avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of Hyperbolic Cosine of a x by Hyperbolic Cosine of p x

:$\ds \int \cosh a x \cosh p x \rd x = \frac {\map \sinh {a + p} x} {2 \paren {a + p} } + \frac {\map \sinh {a - p} x} {2 \paren {a - p} } + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \cosh a x \cosh p x \rd x
      | r = \int \paren {\frac {\map \cosh {a x + p x} + \map \cosh {a x - p x} } 2} \rd x
      | c = Werner Formula for Hyperbolic Cosine by Hyperbolic Cosine
}}
{{eqn | r = \frac 1 2 \int \map \cosh {a + p} x \rd x + \frac 1 2 \int \map \cosh {a - p} x \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 2 \frac {\map \sinh {a + p} x} {a + p} + \frac 1 2 \frac {\map \sinh {a - p} x} {a - p} + C
      | c = Primitive of $\cosh a x$
}}
{{eqn | r = \frac {\map \sinh {a + p} x} {2 \paren {a + p} } + \frac {\map \sinh {a - p} x} {2 \paren {a - p} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Hyperbolic Sine of a x by Sine of p x

:$\ds \int \sinh a x \sin p x \rd x = \frac {a \cosh a x \sin p x - p \sinh a x \cos p x} {a^2 + p^2} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \sinh a x \sin p x \rd x
      | r = \int \paren {\frac {e^{a x} - e^{-a x} } 2} \sin p x \rd x
      | c = {{Defof|Hyperbolic Sine}}
}}
{{eqn | r = \frac 1 2 \int e^{a x} \sin p x \rd x - \frac 1 2 \int e^{- a x} \sin p x \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 2 \paren {\frac {e^{a x} \paren {a \sin p x - p \cos p x} } {a^2 + p^2} } - \frac 1 2 \paren {\frac {e^{-a x} \paren {-a \sin p x - p \cos p x} } {a^2 + p^2} } + C
      | c = Primitive of $e^{a x} \sin b x$
}}
{{eqn | r = \frac 1 {\paren {a^2 + p^2} } \paren {a \paren {\frac {e^{a x} + e^{-a x} } 2} \sin p x - p \paren {\frac {e^{a x} - e^{-a x} } 2} \cos p x} + C
      | c = factoring
}}
{{eqn | r = \frac 1 {\paren {a^2 + p^2} } \paren {a \cosh a x \sin p x - p \paren {\frac {e^{a x} - e^{-a x} } 2} \cos p x} + C
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{eqn | r = \frac 1 {\paren {a^2 + p^2} } \paren {a \cosh a x \sin p x - p \sinh a x \cos p x} + C
      | c = {{Defof|Hyperbolic Sine}}
}}
{{eqn | r = \frac {a \cosh a x \sin p x - p \sinh a x \cos p x} {a^2 + p^2} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Hyperbolic Sine of a x by Cosine of p x

:$\ds \int \sinh a x \cos p x \rd x = \frac {a \cosh a x \cos p x + p \sinh a x \sin p x} {a^2 + p^2} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \sinh a x \cos p x \rd x
      | r = \int \paren {\frac {e^{a x} - e^{- a x} } 2} \cos p x \rd x
      | c = {{Defof|Hyperbolic Sine}}
}}
{{eqn | r = \frac 1 2 \int e^{a x} \cos p x \rd x - \frac 1 2 \int e^{- a x} \cos p x \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 2 \paren {\frac {e^{a x} \paren {a \cos p x + p \sin p x} } {a^2 + p^2} } - \frac 1 2 \paren {\frac {e^{-a x} \paren {-a \cos p x + p \sin p x} } {a^2 + p^2} } + C
      | c = Primitive of $e^{a x} \cos b x$
}}
{{eqn | r = \frac 1 {\paren {a^2 + p^2} } \paren {a \paren {\frac {e^{a x} + e^{-a x} } 2} \cos p x + p \paren {\frac {e^{a x} - e^{-a x} } 2} \sin p x} + C
      | c = factoring
}}
{{eqn | r = \frac 1 {\paren {a^2 + p^2} } \paren {a \cosh a x \cos p x + p \paren {\frac {e^{a x} - e^{-a x} } 2} \sin p x} + C
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{eqn | r = \frac 1 {\paren {a^2 + p^2} } \paren {a \cosh a x \cos p x + p \sinh a x \sin p x} + C
      | c = {{Defof|Hyperbolic Sine}}
}}
{{eqn | r = \frac {a \cosh a x \cos p x + p \sinh a x \sin p x} {a^2 + p^2} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Hyperbolic Cosine of a x by Sine of p x

:$\ds \int \cosh a x \sin p x \rd x = \frac {a \sinh a x \sin p x - p \cosh a x \cos p x} {a^2 + p^2} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \cosh a x \sin p x \rd x
      | r = \int \paren {\frac {e^{a x} + e^{-a x} } 2} \sin p x \rd x
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{eqn | r = \frac 1 2 \int e^{a x} \sin p x \rd x + \frac 1 2 \int e^{- a x} \sin p x \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 2 \paren {\frac {e^{a x} \paren {a \sin p x - p \cos p x} } {a^2 + p^2} } + \frac 1 2 \paren {\frac {e^{-a x} \paren {-a \sin p x - p \cos p x} } {a^2 + p^2} } + C
      | c = Primitive of $e^{a x} \sin b x$
}}
{{eqn | r = \frac 1 {\paren {a^2 + p^2} } \paren {a \paren {\frac {e^{a x} - e^{-a x} } 2} \sin p x - p \paren {\frac {e^{a x} + e^{-a x} } 2} \cos p x} + C
      | c = factoring
}}
{{eqn | r = \frac 1 {\paren {a^2 + p^2} } \paren {a \sinh a x \sin p x - p \paren {\frac {e^{a x} + e^{-a x} } 2} \cos p x} + C
      | c = {{Defof|Hyperbolic Sine}}
}}
{{eqn | r = \frac 1 {\paren {a^2 + p^2} } \paren {a \sinh a x \sin p x - p \cosh a x \cos p x} + C
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{eqn | r = \frac {a \sinh a x \sin p x - p \cosh a x \cos p x} {a^2 + p^2} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Hyperbolic Cosine of a x by Cosine of p x

:$\ds \int \cosh a x \cos p x \rd x = \frac {a \sinh a x \cos p x + p \cosh a x \sin p x} {a^2 + p^2} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \cosh a x \cos p x \rd x
      | r = \int \paren {\frac {e^{a x} + e^{- a x} } 2} \cos p x \rd x
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{eqn | r = \frac 1 2 \int e^{a x} \cos p x \rd x + \frac 1 2 \int e^{- a x} \cos p x \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 2 \paren {\frac {e^{a x} \paren {a \cos p x + p \sin p x} } {a^2 + p^2} } + \frac 1 2 \paren {\frac {e^{-a x} \paren {-a \cos p x + p \sin p x} } {a^2 + p^2} } + C
      | c = Primitive of $e^{a x} \cos b x$
}}
{{eqn | r = \frac 1 {\paren {a^2 + p^2} } \paren {a \paren {\frac {e^{a x} - e^{-a x} } 2} \cos p x + p \paren {\frac {e^{a x} + e^{-a x} } 2} \sin p x} + C
      | c = factoring
}}
{{eqn | r = \frac 1 {\paren {a^2 + p^2} } \paren {a \sinh a x \cos p x + p \paren {\frac {e^{a x} + e^{-a x} } 2} \sin p x} + C
      | c = {{Defof|Hyperbolic Sine}}
}}
{{eqn | r = \frac 1 {\paren {a^2 + p^2} } \paren {a \sinh a x \cos p x + p \cosh a x \sin p x} + C
      | c = {{Defof|Hyperbolic Cosine}}
}}
{{eqn | r = \frac {a \sinh a x \cos p x + p \cosh a x \sin p x} {a^2 + p^2} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of p plus q by Hyperbolic Sine of a x

:$\ds \int \frac {\d x} {p + q \sinh a x}  = \frac 1 {a \sqrt{p^2 + q^2} } \ln \size {\frac {q e^{a x} + p - \sqrt {p^2 + q^2} } {q e^{a x} + p + \sqrt {p^2 + q^2} } } + C$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = e^{a x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d u} {\d x}
      | r = e^{a x} = u
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \d x
      | r = \dfrac {\d u} u
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {p + q \sinh a x}
      | r = \int \frac {\d x} {p + q \paren {\dfrac {e^{a x} - e^{-a x} } 2} }
      | c = {{Defof|Real Hyperbolic Sine}}
}}
{{eqn | r = \int \frac {2 \rd x} {2 p + q \paren {e^{a x} - e^{-a x} } }
      | c = 
}}
{{eqn | r = \int \frac {2 \rd u} {u \paren {2 p + q u - \frac q u} }
      | c = Integration by Substitution: $u = e^{a x}$
}}
{{eqn | r = \int \frac {2 \rd u} {q u^2 + 2 p u - q}
      | c = simplifying
}}
{{end-eqn}}
The discriminant of $q u^2 + 2 p u - q$ is given by:
{{begin-eqn}}
{{eqn | l = \map {\operatorname {Disc} } {q u^2 + 2 p u - q}
      | r = \paren {2 p}^2 + 4 q^2
      | c = 
}}
{{eqn | r = 4 \paren {p^2 + q^2}
      | c = 
}}
{{end-eqn}}
Hence the sign of $\map {\operatorname {Disc} } {q u^2 + 2 p u - q}$ is always positive.
So:
{{begin-eqn}}
{{eqn | l = \int \frac {2 \rd u} {q u^2 + 2 p u - q}
      | r = \dfrac 2 {\sqrt {4 p^2 + 4 q^2} } \ln \size {\dfrac {2 q u + 2 p - \sqrt {4 p^2 + 4 q^2} } {2 q u + 2 p + \sqrt {4 p^2 + 4 q^2} } } + C
      | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$
}}
{{eqn | r = \dfrac 1 {\sqrt {p^2 + q^2} } \ln \size {\dfrac {q u + p - \sqrt {p^2 + q^2} } {q u + p + \sqrt {p^2 + q^2} } } + C
      | c = simplifying
}}
{{eqn | r = \dfrac 1 {\sqrt {p^2 + q^2} } \ln \size {\dfrac {q e^{a x} + p - \sqrt {p^2 + q^2} } {q e^{a x} + p + \sqrt {p^2 + q^2} } } + C
      | c = substituting $u = e^{a x}$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of p plus q by Hyperbolic Cosine of a x

For $p \ne q$:
:$\ds \int \frac {\d x} {p + q \cosh a x} = \begin {cases}
\dfrac 2 {a \sqrt {q^2 - p^2} } \arctan \dfrac {q e^{a x} + p} {\sqrt {q^2 - p^2} } + C & : p^2 < q^2 \\
\dfrac 1 {a \sqrt {p^2 - q^2} } \ln \size {\dfrac {q e^{a x} + p - \sqrt {p^2 - q^2} } {q e^{a x} + p + \sqrt {p^2 - q^2} } } + C & : p^2 > q^2
\end {cases}$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = e^{a x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\d u} {\d x}
      | r = e^{a x} = u
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \d x
      | r = \dfrac {\d u} u
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {p + q \cosh a x}
      | r = \int \frac {\d x} {p + q \paren {\dfrac {e^{a x} + e^{-a x} } 2} }
      | c = {{Defof|Real Hyperbolic Cosine}}
}}
{{eqn | r = \int \frac {2 \rd x} {2 p + q \paren {e^{a x} + e^{-a x} } }
      | c = 
}}
{{eqn | r = \int \frac {2 \rd u} {u \paren {2 p + q u + \frac q u} }
      | c = Integration by Substitution: $u = e^{a x}$
}}
{{eqn | r = \int \frac {2 \rd u} {q u^2 + 2 p u + q}
      | c = simplifying
}}
{{end-eqn}}
The discriminant of $q u^2 + 2 p u + q$ is given by:
{{begin-eqn}}
{{eqn | l = \map {\operatorname {Disc} } {q u^2 + 2 p u + q}
      | r = \paren {2 p}^2 - 4 q^2
      | c = 
}}
{{eqn | r = 4 \paren {p^2 - q^2}
      | c = 
}}
{{end-eqn}}
Hence the result depends on the sign of $p^2 - q^2$.
Let $p^2 < q^2$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {2 \rd u} {q u^2 + 2 p u + q}
      | r = \dfrac 2 {\sqrt {4 q^2 - 4 p^2} } \map \arctan {\dfrac {2 q u + 2 p} {\sqrt {4 q^2 - 4 p ^2} } } + C
      | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$
}}
{{eqn | r = \dfrac 1 {\sqrt {q^2 - p^2} } \map \arctan {\dfrac {q u + p} {\sqrt {q^2 - p ^2} } } + C
      | c = simplifying
}}
{{eqn | r = \dfrac 1 {\sqrt {q^2 - p^2} } \map \arctan {\dfrac {q e^{a x} + p} {\sqrt {q^2 - p^2} } } + C
      | c = substituting $u = e^{a x}$
}}
{{end-eqn}}
Let $p^2 > q^2$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {2 \rd u} {q u^2 + 2 p u + q}
      | r = \dfrac 2 {\sqrt {4 p^2 - 4 q^2} } \ln \size {\dfrac {2 q u + 2 p - \sqrt {4 p^2 - 4 q^2} } {2 q u + 2 p + \sqrt {4 p^2 - 4 q^2} } } + C
      | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$
}}
{{eqn | r = \dfrac 1 {\sqrt {p^2 - q^2} } \ln \size {\dfrac {q u + p - \sqrt {p^2 - q^2} } {q u + p + \sqrt {p^2 - q^2} } } + C
      | c = simplifying
}}
{{eqn | r = \dfrac 1 {\sqrt {p^2 - q^2} } \ln \size {\dfrac {q e^{a x} + p - \sqrt {p^2 - q^2} } {q e^{a x} + p + \sqrt {p^2 - q^2} } } + C
      | c = substituting $u = e^{a x}$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of p plus q by Hyperbolic Tangent of a x

:$\ds \int \frac {\d x} {p + q \tanh a x} = \frac {p x} {p^2 - q^2} - \frac q {a \paren {p^2 - q^2} } \ln \size {q \sinh a x + p \cosh a x} + C$

Proof:
We have:
:$\dfrac \d {\d x} \paren {q \sinh a x + p \cosh a x} = a q \cosh a x + a p \sinh a x$
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {p + q \tanh a x}
      | r = \int \frac {\d x} {p + q \dfrac {\sinh a x} {\cosh a x} }
      | c = {{Defof|Hyperbolic Tangent|index = 2}}
}}
{{eqn | r = \int \frac {\cosh a x \rd x} {p \cosh a x + q \sinh a x}
      | c = 
}}
{{eqn | r = \frac 1 {p^2 - q^2} \int \frac {\paren {p^2 - q^2} \cosh a x \rd x} {p \cosh a x + q \sin a x}
      | c = 
}}
{{eqn | r = \frac 1 {p^2 - q^2} \int \frac {p^2 \cosh a x + p q \sinh a x - p q \sinh a x - q^2 \cosh a x} {p \cosh a x + q \sinh a x} \rd x
      | c = 
}}
{{eqn | r = \frac 1 {p^2 - q^2} \paren {\int \frac {p^2 \cosh a x + p q \sinh a x} {p \cosh a x + q \sinh a x} \rd x - \int \frac {p q \sinh a x + q^2 \cosh a x} {p \cosh a x + q \sinh a x} \rd x}
      | c = 
}}
{{eqn | r = \frac 1 {p^2 - q^2} \paren {\int p \rd x - \frac q a \int \frac {\map \d {p \cosh a x + q \sinh a x} } {p \cosh a x + q \sinh a x} }
      | c = 
}}
{{eqn | r = \frac {p x} {p^2 - q^2} - \frac q {a \paren {p^2 - q^2} } \ln \size {p \cosh a x + q \sinh a x} + C
      | c = Primitive of Constant and Primitive of Reciprocal
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of p plus q by Hyperbolic Cotangent of a x

:$\ds \int \frac {\d x} {p + q \coth a x} = \frac {p x} {p^2 - q^2} - \frac q {a \paren {p^2 - q^2} } \ln \size {p \sinh a x + q \cosh a x} + C$

Proof:
We have:
:$\dfrac \d {\d x} \paren {p \sinh a x + q \cosh a x} = a p \cosh a x + a q \sinh a x$
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {p + q \coth a x}
      | r = \int \frac {\d x} {p + q \dfrac {\cosh a x} {\sinh a x} }
      | c = {{Defof|Hyperbolic Cotangent|index = 2}}
}}
{{eqn | r = \int \frac {\sinh a x \rd x} {p \sinh a x + q \cosh a x}
      | c = 
}}
{{eqn | r = \frac 1 {p^2 - q^2} \int \frac {\paren {p^2 - q^2} \sinh a x \rd x} {p \sinh a x + q \cosh a x}
      | c = 
}}
{{eqn | r = \frac 1 {p^2 - q^2} \int \frac {p^2 \sinh a x + p q \cosh a x - p q \cosh a x - q^2 \sinh a x} {p \sinh a x + q \cosh a x} \rd x
      | c = 
}}
{{eqn | r = \frac 1 {p^2 - q^2} \paren {\int \frac {p^2 \sinh a x + p q \cosh a x} {p \sinh a x + q \cosh a x} \rd x - \int \frac {p q \cosh a x + q^2 \sinh a x} {p \sinh a x + q \cosh a x} \rd x}
      | c = 
}}
{{eqn | r = \frac 1 {p^2 - q^2} \paren {\int p \rd x - \frac q a \int \frac {\map \d {p \sinh a x + q \cosh a x} } {p \sinh a x + q \cosh a x} }
      | c = 
}}
{{eqn | r = \frac {p x} {p^2 - q^2} - \frac q {a \paren {p^2 - q^2} } \ln \size {p \sinh a x + q \cosh a x} + C
      | c = Primitive of Constant and Primitive of Reciprocal
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of q plus p by Hyperbolic Secant of a x

:$\ds \int \frac {\d x} {q + p \sech a x} = \frac x q - \frac p q \int \frac {\d x} {p + q \cosh a x} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {q + p \sech  a x}
      | r = \frac 1 q \int \frac {q \rd x} {q + p \sech a x}
      | c = multiplying top and bottom by $q$
}}
{{eqn | r = \frac 1 q \int \frac {\paren {q + p \sech a x - p \sech a x} \rd x} {q + p \sech a x}
      | c = 
}}
{{eqn | r = \frac 1 q \int \frac {\paren {q + p \sech a x} \rd x} {q + p \sech a x} - \frac p q \int \frac {\sech a x \rd x} {q + p \sech a x}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 q \int \d x - \frac p q \int \frac {\sech a x \rd x} {q + p \sech a x}
      | c = simplifying
}}
{{eqn | r = \frac x q - \frac p q \int \frac {\sech a x \rd x} {q + p \sech a x} + C
      | c = Primitive of Constant
}}
{{eqn | r = \frac x q - \frac p q \int \frac {\d x} {\frac q {\sech a x} + p} + C
      | c = multiplying top and bottom by $\dfrac 1 {\sech a x}$
}}
{{eqn | r = \frac x q - \frac p q \int \frac {\d x} {p + q \cosh a x} + C
      | c = {{Defof|Hyperbolic Secant|index = 2}}
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of q plus p by Hyperbolic Cosecant of a x

:$\ds \int \frac {\d x} {q + p \csch a x} = \frac x q - \frac p q \int \frac {\d x} {p + q \sinh a x} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {q + p \csch a x}
      | r = \frac 1 q \int \frac {q \rd x} {q + p \csch a x}
      | c = multiplying top and bottom by $q$
}}
{{eqn | r = \frac 1 q \int \frac {\paren {q + p \csch a x - p \csch a x} \rd x} {q + p \csch a x}
      | c = 
}}
{{eqn | r = \frac 1 q \int \frac {\paren {q + p \csch a x} \rd x} {q + p \csch a x} - \frac p q \int \frac {\csch a x \rd x} {q + p \csch a x}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 q \int \d x - \frac p q \int \frac {\csch a x \rd x} {q + p \csch a x}
      | c = simplifying
}}
{{eqn | r = \frac x q - \frac p q \int \frac {\csch a x \rd x} {q + p \csch a x} + C
      | c = Primitive of Constant
}}
{{eqn | r = \frac x q - \frac p q \int \frac {\d x} {\frac q {\csch a x} + p} + C
      | c = multiplying top and bottom by $\dfrac 1 {\csch a x}$
}}
{{eqn | r = \frac x q - \frac p q \int \frac {\d x} {p + q \sinh a x} + C
      | c = {{Defof|Hyperbolic Cosecant|index = 2}}
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Square of p plus q by Hyperbolic Sine of a x

:$\ds \int \frac {\d x} {\paren {p + q \sinh a x}^2}  = \frac {-q \cosh a x} {a \paren {p^2 + q^2} \paren {p + q \sinh a x} } + \frac p {p^2 + q^2} \int \frac {\d x} {p + q \sinh a x} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\dfrac {\cosh a x} {p + q \sinh a x} }
      | r = \dfrac {\paren {p + q \sinh a x} \map {\frac \d {\d x} } {\cosh a x} - \cosh a x \map {\frac \d {\d x} } {p + q \sinh a x} } {\paren {p + q \sinh a x}^2}
      | c = Quotient Rule for Derivatives
}}
{{eqn | r = \dfrac {\paren {p + q \sinh a x} \paren {a \sinh a x} - \cosh a x \paren {a q \cosh a x} } {\paren {p + q \sinh a x}^2}
      | c = Derivative of Hyperbolic Cosine Function, Derivative of Hyperbolic Sine Function
}}
{{eqn | r = a \dfrac {p \sinh a x - q \paren {\cosh^2 a x - \sinh^2 a x} } {\paren {p + q \sinh a x}^2}
      | c = simplification
}}
{{eqn | r = a \dfrac {p \sinh a x + q} {\paren {p + q \sinh a x}^2}
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{eqn | r = a \dfrac {p q \sinh a x + q^2} {q \paren {p + q \sinh a x}^2}
      | c = 
}}
{{eqn | r = a \dfrac {p q \sinh a x + q^2 + p^2 - p^2} {q \paren {p + q \sinh a x}^2}
      | c = 
}}
{{eqn | r = a \dfrac {p q \sinh a x + p^2} {q \paren {p + q \sinh a x}^2} - a \dfrac {p^2 + q^2} {q \paren {p + q \sinh a x}^2} 
      | c = 
}}
{{eqn | r = a \dfrac {p \paren {p + q \sinh a x} } {q \paren {p + q \sinh a x}^2} - a \dfrac {p^2 + q^2} {q \paren {p + q \sinh a x}^2}
      | c = 
}}
{{eqn | r = \dfrac {a p} q \dfrac 1 {p + q \sinh a x} - \dfrac {a \paren {p^2 + q^2} } {q \paren {p + q \sinh a x}^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac q {a \paren {p^2 + q^2} } \map {\dfrac \d {\d x} } {\dfrac {\cosh a x} {p + q \sinh a x} }
      | r = \dfrac p {\paren {p^2 + q^2} } \dfrac 1 {p + q \sinh a x} - \dfrac 1 {\paren {p + q \sinh a x}^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac q {a \paren {p^2 + q^2} } {\dfrac {\cosh a x} {p + q \sinh a x} }
      | r = \dfrac p {\paren {p^2 + q^2} } \int \dfrac {\d x} {p + q \sinh a x} - \int \dfrac {\d x} {\paren {p + q \sinh a x}^2}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Primitive of Reciprocal of Square of p plus q by Hyperbolic Cosine of a x

:$\ds \int \frac {\d x} {\paren {p + q \cosh a x}^2}  = \frac {q \sinh a x} {a \paren {q^2 - p^2} \paren {p + q \cosh a x} } - \frac p {q^2 - p^2} \int \frac {\rd x} {p + q \cosh a x} + C$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {\dfrac {\sinh a x} {p + q \cosh a x} }
      | r = \dfrac {\paren {p + q \cosh a x} \map {\frac \d {\d x} } {\sinh a x} - \sinh a x \map {\frac \d {\d x} } {p + q \sinh a x} } {\paren {p + q \cosh a x}^2}
      | c = Quotient Rule for Derivatives
}}
{{eqn | r = \dfrac {\paren {p + q \sinh a x} \paren {a \cosh a x} - \sinh a x \paren {a q \sinh a x} } {\paren {p + q \cosh a x}^2}
      | c = Derivative of Hyperbolic Cosine Function, Derivative of Hyperbolic Sine Function
}}
{{eqn | r = a \dfrac {p \cosh a x + q \paren {\cosh^2 a x - \sinh^2 a x} } {\paren {p + q \cosh a x}^2}
      | c = simplification
}}
{{eqn | r = a \dfrac {p \cosh a x + q} {\paren {p + q \cosh a x}^2}
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{eqn | r = a \dfrac {p q \cosh a x + q^2} {q \paren {p + q \cosh a x}^2}
      | c = 
}}
{{eqn | r = a \dfrac {p q \cosh a x + q^2 + p^2 - p^2} {q \paren {p + q \cosh a x}^2}
      | c = 
}}
{{eqn | r = a \dfrac {q^2 - p^2} {q \paren {p + q \cosh a x}^2} + a \dfrac {p q \cosh a x + p^2} {q \paren {p + q \cosh a x}^2}
      | c = 
}}
{{eqn | r = a \dfrac {q^2 - p^2} {q \paren {p + q \cosh a x}^2} + a \dfrac {p \paren {p + q \cosh a x} } {q \paren {p + q \cosh a x}^2}
      | c = 
}}
{{eqn | r = \dfrac {a \paren {q^2 - p^2} } {q \paren {p + q \cosh a x}^2} + \dfrac {a p} q \dfrac 1 {p + q \cosh a x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac q {a \paren {q^2 - p^2} } \map {\dfrac \d {\d x} } {\dfrac {\sinh a x} {p + q \cosh a x} }
      | r = \dfrac 1 {\paren {p + q \cosh a x}^2} + \dfrac p {\paren {q^2 - p^2} } \dfrac 1 {p + q \cos a x}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac q {a \paren {q^2 - p^2} } {\dfrac {\sinh a x} {p + q \cosh a x} }
      | r = \int \dfrac {\d x} {\paren {p + q \cosh a x}^2} + \dfrac p {\paren {q^2 - p^2} } \int \dfrac {\d x} {p + q \cosh a x}
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Primitive of Reciprocal of p squared plus Square of q by Hyperbolic Cosine of a x/Logarithm Form

:$\ds \int \frac {\d x} {p^2 + q^2 \cosh^2 a x} = \dfrac 1 {2 a p \sqrt {p^2 + q^2} } \ln \size {\dfrac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } }$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {p^2 + q^2 \cosh^2 a x}
      | r = \int \frac {\csch^2 a x \rd x} {p^2 \csch^2 a x + q^2 \coth^2 a x}
      | c = multiplying numerator and denominator by $\csch^2 a x$
}}
{{eqn | r = \int \frac {\csch^2 a x \rd x} {p^2 \paren {\coth^2 a x - 1} + q^2 \coth^2 a x}
      | c = Difference of Squares of Hyperbolic Cotangent and Cosecant
}}
{{eqn | r = \int \frac {\csch^2 a x \rd x} {\paren {p^2 + q^2} \coth^2 a x - p^2}
      | c = simplifying
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \coth a x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \d u
      | r = -a \csch^2 a x \rd x
      | c = Derivative of Hyperbolic Cotangent Function
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {p^2 + q^2 \cosh^2 a x}
      | r = \frac 1 a \int \frac {-\d u} {\paren {p^2 + q^2} u^2 - p^2}
      | c = Integration by Substitution: substituting $u = \tanh a x$
}}
{{eqn | n = 1
      | r = \frac 1 {a \paren {p^2 + q^2} } \int \frac {\rd u} {\frac {p^2} {p^2 + q^2} - u^2}
      | c = rearranging into a standard form
}}
{{end-eqn}}
We have:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {p^2 + q^2 \cosh^2 a x}
      | r = \frac 1 {a \paren {p^2 + q^2} } \int \frac {\rd u} {\paren {\frac p {\sqrt {p^2 + q^2} } }^2 - u^2}
      | c = from $(1)$
}}
{{eqn | r = \frac 1 {a \paren {p^2 + q^2} } \frac {\sqrt {p^2 + q^2} } {2 p} \ln \size {\frac {\frac p {\sqrt {p^2 + q^2} } + u} {\frac p {\sqrt {p^2 + q^2} } - u} } + C
      | c = Primitive of $\dfrac 1 {a^2 - x^2}$
}}
{{eqn | r = \frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \size {\frac {\frac p {\sqrt {p^2 + q^2} } + \coth a x} {\frac p {\sqrt {p^2 + q^2} } - \coth a x} } + C
      | c = substituting $u = \coth a x$
}}
{{eqn | r = \frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \size {\frac {\frac p {\sqrt {p^2 + q^2} } + \frac 1 {\tanh a x} } {\frac p {\sqrt {p^2 + q^2} } - \frac 1 {\tanh a x} } } + C
      | c = {{Defof|Real Hyperbolic Cotangent}}
}}
{{eqn | r = \frac 1 {2 a p \sqrt {p^2 + q^2} } \ln \size {\frac {p \tanh a x + \sqrt {p^2 + q^2} } {p \tanh a x - \sqrt {p^2 + q^2} } } + C
      | c = multiplying numerator and denominator of argument of $\ln$ by $\sqrt {p^2 + q^2} \tanh a x$
}}
{{end-eqn}}
{{qed}}