Theorem: Primitive of Reciprocal of Half Integer Power of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} } = \frac {2 \paren {2 a x + b} } {\paren {2 n - 1} \paren {4 a c - b^2} \paren {a x^2 + b x + c}^{n - \frac 1 2} } + \frac {8 a \paren {n - 1} } {\paren {2 n - 1} \paren {4 a c - b^2} } \int \frac {\d x} {\paren {a x^2 + b x + c}^{n - \frac 1 2} }$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | r = \int \paren {\frac {4 a} {\paren {2 a x + b}^2 + 4 a c - b^2} }^{n + \frac 1 2} \rd x 
      | c = Completing the Square
}}
{{eqn | r = \paren {2 \sqrt a}^{2 n + 1} \int \frac {\d x} {\paren {\paren {2 a x + b}^2 + 4 a c - b^2}^{n + \frac 1 2} } 
      | c = Linear Combination of Primitives
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = \paren {2 a x + b}^2
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 a \cdot 2 \paren {2 a x + b}
      | c = Derivative of Power and Chain Rule for Derivatives
}}
{{eqn | r = 4 a \sqrt z
      | c = 
}}
{{end-eqn}}
Also let $q = 4 a c - b^2$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | r = \paren {2 \sqrt a}^{2 n + 1} \int \frac {\d z} {4 a \sqrt z \paren {z + q}^{n + \frac 1 2} }
      | c = Integration by Substitution
}}
{{eqn | n = 1
      | r = \frac {\paren {2 \sqrt a}^{2 n + 1} } {4 a} \int \frac {\d z} {\paren {z + q}^{n + \frac 1 2} \sqrt z}
      | c = Linear Combination of Primitives
}}
{{end-eqn}}
From Primitive of $\dfrac 1 {\paren {p x + q}^n \sqrt {a x + b} }$:
:$\ds \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} } = \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$
Here $p = 1, a = 1, b = 0$ and $n := n + \dfrac 1 2$:
{{begin-eqn}}
{{eqn | r = \int \frac {\d x} {\paren {a x^2 + b x + c}^{n + \frac 1 2} }
      | o = 
      | c = 
}}
{{eqn | r = \frac {\paren {2 \sqrt a}^{2 n + 1} } {4 a} \int \frac {\d z} {\paren {z + q}^{n + \frac 1 2} \sqrt z}
      | c = from $(1)$
}}
{{eqn | r = \frac {\paren {2 \sqrt a}^{2 n + 1} } {4 a} \paren {\frac {\sqrt z} {\paren {\paren {n + \frac 1 2} - 1} q \paren {z + q}^{\paren {n + \frac 1 2} - 1} } + \frac {2 \paren {n + \frac 1 2} - 3} {2 \paren {\paren {n + \frac 1 2} - 1} q} \int \frac {\d z} {\paren {z + q}^{\paren {n + \frac 1 2} - 1} \sqrt z} }
      | c = Primitive of $\dfrac 1 {\paren {p x + q}^n \sqrt {a x + b} }$
}}
{{eqn | r = \frac {\paren {2 \sqrt a}^{2 n + 1} } {4 a} \paren {\frac {2 \sqrt z} {\paren {2 n - 1} q \paren {z + q}^{n - \frac 1 2} } + \frac {2 \paren {n - 1} } {\paren {2 n - 1} q} \int \frac {\d z} {\paren {z + q}^{n - \frac 1 2} \sqrt z} }
      | c = simplifying
}}
{{eqn | r = \frac {\paren {2 \sqrt a}^{2 n + 1} } {4 a} \paren {\frac {2 \paren {2 a x + b} } {\paren {2 n - 1} q \paren {\paren {2 a x + b}^2 + q}^{n - \frac 1 2} } + \frac {2 \paren {n - 1} } {\paren {2 n - 1} q} \int \frac {4 a \rd x} {\paren {\paren {2 a x + b}^2 + q}^{n - \frac 1 2} } }
      | c = substituting for $z$
}}
{{eqn | r = \frac 2 {2 n - 1} \frac {\paren {2 a x + b} \paren {4 a}^{n - \frac 1 2} } {\paren {4 a c - b^2} \paren {\paren {2 a x + b}^2 + 4 a c - b^2}^{n - \frac 1 2} }
      | c = substituting for $q$
}}
{{eqn | o = 
      | ro= +
      | r = \frac {8 a \paren {n - 1} } {\paren {2 n - 1} \paren {4 a c - b^2} } \int \frac {\paren {4 a}^{n - \frac 1 2} \rd x} {\paren {\paren {2 a x + b}^2 + 4 a c - b^2}^{n - \frac 1 2} }
      | c = and simplifying
}}
{{eqn | r = \frac {2 \paren {2 a x + b} } {\paren {2 n - 1} \paren {4 a c - b^2} \paren {a x^2 + b x + c}^{n - \frac 1 2} } + \frac {8 a \paren {n - 1} } {\paren {2 n - 1} \paren {4 a c - b^2} } \int \frac {\d x} {\paren {a x^2 + b x + c}^{n - \frac 1 2} }
      | c = Completing the Square
}}
{{end-eqn}}
{{qed}}