Theorem: Primitive of Reciprocal of x cubed by a x + b cubed

:$\ds \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} - \frac {4 a^3 x} {b^5 \paren {a x + b} } - \frac {\paren {a x + b}^2} {2 b^5 x^2} + \frac {4 a} {b^4 x} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C$

Proof:
A partial fraction expansion of the integrand gives:
:$\dfrac 1 {x^3 \paren {a x + b}^3} = \dfrac {6 a^2} {b^5 x} - \dfrac {3 a} {b^4 x^2} + \dfrac 1 {b^3 x^3} - \dfrac {6 a^3} {b^5 \paren {a x + b} } - \dfrac {3 a^3} {b^4 \paren {a x + b}^2} - \dfrac {a^3} {b^3 \paren {a x + b}^3}$
From Linear Combination of Primitives it follows that the expression can be rendered as:
:$\ds \int \frac {\d x} {x^3 \paren {a x + b}^3} = \frac {6 a^2} {b^5} \int \frac {\d x} x - \frac {3 a} {b^4} \int \frac {\d x} {x^2} + \frac 1 {b^3} \int \frac {\d x} {x^3} - \frac {6 a^3} {b^5} \int \frac {\d x} {\paren {a x + b} } - \frac {3 a^3} {b^4} \int \frac {\d x} {\paren {a x + b}^2} - \frac {a^3} {b^3} \int \frac {\d x} {\paren {a x + b}^3}$
This can be integrated termwise as follows (ignoring for now the arbitrary constant):
{{begin-eqn}}
{{eqn | l = \frac {6 a^2} {b^5} \int \frac {\d x} x 
      | r = \frac {6 a^2} {b^5} \ln \size x
      | c = Primitive of Reciprocal
}}
{{eqn | l = -\frac {3 a} {b^4} \int \frac {\d x} {x^2}
      | r = \frac {3 a} {b^4 x}
      | c = Primitive of Power
}}
{{eqn | l = \frac 1 {b^3} \int \frac {\d x} {x^3}
      | r = -\frac 1 {2 b^3 x^2}
      | c = Primitive of Power
}}
{{eqn | l = -\frac {6 a^3} {b^5} \int \frac {\d x} {\paren {a x + b} }
      | r = -\frac {6 a^2} {b^5} \ln \size {a x + b}
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{eqn | l = - \frac {3 a^3} {b^4} \int \frac {\d x} {\paren {a x + b}^2}
      | r = \frac {3 a^3} {b^4 a \paren {a x + b} }
      | c = Primitive of $\dfrac 1 {\paren {a x + b}^2}$
}}
{{eqn | l = -\frac {a^3} {b^3} \int \frac {\d x} {\paren {a x + b}^3}
      | r = \frac {a^2} {2 b^3 \paren {a x + b}^2}
      | c = Primitive of $\dfrac 1 {\paren {a x + b}^3}$
}}
{{end-eqn}}
The terms with the logarithm can be combined thus:
{{begin-eqn}}
{{eqn | l = \frac {6 a^2} {b^5} \ln \size x - \frac {6 a^2} {b^5} \ln \size {a x + b}
      | r = \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} }
      | c = Difference of Logarithms
}}
{{end-eqn}}
and so the solution to the integral can be expressed as:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^3 \paren {a x + b}^3}
      | r = \frac {3 a} {b^4 x} - \frac 1 {2 b^3 x^2} + \frac {3 a^2} {b^4 \paren {a x + b} } + \frac {a^2} {2 b^3 \paren {a x + b}^2} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C
}}
{{end-eqn}}
While this would usually be considered as an acceptable form to leave such an expression, there is some way to go to obtain the result requested.
So, placing the non-logarithmic terms over a common denominator and rearranging, with a view to our destination:
{{begin-eqn}}
{{eqn | r = \frac {3 a} {b^4 x} - \frac 1 {2 b^3 x^2} + \frac {3 a^2} {b^4 \paren {a x + b} } + \frac {a^2} {2 b^3 \paren {a x + b}^2}
      | o = 
}}
{{eqn | r = \frac {6 a x \paren {a x + b}^2} {2 b^4 x^2 \paren {a x + b}^2} - \frac {b \paren {a x + b}^2} {2 b^4 x^2 \paren {a x + b}^2} + \frac {6 a^2 x^2 \paren {a x + b} } {2 b^4 x^2 \paren {a x + b}^2} + \frac {a^2 b x^2} {2 b^4 x^2 \paren {a x + b}^2}
}}
{{eqn | r = \frac {12 a^3 x^3 + 18 a^2 b x^2 + 4 a b^2 x - b^3} {2 b^4 \paren {a x + b}^2 x^2}
}}
{{eqn | r = \frac {a^4 x^4 + 16 a^3 b x^3 + 24 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {a^4 x^4 + 4 a^3 b x^3 + 6 a^2 b^2 x^2 + 4 a b^3 x + b^4} {2 b^5 \paren {a x + b}^2 x^2}
      | c = 
}}
{{eqn | r = \frac {a^4 x^4 + 16 a^3 b x^3 + 24 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^4} {2 b^5 \paren {a x + b}^2 x^2}
      | c = 
}}
{{eqn | r = \frac {a^4 x^4 + 16 a^3 b x^3 + 24 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
      | c = 
}}
{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {16 a^3 b x^3 + 24 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
      | c = 
}}
{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {8 a^3 b x^3 + 8 a^2 b^2 x^2} {2 b^5 \paren {a x + b}^2 x^2} + \frac {8 a^3 b x^3 + 16 a^2 b^2 x^2 + 8 a b^3 x} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
      | c = 
}}
{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {8 a^2 b x^2 \paren {a x + b} } {2 b^5 \paren {a x + b}^2 x^2} + \frac {8 a b x \paren {a x + b}^2} {2 b^5 \paren {a x + b}^2 x^2} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
      | c = 
}}
{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {4 a^2} {b^4 \paren {a x + b} } + \frac {4 a} {b^4 x} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
      | c = 
}}
{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {4 a^2 b} {b^5 \paren {a x + b} } + \frac {4 a} {b^4 x} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
      | c = 
}}
{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {4 a^2 \paren {a x + b} } {b^5 \paren {a x + b} } - \frac {4 a^3 x} {b^5 \paren {a x + b} } + \frac {4 a} {b^4 x} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
      | c = 
}}
{{eqn | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} + \frac {4 a^2} {b^5} - \frac {4 a^3 x} {b^5 \paren {a x + b} } + \frac {4 a} {b^4 x} - \frac {\paren {a x + b}^2} {2 b^5 x^2}
      | c = 
}}
{{end-eqn}}
Thus the full solution to the integral can be assembled as follows, where $\dfrac {4 a^2} {b^5}$ is subsumed into the arbitrary constant:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^3 \paren {a x + b}^3}
      | r = \frac {a^4 x^2} {2 b^5 \paren {a x + b}^2} - \frac {4 a^3 x} {b^5 \paren {a x + b} } - \frac {\paren {a x + b}^2} {2 b^5 x^2} + \frac {4 a} {b^4 x} + \frac {6 a^2} {b^5} \ln \size {\frac x {a x + b} } + C
}}
{{end-eqn}}
{{qed}}