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

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {x^3 \left({a x + b}\right)^3} = \frac {a^4 x^2} {2 b^5 \left({a x + b}\right)^2} - \frac {4 a^3 x} {b^5 \left({a x + b}\right)} - \frac {\left({a x + b}\right)^2} {2 b^5 x^2} + \frac {4 a} {2 b^4 x} + \frac {6 a^2} {b^5} \ln \left\vert{\frac x {a x + b} }\right\vert + C$

Proof
A partial fraction expansion of the integrand gives:
 * $\dfrac 1 {x^3 \left({a x + b}\right)^3} = \dfrac {6 a^2} {b^5 x} - \dfrac {3 a} {b^4 x^2} + \dfrac 1 {b^3 x^3} - \dfrac {6 a^3} {b^5 \left({a x + b}\right)} - \dfrac {3 a^3} {b^4 \left({a x + b}\right)^2} - \dfrac {a^3} {b^3 \left({a x + b}\right)^3}$

From Linear Combination of Integrals it follows that the expression can be rendered as:


 * $\displaystyle \int \frac {\mathrm d x} {x^3 \left({a x + b}\right)^3} = \frac {6 a^2} {b^5} \int \frac {\mathrm d x} x - \frac {3 a} {b^4} \int \frac {\mathrm d x} {x^2} + \frac 1 {b^3} \int \frac {\mathrm d x} {x^3} - \frac {6 a^3} {b^5} \int \frac {\mathrm d x} {\left({a x + b}\right)} - \frac {3 a^3} {b^4} \int \frac {\mathrm d x} {\left({a x + b}\right)^2} - \frac {a^3} {b^3} \int \frac {\mathrm d x} {\left({a x + b}\right)^3}$

This can be integrated termwise as follows (ignoring for now the arbitrary constant):

The terms with the logarithm can be combined thus:

and so the solution to the integral can be expressed as:

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:

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: