Primitive of x cubed over a x + b squared

Theorem

 * $\ds \int \frac {x^3 \rd x} {\paren {a x + b}^2} = \frac {\paren {a x + b}^2} {2 a^4} - \frac {3 b \paren {a x + b} } {a^4} + \frac {b^3} {a^4 \paren {a x + b} } + \frac {3 b^2} {a^4} \ln \size {a x + b} + C$

Proof
Put $u = a x + b$.

Then:

Then: