Primitive of x cubed over a x + b

Theorem

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

Proof
Put $u = a x + b$.

Then:

Then: