Primitive of x squared by Power of Root of a x + b

Theorem

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

Proof
Let:

Then: