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

Theorem

 * $\ds \int x^2 \paren {\sqrt {a x + b} }^m \rd x = \frac {2 \paren {\sqrt {a x + b} }^{m + 6} } {a^3 \paren {m + 6} } - \frac {4 b \paren {\sqrt {a x + b} }^{m + 4} } {a^3 \paren {m + 4} } + \frac {2 b^2 \paren {\sqrt {a x + b} }^{m + 2} } {a^3 \paren {m + 2} } + C$

Proof
Let:

Then: