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

Theorem

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

Proof
Let:

Then: