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

Theorem

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

Proof
From Reduction Formula for Primitive of Power of $x$ by Power of $a x + b$: Decrement of Power of $a x + b$:


 * $\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^n} {m + n + 1} + \frac {n b} {m + n + 1} \int x^m \paren {a x + b}^{n - 1} \rd x$

Putting $n := \dfrac m 2$ and $m := -1$: