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

Theorem

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

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


 * $\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$

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