Primitive of Root of a x + b over Power of x/Formulation 2

Theorem

 * $\ds \int \frac {\sqrt{a x + b} } {x^m} \rd x = -\frac {\paren {\sqrt{a x + b} }^3} {\paren {m - 1} b x^{m - 1} } - \frac {\paren {2 m - 5} a} {\paren {2 m - 2} b} \int \frac {\sqrt {a x + b} } {x^{m - 1} } \rd x$

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