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

Theorem

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

Proof
From Primitive of Power of $a x + b$ over Power of $p x + q$: Formulation 3:


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

Setting $m := \dfrac 1 2$: