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

Theorem

 * $\displaystyle \int \left({p x + q}\right)^n \sqrt{a x + b} \ \mathrm d x = \frac {2 \left({p x + q}\right)^{n+1} \sqrt{a x + b} } {\left({2 n + 3}\right) p} + \frac {b p - a q} {\left({2 n + 3}\right) p} \int \frac {\left({p x + q}\right)^n} {\sqrt{a x + b} } \ \mathrm d x$

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


 * $\displaystyle \int \left({a x + b}\right)^m \left({p x + q}\right)^n \ \mathrm d x = \frac {\left({a x + b}\right)^m \left({p x + q}\right)^{n+1}} {\left({m + n + 1}\right) p} + \frac {m \left({b p - a q}\right)} {\left({m + n + 1}\right) p} \int \left({a x + b}\right)^{m-1} \left({p x + q}\right)^n \ \mathrm d x$

Setting $m := \dfrac 1 2$: