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

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

Setting $m := \dfrac 1 2$:

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

$\blacksquare$