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

Theorem

 * $\displaystyle \int \sqrt{\left({a x + b}\right) \left({p x + q}\right)} \ \mathrm d x = \frac {2 a p x + b p + a q} {4 a p} \sqrt{\left({a x + b}\right) \left({p x + q}\right)} - \frac {\left({b p - a q}\right)^2} {8 a p} \int \frac {\mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} }$

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

Putting $n = \frac 1 2$: