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

Theorem

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

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

Putting $n = \dfrac 1 2$: