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

Theorem

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

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

Putting $n = \dfrac 3 2$: