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

Theorem

 * $\displaystyle \int \frac {\d x} {\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} } } = \frac {2 \sqrt{a x + b} } {\paren {a q - b p} \sqrt {p x + q} } + C$

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

Putting $n = \dfrac 3 2$: