Primitive of Reciprocal of p x + q by Root of a x + b/p (b p - a q) greater than 0/Also presented as

Primitive of $\frac 1 {\paren {p x + q} \sqrt {a x + b} }$ where $p \paren {b p - a q} > 0$: Also presented as
This result can also be seen presented as:


 * $\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \frac 1 {\sqrt {b p - a q} \sqrt p} \ln \size {\frac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C$

but this presupposes both that $p > 0$ and $b p - a q > 0$.