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

Theorem
Let $a, b, p, q \in \R$ such that $a p \ne b q$ and such that $p \ne 0$.

Then:


 * $\ds \int \frac {\d x} {\paren {p x + q} \sqrt {a x + b} } = \begin {cases}

\dfrac 1 {\sqrt {p \paren {b p - a q} } } \ln \size {\dfrac {\sqrt {p \paren {a x + b} } - \sqrt {b p - a q} } {\sqrt {p \paren {a x + b} } + \sqrt {b p - a q} } } + C & : p \paren {b p - a q} > 0 \\ \dfrac 2 {\sqrt {p \paren {a q - b p} } } \arctan \sqrt {\dfrac {p \paren {a x + b} } {a q - b p} } + C & : p \paren {b p - a q} < 0 \\ \end {cases}$