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

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

Then:


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

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

Proof
Let:

Then:

Let $c^2 > 0$.

Then:

Let $c^2 < 0$.

Then: