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

Theorem

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

\dfrac 2 {\sqrt {a p} } \ln \left({\sqrt {p \left({a x + b}\right)} + \sqrt {a \left({p x + q}\right)} }\right) + C & : \dfrac {b p - a q} p > 0 \\ \dfrac 2 {\sqrt {a p} } \sinh^{-1} \sqrt {\dfrac {p \left({a x + b}\right)} {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: