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

Theorem

 * $\displaystyle \int \frac {x \ \mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } = \frac {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} } {a p} - \frac {b p + a q} {2 a p} \int \frac {\mathrm d x} {\sqrt{\left({a x + b}\right) \left({p x + q}\right)} }$

Proof
Let:

Let $c^2 = \dfrac {b p - a q} p$.

Then: