Primitive of Reciprocal of Root of a x + b by Root of p x + q/Lemma 1

Lemma for Primitive of $\frac 1 {\sqrt {\paren {a x + b} \paren {p x + q} } }$
Let $u = \sqrt {a x + b}$.

Then:
 * $\ds \sqrt {p x + q} = \sqrt {\paren {\frac p a} \paren {u^2 - \paren {\frac {b p - a q} p} } }$

Proof
We have: