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:

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

We have:

$\ds u$

$=$

$\ds \sqrt {a x + b}$

$\ds \leadsto \ \ $

$\ds x$

$=$

$\ds \frac {u^2 - b} a$

$\ds \leadsto \ \ $

$\ds \sqrt {p x + q}$

$=$

$\ds \sqrt {p \paren {\frac {u^2 - b} a} + q}$

$\ds $

$=$

$\ds \sqrt {\frac {p \paren {u^2 - b} + a q} a}$

$\ds $

$=$

$\ds \sqrt {\frac {p u^2 - b p + a q} a}$

$\ds $

$=$

$\ds \sqrt {\paren {\frac p a} \paren {u^2 - \paren {\frac {b p - a q} p} } }$

$\blacksquare$