Primitive of Reciprocal of Root of a x + b by Root of p x + q/Lemma 2

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

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

\ds \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 - \paren {\frac {b p - a q} p} } } & : a p > 0 \\ \ds \frac 2 {\sqrt {-a p} } \int \frac {\d u} {\sqrt {\paren {\frac {b p - a q} p} - u^2} } & : a p < 0 \end {cases}$

where:

$u := \sqrt {a x + b}$

Proof

Let us make the substitution:

$u = \sqrt {a x + b}$

Lemma

Let $u = \sqrt {a x + b}$.

Then:

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

$\Box$

Case $1

\quad a p > 0$

\(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\)

\(=\)

\(\ds \int \frac {2 u \rd u} {a \sqrt {\frac p a} \sqrt {u^2 - \paren {\frac {b p - a q} p} } u}\)

Primitive of Function of $\sqrt {p x + q}$

\(\ds \)

\(=\)

\(\ds \frac 2 {\sqrt {a p} } \int \frac {\d u} {\sqrt {u^2 - \paren {\frac {b p - a q} p} } }\)

Primitive of Constant Multiple of Function

$\Box$

Case $2

\quad a p < 0$

We have:

\(\ds a p\)

\(<\)

\(\ds 0\)

by hypothesis

\(\ds \leadsto \ \ \)

\(\ds \dfrac p a\)

\(<\)

\(\ds 0\)

Then:

\(\ds \sqrt {p x + q}\)

\(=\)

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

from $(1)$

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

Then:

\(\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }\)

\(=\)

\(\ds \dfrac 2 p \int \frac {u \rd u} {\sqrt {-\frac p a} \paren {\sqrt {\paren {\frac {b p - a q} p} - u^2} } u}\)

Primitive of Function of $\sqrt {p x + q}$

\(\ds \)

\(=\)

\(\ds \frac 2 {\sqrt {-a p} } \int \frac {\d u} {\sqrt {\paren {\frac {b p - a q} p} - u^2} }\)

Primitive of Constant Multiple of Function and simplifying

$\blacksquare$