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

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Theorem

Let $a, b, p, q \in \R$ such that $a p \ne b q$.

Then:

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

\dfrac 2 {\sqrt {-a p} } \map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } + C & : a p < 0 \\ \dfrac {-1} {\sqrt {-a p} } \map \arcsin {\dfrac {2 a p x + b p + a q} {a q - b p} } + C & : a > 0, p < 0 \\ \ds \dfrac 2 {\sqrt {a p} } \ln \size {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C & : a p > 0 \\ \end {cases}$

for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.


Case where $a p = b q$

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

Proof

Case $1$: $a p < 0$

Let $a p < 0$.

Then:

$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \dfrac 2 {\sqrt {-a p} } \map \arctan {\sqrt {\dfrac {-p \paren {a x + b} } {a \paren {p x + q} } } } + C$

for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.

$\Box$


Case $2$: $a > 0, p < 0$

Let $a > 0$ and $p < 0$.

Then:

$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \dfrac {-1} {\sqrt {-a p} } \map \arcsin {\dfrac {2 a p x + b p + a q} {a q - b p} } + C$

for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.

$\Box$


Case $3$: $a p > 0$

Let $a, b, p, q \in \R$ such that $a p \ne b q$.

Let $a p > 0$.

Then:

$\ds \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \dfrac 2 {\sqrt {a p} } \ln \size {\sqrt {p \paren {a x + b} } + \sqrt {a \paren {p x + q} } } + C$

for all $x \in \R$ such that $\paren {a x + b} \paren {p x + q} > 0$.

$\blacksquare$


Sources

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