Primitives of Functions involving Root of a x + b and Root of p x + q

Theorem

This page gathers together the primitives of some functions involving $\sqrt {a x + b}$ and $\sqrt {p x + q}$.

Primitive of Reciprocal of $\sqrt {a x + b} \sqrt {p x + q}$

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 $\paren {a x + b} \paren {p x + q} > 0$.

Primitive of $x$ over $\sqrt {a x + b} \sqrt {p x + q}$

$\ds \int \frac {x \rd x} {\sqrt {\paren {a x + b} \paren {p x + q} } } = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } {a p} - \frac {b p + a q} {2 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$

Primitive of $\sqrt {a x + b} \sqrt {p x + q}$

$\ds \int \sqrt {\paren {a x + b} \paren {p x + q} } \rd x = \frac {2 a p x + b p + a q} {4 a p} \sqrt {\paren {a x + b} \paren {p x + q} } - \frac {\paren {b p - a q}^2} {8 a p} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$

Primitive of $\sqrt {p x + q}$ over $\sqrt {a x + b}$

$\ds \int \sqrt {\frac {p x + q} {a x + b} } \rd x = \frac {\sqrt {\paren {a x + b} \paren {p x + q} } } a + \frac {a q - b p} {2 a} \int \frac {\d x} {\sqrt {\paren {a x + b} \paren {p x + q} } }$

Primitive of Reciprocal of $\paren {p x + q} \sqrt {\paren {a x + b} \paren {p x + q} }$

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