Primitive of Reciprocal of a x squared + b by Root of c x squared + d

Theorem

Let $a, b, c, d \in \R$ be real numbers such that $a d \ne b c$.

Then:

$\ds \int \dfrac {\d x} {\paren {a x^2 + b} \sqrt {c x^2 + d} } = \begin {cases}

\dfrac 1 {\sqrt {b \paren {a d - b c} } } \arctan \dfrac {x \sqrt {a d - b c} } {\sqrt {b \paren {c x^2 + d} } } + C & : a d > b c \\ \dfrac 1 {2 \sqrt {b \paren {a d - b c} } } \ln \size {\dfrac {\sqrt {b \paren {c x^2 + d} } + x \sqrt {b c - a d} } {\sqrt {b \paren {c x^2 + d} } - x \sqrt {b c - a d} } } + C & : b c > a d \end {cases}$

Proof

Case $1$: $a d > b c$

$\ds \int \dfrac {\d x} {\paren {a x^2 + b} \sqrt {c x^2 + d} } = \dfrac 1 {\sqrt {b \paren {a d - b c} } } \arctan \dfrac {x \sqrt {a d - b c} } {\sqrt {b \paren {c x^2 + d} } } + C$

$\Box$

Case $2$: $b c > a d$

$\ds \int \dfrac {\d x} {\paren {a x^2 + b} \sqrt {c x^2 + d} } = \dfrac 1 {2 \sqrt {b \paren {a d - b c} } } \ln \size {\dfrac {\sqrt {b \paren {c x^2 + d} } + x \sqrt {b c - a d} } {\sqrt {b \paren {c x^2 + d} } - x \sqrt {b c - a d} } } + C$

$\blacksquare$

Sources

• 1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Irrational Algebraic Functions