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

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

Let $a d > b c$.

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