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

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

Then:
 * $\ds \int \dfrac {\d x} {\paren {a x^2 + b} \sqrt {c x^2 + d} } = \dfrac {\sqrt c} {2 a} \int \dfrac {\d u} {\paren {\sqrt {\paren {u - \frac d 2}^2 - \frac d 4} } \paren {u - \frac {a d + b c} a} }$

where $u := c x^2 + d$.

Proof
Hence: