Primitive of Reciprocal of p squared minus square of q by Sine of a x

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {p^2 - q^2 \sin^2 a x} = \begin{cases}

\displaystyle \frac 1 {a p \sqrt{p^2 - q^2} } \arctan \frac {\sqrt{p^2 - q^2} \tan a x} p & : p^2 > q^2 \\ \displaystyle \frac 1 {2 a p \sqrt{q^2 - p^2} } \ln \left\vert{\frac {\sqrt{q^2 - q^p} \tan a x + p} {\sqrt{q^2 - q^p} \tan a x - p} }\right\vert & : p^2 < q^2 \end{cases}$