Primitive of Reciprocal of p squared plus Square of q by Hyperbolic Sine of a x

Theorem

 * $\ds \int \frac {\d x} {p^2 + q^2 \sinh^2 a x} = \begin {cases}

\dfrac 1 {a p \sqrt{q^2 - p^2} } \arctan \dfrac {\sqrt {q^2 - p^2} \tanh a x} p + C & : p^2 < q^2 \\ \dfrac 1 {2 a p \sqrt{p^2 - q^2} } \ln \size {\dfrac {p + \sqrt {p^2 - q^2} \tanh a x} {p - \sqrt {p^2 - q^2} \tanh a x} } + C & : p^2 > q^2 \\ \end {cases}$

Proof
Then:

There are two cases to address: $q^2 - p^2 > 0$ and $q^2 - p^2 < 0$.

First suppose that $q^2 - p^2 > 0$.

Then:

Now suppose that $q^2 - p^2 < 0$.

Then:

Also see

 * Primitive of $\dfrac 1 {p^2 + q^2 \cosh^2 a x}$