Primitive of Reciprocal of Square of p plus q by Hyperbolic Cosine of a x

Theorem

 * $\ds \int \frac {\d x} {\paren {p + q \cosh a x}^2} = \frac {q \sinh a x} {a \paren {q^2 - p^2} \paren {p + q \cosh a x} } - \frac p {q^2 - p^2} \int \frac {\rd x} {p + q \cosh a x} + C$

Proof
Hence the result.

Also see

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