Primitive of Reciprocal of a x + b squared by p x + q

Theorem

 * $\displaystyle \int \frac {\mathrm d x} {\left({a x + b}\right)^2 \left({p x + q}\right)} = \frac 1 {b p - a q} \left({\frac 1 {a x + b} + \frac p {b p - a q} \ln \left\vert{\frac {p x + q} {a x + b} }\right\vert}\right) + C$