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

Theorem

 * $\ds \int \frac {\d x} {\paren {a x + b}^2 \paren {p x + q} } = \frac 1 {b p - a q} \paren {\frac 1 {a x + b} + \frac p {b p - a q} \ln \size {\frac {p x + q} {a x + b} } } + C$