Primitive of Reciprocal of x squared by a x + b cubed

Theorem

 * $\ds \int \frac {\d x} {x^2 \paren {a x + b}^3} = \frac {-a} {2 b^2 \paren {a x + b}^2} - \frac {2 a} {b^3 \paren {a x + b} } - \frac 1 {b^3 x} + \frac {3 a} {b^4} \ln \size {\frac {a x + b} x} + C$