Primitive of Reciprocal of x by square of a x squared plus b x plus c

Theorem
Let $a \in \R_{\ne 0}$.

Then:
 * $\displaystyle \int \frac {\mathrm d x} {x \left({a x^2 + b x + c}\right)^2} = \frac 1 {2 c \left({a x^2 + b x + c}\right)} - \frac b {2 c} \int \frac {\mathrm d x} {\left({a x^2 + b x + c}\right)^2} - \frac 1 c \int \frac {\mathrm d x} {x \left({a x^2 + b x + c}\right)}$