Primitive of x squared over Cube of Root of a x squared plus b x plus c

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

Then:
 * $\displaystyle \int \frac {x^2 \ \mathrm d x} {\left({\sqrt {a x^2 + b x + c} }\right)^3} = \frac {\left({2 b^2 - 4 a c}\right) x + 2 b c} {a \left({4 a c - b^2}\right) \sqrt {a x^2 + b x + c} } + \frac 1 a \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }$