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

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

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

Proof
First:

Then: