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

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

Then:
 * $\displaystyle \int \frac {x^2 \ \mathrm d x} {\sqrt {a x^2 + b x + c} } = \frac {2 a x - 3 b} {4 a^2} \sqrt {a x^2 + b x + c} + \frac {3 b^2 - 4 a c} {8 a^2} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }$

Proof
First:

Then: