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

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

Then:
 * $\displaystyle \int x^2 \sqrt {a x^2 + b x + c} \ \mathrm d x = \frac {6 a x - 5 b} {24 a^2} \left({\sqrt {a x^2 + b x + c} }\right)^3 + \frac {5 b^2 - 4 a c} {16 a^2} \int \sqrt {a x^2 + b x + c} \ \mathrm d x$