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

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

Then:
 * $\displaystyle \int x \sqrt {a x^2 + b x + c} \ \mathrm d x = \frac {\left({\sqrt {a x^2 + b x + c} }\right)^3} {3 a} - \frac {b \left({2 a x + b}\right) \sqrt {a x^2 + b x + c} } {8 a^2} - \frac {b \left({4 a c - b^2}\right)} {16 a^2} \int \frac {\mathrm d x} {\sqrt {a x^2 + b x + c} }$

Proof
Let:

So: