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

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

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

Proof
Let:

Then: