Primitive of Half Integer Power of a x squared plus b x plus c

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

Then:
 * $\displaystyle \int \left({a x^2 + b x + c}\right)^{n + \frac 1 2} \ \mathrm d x = \frac {\left({2 a x + b}\right) \left({a x^2 + b x + c}\right)^{n + \frac 1 2} } {4 a \left({n + 1}\right)} + \frac {\left({2 n + 1}\right) \left({4 a c - b^2}\right)} {8 a \left({n + 1}\right)} \int \left({a x^2 + b x + c}\right)^{n - \frac 1 2} \ \mathrm d x$

Proof
Let:

Also let $q = 4 a c - b^2$.

Then:

From Primitive of $\dfrac {\left({p x + q}\right)^n} {\sqrt {a x + b} }$:
 * $\displaystyle \int \frac {\left({p x + q}\right)^n} {\sqrt{a x + b} } \ \mathrm d x = \frac {2 \left({p x + q}\right)^n \sqrt{a x + b} } {\left({2 n + 1}\right) a} + \frac {2 n \left({a q - b p}\right)} {\left({2 n + 1}\right) a} \int \frac {\left({p x + q}\right)^{n-1} } {\sqrt{a x + b} } \ \mathrm d x$

Here $p = 1, a = 1, b = 0$ and $n := n + \dfrac 1 2$: