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

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

Then:
 * $\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \begin {cases}

\dfrac 1 {\sqrt a} \ln \size {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C & : b^2 - 4 a c > 0 \\ \dfrac 1 {\sqrt a} \map {\sinh^{-1} } {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \text { where } a > 0 \\ \dfrac {-1} {\sqrt {-a} } \map \arcsin {\dfrac {2 a x + b} {\sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c < 0 \text { where } a < 0 \\ \dfrac 1 {\sqrt a} \ln \size {2 a x + b} + C & : b^2 - 4 a c = 0 \end {cases}$

Proof
First:

Put:

Let $D = b^2 - 4 a c$.

Thus: