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

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

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

Proof
Let:

Then:

Let $u = z^2$.

Let:

Recall the result Primitive of $\dfrac 1 {\left({p x + q}\right)^n \sqrt{a x + b} }$:
 * $\displaystyle \int \frac {\mathrm d x} {\left({p x + q}\right)^n \sqrt{a x + b} } = \frac {\sqrt{a x + b} } {\left({n - 1}\right) \left({a q - b p}\right) \left({p x + q}\right)^{n-1} } + \frac {\left({2 n - 3}\right) a} {2 \left({n - 1}\right) \left({a q - b p}\right)} \int \frac {\mathrm d x} {\left({p x + q}\right)^{n-1} \sqrt{a x + b} }$

Let:

Then: