Reduction formula for integral of $\int \frac{1}{x^2 \sqrt{ax^2+bx+c}} dx $ The following reduction formula is taken from http://www.sosmath.com/tables/integral/integ15/integ15.html:
$$\int \frac{1}{x^2 \sqrt{ax^2+bx+c}} dx = -\frac{\sqrt{ax^2+bx+c}}{cx} - \frac{b}{2c} \int \frac{1}{x \sqrt{ax^2+bx+c}} dx$$


*

*I've been trying to derive this reduction formula myself, but without success. Can anyone point me in the right direction?

*Are there similar formulae for higher exponents of the $x$ in the denominator? Or even better, for general integer exponents $n$?

 A: One may observe that
$$
\begin{align}
\left(-\frac{\sqrt{ax^2+bx+c}}{cx}\right)'&=\frac{\sqrt{ax^2+bx+c}}{c x^2}-\frac{2ax+b}{2cx\sqrt{ax^2+bx+c}}
\\\\&=\frac{2(ax^2+bx+c)-(2ax+b)x}{2cx^2 \sqrt{ax^2+bx+c}}
\\\\&=\frac{bx+2c}{2cx^2 \sqrt{ax^2+bx+c}}
\\\\&=\frac{b}{2c}\cdot\frac{1}{x \sqrt{ax^2+bx+c}}+\frac{1}{x^2 \sqrt{ax^2+bx+c}}
\end{align}
$$ which yields the first result.
This might be generalized to get 
$$
\begin{align}
\left(-\frac{\sqrt{ax^2+bx+c}}{cx^{\color{red}{n}+1}}\right)'&=\frac{a\cdot \color{red}{n}}{cx^{\color{red}{n}}\sqrt{ax^2+bx+c}}+\frac{b}{2c}\cdot\frac{(2\color{red}{n}+1)}{x^{\color{red}{n}+1} \sqrt{ax^2+bx+c}}+\frac{\color{red}{n}+1}{x^{\color{red}{n}+2}\sqrt{ax^2+bx+c}}.
\end{align}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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It's convenient to rewrite the original question as
  $$
\int{\dd x \over x^{2}\root{ax^{2} + bx + c}} =
{1 \over \root{a}}
\color{#f00}{\int{\dd x \over x^{2}\root{x^{2} + 2px + q}}}\,,\qquad
p \equiv {b \over 2a}\,,\quad q \equiv {c \over a}
$$

Lets
$\ds{t = \root{x^{2} + 2px + q} - x \iff
x = -\,{1 \over 2}\,{t^{2} - q \over t - p}}$:
\begin{align}
&\color{#f00}{\int{\dd x \over x^{2}\root{x^{2} + 2px + q}}}  =
-4\int{t - p \over \pars{t^{2} - q}^{2}}\,\dd t =
-\,{2 \over q}\,{pt - q \over t^{2} - q} + {2p \over q^{3/2}}\,
\,\mrm{arctanh}\pars{t \over \root{q}}
\end{align}
Replace $\ds{t = \root{x^{2} + 2px + q} - x}$ in the right hand side.
