Reduction formula for $\int\frac{dx}{(ax^2+b)^n}$ I recently stumbled upon the following reduction formula on the internet which I am so far unable to prove.
$$I_n=\int\frac{\mathrm{d}x}{(ax^2+b)^n}\\I_n=\frac{x}{2b(n-1)(ax^2+b)^{n-1}}+\frac{2n-3}{2b(n-1)}I_{n-1}$$
I tried the substitution $x=\sqrt{\frac ba}t$, and it gave me 
$$I_n=\frac{b^{1/2-n}}{a^{1/2}}\int\frac{\mathrm{d}t}{(t^2+1)^n}$$
To which I applied $t=\tan u$:
$$I_n=\frac{b^{1/2-n}}{a^{1/2}}\int\cot^{n-1}u\ \mathrm{d}u$$
I then used the $\cot^nu$ reduction formula to find 
$$I_n=\frac{-b^{1/2-n}}{a^{1/2}}\bigg(\frac{\cot^{n-2}u}{n-2}+\int\cot^{n-3}u\ \mathrm{d}u\bigg)$$
$$I_n=\frac{-b^{1/2-n}\cot^{n-2}u}{a^{1/2}(n-2)}-b^2I_{n-2}$$
Which is a reduction formula, but not the reduction formula. 
Could someone provide a derivation of the reduction formula? Thanks.

To prove that the integral $$I_n=\int\frac{\mathrm{d}x}{(ax^2+b)^n} $$
has a reduction formula
$$I_n=\frac{x}{2b(n-1)(ax^2+b)^{n-1}}+\frac{2n-3}{2b(n-1)}I_{n-1},$$
which is equivalent to $$
2 b(n-1) J_{n} -(2 n-3) J_{n-1} =\frac{x}{\left(a x^{2}+b\right)^{n-1}}
\begin{aligned}
\Leftrightarrow & \frac{d}{d x}\left[\frac{x}{\left(a x^{2}+b\right)^{n-1}}\right] =\frac{b-(2 n-3) a x^{2}}{\left(a x^{2}+b\right) ^n}
\end{aligned}
$$
In fact, using quotient rule prove that
$$
\begin{aligned}
& \frac{d}{d x}\left[\frac{x}{\left(a x^{2}+b\right)^{n-1}}\right] \\
=& \frac{\left(a x^{2}+b\right)^{n-1}-x(n-1)\left(a x^{2}+b\right)^{n-2}(2 a x)}{\left(a x^{2}+b\right)^{2 n-2}} \\
=& \frac{\left(a x^{2}+b\right)-x(n-1)(2 a x)}{\left(a x^{2}+b\right)^{n}} \\
=& \frac{b-(2 n-3) a x^{2}}{\left(a x^{2}+b\right)^{n}},
\end{aligned}
$$
which complete the proof.
