How to find reduction formula for $I_n=\int\frac{(px+q)^n}{\sqrt{ax+b}}dx$ I have to find the reduction formula for the following :
$$\int\frac{(px+q)^n}{\sqrt{ax+b}}dx$$
I took this integral from wikipedia. By using parts this is what I got.
${2a^{-1}(px+q)^{n}\sqrt{ax+b}-2pna^{-1}\int\{px+q}^{n-1}\sqrt{ax+b}dx$
The answer is in this link in reduction formula table:
https://en.m.wikipedia.org/wiki/Integration_by_reduction_formulae
Any help is appreciated. Thanks in advance.
 A: Hint: Perform integration-by-parts suggested below
$$I_n= \int\frac{(px+q)^n}{\sqrt{ax+b}}dx
=\frac{1}{a(n+\frac12)} \int \left( \frac{px+q}{ax+b}\right)^n d\left( 
(ax+b)^{n+\frac12}\right)
$$
A: We first perform integration by parts on $\sqrt{ax+b}$.
\begin{aligned}
I_{n} &=\int \frac{(p x+q)^{n}}{\sqrt{a x+b}} d x \\
&=\frac{2}{a} \int(p x+q)^{n} d(\sqrt{a x+b}) \\
&=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a} \int(p x+q)^{n-1} \sqrt{a x+b} d x
\end{aligned}
Rationalization gives back our integrals. $$
\begin{aligned}
I_{n} &=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a} \int \frac{(p x+q)^{n-1}(a x+b)}{\sqrt{a x+b}} d x \\
&=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a} \int \frac{(p x+q)^{n-1}\left[\frac{a}{p}(p x+q)+b-\frac{a q}{p}\right]}{\sqrt{a x+b}} d x \\
&=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}-\frac{2 pn}{a}\left(\frac{a}{p} I_{n}+\frac{b p-a q}{p} I_{n-1}\right)
\end{aligned}
$$
Rearranging gives
$$
\begin{aligned}
\left(2n+1\right) I_{n} &=\frac{2(p x+q)^{n}}{a \sqrt{a x+b}}+\frac{2 n(a q-bp)}{a } I_{n-1} \\
I_{n} &=\frac{2 p(p x+q)^{n}}{(2 n+1)\sqrt{a x+b}}+\frac{2 n(aq-b p)}{2 n+1} I_{n-1},
\end{aligned}
$$
which is the reduction formula of $I_n$.
