How to find reduction formula for this integral? So, I'm practicing  reduction formula and so I searched wikipedia for some integrals. I came across this one and even after trying it for hours,I couldn't find a reduction formula for it.
$$\int\frac{dx}{{[{ax+b}]^{m}}{{[p x+q}]^{n}}}$$
I have no clue how to do it. I tried integration by parts but it was useless.
Answer is in this link https://en.m.wikipedia.org/wiki/Integration_by_reduction_formulae in the tables of integral reduction. I would've  pasted the answer but couldn't because my phone is not supporting it. Any hint is appreciated. Thanks in advance.
 A: WLOG, $a=c=1$. By parts,
$$\int\frac{dx}{(x+b)^m(x+d)^n}=
\\-\frac1{(m-1)(x+b)^{m-1}(x+d)^n}-\frac{n}{m-1}\int\frac{dx}{(x+b)^{m-1}(x+d)^{n+1}}.$$
Notice that $m$ decreases. When it reaches $1$, you have
$$\int\frac{dx}{(x+b)(x+d)^k}=\frac1{b-d}\int\frac{(x+b-x-d)dx}{(x+b)(x+d)^k}=\\\frac1{b-d}\int\frac{dx}{(x+d)^k}-\frac1{b-d}\int\frac{dx}{(x+b)(x+d)^{k-1}}.$$
The first terms integrates immediately and the second is reduced.
A: Proceed with IBP as follows
\begin{align}
I_{m,n}=& \int\frac{dx}{{({ax+b})^{m}}{{(p x+q})^{n}}}\\
=& \frac{1}{(m-1)(bp-aq)}\int \frac1{(p x+q)^{n+m-2}}d\left( \frac{(px+q)^{m-1}}{({ax+b})^{m-1}}\right)\\
= & \frac{1}{(m-1)(bp-aq)}\left( \frac{1}{(ax+b)^{m-1}{{(p x+q})^{n-1}}}-\int \frac{(px+q)^{m-1}}{({ax+b})^{m-1}} d\left(\frac1{(p x+q)^{n+m-2}}\right)\right)\\
=  & \frac{1}{(m-1)(bp-aq)}\left( \frac{1}{(ax+b)^{m-1}{{(p x+q})^{n-1}}}+p(m+n-2)I_{m-1, n} \right)\\
\end{align}
Alternatively
$$I_{m,n} =\frac{1}{(n-1)(aq-bp)}\left( \frac{1}{(ax+b)^{m-1}{{(p x+q})^{n-1}}}+a(m+n-2)I_{m, n-1} \right)
$$
A: You know the form of the formula you are trying to get:
$$
\int\frac{dx}{{[{ax+b}]^{m}}{{[p x+q}]^{n}}} = \frac{\alpha}{[ax+b]^\beta [px+q]^\gamma} + \delta\int\frac{dx}{{[{ax+b}]^{m-1}}{{[p x+q}]^{n}}}.
$$
You differentiate it:
$$
\frac1{[{ax+b}]^{m}{[p x+q}]^{n}} =\\ -\frac{\alpha\beta a}{[ax+b]^{\beta+1} [px+q]^\gamma}
-\frac{\alpha\gamma p}{[ax+b]^{\beta} [px+q]^{\gamma+1}} + \frac{\delta}{[{ax+b}]^{m-1}{[p x+q}]^{n}},\\
1 =\delta(ax+b) - \alpha\beta a (ax+b)^{m-\beta-1}(px+q)^{n-\gamma}
- \alpha\gamma p (ax+b)^{m-\beta}(px+q)^{n-\gamma-1}.
$$
The smallest order of the polynomial we can do is linear. By setting $m=\beta+1$ and $n=\gamma+1$:
$$
1 = \delta(ax+b) - \alpha(m-1)a(px+q) - \alpha(n-1)p(ax+b).
$$
You know that the polynomial on the right has two parameters $\alpha,\delta$ and should be equal to $1+0x$. It allows you to find these parameters.
