Context
For curiosity I wanted to calculate this integral: $$\displaystyle \int\frac{x^m}{(a^n\pm x^n)^p}\mathrm{d}x\qquad m\in\mathbb{N},\ n,p\in\mathbb{N}^{+}, \ a\in\mathbb{R}$$
For brevity:
- $m\mapsto m-1$
- $a=1$ (it's a simple scaling)
My work
I omit the intermediate steps because they are long and now I don't feel like re-transcribing them all in LaTeX, however I leave a Desmos link for anyone who wants to verify its accuracy at least from a graphic point of view
For $m,n,p\in\mathbb{N}^{+}$
$$\bbox[15px,#E0EFFF,border:5px groove #0058B3]{\int_0^x \frac{t^{m-1}}{(1\pm t^n)^p}\mathrm{d}t=\frac{x^{m}}{n}\sum_{j=1}^{p-1}\frac{\displaystyle\left[\prod_{k=j+1}^{p-1}\frac{nk-m}{nk}\right]}{j\left(1\pm x^{n}\right)^{j}}+(I(x)-J(x))\left[\prod_{k=1}^{p-1}\frac{nk-m}{nk}\right]}$$ Where: $$\bbox[2px,#E0EFFF,border:5px groove #0058B3]{\begin{matrix} \displaystyle I(x)=\frac{1}{n}\sum_{k=1}^{n}\left[\sin(m\theta)\operatorname{tan^{-1}}\left(x\sin(\theta),1\!-\!x\cos(\theta)\right)\!-\!\cos(m\theta)\ln\left(\sqrt{x^{2}\!-\!2\cos(\theta)x\!+\!1}\right)\right]\\ \begin{cases}\text{When }+:&J(x)=\displaystyle\sum_{k=1}^{\left\lfloor{\frac{m-1}{n}}\right\rfloor}(-1)^{k}\frac{x^{m-nk}}{m-nk}\qquad \theta:=\frac{2k-1}{n}\pi\\ \text{When }-:&\displaystyle J(x)=\sum_{k=1}^{\left\lfloor{\frac{m-1}{n}}\right\rfloor}\frac{x^{m-nk}}{m-nk}\quad\qquad\qquad \theta:=\frac{2k}{n}\pi \end{cases}\end{matrix}}$$
Question
Given that:
$$\displaystyle \int\frac{x^{m-1}}{(1\pm x^n)^p}\mathrm{d}x=\frac{x^m}{m}{}_2F_1\left(\left.{\frac{m}{n},p\atop1+\frac{m}{n}}\right|\mp x^n\right)\color{gray}{+C}\qquad \forall m$$ and $$\displaystyle \int\frac{1}{x^{m-1}(1\pm x^n)^p}\mathrm{d}x=-\frac{1}{mx^m}{}_2F_1\left(\left.{-\frac{m}{n},p\atop1-\frac{m}{n}}\right|\mp x^n\right)\color{gray}{+C}\qquad \forall m$$ it is possible to calculate the integral as well $$\displaystyle \int\frac{1}{x^{m-1}(1\pm x^n)^p}\mathrm{d}x\qquad \text{for }m\in\mathbb{N}^{+}$$ In terms of the functions already used above?
Essentially if there is a relation that connects the
$${}_2F_1\left(\left.{a_1,a_2\atop 1+a_1}\right| z\right)\quad\text{with}\quad {}_2F_1\left(\left.{-a_1,a_2\atop 1-a_1}\right| z\right)$$