Evaluate the Integral \(\int x\left(1+ax^{-k}\right)^{-m} \mathrm{d}x\)

How to evaluate
\int x\left(1+\frac{a}{x^{k}}\right)^{-m} \mathrm{d}x
was part of a question at Mathematics Stack Exchange. Here is my solution.

Let \(y = -\frac{a}{x^{k}}\)
\int x\left(1+\frac{a}{x^{k}}\right)^{-m} \mathrm{d}x &=
\frac{1}{k} (-1)^{1+\frac{2}{k}} a^{\frac{2}{k}} \int (1-y)^{-m} y^{-1-\frac{2}{k}} \mathrm{d}y \\
&= \frac{1}{k} (-1)^{1+\frac{2}{k}} a^{\frac{2}{k}} \mathrm{B}_{y} \left(-\frac{2}{k},1-m \right) \\
&= \frac{1}{k} (-1)^{1+\frac{2}{k}} a^{\frac{2}{k}} y^{-\frac{2}{k}} (-1) \frac{k}{2}
\,{}_{2}\mathrm{F}_{1} \left(-\frac{2}{k},m;1-\frac{2}{k};y \right) \\
&= \frac{x^{2}}{2} \,{}_{2}\mathrm{F}_{1} \left(-\frac{2}{k},m;1-\frac{2}{k};-\frac{a}{x^{k}} \right)

\mathrm{B}_{z}(p,q) = \int_{0}^{z} t^{p-1} (1-t)^{q-1} \mathrm{d}t
is the incomplete beta function.
\mathrm{B}_{z}(p,q) = \frac{z^{p}}{p} \,{}_{2}\mathrm{F}_{1}(p,1-q;p+1;z)
is the incomplete beta function in terms of Gauss’s hypergeometric function.

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