calculus - Understanding the solution of $intleft(1-x^{p}right)^{frac{n-1}{p}}logleft(1-x^{p}right)dx$

I would like to solve the integral

$$\int\left(1-x^{p}\right)^{\frac{n-1}{p}}\log\left(1-x^{p}\right)dx.$$

My problem is that I arrived at a solution via wolfram alpha, but I would like to understand how one would arrive there by hand.

What I did is to use the substitution $x=\left(1-\exp(z)\right)^{\frac{1}{p}}$. This yields
$$-\frac{1}{p}\int\left(1-\exp(z)\right)^{\frac{1}{p}-1}\exp\left(\frac{n-1+p}{p}z\right)zdz$$ for which wolfram alpha yields the integral

\begin{align}
&\frac{\exp\left(z\frac{(n+p-1)}{p}\right)}{\left(n+p-1\right)^{2}}p\,_{3}F_{2}\left(1-\frac{1}{p},\frac{n}{p}-\frac{1}{p}+1,\frac{n}{p}-\frac{1}{p}+1;\frac{n}{p}-\frac{1}{p}+2,\frac{n}{p}-\frac{1}{p}+2;\exp(z)\right)\\&-z\frac{\exp\left(z\frac{(n+p-1)}{p}\right)}{\left(n+p-1\right)}\,_{2}F_{1}\left(\frac{p-1}{p},\frac{n+p-1}{p};\frac{n+2p-1}{p};\exp z\right).
\end{align}

I would really like to understand how it arrived at the generalized hypergeometric function in the solution because I think it would be useful to spot it in an integral (like spotting possible solutions in terms of a gamma function or beta function in integrals).

Does anybody have a hint how it arrived at the solution?

Answer

You made a substitution $u=1-x^p$, transforming the integral as
$$
\int \left(1-x^p\right)^{(n-1)/p} \log\left(1-x^p\right) \mathrm{d} x = - \frac{1}{p} \int \left(1-u\right)^a u^b \log\left(u\right) \mathrm{d} u
$$
where $a = \frac{1}{p}-1$ and $b=\frac{n-1}{p}$. In order to integrate this we shall take advantage of the following differentiation property of the Gauss hypergeometric function:
$$

\left( \frac{1}{\alpha_1} z \frac{\mathrm{d}}{\mathrm{d} z} + 1 \right) {}_2 F_1\left( \left. \begin{array}{cc} \alpha_1 & \alpha_2 \cr \beta&- \end{array} \right| z \right) = {}_2 F_1\left( \left. \begin{array}{cc} \alpha_1 +1 & \alpha_2 \cr \beta& - \end{array} \right| z \right)
$$
Choosing $\beta=\alpha_1 + 1$ we get
$$
\left( \frac{1}{\alpha_1} z \frac{\mathrm{d}}{\mathrm{d} z} + 1 \right) {}_2 F_1\left( \left. \begin{array}{cc} \alpha_1 & \alpha_2 \cr &\alpha_1+1& \end{array} \right| z \right) = {}_1 F_0\left( \left. \begin{array}{c} \alpha_2 \\ - \end{array} \right| z \right) = \left(1-z\right)^{-\alpha_2} \tag{1}
$$
Now, using
$$\left( \frac{1}{\alpha_1} z \frac{\mathrm{d}}{\mathrm{d} z} + 1 \right) f(z) = \frac{1}{\alpha_1} z^{1-\alpha_1} \frac{\mathrm{d}}{\mathrm{d} z} \left( z^{\alpha_1} f(z) \right) \tag{2}
$$
Combining eqs. $(1)$ and $(2)$ and choosing $\alpha_2 = -a$ and $\alpha_1 = b +1$ we arrive at

$$
\int \left(1-u\right)^a u^b \mathrm{d}u = \frac{u^{b+1}}{b+1} {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) \tag{3}
$$
We can now integrate by parts:
$$ \begin{eqnarray}
\int \left(1-u\right)^a u^b \log(u) \mathrm{d}u &=& \frac{u^{b+1}}{b+1} \log\left(u \right) \cdot {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) \\ && - \frac{1}{b+1} \int u^b \cdot {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) \mathrm{d}u
\end{eqnarray}
$$
The latter integral is integrated in a similar way by using differentiation properties of generalized hypergeometric function:
$$

\frac{\mathrm{d}}{\mathrm{d} z} \left( z^{\alpha_1} \cdot {}_3 F_2\left( \left. \begin{array}{ccc} \alpha_1 & \alpha_2 & \alpha_3 \cr \beta_1& \beta_2 & \end{array} \right| z \right) \right) = \alpha_1 z^{\alpha_1-1} \cdot {}_3 F_2\left( \left. \begin{array}{ccc} \alpha_1+1 & \alpha_2 & \alpha_3 \cr \beta_1& \beta_2 & \end{array} \right| z \right)
$$
we get
$$\begin{eqnarray}
\int \left(1-u\right)^a u^b \log(u) \, \mathrm{d} u &=& \frac{u^{b+1}}{b+1} \log\left(u \right) \cdot {}_2 F_1\left( \left. \begin{array}{cc} -a & b+1 \cr &b+2& \end{array} \right| u \right) \\ && - \frac{u^{b+1}}{(b+1)^2} \cdot {}_3 F_2\left( \left. \begin{array}{ccc} -a & b+1 & b+1 \cr b+2& b+2 & \end{array} \right| u \right)
\end{eqnarray} \tag{4}
$$