Do you guys have any ideas on solving the following integral: $$ \int_{0}^{D} z \cdot (\sqrt{1+z^{a}})^{b} \cdot \ln(\sqrt{1+z^{a}})\; \mathrm dz $$ with $a, b$ being constants.
I was trying to use the formula (4.253.1) (shown below) of the book Table of Integrals, Series, and Products, 8th edition, to solve the integral by applying change of random variable, but it seems not to work.
Thanks in advance.
Concerning the antiderivative $$I=\int z \, (\sqrt{1+z^{a}})^{b} \, \ln(\sqrt{1+z^{a}})\, dz$$ let $$\sqrt{1+z^{a}}=t \implies z=\left(t^2-1\right)^{\frac{1}{a}}\implies dz=\frac{2 t \left(t^2-1\right)^{\frac{1}{a}-1}}{a}dt$$ which makes $$I=\frac 2a \int \left(t^2-1\right)^{\frac{2}{a}-1} t^{b+1} \log (t)\,dt$$ and then $$I=\frac{2 }{a \,(2+b)^2}\left(1-t^2\right)^{-2/a} \left(t^2-1\right)^{2/a} t^{b+2}\,J$$where $$J= \, _3F_2\left(1-\frac{2}{a},\frac{b}{2}+1,\frac{b}{2}+1;\frac{b}{2}+2,\frac{b}{2}+2 ;t^2\right)-$$ $$(b+2) \log (t) \, _2F_1\left(1-\frac{2}{a},\frac{b}{2}+1;\frac{b}{2}+2;t^2\right)$$
where appear hypergeometric functions.
When $z\to 0$, $t \to 1$ and $$\lim_{t\to 1} \, I=-\frac{(-1)^{-2/a} \Gamma \left(\frac{2}{a}\right) \Gamma \left(\frac{b}{2}+2\right) \left(\psi ^{(0)}\left(\frac{b}{2}+1\right)-\psi ^{(0)}\left(\frac{b}{2}+\frac{2}{a}+1\right)\right)}{a (b+2) \Gamma \left(\frac{b}{2}+\frac{2}{a}+1\right)}$$