How do we evaluate
$$\int_{0}^{\infty}x^n\ln\left(ae^x-b\over ae^x+b\right)\mathrm dx\tag1$$
$a\ge b$ and $n\ge0$
$$\int_{0}^{\infty}x^n\ln(ae^x-b)\mathrm dx-\int_{0}^{\infty}x^n\ln(ae^x+b)\mathrm dx\tag2$$
$$I_n=\int x^n\ln(ae^x-b)\mathrm dx\tag3$$
$$I_n={x^{n+1}\over n+1}\ln(ae^x-b)-{a\over n+1}\int {x^{n+1}\over ae^x-b}\mathrm dx\tag4$$
$$u={ae^x-b\over ae^x+b}$$
$$\mathrm dx={(ae^x+b)^2\over 2ab}\mathrm du$$
$${1\over 2ab}\int {(ae^x+b)^2\over e^x}\ln^n\left({b\over a}\cdot {1+u\over 1-u}\right)\mathrm du\tag5$$
Let $t=b/a$, where $0<t<1$. Then, write the logarithm term as
$$\log\left( \frac{ae^x-b}{ae^x+b}\right)=\log(1-te^{-x})-\log(1+te^{-x})\tag 1 $$
Expanding the right-hand side of $(1)$ in a Taylor series, we obtain
$$\log\left( \frac{ae^x-b}{ae^x+b}\right)=-\sum_{k=1}^\infty \frac{(te^{-x})^k}k +\sum_{k=1}^\infty \frac{(-te^{-x})^k}k \tag 2 $$
Using $(2)$, we find that
$$\begin{align} \int_0^\infty x^n \log\left( \frac{ae^x-b}{ae^x+b}\right)\,dx&=\int_0^\infty x^n\left(-\sum_{k=1}^\infty \frac{(te^{-x})^k}k +\sum_{k=1}^\infty \frac{(-te^{-x})^k}k \right)\,dx\\\\ &=-\sum_{k=1}^\infty \frac{t^k}{k}\int_0^\infty x^ne^{-kx}\,dx+\sum_{k=1}^\infty \frac{(-t)^k}{k}\int_0^\infty x^ne^{-kx}\,dx\\\\ &=-\sum_{k=1}^\infty \frac{t^k}{k^{n+2}}\underbrace{\int_0^\infty x^ne^{-x}\,dx}_{=n!}+\sum_{k=1}^\infty \frac{(-t)^k}{k^{n+2}}\underbrace{\int_0^\infty x^ne^{-x}\,dx}_{=n!}\\\\ &=n!\left(\underbrace{\sum_{k=1}^\infty \frac{(-t)^k}{k^{n+2}}}_{=\text{Li}_{n+2}(-t)}-\underbrace{\sum_{k=1}^\infty \frac{t^k}{k^{n+2}}}_{=\text{Li}_{n+2}(t)}\right)\\\\ &=n!\left(\text{Li}_{n+2}\left(-t\right)-\text{Li}_{n+2}\left(t\right)\right)\\\\ &=n!\left(\text{Li}_{n+2}\left(-\frac ba\right)-\text{Li}_{n+2}\left(\frac ba\right)\right) \end{align}$$