Is it hard to evaluate the integral $\int_{0}^{1} \frac{x^{3} \ln \left(\frac{1+x}{1-x}\right)}{\sqrt{1-x^{2}}} d x$?

When I encountered this question, I just wondered whether there is a simple way to tackle the integral. After trying many substitutions, I still failed. Suddenly, integration by parts come to my mind and helped me solve the problem. I first integrate the part $$\dfrac{x^{3}}{\sqrt{1-x^{2}}}$$ and then evaluate it via integration by parts.

Are there any alternate methods other than integration by parts?

$ \text{We are going to evaluate the integral}$ $\displaystyle I=\int_{0}^{1} \frac{x^{3} \ln \left(\frac{1+x}{1-x}\right)}{\sqrt{1-x^{2}}} d x. \tag*{}$ $\textrm{using integration by parts only.}$

$\textrm{Noting that}$ $\displaystyle \quad \frac{d}{d x} \ln \left(\frac{1+x}{1-x}\right)=\frac{2}{1-x^{2}},\tag*{}$ $\textrm{we prefer to using integration by parts.}$ $\begin{align*} \displaystyle I&= \quad \int \frac{x^{3}}{\sqrt{1-x^{2}}} d x \\&=\displaystyle \int \frac{x-x\left(1-x^{2}\right)}{\sqrt{1-x^{2}}} d x \\&=\displaystyle \int \frac{x}{\sqrt{1-x^{2}}} d x-\int x \sqrt{1-x^{2}} d x \\&=\displaystyle -\sqrt{1-x^{2}}+\frac{\left(1-x^{2}\right)^{\frac{3}{2}}}{3}+c \\\displaystyle &=-\frac{\sqrt{1-x^{2}}}{3}\left(2+x^{2}\right)+c\end{align*} \tag*{} $ $\textrm{We can now start the evaluation.}$ $ \displaystyle \begin{aligned}I\displaystyle =&\int_{0}^{1} \ln \left(\frac{1+x}{1-x}\right) d\left(-\frac{\sqrt{1-x^{2}}}{3}\left(2+x^{2}\right)\right) \\\displaystyle =&-\left[\frac{\sqrt{1-x^{2}}}{3}\left(2+x^{2}\right)\ln \left(\frac{1+x}{1-x}\right)\right]_{0}^{1}+\int_{0}^{1} \frac{\sqrt{1-x^{2}}}{3}\left(2+x^{2}\right)\left(\frac{2}{1-x^{2}}\right) d x \\ \displaystyle =&\frac{2}{3} \int_{0}^{1} \frac{2+x^{2}}{\sqrt{1-x^{2}}} d x\end{aligned} \tag*{}$

$\text {As usual, we let } x=\sin \theta ,$ $\displaystyle \begin{aligned} \int_{0}^{1} \frac{2+x^{2}}{\sqrt{1-x^{2}}} d x=&\int_{0}^{\frac{\pi}{2}} \frac{2+\sin ^{2} \theta}{\sqrt{1-\sin ^{2} \theta}} \cdot \cos \theta d \theta \\&=[2 \theta]_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}} \frac{1-\cos 2 \theta}{2} d \theta \\&=\pi+\frac{1}{2}\left[\theta-\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{2}} \\&=\frac{5 \pi}{4}\end{aligned} \tag*{} $

We can now conclude that $\displaystyle \boxed{\int_{0}^{1} \frac{x^{3} \ln \left(\frac{1+x}{1-x}\right)}{\sqrt{1-x^{2}}} d x =\frac{2}{3}\cdot \frac{5 \pi}{4}=\frac{5 \pi}{6}}\tag*{} $