An exercise from my brother: $\int_{-1}^1\frac{\ln (2x-1)}{\sqrt[\large 6]{x(1-x)(1-2x)^4}}\,dx$

My brother asked me to calculate the following integral before we had dinner and I have been working to calculate it since then ($\pm\, 4$ hours). He said, it has a beautiful closed form but I doubt it and I guess he has tried to trick me again (as usual). So I am curious, what is the closed form (if any) of the following integral:

\begin{equation} \int_{-1}^1\frac{\ln (2x-1)}{\sqrt[\large 6]{x(1-x)(1-2x)^4}}\,dx \end{equation}

I have tried by parts method, partial fractions (stupid idea), converting into series (nothing familiar), many substitutions such as: $u=2x-1$, $u=1-x$, $x=\cos^2\theta$, etc, but I failed and got nothing. Wolfram Alpha also doesn't give an answer. Either he is lying to me or telling the truth, I don't know. Could anyone here please help me to obtain the closed form of the integral with any methods (whatever it takes)? Any help would be greatly appreciated. Thank you.

Too long for a comment and this is not the answer too, but following @David H's comment. Perhaps your brother means

$$\int_{\color{red}{\large\frac12}}^1\frac{\ln (2x-1)}{\sqrt[\large 6]{x(1-x)(1-2x)^4}}\ dx.\tag1$$

If so, then setting $u=2x-1$ to $(1)$ yields $$ \frac1{\sqrt[\large 3]{4}}\int_{0}^1\frac{\ln u}{\sqrt[\large 6]{(1+u)(1-u)u^4}}\ du=\frac1{\sqrt[\large 3]{4}}\int_{0}^1\frac{\ln u}{\sqrt[\large 6]{(1-u^2)u^4}}\ du.\tag2 $$ Setting $t=u^2$ yields $$ \frac1{\sqrt[\large 3]{128}}\int_{0}^1\frac{\ln t}{(1-t)^{\large\frac16}t^{\large\frac56}}\ dt=\frac1{\sqrt[\large 3]{128}}\int_{0}^1(1-t)^{-\large\frac16}t^{\large-\frac56}\ln t\ dt.\tag3 $$

Now, consider beta function $$ \text{Beta}(x,y)=\int_{0}^1t^{\large x-1}(1-t)^{\large y-1}\ dt.\tag4 $$ Differentiating $(4)$ with respect to $x$ yields \begin{align} \frac{\partial}{\partial x}\text{B}(x,y)&=\int_0^1\frac{\partial}{\partial x}\left(t^{\large x-1}(1-t)^{\large y-1}\right)\ dt\\ (\psi(x)-\psi(x+y))\text{B}(x,y)&=\int_0^1 t^{\large x-1}(1-t)^{\large y-1}\ln t\ dt,\tag5 \end{align} where $\psi(\cdot)$ is the digamma function.

Using $(5)$ then $(3)$ turns out to be $$ \frac1{\sqrt[\large 3]{128}}\left(\psi\left(\frac16\right)-\psi\left(1\right)\right)\text{B}\left(\frac16,\frac56\right)=\large\color{blue}{-\frac\pi{\sqrt[\large 3]{256}}\left(\pi\sqrt{3}+4\ln2+3\ln3\right)}. $$ Either your brother is teasing you or he just misplaces the interval of integral.

I highly doubt that one can evaluate that integral by hand, however the integral does have a closed form found by Mathematica to be: $$\frac{2 \left(4 (-1)^{5/6} \, _4F_3\left(1,1,1,\frac{7}{6};2,2,2;\frac{1}{9}\right)+8 (-1)^{5/6} \log (3) \, _3F_2\left(1,1,\frac{7}{6};2,2;\frac{1}{9}\right)+432 (-1)^{5/6} \log ^2(2)-189 (-1)^{5/6} \log ^2(3)+648 (-1)^{5/6} \log (3) \log (2)-54 i \psi ^{(1)}\left(\frac{5}{6}\right)+54 \sqrt{3} \psi ^{(1)}\left(\frac{5}{6}\right)\right)+\pi ^2 \left(-963 \sqrt{3}+1827 i\right)-54 i \left(\sqrt{3}-13 i\right) \pi \log (432)}{1728\ 2^{2/3}},$$ where $pF_q(a;b;z)$ is the Generalized Hypergeometric function, and $\psi^{(n)}$ is the Polygamma function of order $n$.