Show that $\int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx = 0$ I am looking for a nice slick way to show

$$\int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx = 0.$$

So far I can only show the result using brute force as follows.
Let 
$$I = \int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx.$$
Since
$$\tanh^{-1} x = \frac{1}{2} \ln \left (\frac{1 + x}{1 - x} \right ),$$
the above integral, after rearranging, can be rewritten as
$$I = \frac{3}{2} \int^1_0 \frac{\ln^2 (1 + x)}{x} \, dx - \int^1_0 \frac{\ln (1 - x) \ln (1 + x)}{x} \, dx - \frac{1}{2} \int^1_0 \frac{\ln^2 (1 - x)}{x} \, dx.\tag1$$
Each of the above three integrals can be found. The results are:
$$\int^1_0 \frac{\ln^2 (1 + x)}{x} \, dx = \frac{1}{4} \zeta (3).$$ For a proof, see here or here.
$$\int^1_0 \frac{\ln (1 - x) \ln (1 + x)}{x} \, dx = -\frac{5}{8} \zeta (3).$$
For a proof, see here. And
$$\int^1_0 \frac{\ln^2 (1 - x)}{x} \, dx = 2 \zeta (3).$$
For a proof of this last one, see here.
Thus (1) becomes
$$\int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx = \frac{3}{8 } \zeta (3) + \frac{5}{8} \zeta (3) - \zeta (3) = 0,$$
as expected. 
 A: \begin{align}J=\int^1_0 \frac{\tanh^{-1} x}{x} \ln [(1 + x)^3 (1 - x)] \, dx\end{align}
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
\begin{align}J&=\int^1_0\frac{\ln\left(\frac{16x}{(1+x)^4}\right)\ln x}{1-x^2}\, dx\\
&=4\ln 2\int_0^1\frac{\ln x}{1-x^2}\,dx+\int_0^1\frac{\ln^2 x}{1-x^2}\,dx-4\int_0^1\frac{\ln x\ln(1+x)}{1-x^2}\,dx\\
\end{align}
Define on $[0;1]$ the function $R$ by,
\begin{align}R(x)&=\int_0^x \frac{\ln t}{1-t^2}\,dt\\
&=\int_0^1 \frac{x\ln(tx)}{1-t^2x^2}\,dt
\end{align}
Therefore,
\begin{align}K&=\int_0^1\frac{\ln x\ln(1+x)}{1-x^2}\,dx\\
&=\Big[R(x)\ln(1+x)\Big]_0^1-\int_0^1\int_0^1 \frac{x\ln(tx)}{(1-t^2x^2)(1+x)}\,dt\,dx\\
&=\int_0^1 \frac{\ln 2\ln t}{1-t^2}\,dt-\int_0^1\left(\int_0^1 \frac{x\ln t}{(1-t^2x^2)(1+x)}\,dx\right)\,dt-\int_0^1\left(\int_0^1 \frac{x\ln x}{(1-t^2x^2)(1+x)}\,dt\right)\,dx\\
&=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt-\\
&\frac{1}{2}\left(\int_0^1 \frac{\ln t\ln(1+t)}{1-t}\,dt-\int_0^1 \frac{\ln t\ln\left(\frac{1-t}{1+t}\right)}{t}\,dt-\int_0^1 \frac{2\ln 2\ln t}{1-t^2}\,dt+\int_0^1 \frac{\ln(1-t)\ln t}{1+t}\,dt\right)-\\
&\frac{1}{2}\left(\int_0^1 \frac{\ln x\ln(1+x)}{1+x}\,dx-\int_0^1 \frac{\ln x\ln(1-x)}{1+x}\,dx\right)\\
&=2\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt-K+\frac{1}{2}\int_0^1 \frac{\ln t\ln\left(\frac{1-t}{1+t}\right)}{t}\,dt
\end{align}
Therefore,
\begin{align}K&=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{1}{4}\int_0^1 \frac{\ln t\ln\left(\frac{1-t}{1+t}\right)}{t}\,dt\\
&=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{1}{8}\left[\ln^2 t\ln\left(\frac{1-t}{1+t}\right)\right]_0^1+\frac{1}{4}\int_0^1 \frac{\ln^2 t}{1-t^2}\,dt\\
&=\ln 2\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{1}{4}\int_0^1 \frac{\ln^2 t}{1-t^2}\,dt\\
\end{align}
Therefore,
\begin{align}\boxed{J=0}\end{align}
NB:
It's easy to deduce that,
\begin{align}\int_0^1\frac{\ln x\ln(1+x)}{1-x^2}\,dx=\frac{7}{16}\zeta(3)-\frac{1}{8}\pi^2\ln 2\end{align}
