Consider the generalized parametric case where $$0 < z \le 1$$:$$\mathcal{I}(z)=\int_0^z\frac{\log(1+x)\log(1-x)}{(1+x)}\,dx$$Substitute $$y=1+x$$ to obtain:$$\int_1^{1+z}\frac{\log y\log[1-(y-1)]}{y}\,dy=\int_1^{1+z}\frac{\log y\log(2-y)}{y}\,dy=$$$$\int_1^{1+z}\frac{\log y\log\left[2 \left(1-\frac{y}{2} \right) \right]}{y}\,dy=$$$$\log 2\, \int_1^{1+z}\frac{\log y}{y}\,dy+\int_1^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy$$Performing an integration by parts on that first integral gives:$$\frac{1}{2}\log 2\, (\log y)^2\, \Bigg|_1^{1+z}=\frac{1}{2}\log 2\log^2(1+z)$$So$$\mathcal{I}(z)=\frac{1}{2}\log 2\log^2(1+z)+\int_1^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy$$Next, we split that last integral into two:$$\int_1^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy=
\int_0^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy-\int_0^1\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy$$For $$0 < z \le 1$$, Polylogarithms of order $$m \ge 1$$ have the integral representation:$$\text{Li}_m(z)=\frac{(-1)^{m-1}}{(m-2)!}\int_0^1\frac{(\log x)^{m-2}\log(1-zx)}{x}\,dx$$Hence$$\int_0^1\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy=\text{Li}_3\left( \tfrac{1}{2}\right) $$So$$\mathcal{I}(z)=\frac{1}{2}\log 2\log^2(1+z)-\text{Li}_3\left( \tfrac{1}{2}\right) +\int_0^{1+z}\frac{\log y\log\left( 1- \frac{y}{2} \right)}{y}\,dy$$For that final integral, apply the substitution $$y=(1+z)\, x$$ to change it into:$$\int_0^1\frac{\log[(1+z)\, x]\, \log\left(1-\frac{(1+z)}{2}\, x \right)}{x}\,dx=$$$$\log(1+z)\, \int_0^1\frac{\log\left(1-\frac{(1+z)}{2}\, x \right)}{x}\,dx+\int_0^1\frac{\log x\, \log\left(1-\frac{(1+z)}{2}\, x \right)}{x}\,dx$$By the integral representation for arbitrary (non-zero) order Polylogs given above, this equates to:$$-\log(1+z) \, \text{Li}_2\left(\frac{1+z}{2}\right)+ \text{Li}_3 \left(\frac{1+z}{2}\right)$$
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General parametric solution:
$$\int_0^z\frac{\log(1+x)\log(1-x)}{(1+x)}\,dx=$$$$\frac{1}{2}\log 2\log^2(1+z)-\text{Li}_3\left( \tfrac{1}{2}\right) - \log(1+z) \, \text{Li}_2\left(\frac{1+z}{2}\right)+ \text{Li}_3 \left(\frac{1+z}{2}\right)$$For $$\, \, 0 < z \le 1$$