Evaluating $\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x$ While I was working on computing $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}$, I came across the integral:
$$I=\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x.$$
I tried $x=\sin(u)$ then $\tan(u/2)=t$ and got
$$I=\int_0^{\pi/2}\cot(u)\ln(\sin u)\ln(1-\sin u)\ln(1+\cos u)\mathrm{d}u$$
$$=\int_0^1\frac{1-t^2}{t(1+t^2)}\ln\left(\frac{2t}{1+t^2}\right)\ln\left(\frac{(1-t)^2}{1+t^2}\right)\ln\left(\frac{2}{1+t^2}\right)\mathrm{d}t.$$
Any idea?
Thanks
 A: Bonus: Differentiate both sides of $\int_0^1 x^{2n-1}\ln(1-x)dx=-\frac{H_{2n}}{2n}$, we get
$$\int_0^1 x^{2n-1}\ln(x)\ln(1-x)dx=\frac{H_{2n}}{4n^2}+\frac{H_{2n}^{(2)}}{2n}-\frac{\zeta(2)}{2n}.$$
Multiply both sides by $\frac{2n\choose n}{4^nn}$ then sum up from $n=1$ to $\infty$, we obtain
$$\frac14\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}}{4^n n^3}+\frac12\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}-\zeta(2)\sum_{n=1}^\infty\frac{{2n\choose n}}{4^n n^2}=\int_0^1\frac{\ln(x)\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{{2n\choose n}x^{2n}}{4^n n}\right)dx$$
$$=\int_0^1\frac{\ln(x)\ln(1-x)}{x}\left(2\ln\left(\frac{2}{1+\sqrt{1-x^2}}\right)\right)dx$$
$$=2\ln(2)\int_0^1\frac{\ln(x)\ln(1-x)}{x}dx-2\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}dx$$
$$=2\ln(2)\zeta(3)-2I.$$
The sum $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}$ is calculated by Jorge above and the sum $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}}{4^n n^3}$ is calculated here. Collecting these two results along with using $\sum_{n=1}^\infty\frac{{2n\choose n}}{4^n n^2}=\zeta(2)-2\ln^2(2)$, we get
$$I = -2 \pi  \Im \left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}-\frac12\text{Li}_4\left(\frac{1}{2}\right)+\frac{281 }{64}\zeta(4)+\frac{7}{16} \ln (2)\zeta (3)-\frac{5}{8} \ln ^2(2)\zeta(2)-\frac{1}{48} \ln ^4(2).$$
A: Since you are primarily interested in that series I'll evaluate it.
$$\sum _{k=1}^{\infty }\frac{H_{2k}^{\left(2\right)}}{4^kk^2}\binom{2k}{k}$$
Consider the well-known expansion:
$$2\sum _{k=1}^{\infty }\frac{\left(H_{2k}^{\left(2\right)}-\frac{1}{4}H_k^{\left(2\right)}\right)}{4^k}\binom{2k}{k}x^{2k}=\frac{\arcsin ^2\left(x\right)}{\sqrt{1-x^2}}$$
$$-4\sum _{k=1}^{\infty }\frac{\left(H_{2k}^{\left(2\right)}-\frac{1}{4}H_k^{\left(2\right)}\right)}{4^k}\binom{2k}{k}\int _0^1x^{2k-1}\ln \left(x\right)\:dx=-2\int _0^1\frac{\ln \left(x\right)\arcsin ^2\left(x\right)}{x\sqrt{1-x^2}}\:dx$$
$$\sum _{k=1}^{\infty }\frac{H_{2k}^{\left(2\right)}}{4^kk^2}\binom{2k}{k}=-2\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\ln \left(\sin \left(x\right)\right)\:dx+\frac{1}{4}\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{4^kk^2}\binom{2k}{k}$$
You can find that integral here. I can also provide a new way to calculate it but it's lenghty.

For the remaining sum consider:
$$\int _0^1x^{k-1}\ln ^2\left(1-x\right)\:dx=\frac{H_k^2+H_k^{\left(2\right)}}{k}$$
$$\int _0^1\left(\sum _{k=1}^{\infty }\frac{1}{4^kk}\binom{2k}{k}x^k\right)\frac{\ln ^2\left(1-x\right)}{x}\:dx=\sum _{k=1}^{\infty }\frac{H_k^2}{4^kk^2}\binom{2k}{k}+\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{4^kk^2}\binom{2k}{k}$$
$$\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{4^kk^2}\binom{2k}{k}=-16\underbrace{\int _0^1\frac{x\ln ^2\left(x\right)\ln \left(1+x\right)}{1-x^2}\:dx}_{I}+2\ln \left(2\right)\int _0^1\frac{\ln ^2\left(1-x\right)}{x}\:dx-\underbrace{\sum _{k=1}^{\infty }\frac{H_k^2}{4^kk^2}\binom{2k}{k}}_{S}$$
$$\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{4^kk^2}\binom{2k}{k}=3\zeta \left(4\right)-3\ln \left(2\right)\zeta \left(3\right)$$
$I$ transforms into known integrals after integration by parts and $S$ can be calculated by simple means as shown in the OP's book, page $\#297$

And therefore:
$$\sum _{k=1}^{\infty }\frac{H_{2k}^{\left(2\right)}}{4^kk^2}\binom{2k}{k}=-\frac{129}{8}\zeta \left(4\right)+\frac{25}{4}\ln \left(2\right)\zeta \left(3\right)-\frac{3}{2}\ln ^2\left(2\right)\zeta \left(2\right)+8\pi \operatorname{\mathfrak{I}} \left\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
