Integral ${\large\int}_0^\infty\frac{\ln x}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}\ \mathrm dx$ [closed]

With some help from a CAS I got this result with only one dilogarithm term: $$I=\frac{\sqrt[4]2\sqrt{2-\sqrt2}}{16}\\\left\{\left[16\ln2+\left(1+\sqrt2\right)\left(4\pi-16\arctan\sqrt{\sqrt2-1}\right)\right]\ln\left(1+\sqrt2+\sqrt{2+2\sqrt2}\right)\\-\Big(32\left(1+\sqrt2\right)\ln2+4\pi\Big)\arctan\sqrt{\sqrt2-1}-3\pi\arctan\frac{2\sqrt{2+10\sqrt2}}7\\-16\arctan^2\sqrt{\sqrt2-1}+4\ln^2\!\left(1+\sqrt2+\sqrt{2+2\sqrt2}\right)\\+3\pi^2-\frac{32}{\sqrt{2-\sqrt2}}\,\Re\left[\left(-1\right)^{5/8}\operatorname{Li}_2\!\left(2i\left( 1+\sqrt{1+i}\right)\right)\right]\right\}.$$

Update: It was suggested in comments that this expression gives a numerically incorrect result when evaluated with Maple. Here is what I got with Maple 18:

This value numerically matches the integral. I also checked it with Mathematica 10 and it gave the same numeric result.

\begin{align} -\frac{1}{4}\int^\infty_0\frac{\ln{x}}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}{\rm d}x &=-\frac{1}{4}\int^\infty_0\frac{\ln{\sinh{x}}}{1+\sinh{x}}e^{x/2}{\rm d}x\\ &=-\frac{1}{2}\int^\infty_1\frac{\ln\left(\frac{x^2-x^{-2}}{2}\right)}{1+\frac{x^2-x^{-2}}{2}}{\rm d}x\\ &=\int^1_0\frac{\ln(1-x^4)-\ln(2x^2)}{x^4-2x^2-1}{\rm d}x\\ \end{align} For simplicity's sake, let $\displaystyle\frac{1}{x^4-2x^2-1}=\sum^4_{k=1}\frac{c_k}{x-r_k}$. The integral becomes $$I=-4\sum^4_{k=1}\int^1_0\frac{c_k\ln(1-x^4)-c_k\ln{2}-2c_k\ln{x}}{x-r_k}{\rm d}x$$ The second integral is \begin{align} -c_k\ln{2}\int^1_0\frac{1}{x-r_k}{\rm d}x=-c_k\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right) \end{align} The third integral is \begin{align} -2c_k\int^1_0\frac{\ln{x}}{x-r_k}{\rm d}x &=2c_k\int^{1/r_k}_0\frac{\ln(r_kx)}{1-x}{\rm d}x\\ &=-2c_k{\rm Li}_2\left(\frac{1}{r_k}\right) \end{align} Pluck these results back in. $$I=4\sum^4_{k=1}c_k\left[\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)+2{\rm Li}_2\left(\frac{1}{r_k}\right)-\int^1_0\frac{\ln(1-x^4)}{x-r_k}{\rm d}x\right]$$ The remaining integral is \begin{align} &\ \ \ \ \ \int^1_0\frac{\ln(1-x^4)}{x-r_k}{\rm d}x\\ &=\sum_{j=1,-1,i,-i}\int^1_0\frac{\ln(1+jx)}{x-r_k}{\rm d}x\\ &=-\sum_{j=1,-1,i,-i}\int^{\frac{\lambda}{1-r_k}}_{-\frac{\lambda}{r_k}}\ln\left(1+\frac{j\lambda}{y}+jr_k\right)\frac{{\rm d}y}{y}\\ &=-\sum_{j=1,-1,i,-i}\int^{\frac{\lambda}{1-r_k}}_{-\frac{\lambda}{r_k}}\left[\ln\left(1+\frac{1+jr_k}{j\lambda}y\right)-\ln\left(\frac{y}{j\lambda}\right)\right]\frac{{\rm d}y}{y}\\ &=-\sum_{j=1,-1,i,-i}\int^{\frac{1+jr_k}{j-jr_k}}_{-\frac{1+jr_k}{jr_k}}\frac{\ln(1+y)}{y}-\frac{1}{y}\ln\left(\frac{y}{1+jr_k}\right){\rm d}y\\ &=-\sum_{j=1,-1,i,-i}\left[{\rm Li}_2\left(\frac{1+jr_k}{jr_k}\right)-{\rm Li}_2\left(\frac{1+jr_k}{jr_k-j}\right)+\frac{1}{2}\ln^2\left(-jr_k\right)-\frac{1}{2}\ln^2\left(j-jr_k\right)\right] \end{align} Final Result: \begin{align} \color\purple{\int^\infty_0\frac{\ln{x}}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}{\rm d}x =4\sum^4_{k=1}c_k\left[\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)+2{\rm Li}_2\left(\frac{1}{r_k}\right)+\sum_{j=1,-1,i,-i}\left[{\rm Li}_2\left(\frac{1+jr_k}{jr_k}\right)-{\rm Li}_2\left(\frac{1+jr_k}{jr_k-j}\right)+\frac{1}{2}\ln^2\left(-jr_k\right)-\frac{1}{2}\ln^2\left(j-jr_k\right)\right]\right]} \end{align} where \begin{align} \frac{c_1}{x-r_1}&=\frac{1}{4\sqrt{2+2\sqrt{2}}\left(x-\sqrt{1+\sqrt{2}}\right)}\\ \frac{c_2}{x-r_2}&=-\frac{1}{4\sqrt{2+2\sqrt{2}}\left(x+\sqrt{1+\sqrt{2}}\right)}\\ \frac{c_3}{x-r_3}&=-\frac{1}{4\sqrt{2}\left(x+i\sqrt{\sqrt{2}-1}\right)}\\ \frac{c_4}{x-r_4}&=-\frac{1}{4\sqrt{2}\left(x-i\sqrt{\sqrt{2}-1}\right)}\\ \end{align}