$$I:=　\int _{0}^{\infty }\frac {\ln (x)\ln (1+x)}{(1+x)(1+x^{2})}dx$$

$$\begin {aligned} f(z):=\frac {\log ^{2}(z)\log (1+z)}{(1+z)(1+z^{2})} \end {aligned}$$

$$\begin {aligned} L_1&:z=x+0i\quad (x:R\to r_1)\\ L_2&:z=x-0i\quad (x:r_1\to R)\\ M_1&:z=x+0i\quad (x:-r_1\to -1+r_2)\\ M_2&:z=x-0i\quad (x:-1+r_2\to -r_12)\\ C_1&:z=Re^{i\theta}\quad (\theta :2\pi \to 0)\\ C_2&:z=-1+r_1e^{i\theta}\quad (\theta:0\to 2\pi)\\ C_3&:z=r_2e^{i\theta}\quad (\theta:0\to \pi)\\ C_4&:z=r_1e^{i\theta}\quad (\theta:\pi \to 2\pi)\\ \\ L&:=L_1+L_2\\ \Gamma &:=M_1+C_2+M_2 \end {aligned}$$

$$\displaystyle{ \int_{C_1},\int_{C_3},\int_{C_4} = 0 }$$

$$\begin {aligned} \int_{\Gamma}+\int_{L}&=-2\pi i\Res{z=\pm i}f(z)\\ \frac 1{2\pi }\Im \left (\int _{\Gamma }+\int _{L}\right )&=-\Re \left (\Res{z=\pm i}f(z)\right ) \quad\quad (33.1) \end {aligned}$$

$$\begin {aligned} -\Re \left (\Res{z=i}f(z)+\Res{z=-i}f(z)\right ) &=-\Re \left (\frac {\log ^{2}(i)\log (1+i)}{(1+i)2i}-\frac {\log ^{2}(-i)\log (1-i)}{(1-i)2i}\right )\\ &=-\Re\left (\frac {\pi ^{2}}{16}(1+i)\left (\frac {\ln 2}2+\frac {\pi i}4\right )+\frac {9\pi ^{2}}{16}(1-i)\left (\frac {\ln 2}2+\frac {7\pi i}4\right )\right )\\ &=-\left (\frac {\pi ^{2}\log 2}{32}-\frac {\pi ^{3}}{64}+\frac {9\ln 2}{32}+\frac {63\pi ^{3}}{64}\right )\\ &=-\frac {31\pi ^{3}}{32}-\frac {5\pi ^{2}\ln 2}{16} \quad\quad(33.2) \end {aligned}$$

$$\begin {aligned} \frac {1}{2\pi }\Im \int _{L} &=\frac {1}{2\pi }\Im \left (\int _{L_1}+\int _{L_2}\right )\\ &=\frac {1}{2\pi }\Im \int _{L_2}\\ &=\frac {1}{2\pi }\Im\int _{0}^{\infty }\frac {(\ln x+2\pi i)^{2}(\ln (1+x)+2\pi i)}{(1+x)(1+x^{2})}dx\\ &=\frac {1}{2\pi }\int _{0}^{\infty }\Im \frac {(\ln ^{2}x-4\pi ^{2}+4\pi i\ln x)(\ln (1+x)+2\pi i)}{(1+x)(1+x^{2})}dx\\ &=\frac {1}{2\pi }\int _{0}^{\infty }\frac {(\ln ^{2}x-4\pi ^{2})2\pi +4\pi \ln (x)\ln (1+x)}{(1+x)(1+x^2)}dx\\ &=\int _{0}^{\infty }\frac {\ln ^{2}x}{(1+x)(1+x^{2})}dx-4\pi ^{2}\int _{0}^{\infty }\frac {dx}{(1+x)(1+x^{2})}+2I\\ &=\int _{0}^{1}\frac {\ln ^{2}x}{1+x^{2}}dx-4\pi ^{2}\int _{0}^{1}\frac {dx}{1+x^{2}}+2I\\ &=\frac {\pi ^{3}}{16}-\pi ^{3}+2I\\ &=-\frac {15\pi ^{3}}{16}+2I \quad\quad(33.3) \end {aligned}$$

$$\begin {aligned} \int _{C_2}f(z)dz&=\int _{C}\left (\frac {\log ^{2}(-1)}{2(1+z)}+O(z)\right )\log (1+z)dz\\ &=\left [-\frac {\pi ^{2}}4\log ^{2}(1+z)\right ]_{-1+r_1e^0}^{-1+r_1e^{2\pi i}}+O(r\ln r)\\ &=-\frac {\pi ^{2}}4\left \{ \left (\ln r+2\pi i\right )^2-\ln ^{2}r\right \} +O(r\ln r)\\ &=-\frac {\pi ^{2}}4\left (4\pi i\ln r-4\pi ^{2}\right )+O(r\ln r)\\ \frac {1}{2\pi }\Im \int _{C_2}&=-\frac {\pi ^2}2\ln r+O(r\ln r) \end {aligned}$$

$$\begin {aligned} \frac {1}{2\pi }\Im\int _{M_1+M_2}f(z)dz &=\frac {1}{2\pi }\Im \int _{0}^{1-r_1}(f(-x-0i)-f(-x+0i))dx\\ &=\frac {1}{2\pi }\Im \int _{0}^{1-r_1}\frac {\log ^{2}(-x)}{(1-x)(1+x^{2})}\left ((\ln (1-x)+2\pi i)-\ln (1-x)\right )dx\\ &=\Re \int _{0}^{1-r_1}\frac {(\ln x+\pi i)^{2}}{(1-x)(1-x^{2})}dx\\ &=\int _{0}^{1-r_1}\frac {\ln ^{2}x}{(1-x)(1+x^{2})}dx-\pi ^{2}\int _{0}^{1-r}\frac {dx}{(1-x)(1+x^{2})}dx \end {aligned}$$

$$\begin {aligned} \frac {1}{2\pi }\Im\int _{\Gamma } &=\frac {1}{2\pi }\Im \int _{M_1+M_2}+\frac {1}{2\pi }\Im \int _{C_2}\\ &=\int _{0}^{1-r}\frac {\ln ^{2}x}{(1-x)(1+x^{2})}dx-\pi ^{2}\int _{0}^{1-r}\frac {dx}{(1-x)(1+x^{2})}-\frac {\pi ^{2}}2\ln r+O(r\ln r)\\ &\xrightarrow{r\to 0}\int _{0}^{1}\frac {\ln ^{2}x}{(1-x)(1+x^{2})}dx-\pi ^{2}\left (\frac {\pi }8+\frac {\ln 2}4\right )\\ &=\frac {35}{32}\zeta (3)+\frac {\pi ^{3}}{32}-\frac {\pi ^{3}}8-\frac {\pi ^{2}\ln 2}4 \\ &=\frac {35}{32}\zeta (3)-\frac {3\pi ^{2}}{32}-\frac {\pi ^{2}\ln 2}4 \quad\quad(33.4) \end {aligned}$$

$$\begin {aligned} \frac {35}{32}\zeta (3)-\frac {3\pi ^{3}}{32}-\frac {\pi ^{2}\ln 2}4-\frac {15\pi ^{3}}{16}+2I &=-\frac {31\pi ^{3}}{32}-\frac {5\pi ^{2}\ln 2}{16}\\ 2I&=-\frac {35}{32}\zeta (3)+\frac {\pi ^{3}}{16}-\frac {\pi ^{2}\ln 2}{16}\\ I&=\textcolor{blue}{-\frac {35}{64}\zeta (3)+\frac {\pi ^{3}}{32}-\frac {\pi ^{2}\ln 2}{32}}. \end {aligned}$$