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$$\begin{array}{rl}& {\ln }^2\left(\frac{2}{1+x}\right){\ln }^3\left(\frac{2x}{1+x}\right)=-{\ln }^2\text{2}{\ln }^3(1+x)+\text{2}\ln \text{2}{\ln }^4(1+x)-{\ln }^5(1+x)\\ & +\text{3}{\ln }^2\text{2}{\ln }^2(1+x)\ln \left(2x\right)-\text{6}\ln \text{2}{\ln }^3(1+x)\ln \left(2x\right)+\text{3}{\ln }^4(1+x)\ln \left(2x\right)\\ & -\text{3}{\ln }^2\text{2}\ln (1+x){\ln }^2\left(2x\right)+\text{6}\ln \text{2}{\ln }^2(1+x){\ln }^2\left(2x\right)-\text{3}{\ln }^3(1+x){\ln }^2\left(2x\right)\\ & +{\ln }^2\text{2}{\ln }^3\left(2x\right)-\text{2}\ln \text{2}\ln (1+x){\ln }^3\left(2x\right)+{\ln }^2(1+x){\ln }^3\left(2x\right)\end{array}$$

$$\begin{array}{rl}{I}_4& =\text{3}{\ln }^{2}2{\int }_0^1\frac{{\ln }^2(1+x)\ln \left(2x\right)}{{(1+x)}^2}\text{d}x\\ & =\text{3}{\ln }^{2}2{\int }_0^1\frac{{\ln }^2(1+x)\ln x}{{(1+x)}^2}\text{d}x+\text{3}{\ln }^{3}2{\int }_0^1\frac{{\ln }^2(1+x)}{{(1+x)}^2}\text{d}x\end{array}$$

$$\begin{array}{rl}{I}_5& =-\text{6}\ln 2{\int }_0^1\frac{{\ln }^3(1+x)\ln \left(2x\right)}{{(1+x)}^2}\text{d}x\\ & =-\text{6}\ln 2{\int }_0^1\frac{{\ln }^3(1+x)\ln x}{{(1+x)}^2}\text{d}x-\text{6}{\ln }^{2}2{\int }_0^1\frac{{\ln }^3(1+x)}{{(1+x)}^2}\text{d}x\end{array}$$

$$\begin{array}{rl}{I}_{51}& ={\int }_0^1\frac{{\ln }^3(1+x)\ln x}{{(1+x)}^2}\text{d}x={\int }_0^1\frac{{\ln }^3(1+x)}{x(1+x)}\text{d}x+3{\int }_0^1\frac{\ln x{\ln }^2(1+x)}{{(1+x)}^2}\text{d}x\\ & ={\int }_0^1\frac{{\ln }^3(1+x)}{x}\text{d}x-{\int }_0^1\frac{{\ln }^3(1+x)}{1+x}\text{d}x+3{\int }_0^1\frac{\ln x{\ln }^2(1+x)}{{(1+x)}^2}\text{d}x\\ & =[-{\text{6Li}}_4\left(\frac{1}{2}\right)-\frac{21}{4}\ln 2\zeta \left(3\right)+\frac{1}{15}{\pi }^4-\frac{1}{4}{\ln }^{4}2+\frac{1}{4}{\pi }^2{\ln }^{2}2]-\frac{{\ln }^{4}2}{4}\\ & +3[\frac{1}{4}\zeta \left(3\right)+\frac{{\pi }^2}{6}-\frac{1}{3}{\ln }^{3}2-{\ln }^{2}2-\text{2}\ln 2]\\ & =-{\text{6Li}}_4\left(\frac{1}{2}\right)-\frac{21}{4}\ln 2\zeta \left(3\right)+\frac{1}{15}{\pi }^4-\frac{1}{2}{\ln }^{4}2-{\ln }^{3}2+\frac{1}{4}{\pi }^2{\ln }^{2}2\\ & +\frac{3}{4}\zeta \left(3\right)+\frac{{\pi }^2}{2}-\text{3}{\ln }^{2}2-\text{6}\ln 2\end{array}$$

$$\begin{array}{rl}{I}_{61}& ={\int }_0^1\frac{{\ln }^4(1+x)\ln x}{{(1+x)}^2}\text{d}x={\int }_0^1\frac{{\ln }^4(1+x)}{x(1+x)}\text{d}x+4{\int }_0^1\frac{\ln x{\ln }^3(1+x)}{{(1+x)}^2}\text{d}x\\ & ={\int }_0^1\frac{{\ln }^4(1+x)}{x}\text{d}x-{\int }_0^1\frac{{\ln }^4(1+x)}{1+x}\text{d}x+4{\int }_0^1\frac{\ln x{\ln }^3(1+x)}{{(1+x)}^2}\text{d}x\\ & =[-{\text{24Li}}_5\left(\frac{1}{2}\right)-\text{24}\ln {\text{2Li}}_4\left(\frac{1}{2}\right)+24\zeta \left(5\right)-\frac{21}{2}{\ln }^{2}2\zeta \left(3\right)-\frac{4}{5}{\ln }^{5}2+\frac{2}{3}{\pi }^2{\ln }^{3}2]-\frac{1}{5}{\ln }^{5}2\\ & +4[-{\text{6Li}}_4\left(\frac{1}{2}\right)-\frac{21}{4}\ln 2\zeta \left(3\right)+\frac{1}{15}{\pi }^4-\frac{1}{2}{\ln }^{4}2-{\ln }^{3}2+\frac{1}{4}{\pi }^2{\ln }^{2}2+\frac{3}{4}\zeta \left(3\right)+\frac{{\pi }^2}{2}-\text{3}{\ln }^{2}2-\text{6}\ln 2]\\ & =-{\text{24Li}}_5\left(\frac{1}{2}\right)-{\text{24Li}}_4\left(\frac{1}{2}\right)-\text{24}\ln {\text{2Li}}_4\left(\frac{1}{2}\right)+24\zeta \left(5\right)-\frac{21}{2}{\ln }^{2}2\zeta \left(3\right)-\text{21}\ln 2\zeta \left(3\right)-{\ln }^{5}2+\frac{4}{15}{\pi }^4\\ & -\text{2}{\ln }^{4}2-\text{4}{\ln }^{3}2+\frac{2}{3}{\pi }^2{\ln }^{3}2+{\pi }^2{\ln }^{2}2+3\zeta \left(3\right)+2{\pi }^2-\text{12}{\ln }^{2}2-\text{24}\ln 2\end{array}$$

$$\begin{array}{rl}{I}_7& =-\text{3}{\ln }^{2}2{\int }_0^1\frac{\ln (1+x){\ln }^2\left(2x\right)}{{(1+x)}^2}\text{d}x=-\text{3}{\ln }^{2}2{\int }_0^1\frac{\ln (1+x){\ln }^{2}x}{{(1+x)}^2}\text{d}x\\ & -\text{3}{\ln }^{4}2{\int }_0^1\frac{\ln (1+x)}{{(1+x)}^2}\text{d}x-\text{6}{\ln }^{3}2{\int }_0^1\frac{\ln (1+x)\ln x}{{(1+x)}^2}\text{d}x\end{array}$$

$$\begin{array}{rl}{I}_{71}& ={\int }_0^1\frac{\ln (1+x){\ln }^{2}x}{{(1+x)}^2}\text{d}x={\int }_0^1\frac{{\ln }^{2}x}{{(1+x)}^2}\text{d}x+2{\int }_0^1\frac{\ln (1+x)\ln x}{x(1+x)}\text{d}x\\ & ={\int }_0^1\frac{{\ln }^{2}x}{{(1+x)}^2}\text{d}x+2{\int }_0^1\frac{\ln (1+x)\ln x}{x}\text{d}x-2{\int }_0^1\frac{\ln (1+x)\ln x}{1+x}\text{d}x\\ & =\frac{1}{6}{\pi }^2+2[-\frac{3}{4}\zeta \left(3\right)]-2[-\frac{1}{8}\zeta \left(3\right)]=-\frac{5}{4}\zeta \left(3\right)+\frac{1}{6}{\pi }^2\end{array}$$

$$\begin{array}{rl}{I}_8& =\text{6}\ln 2{\int }_0^1\frac{{\ln }^2(1+x){\ln }^2\left(2x\right)}{{(1+x)}^2}\text{d}x=\text{6}\ln 2{\int }_0^1\frac{{\ln }^2(1+x){\ln }^{2}x}{{(1+x)}^2}\text{d}x\\ & +\text{6}{\ln }^{3}2{\int }_0^1\frac{{\ln }^2(1+x)}{{(1+x)}^2}\text{d}x+\text{12}{\ln }^{2}2{\int }_0^1\frac{{\ln }^2(1+x)\ln x}{{(1+x)}^2}\text{d}x\end{array}$$

$$\begin{array}{rl}{I}_{81}& ={\int }_0^1\frac{{\ln }^2(1+x){\ln }^{2}x}{{(1+x)}^2}\text{d}x=2{\int }_0^1\frac{\ln (1+x){\ln }^{2}x}{{(1+x)}^2}\text{d}x+2{\int }_0^1\frac{{\ln }^2(1+x)\ln x}{x(1+x)}\text{d}x\\ & =2{\int }_0^1\frac{{\ln }^{2}x}{{(1+x)}^2}\text{d}x+4{\int }_0^1\frac{\ln (1+x)\ln x}{x(1+x)}\text{d}x+2{\int }_0^1\frac{{\ln }^2(1+x)\ln x}{x}\text{d}x-2{\int }_0^1\frac{{\ln }^2(1+x)\ln x}{1+x}\text{d}x\\ & =2{\int }_0^1\frac{{\ln }^{2}x}{{(1+x)}^2}\text{d}x+4{\int }_0^1\frac{\ln (1+x)\ln x}{x}\text{d}x-4{\int }_0^1\frac{\ln (1+x)\ln x}{1+x}\text{d}x+2{\int }_0^1\frac{{\ln }^2(1+x)\ln x}{x}\text{d}x-2{\int }_0^1\frac{{\ln }^2(1+x)\ln x}{1+x}\text{d}x\\ & =2\cdot \frac{1}{6}{\pi }^2+4[-\frac{3}{4}\zeta \left(3\right)]-4[-\frac{1}{8}\zeta \left(3\right)]+2[-{\text{4Li}}_4\left(\frac{1}{2}\right)-\frac{7}{2}\ln 2\zeta \left(3\right)+\frac{1}{24}{\pi }^4-\frac{1}{6}{\ln }^{4}2+\frac{1}{6}{\pi }^2{\ln }^{2}2]\\ & -2[{\text{2Li}}_4\left(\frac{1}{2}\right)+\frac{7}{4}\ln 2\zeta \left(3\right)-\frac{1}{45}{\pi }^4+\frac{1}{12}{\ln }^{4}2-\frac{1}{12}{\pi }^2{\ln }^{2}2]\\ & =-{\text{12Li}}_4\left(\frac{1}{2}\right)-\frac{21}{2}\ln 2\zeta \left(3\right)-\frac{5}{2}\zeta \left(3\right)+\frac{23}{180}{\pi }^4-\frac{1}{2}{\ln }^{4}2+\frac{1}{2}{\pi }^2{\ln }^{2}2+\frac{1}{3}{\pi }^2\end{array}$$

$$\begin{array}{rl}{I}_{83}& ={\int }_0^1\frac{{\ln }^2(1+x)\ln x}{{(1+x)}^2}\text{d}x={\int }_0^1\frac{{\ln }^2(1+x)}{x(1+x)}\text{d}x+2{\int }_0^1\frac{\ln x\ln (1+x)}{{(1+x)}^2}\text{d}x\\ & ={\int }_0^1\frac{{\ln }^2(1+x)}{x}\text{d}x-{\int }_0^1\frac{{\ln }^2(1+x)}{1+x}\text{d}x+2{\int }_0^1\frac{\ln x}{{(1+x)}^2}\text{d}x+2{\int }_0^1\frac{\ln (1+x)}{x(1+x)}\text{d}x\\ & ={\int }_0^1\frac{{\ln }^2(1+x)}{x}\text{d}x-{\int }_0^1\frac{{\ln }^2(1+x)}{1+x}\text{d}x+2{\int }_0^1\frac{\ln x}{{(1+x)}^2}\text{d}x+2{\int }_0^1\frac{\ln (1+x)}{x}\text{d}x-2{\int }_0^1\frac{\ln (1+x)}{1+x}\text{d}x\\ & =\frac{1}{4}\zeta \left(3\right)-\frac{{\ln }^{3}2}{3}+2(-\ln 2)+2\cdot \frac{{\pi }^2}{12}-2\cdot \frac{{\ln }^{2}2}{2}=\frac{1}{4}\zeta \left(3\right)+\frac{{\pi }^2}{6}-\frac{1}{3}{\ln }^{3}2-{\ln }^{2}2-\text{2}\ln 2\end{array}$$

$$\begin{array}{rl}{I}_9& =-3{\int }_0^1\frac{{\ln }^3(1+x){\ln }^2\left(2x\right)}{{(1+x)}^2}\text{d}x=-3{\int }_0^1\frac{{\ln }^3(1+x){\ln }^{2}x}{{(1+x)}^2}\text{d}x\\ & -\text{3}{\ln }^{2}2{\int }_0^1\frac{{\ln }^3(1+x)}{{(1+x)}^2}\text{d}x-\text{3}\ln 2{\int }_0^1\frac{{\ln }^3(1+x)\ln x}{{(1+x)}^2}\text{d}x\end{array}$$

$$\begin{array}{rl}{I}_{91}& ={\int }_0^1\frac{{\ln }^3(1+x){\ln }^{2}x}{{(1+x)}^2}\text{d}x=2{\int }_0^1\frac{\ln x{\ln }^3(1+x)}{x(1+x)}\text{d}x+3{\int }_0^1\frac{{\ln }^{2}x{\ln }^2(1+x)}{{(1+x)}^2}\text{d}x\\ & =2{\int }_0^1\frac{\ln x{\ln }^3(1+x)}{x}\text{d}x-2{\int }_0^1\frac{\ln x{\ln }^3(1+x)}{1+x}\text{d}x+3{\int }_0^1\frac{{\ln }^{2}x{\ln }^2(1+x)}{{(1+x)}^2}\text{d}x\\ & =2[-{\text{12Li}}_5\left(\frac{1}{2}\right)-\text{12}\ln {\text{2Li}}_4\left(\frac{1}{2}\right)+\frac{1}{2}{\pi }^2\zeta \left(3\right)+\frac{99}{16}\zeta \left(5\right)-\frac{2}{5}{\ln }^{5}2+\frac{1}{3}{\pi }^2{\ln }^{3}2-\frac{21}{4}{\ln }^{2}2\zeta \left(3\right)]\\ & -2[{\text{6Li}}_5\left(\frac{1}{2}\right)+\text{6}\ln {\text{2Li}}_4\left(\frac{1}{2}\right)-6\zeta \left(5\right)+\frac{21}{8}{\ln }^{2}2\zeta \left(3\right)+\frac{1}{5}{\ln }^{5}2-\frac{1}{6}{\pi }^2{\ln }^{3}2]\\ & +3[-{\text{12Li}}_4\left(\frac{1}{2}\right)-\frac{21}{2}\ln 2\zeta \left(3\right)-\frac{5}{2}\zeta \left(3\right)+\frac{23}{180}{\pi }^4-\frac{1}{2}{\ln }^{4}2+\frac{1}{2}{\pi }^2{\ln }^{2}2+\frac{1}{3}{\pi }^2]\\ & =-{\text{36Li}}_5\left(\frac{1}{2}\right)-{\text{36Li}}_4\left(\frac{1}{2}\right)-\text{36}\ln {\text{2Li}}_4\left(\frac{1}{2}\right)+{\pi }^2\zeta \left(3\right)+\frac{195}{8}\zeta \left(5\right)-\frac{15}{2}\zeta \left(3\right)-\frac{6}{5}{\ln }^{5}2+{\pi }^2{\ln }^{3}2\\ & -\frac{63}{4}{\ln }^{2}2\zeta \left(3\right)-\frac{63}{2}\ln 2\zeta \left(3\right)+\frac{23}{60}{\pi }^4-\frac{3}{2}{\ln }^{4}2+\frac{3}{2}{\pi }^2{\ln }^{2}2+{\pi }^2\end{array}$$

$$\begin{array}{rl}{I}_{93}& =I_{51}={\int }_0^1\frac{{\ln }^3(1+x)\ln x}{{(1+x)}^2}\text{d}x\\ & =-{\text{6Li}}_4\left(\frac{1}{2}\right)-\frac{21}{4}\ln 2\zeta \left(3\right)+\frac{1}{15}{\pi }^4-\frac{1}{2}{\ln }^{4}2-{\ln }^{3}2+\frac{1}{4}{\pi }^2{\ln }^{2}2+\frac{3}{4}\zeta \left(3\right)+\frac{{\pi }^2}{2}-\text{3}{\ln }^{2}2-\text{6}\ln 2\end{array}$$

$$\begin{array}{rl}{I}_{11}& =-\text{2}\ln 2{\int }_0^1\frac{\ln (1+x){\ln }^3\left(2x\right)}{{(1+x)}^2}\text{d}x=-\text{2}{\ln }^{4}2{\int }_0^1\frac{\ln (1+x)}{{(1+x)}^2}\text{d}x-\text{6}{\ln }^{3}2{\int }_0^1\frac{\ln (1+x)\ln x}{{(1+x)}^2}\text{d}x\\ & -\text{6}{\ln }^{2}2{\int }_0^1\frac{\ln (1+x){\ln }^{2}x}{{(1+x)}^2}\text{d}x-\text{2}\ln 2{\int }_0^1\frac{\ln (1+x){\ln }^{3}x}{{(1+x)}^2}\text{d}x\end{array}$$

$$\begin{array}{rl}{I}_{114}& ={\int }_0^1\frac{\ln (1+x){\ln }^{3}x}{{(1+x)}^2}\text{d}x={\int }_0^1\frac{{\ln }^{3}x}{{(1+x)}^2}\text{d}x+3{\int }_0^1\frac{\ln (1+x){\ln }^{2}x}{x(1+x)}\text{d}x\\ & ={\int }_0^1\frac{{\ln }^{3}x}{{(1+x)}^2}\text{d}x+3{\int }_0^1\frac{\ln (1+x){\ln }^{2}x}{x}\text{d}x-3{\int }_0^1\frac{\ln (1+x){\ln }^{2}x}{1+x}\text{d}x\\ & =-\frac{9}{2}\zeta \left(3\right)+3\cdot \frac{7{\pi }^4}{360}-3[{\text{4Li}}_4\left(\frac{1}{2}\right)+\frac{7}{2}\ln 2\zeta \left(3\right)-\frac{1}{24}{\pi }^4+\frac{1}{6}{\ln }^{4}2-\frac{1}{6}{\pi }^2{\ln }^{2}2]\\ & =-{\text{12Li}}_4\left(\frac{1}{2}\right)-\frac{21}{2}\ln 2\zeta \left(3\right)-\frac{9}{2}\zeta \left(3\right)+\frac{11}{60}{\pi }^4-\frac{1}{2}{\ln }^{4}2+\frac{1}{2}{\pi }^2{\ln }^{2}2\end{array}$$

$$\begin{array}{rl}{I}_{12}& ={\int }_0^1\frac{{\ln }^2(1+x){\ln }^3\left(2x\right)}{{(1+x)}^2}\text{d}x={\ln }^{3}2{\int }_0^1\frac{{\ln }^2(1+x)}{{(1+x)}^2}\text{d}x+\text{3}{\ln }^{2}2{\int }_0^1\frac{{\ln }^2(1+x)\ln x}{{(1+x)}^2}\text{d}x\\ & +\text{3}\ln 2{\int }_0^1\frac{{\ln }^2(1+x){\ln }^{2}x}{{(1+x)}^2}\text{d}x+{\int }_0^1\frac{{\ln }^2(1+x){\ln }^{3}x}{{(1+x)}^2}\text{d}x\end{array}$$

$$\begin{array}{rl}{I}_{124}& ={\int }_0^1\frac{{\ln }^2(1+x){\ln }^{3}x}{{(1+x)}^2}\text{d}x=2{\int }_0^1\frac{\ln (1+x){\ln }^{3}x}{{(1+x)}^2}\text{d}x+3{\int }_0^1\frac{{\ln }^2(1+x){\ln }^{2}x}{x(1+x)}\text{d}x\\ & =2{\int }_0^1\frac{\ln (1+x){\ln }^{3}x}{{(1+x)}^2}\text{d}x+3{\int }_0^1\frac{{\ln }^2(1+x){\ln }^{2}x}{x}\text{d}x-3{\int }_0^1\frac{{\ln }^2(1+x){\ln }^{2}x}{1+x}\text{d}x\\ & =2[-{\text{12Li}}_4\left(\frac{1}{2}\right)-\frac{21}{2}\ln 2\zeta \left(3\right)-\frac{9}{2}\zeta \left(3\right)+\frac{11}{60}{\pi }^4-\frac{1}{2}{\ln }^{4}2+\frac{1}{2}{\pi }^2{\ln }^{2}2]+3[\frac{1}{3}{\pi }^2\zeta \left(3\right)-\frac{29}{5}\zeta \left(5\right)]\\ & -3[{\text{8Li}}_5\left(\frac{1}{2}\right)+\text{8}\ln {\text{2Li}}_4\left(\frac{1}{2}\right)-\frac{33}{8}\zeta \left(5\right)-\frac{1}{3}{\pi }^2\zeta \left(3\right)+\frac{7}{2}{\ln }^{2}2\zeta \left(3\right)+\frac{4}{15}{\ln }^{5}2-\frac{2}{9}{\pi }^2{\ln }^{3}2]\\ & =-{\text{24Li}}_5\left(\frac{1}{2}\right)-{\text{24Li}}_4\left(\frac{1}{2}\right)-\text{24}\ln {\text{2Li}}_4\left(\frac{1}{2}\right)-\frac{201}{40}\zeta \left(5\right)-\frac{21}{2}{\ln }^{2}2\zeta \left(3\right)-\text{21}\ln 2\zeta \left(3\right)+2{\pi }^2\zeta \left(3\right)\\ & -9\zeta \left(3\right)+\frac{11}{30}{\pi }^4-{\ln }^{4}2+{\pi }^2{\ln }^{2}2-\frac{4}{5}{\ln }^{5}2+\frac{2}{3}{\pi }^2{\ln }^{3}2\end{array}$$

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