Calculating $\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx$
$$I:=\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx\overset{\large x\to\frac{1-x}{1+x}}=\int_0^1 \frac{1}{1+x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx$$ $$J:=\int_0^1 \frac{1}{1-x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx$$ $$I=\frac12\left((I+J)+(I-J)\right)=\boxed{\int_0^1 \frac{1}{1+x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx=\frac{13}{8}\zeta(2)\ln 2-\frac{33}{32}\zeta(3)}$$ $$J=\frac12\left((I+J)-(I-J)\right)=\boxed{\int_0^1 \frac{1}{1-x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx=\frac58\zeta(2)\ln 2+\frac{19}{32}\zeta(3)}$$
$$I-J=-2\int_0^1\frac{x}{1-x^2}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx\overset{\large \frac{1-x^2}{1+x^2}\to x}=\color{blue}{\int_0^1 \frac{\operatorname{Li}_2(x)}{1+x}dx}-\color{red}{\int_0^1 \frac{\operatorname{Li}_2(x)}{x}dx}$$ $$\overset{\color{blue}{IBP}}=\color{blue}{\zeta(2)\ln 2+\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx}-\color{red}{\sum_{n=1}^\infty \frac{1}{n^2}\int_0^1 x^{n-1}dx}=\boxed{\zeta(2)\ln 2-\frac{13}{8}\zeta(3)}$$ See here for the first integral.
$$I+J=2\int_0^1 \frac{1}{1-x^2}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx\overset{\large \frac{1-x^2}{1+x^2}\to x}=\int_0^1 \frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}dx$$ Now we're going to use the following result: $$\int_0^1 \frac{\ln t}{t-\frac{1}{x}}dt=- \sum_{n=1}^\infty x^n\int_0^1 t^{n-1}\ln t\,dt=-\sum_{n=1}^\infty x^n \left(\frac{1}{n}\right)'=\sum_{n=1}^\infty \frac{x^n}{n^2}=\operatorname{Li}_2(x)$$ $$\Rightarrow I+J=- \int_0^1\int_0^1 \frac{\ln t}{(1-tx)\sqrt{1-x^2}}dx dt\overset{x\to \sin x}=- \int_0^1\ln t\int_0^\frac{\pi}{2} \frac{1}{1-t\sin x}dxdt$$ $$=-\int_0^1\ln t \ \frac{\frac{\pi}{2}+\arcsin t}{\sqrt{1-t^2}} dt\overset{t=\sin x}=-\int_0^\frac{\pi}{2}\left(\frac{\pi}{2}+x\right)\ln (\sin x) dx=\boxed{\frac94\zeta(2)\ln 2 -\frac7{16}\zeta(3)}$$ See here and here for the above integrals.
Also to prove a leftover integral from above we will consider: $$f(a)=\int_0^\pi \frac{dx}{1+\sin a\sin x}\overset{\tan\frac{x}{2}=y}=\int_0^\infty \frac{dy}{1+y^2+2\sin a y}$$ $$=\int_0^\infty \frac{dy}{(\sin a+y)^2+\cos^2 a}=\frac{1}{\cos a}\arctan\left(\frac{\sin a+y}{\cos a}\right)\bigg|_0^\infty=\frac{\frac{\pi}{2}-a}{\cos a}$$ $$\Rightarrow \int_0^\frac{\pi}{2}\frac{1}{1-t\sin x}dx=\frac12 f(-\arcsin t)=\frac{\frac{\pi}{2}+\arcsin t}{\sqrt{1-t^2}}$$
Starting with the identity
$$4\int_0^x\frac{t\ln(1-t)}{1+t^2}\ dt=\frac12\ln^2(1+x^2)-2\operatorname{Li}_2(x)+\frac12\operatorname{Li}_2(-x^2)+\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)\tag1$$
Divide both sides by $1+x$ then integrate from $x=0$ to $1$ we get
$$4\underbrace{\int_0^1\int_0^x\frac{t\ln(1-t)}{1+t^2}\ dtdx}_{\mathcal{\Large I_1}}$$ $$\small{=\frac12\underbrace{\int_0^1\frac{\ln^2(1+x^2)}{1+x}dx}_{\mathcal{\Large I_2}}-2\underbrace{\int_0^1\frac{\operatorname{Li}_2(x)}{1+x}dx}_{\mathcal{\Large I_3}}+\frac12\underbrace{\int_0^1\frac{\operatorname{Li}_2(-x^2)}{1+x}dx}_{\mathcal{\Large I_4}}+\underbrace{\int_0^1\frac1{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx}_{\mathcal{\Large I}}}\tag2$$
$$\mathcal{I}_1=\int_0^1\int_0^x\frac{t\ln(1-t)}{1+t^2}\ dtdx=\int_0^1\frac{t\ln(1-t)}{1+t^2}\left(\int_t^1\frac{dx}{1+x}\right)dt$$
$$=\int_0^1\frac{t\ln(1-t)}{1+t^2}(\ln2-\ln(1+t))dt$$ $$=\ln2\underbrace{\int_0^1\frac{t\ln(1-t)}{1+t^2}dt}_{\mathcal{\Large I_1'}}-\underbrace{\int_0^1\frac{t\ln(1-t)\ln(1+t)}{1+t^2}dt}_{\mathcal{\Large I_1''}}$$
By using $(1)$ we have
$$\mathcal{I_1'}=\frac18\ln^22-\frac5{16}\zeta(2)$$
For $\mathcal{I_1''}$ , we use $ab=\frac14(a+b)^2-\frac14(a-b)^2$
$$\mathcal{I_1''}=\frac14\underbrace{\int_0^1\frac{t\ln^2(1-t^2)}{1+t^2}dt}_{t^2=x}-\frac14\underbrace{\int_0^1\frac{t\ln^2\left(\frac{1-t}{1+t}\right)}{1+t^2}dt}_{(1-t)/(1+t)=x}$$
$$=\frac18\int_0^1\frac{\ln^2(1-x)}{1+x}dx-\frac14\int_0^1\frac{\ln^2x}{1+x}dx+\frac14\underbrace{\int_0^1\frac{x\ln^2x}{1+x^2}dx}_{x^2\mapsto x}$$
$$=\frac18\int_0^1\frac{\ln^2(1-x)}{1+x}dx-\frac7{32}\int_0^1\frac{\ln^2x}{1+x}dx$$
$$=\frac18\left(\frac13\ln^32-\ln2\zeta(2)+\frac74\zeta(3)\right)-\frac{7}{32}\left(\frac32\zeta(3)\right)$$
$$=\frac1{24}\ln^32-\frac18\ln2\zeta(2)-\frac{7}{64}\zeta(3)$$ combine $\mathcal{I_1'}$ and $\mathcal{I_1''}$ to get
$$\boxed{\mathcal{I}_1=\frac1{12}\ln^32-\frac3{16}\ln2\zeta(2)+\frac7{64}\zeta(3)}$$
Evaluation of $\mathcal{I}_2$: Its already calculated here
$$\boxed{\mathcal{I}_2=\frac23\ln^32-\pi G-\frac14\ln2\zeta(2)+\frac52\zeta(3)}$$
Evaluation of $\mathcal{I}_3$
$$\mathcal{I}_3\overset{IBP}{=}\ln(2)\zeta(2)+\int_0^1\frac{\ln(1+x)\ln(1-x)}{x}dx=\boxed{\ln2\zeta(2)-\frac58\zeta(3)}$$
where the last result follows from this solution.
Evaluation of $\mathcal{I}_4$
$$\mathcal{I}_4\overset{IBP}{=}-\frac12\ln(2)\zeta(2)+2\int_0^1\frac{\ln(1+x)\ln(1+x^2)}{x}dx$$
$$=-\frac12\ln(2)\zeta(2)-2\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1 x^{2n-1}\ln(1+x)dx$$
$$=-\frac12\ln(2)\zeta(2)-2\sum_{n=1}^\infty\frac{(-1)^n}{n}\left(\frac{H_{2n}-H_n}{2n}\right)$$
$$=-\frac12\ln(2)\zeta(2)-4\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^2}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}$$
$$=-\frac12\ln(2)\zeta(2)-4\Re\sum_{n=1}^\infty\frac{(i)^nH_{n}}{n^2}+\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}$$
$$\boxed{=\pi G-\frac12\ln2\zeta(2)-\frac{33}{16}\zeta(3)}$$
where the last result follows from using the generating function
$$\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$
By substituting the boxed results in$(2)$ we obtain the claimed closed form.
Actually this approach is very similar to my approach in the question body as we see common integrals but I think its more interesting to to use the identity in $(1)$.