Recall an integral, which I already confirmed that $$ \int_{0}^{1} \frac{\arcsin x\ln x}{1+x} \,\mathrm{d}x = -\frac{5\pi^3}{48}-2G\ln2-\frac{\pi}{2}\ln^22 +4\Im\operatorname{Li}_3(1+i) $$ whose $\arctan$ analog may be more common in typical logarithmic integral, yet I can not find any old post on this one.
We can obtain this result by establishing a clear identity $$ \int_{0}^{1} \frac{\arcsin x\ln x}{1+x} \,\mathrm{d}x = \frac{\pi^3}{48} - \frac{\pi}{4}\ln^22 - \frac{4}{3}\int_{0}^{1} \frac{\arctan x\ln(1+x)}{x} \,\mathrm{d}x \tag{*} $$ where recall a typical identity from here with its equivalent Euler Sum $$ \frac{2}{3}\int_{0}^{1} {\frac{\arctan x \ln(1+x)}{x} \,\mathrm{d}x} = \int_{0}^{1} {\frac{\arctan x \ln(1+x^2)}{x} \,\mathrm{d}x} = 2\sum_{n=1}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{(2n+1)^2}} $$ $$ \sum_{n=0}^{\infty} {\frac{(-1)^{n+1} H_{2n+1}}{(2n+1)^2}} = -\Im\sum_{n=1}^{\infty} {\frac{i^{n} H_{n}}{n^2}} = \frac{\pi}{16}\ln^2\!2 + \frac{G}{2}\ln2 - \Im\operatorname{Li}_{3}(1+i) $$ which give us the result from above, note that the tricky yet only polylog argument comes from this Euler Sum.
However, the path I get to $(*)$ is very tedious and inconvenient, which briefly represented here, integrating by parts $$ \begin{aligned} \int_{0}^{1} {\frac{\arcsin x \ln x}{1+x} \,\mathrm{d}x} = &\; \frac{\pi}{2}\int_{0}^{1} {\frac{\ln t}{1+t} \,\mathrm{d}t} - \int_{0}^{1} {\frac{1}{\sqrt{1-x^2}}\left( \int_{0}^{x} {\frac{\ln t}{1+t} \,\mathrm{d}t} \right) \mathrm{d}x}\\ = & -\frac{\pi^3}{24} - \int_{0}^{1} {\left( \int_{0}^{1} {\frac{x\ln(tx)}{(1+tx)\sqrt{1-x^2}} \,\mathrm{d}t} \right) \mathrm{d}x}\\ = & -\frac{\pi^3}{24} - \int_{0}^{1} {\ln t\left( \int_{0}^{1} {\frac{x\,\mathrm{d}x}{(1+tx)\sqrt{1-x^2}}} \right) \mathrm{d}t} \\ & - \int_{0}^{1} {\frac{\ln x}{\sqrt{1-x^2}}\left( \int_{0}^{1} {\frac{x\,\mathrm{d}t}{1+tx}} \right) \mathrm{d}x}\\ = & -\frac{\pi^3}{24} - \frac{\pi}{2}\int_{0}^{1} {\left(1-\frac{1}{\sqrt{1-t^2}}\right)\frac{\ln t}{t}\,\mathrm{d}t} -\int_{0}^{1} {\frac{\arcsin t\ln t}{t\sqrt{1-t^2}} \,\mathrm{d}t} \\ & - \int_{0}^{1} {\frac{\ln x\ln(1+x)}{\sqrt{1-x^2}} \,\mathrm{d}x} \\ \end{aligned} $$ where the first integral is relatively easy to solve $$ \int_{0}^{1} {\left(1-\frac{1}{\sqrt{1-t^2}}\right)\frac{\ln t}{t}\,\mathrm{d}t} = \frac{\pi^2}{24} - \frac1{2}\ln^22 $$ the second one can be find in this post, with $t=\sin x$ and Weierstrass subs we could have $$ \begin{aligned} \int_{0}^{1} {\frac{\arcsin t\ln t}{t\sqrt{1-t^2}} \,\mathrm{d}t} & = \int_{0}^{\pi/2} \frac{x\ln(\sin x)}{\sin x} \,\mathrm{d}x = -4\int_{0}^{1} \frac{\ln u\ln(1-u)}{1+u^2}\,\mathrm{d}u \\ & = -2\int_{0}^{1} \frac{\ln u}{1+u^2}\ln\left(\frac{(1-u)^2}{1+u^2}\right)\mathrm{d}u - 2\int_{0}^{1} \frac{\ln u\ln(1+u^2)}{1+u^2}\,\mathrm{d}u\\ & = -2G\ln2 - 2\int_{0}^{1} \frac{\ln u\ln(1+u^2)}{1+u^2}\,\mathrm{d}u \end{aligned} $$ note that the Catalan term is not trivial, I find no easy way. Then the third one can also deal with Weierstrass subs $$ \begin{aligned} \int_{0}^{1} {\frac{\ln x\ln(1+x)}{\sqrt{1-x^2}} \,\mathrm{d}x} = &\; \int_{0}^{\pi/2} \ln(\sin x)\ln(1+\sin x) \,\mathrm{d}x \\ = & -\frac{\pi^3}{12} + 2G\ln 2 + \frac{\pi}{2}\ln^2\!2 + 2\int_{0}^{1} \frac{\ln u\ln(1+u^2)}{1+u^2}\,\mathrm{d}u \\ & + \frac{4}{3}\int_{0}^{1} \frac{\arctan u\ln(1+u)}{u}\,\mathrm{d}u \end{aligned} $$ This result takes me lots of step to get (as well, can not find post on it), where the most tricky part is the last integral, comes from those typical high-order logarithmic integral like $$ \int_{0}^{1} \frac{\ln^2(1+u)}{1+u^2}\,\mathrm{d}u, \quad \int_{0}^{1} \frac{\ln u\ln(1+u)}{1+u^2}\,\mathrm{d}u $$ yet, except this component, other constants and parts in this prove canceled in a really coincidence way, makes me doubt that there maybe an easier way I missed to reach identity $(*)$, hope somebody else could discover.
Thanks in advance for any suggestion.
