How to prove the following identity $$ \operatorname{Li}_{3}\left(\frac{1}{2} \right) = \sum_{n=1}^{\infty}\frac{1}{2^n n^3}= \frac{1}{24} \left( 21\zeta(3)+4\ln^3 (2)-2\pi^2 \ln2\right)\,?$$
Where $\operatorname{Li}_3 (x)$ is the trilogarithm also the result from above can be found here in $(2)$.
In particular for $x=\frac12$ we have:$$\operatorname{Li}_3\left(\frac12\right)=\sum_{n=1}^\infty \frac{1}{n^3}\frac{1}{2^n}=\frac14 \sum_{n=1}^\infty \frac{1}{2^{n-1}} \int_0^\frac12 x^{n-1}\ln^2 xdx=\frac14 \int_0^1 \frac{\ln^2 x}{1-\frac{x}{2}}dx$$
First, evaluate \begin{aligned} \operatorname{Li}_2\left(\frac12\right)&=-\int_0^{1/2}\frac{\ln(1-x)}x{d}x \overset{x\to 1-x}=\int_0^{1/2}\frac{\ln x}{1-x}\overset{ibp}{dx} -\int_0^1\frac{\ln x}{1-x}{d}x\\ &=-\operatorname{Li}_2\left(\frac12\right) -\ln^22 -\left(-\frac{\pi^2}6 \right)=-\frac12\ln^22+\frac{\pi^2}{12} \end{aligned}
and then express \begin{align} \operatorname{Li}_3\left(\frac12\right)&= \int_0^{1/2}\frac{\operatorname{Li}_2(x)}x{d}x \overset{ibp}=-\ln2 \operatorname{Li}_2\left(\frac12\right)+\int_0^{1/2}\frac{\ln x\ln(1-x)}{x}\>\overset{ibp}{dx} \\ &=-\frac{\pi^2}{12}\ln2 + \frac12\int_0^{1/2}\frac{\ln^2 x}{1-x} dx \end{align} where \begin{aligned} \int_0^{1/2}\frac{\ln^2 x}{1-x} dx& \overset{x=\frac t{1+t}}=\int_0^{1}\frac{\ln^2(1+t)}{1+t}dt +\int_0^{1}\frac{\ln^2t}{1+t}dt-2\int_0^{1}\frac{\ln t\ln (1+t)}{1+t}\>\overset{ibp}{dt}\\ &=\frac13\ln^32+ \left( \int_0^{1}\frac{\ln^2t}{1-t}dt - \int_0^{1}\frac{2t\ln^2t}{1-t^2}\overset{t^2\to t}{dt}\right)+\int_0^{1}\frac{\ln^2(1+t)}{t}\> \overset{t\to\frac t{1-t}}{dt}\\ &= \frac13\ln^32+ \frac34 \int_0^{1}\frac{\ln^2t}{1-t}dt+\int_{1/2}^{1}\frac{\ln^2t}{1-t}dt+ \int_{1/2}^{1}\frac{\ln^2t}{t} dt\\ &= \frac23\ln^32+ \frac74\int_0^{1}\frac{\ln^2t}{1-t}dt-\int_{0}^{1/2}\frac{\ln^2t}{1-t}dt\\ &= \frac13 \ln^32+ \frac78 \int_0^{1}\frac{\ln^2t}{1-t}dt = \frac13 \ln^32+ \frac74\zeta(3) \end{aligned} As a result $$\operatorname{Li}_3\left(\frac12\right) = -\frac{\pi^2}{12}\ln2 + \frac16 \ln^32+ \frac78 \zeta(3) $$