Conjectured closed form for $\operatorname{Li}_2\!\left(\sqrt{2-\sqrt3}\cdot e^{i\pi/12}\right)$ There are few known closed form for values of the dilogarithm at specific points. Sometimes only the real part or only the imaginary part of the value is known, or a relation between several different values is known: [1][2][3][4][5][6]. Discovering a new identity of this sort is always of a great interest.
I numerically discovered the following conjectured closed form and now am looking how to prove it:
$$\operatorname{Li}_2\!\left(\sqrt{2-\sqrt3}\cdot e^{i\pi/12}\right)=\operatorname{Li}_2\!\left(\tfrac12+i\left(1-\tfrac{\sqrt3}2\right)\right)\stackrel{\color{gray}?}=\\\frac{23\pi^2}{288}-\frac18\ln^2\!\left(2+\sqrt3\right)+i\left[\frac{2\;\!G}3+\frac{\pi^2}{12\sqrt3}-\frac\pi{24}\ln\left(2+\sqrt3\right)-\frac1{8\sqrt3}\psi^{\small(1)}\!\left(\tfrac13\right)\right]$$
If you have got any ideas please share them.
$G$ is the Catalan constant, and $\psi^{\small(1)}(z)$ is the trigamma function.

To respond questions in comments: my approach to find closed forms like this is to evaluate the expression with hundreds or sometimes thousands digits of precision and then use integer relation algorithms to find a matching linear combination with rational coefficients from a pool of candidate terms. The tricky part is that the pool cannot be too big, otherwise the algorithms begin to work extremely slowly or fail and produce false positives, so I need some intuition to select a restricted set of candidates that are likely to appear in a closed form I'm looking for. I usually select candidates from known closed forms of similar expressions, or from known closed forms of integrals that contain the expression in question along with other simpler terms (its "siblings"). Sometimes I also add some variations of existing terms, e.g. multiply them by a simple irrational coefficient like $\sqrt3$, or replace occurences of $\ln2$ with $\ln3$. When I find a matching linear combination, I verify it with much higher precision (usually tens or hundreds of thousand of decimal digits) to reduce chances that it is an accidental close but non-exact expression.
 A: We use the main three steps from my answer to one of your previous questions.

$$\operatorname{Li}_2(z) = -\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac{1}{2}\ln^2(1-z), \quad z \notin (1,\infty).\tag{$\diamondsuit$}$$
  $$\operatorname{Li}_2\left(e^{i\theta}\right) = \operatorname{Sl}_2(\theta)+i\operatorname{Cl}_2(\theta), \quad \theta \in [0,2\pi).\tag{$\heartsuit$}
$$
  $$
\operatorname{Sl}_2(\theta) = \frac{\pi^2}{6}-\frac{\pi\theta}{2}+\frac{\theta^2}{4}, \quad \theta \in [0,2\pi).\tag{$\spadesuit$}
$$
  For definitions and notations see my previous answer.


Let $z:=\tfrac{1}{2}+i\left(1-\tfrac{\sqrt3}{2}\right).$ Note that
$$
\frac{z}{z-1} = -\frac{\sqrt 3}{2} - \frac{i}{2},
$$
and therefore $\left|\frac{z}{z-1}\right|=1$.
The equation
$$
e^{i\theta} = \frac{z}{z-1} = -\frac{\sqrt 3}{2} - \frac{i}{2}
$$
has the only solution $\theta=\tfrac76 \pi$ in $[0,2\pi)$.
Because of $(\diamondsuit)$ and $(\heartsuit)$ we have
$$
\operatorname{Li}_2(z) = -\color{red}{\operatorname{Sl}_2(\theta)} - i \color{green}{\operatorname{Cl}_2(\theta)} - \color{blue}{\frac{1}{2}\ln^2(1-z)},
$$
for $z=\tfrac{1}{2}+i\left(1-\tfrac{\sqrt3}{2}\right)$ and $\theta =\tfrac76 \pi$.
For the logarithm term we get
$$
\Re{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{1}{8}\ln^2\left(2-\sqrt{3}\right)-\frac{\pi^2}{288}
$$
and
$$
\Im{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = -\frac{\pi}{24}\ln\left(2-\sqrt{3}\right).
$$
We know that $\color{red}{\operatorname{Sl}_2(\theta)}$ and $\color{green}{\operatorname{Cl}_2(\theta)}$ are real quantities. By using $(\spadesuit)$ for the SL-type Clausen term we get
$$
\color{red}{\operatorname{Sl}_2(\theta)} =-\frac{11\pi^2}{144}.
$$
Now we could obtain your conjectured closed-form for the real part:
$$\Re\left[\operatorname{Li}_2(z)\right] = -\color{red}{\operatorname{Sl}_2(\theta)} - \Re{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} = \frac{23\pi^2}{288}-\frac{1}{8}\ln^2\left(2-\sqrt{3}\right).$$
For the imaginary part we have
$$\begin{align}\Im\left[\operatorname{Li}_2(z)\right] &= -\color{green}{\operatorname{Cl}_2(\theta)} - \Im{\left[\color{blue}{\frac{1}{2}\ln^2(1-z)}\right]} \\ &= -\operatorname{Cl}_2\left(\tfrac{7}{6}\pi\right)+\frac{\pi}{24}\ln\left(2-\sqrt{3}\right).\end{align}$$
Now by using the relationship between Clausen function and polygamma function and by using some polygamma tricks, we can get that
$$\operatorname{Cl}_2 \left(\frac{7}{6}\pi\right) = -\frac{2}{3}G - \frac{\pi^2}{12\sqrt{3}} + \frac{1}{8\sqrt{3}} \psi ^{(1)}\left(\frac{1}{3}\right).$$
This completes the proof.
