Here in my answer I got the real part for the polylogarithm function at $1+i$ for natural $n$

$$ \Re\left(\text{Li}_n(1+i)\right)=\left(\frac{-1}{4}\right)^{n+1}A_n-B_n $$

where $$ B_n=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \frac{2\eta(2k)}{(n-2k)!}\operatorname{Re}\left[\left(\frac{\ln2}{2}-\frac{3\pi}{4}i\right)^{n-2k}\right] $$ which is easy to evaluate the closed form by using Dirichlet eta function.

But $A_n$ is defined as
$$ A_n=4 \text{Li}_n \left(\frac{-1}{4} \right)-\Phi \left(-\frac{1}{4},n,\frac{3}{4}\right)+2\Phi \left(-\frac{1}{4},n,\frac{1}{4}\right) $$
by using Lerch zeta function, so it has an integral representation
$$ A_n=\frac{(-4)^{n+1}}{(n-1)!} \int_0^1 \frac{x}{x^2+1} \left(\ln(1-x) \right)^{n-1}dx.$$
I got closed form of this integral for $n=2,3,4$: precisely
$$
\begin{split}
A_2 &=\frac{10\pi^2}{3}-8\ln^22,\\
A_3 &=140\zeta(3)+\frac{16}{3}\ln^32-\frac{20\pi^2}{3} \ln2\\
A_4 &=320 \text{Li}_4\left(\frac{1}{2}\right)+\frac{343}{90}\pi^4+\frac{32}{3}\ln^42-\frac{20\pi^2}{3}\ln^22
\end{split}$$
**Question**: is it possible to find a closed form expression for $A_n$ for every natural $n$?

**Note**: by closed form I meant way to simplify the Lerch $\Phi$ function and use only the polylogarithm function (without complex numbers) or Clausen function, because some values of this functions are known as constants and it cant be express simply as constants just like $Li_4\left(\frac{1}{2}\right)$.

this question asked on MSE.

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