Closed Form for the Imaginary Part of $\text{Li}_3\Big(\frac{1+i}2\Big)$ 
$\qquad\qquad$ Is there any closed form expression for the imaginary part of $~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)$ ?


Motivation: We already know that $~\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3)$, 
so I was wondering whether a similar closed form expression might also exist for its
imaginary part as well. Thank you !

Apparently, $~\Im~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)~+~\Im~\text{Li}_3(1+i)~=~\dfrac7{128}\cdot\pi^3~+~\dfrac3{32}\cdot\pi\cdot\ln^22,~$ so the question
is equivalent to asking for the closed form of the imaginary part of $~\Im~\text{Li}_3(1+i).$

I'm seriously late for this party, but to complement Cleo's answer, if we consider a hypergeometric as a closed-form, then using a different hypergeometric, the family is,
$$\Im\left[\operatorname{Li}_\color{red}2\big(1+i\big)\right] =\sqrt2 \;{_3F_2}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)$$
$$\Im\left[\operatorname{Li}_\color{red}2\left(\frac{1+i}2\right)\right] =\frac{-1}{\sqrt2} \;{_3F_2}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)+\frac{3K}2$$
with Catalan's constant $K=\Im\left[\operatorname{Li}_\color{blue}2(i)\right]$ and,
$$\Im\left[\operatorname{Li}_\color{red}3\big(1+i\big)\right] =\sqrt2 \;{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)-\frac{\pi^3}{192}$$
$$\Im\left[\operatorname{Li}_\color{red}3\left(\frac{1+i}2\right)\right] =-\sqrt2 \;{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)+\frac{23\pi^3}{384}+\frac{3\pi\ln^2 2}{32}$$
where $\frac{\pi^3}{32}=\Im\left[\operatorname{Li}_\color{blue}3(i)\right]$. It is tempting to speculate what comes next.

Edit: July 2019. With insights from this answer, it turns out,
$$\Im\left[\operatorname{Li}_\color{red}4\big(1+i\big)\right] =\sqrt2 \;{_5F_4}\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\;\tfrac12\right)-\frac{\pi^3\ln 2}{382}-\frac{\pi\zeta(3)}8+\frac{\Im\left[\operatorname{Li}_\color{blue}4(i)\right]}4$$
P.S. The first two hypergeometrics actually have a closed-form, but I didn't want to clutter my answer.
