Simplifying real part of hypergeometric function with complex parameters I am looking for a simpler representation of the following hypergeometric function with complex parameters in terms of more basic functions and manifestly real parameters:

$$\mathcal{I}{\left(z\right)}:=\Re{\left[{_2F_1}{\left(i,1;1+i;-z\right)}\right]}=???\tag{1}$$

where $0<z<1$.
This question is based on this response of mine that I began almost a year ago but was never able to complete. The final step required the evaluation of the series
$$\sum_{n=0}^{\infty}\frac{r^n}{n^2+1}$$
at the particular value $r=-\sqrt[3]{2}+\sqrt{2^{2/3}-1}$. A closed form value to $(1)$ would allow me to complete that solution.

My attempt:
My best idea for a first step was to find an explicit integral representation. I use the following integral representations for the Gauss hypergeometric function ${_2F_1}$,
$${_2F_1}{\left(a,b;c;z\right)}=\frac{1}{\Gamma{\left(b\right)}}\int_{0}^{\infty}e^{-t}t^{b-1}\,{_1F_1}{\left(a;c;zt\right)}\,\mathrm{d}t;~~~\small{\left[Re{(b)}>0\right]}\tag{2}$$
the Kummer confluent hypergeometric function ${_1F_1}$,
$${_1F_1}{\left(a;b;z\right)}=\frac{1}{\Gamma{\left(a\right)}}\int_{0}^{\infty}e^{-t}t^{a-1}\,{_0F_1}{\left(;b;zt\right)}\,\mathrm{d}t;~~~\small{\left[\Re{(a)}>0\right]}\tag{3}$$
and the confluent hypergeometric function ${_0F_1}$,
$${_0F_1}{\left(;b;z\right)}=\frac{\Gamma{\left(b\right)}}{\sqrt{\pi}\,\Gamma{\left(b-\frac12\right)}}\int_{-1}^{1}\frac{\left(1-t^2\right)^{b-\frac32}}{e^{2t\sqrt{z}}}\,\mathrm{d}t;~~~\small{\left[\Re{(b)}>\frac12\right]}.\tag{4}$$
I also use the following relation for Kummer confluent hypergeometric functions:
$${_1F_1}{\left(b-a;b;z\right)}=e^{z}\,{_1F_1}{\left(a;b;-z\right)}.\tag{5}$$
Denote the function ${_2F_1}{\left(i,1;1+i;-z\right)}$ by $g{(z)}$. For $z\in\mathbb{R}^{+}$, we have
$$\begin{align}
g{\left(z\right)}
&={_2F_1}{\left(i,1;1+i;-z\right)}\\
&=\int_{0}^{\infty}e^{-y}\,{_1F_1}{\left(i;1+i;-zy\right)}\,\mathrm{d}y\\
&=\int_{0}^{\infty}e^{-y}e^{-zy}\,{_1F_1}{\left(1;1+i;zy\right)}\,\mathrm{d}y\\
&=\int_{0}^{\infty}e^{-\left(1+z\right)y}\,{_1F_1}{\left(1;1+i;zy\right)}\,\mathrm{d}y\\
&=\int_{0}^{\infty}\mathrm{d}y\,e^{-\left(1+z\right)y}\int_{0}^{\infty}\mathrm{d}x\,e^{-x}\,{_0F_1}{\left(;1+i;zyx\right)}\\
&=\int_{0}^{\infty}\mathrm{d}y\,e^{-\left(1+z\right)y}\int_{0}^{\infty}\mathrm{d}x\,e^{-x}\frac{\Gamma{\left(1+i\right)}}{\sqrt{\pi}\,\Gamma{\left(\frac12+i\right)}}\int_{-1}^{1}\mathrm{d}w\,\frac{\left(1-w^2\right)^{-\frac12+i}}{e^{2w\sqrt{zyx}}}\\
&=\frac{\Gamma{\left(1+i\right)}}{\sqrt{\pi}\,\Gamma{\left(\frac12+i\right)}}\int_{0}^{\infty}\mathrm{d}y\,e^{-\left(1+z\right)y}\int_{0}^{\infty}\mathrm{d}x\,e^{-x}\int_{-1}^{1}\mathrm{d}w\,\frac{e^{i\ln{\left(1-w^2\right)}}}{e^{2w\sqrt{zyx}}\sqrt{1-w^2}}.\\
\end{align}$$
The inner integral looks similar to representations of Bessel functions, but I'm not too familiar with those. Does anybody have any ideas how to proceed?
 A: Some notes: 
\begin{align}
{}_{2}F_{1}(i, 1; 1+i; x) = \sum_{n=0}^{\infty} \frac{(i)_{n} \, x^{n}}{(1+i)_{n}}.
\end{align}
Now,
\begin{align}
\frac{(i)_{n}}{(1+i)_{n}} = \frac{\Gamma(i+1) \, \Gamma(n+i)}{\Gamma(i) \, \Gamma(n+i+1)} = \frac{i}{n+i} = \frac{1}{1-i n} = \frac{1+i n}{n^{2}+1}
\end{align}
which leads to
\begin{align}
{}_{2}F_{1}(i, 1; 1+i; x) = \sum_{n=0}^{\infty} \frac{(1 + in) \, x^{n}}{n^{2} + 1}
\end{align}
and, $S(x)$ is the real part of,
\begin{align}\tag{1}
S(x) = \sum_{n=0}^{\infty} \frac{x^{n}}{n^{2}+1}.
\end{align}
It can be quickly determined that $S(x)$ satisfies the differential equation
\begin{align}
x^{2} S'' + x S' + S = \frac{1}{1-x}
\end{align}
with the conditions $S(0) =1$ and $S'(0) = 0$. The solution of which is of the form
\begin{align}
S(x) = A \cos(\ln x) + B \sin(\ln x) + f(x).
\end{align}
The differential equation seems to lead to more problems. Another form can be obtained and is given by
\begin{align}
S(x) = \int_{0}^{\infty} \frac{\sin t \, dt}{1 - x e^{-t}}.
\end{align}
It is evident that complications arise when $x=0$. 
