Evaluate $2i\int_{-1}^1\frac{\operatorname{Li}_2(-ix^2)-\operatorname{Li}_2(ix^2)}{x^2}dx$, where $\operatorname{Li}_2(s)$ is the polylogarithm Let $\sqrt{-1}=i$ the imaginary unit. 
Playing with Wolfram Alpha online calculator I know that $$I:=\int_{-1}^1\left(\int_{-1}^1\left(\int_{-1}^1\frac{1}{1-ix^2yz}dy\right)dz\right)dx=\int_{-1}^1\left(\int_{-1}^1\frac{2\arctan(x^2z)}{x^2z}dz\right)dx,\tag{1}$$
and this RHS equals to $$\int_{-1}^1\frac{2i(\operatorname{Li}_2(-ix^2)-\operatorname{Li}_2(ix^2))}{x^2}dx\tag{2}$$ where $\operatorname{Li}_2(s)$ is the polylogarithm (see if you need the definition from this MathWorld).

Question. I would like to get, if it is possible a closed-form for $I$, thus I need to evaluate 
  $$\int_{-1}^1\frac{2i(\operatorname{Li}_2(-ix^2)-\operatorname{Li}_2(ix^2))}{x^2}dx.$$
  Can you finish this task providing a simplified form of the result? Many thanks.

As summary, as I've said using Wolfram Alpha online calculator I've calculated the definite integrals
int 1/(1-i x^2 y z)dy, from y=-1 to 1
and 
int (2 tan^(-1)(x^2 z))/(x^2 z)dz, from z=-1 to 1
Also we know the indefinite integral
int  (2 i (Polylog[2,-i x^2] - Polylog[2,i x^2]))/x^2 dx
I believe also that
$$2i(\frac{-\operatorname{Li}_2(-ix^2)+\operatorname{Li}_2(ix^2)}{x} \Big|_{-1}^1=-8C,$$ where $C$ is the Catalan's constant, and $$(\frac{2i}{x}2(-\log(1-ix^2)+\log(1+ix^2))\Big|_{-1}^1=-4\pi.$$
Feel free to add a different approach than previous to evaluate $I$, or our integral in $(2)$.
 A: Hint. By integrating
$$
\frac{2i(\operatorname{Li}_2(-ix^2)-\operatorname{Li}_2(ix^2))}{x^2}
$$ termwise, which is allowed here, one gets
$$
\int_{-1}^1\frac{2i(\operatorname{Li}_2(-ix^2)-\operatorname{Li}_2(ix^2))}{x^2}dx=4i\cdot\sum_{n=1}^\infty\frac{i^n\left((-1)^n-1\right)}{(2n-1)n^2}
$$ by using a partial fraction decomposition, the preceding series admits a closed form, one gets
$$
\int_{-1}^1\frac{2i(\operatorname{Li}_2(-ix^2)-\operatorname{Li}_2(ix^2))}{x^2}dx=4\pi\sqrt{2}+4\sqrt{2}\ln\left(3+2\sqrt{2}\right)-4\pi-8 C
$$ where $C$ is the Catalan constant.
A: There is an additional approximate solution for this integral. 
For $-1<x<1$ the  expression $i(Li_2(-ix^2) - Li_2(ix^2))$ can be approximated with $2x^2$.
As a consequence, we can write $\int_{-1}^1\frac{2 (2x^2)}{x^2}dx=\int_{-1}^14dx=8$. This comes out with the cost of an inaccuracy, that is this approximation calculates a number of 8 instead of 7.84904 which is calculated using numerical integration with mathematica.
A: In the same spirit as Dimitris U., let us consider the Taylor series built around $x=0$.
You will have $$\frac{2 i \left(\text{Li}_2\left(-i x^2\right)-\text{Li}_2\left(i
   x^2\right)\right)}{x^2}=4 \sum_{k=0}^\infty (-1)^k \frac{x^{4k}}{(2k+1)^2}$$ leading to $$\int_{-1}^1\frac{2 i \left(\text{Li}_2\left(-i x^2\right)-\text{Li}_2\left(i
   x^2\right)\right)}{x^2}\,dx=8 \sum_{k=0}^\infty  \frac{ (-1)^k}{(2 k+1)^2\, (4 k+1)}$$ and, as Olivier Oloa answered, take into account the fact that $$\frac{1}{(2 k+1)^2 (4 k+1)}=\frac{4}{4 k+1}-\frac{2}{2 k+1}-\frac{1}{(2 k+1)^2}$$ Using the summation, it would required many terms even for a moderate accuracy.
