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 $(2)$.