Triple integral of a long expression

I have to integrate this:

$$\int_{-1}^1 \int_{0}^{2} \int_{-2}^2 [1+x^2 z^2 + x y \cos(x^4 + y^4 + z^4) ]\;\operatorname d\!z \;\operatorname d\!y \operatorname d\!x$$

I don't know how to do it. I tried with $u = x^4 + y^4 + z^4$, but it doesn't work.

Could someone point me in the right direction?

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The function $xy\cos(x^4+y^4+z^4)$ is odd w.r.t. $x$. Thus,
$\displaystyle\int\limits_{-1}^0 \int\limits_{0}^{2} \int\limits_{-2}^2 xy\cos(x^4+y^4+z^4) \,dz\,dy\,dx = - \int\limits_{0}^1 \int\limits_{0}^{2} \int\limits_{-2}^2 xy\cos(x^4+y^4+z^4) \,dz\,dy\,dx$.
Move everything to the left side to get, $\displaystyle\int\limits_{-1}^1 \int\limits_{0}^{2} \int\limits_{-2}^2 xy\cos(x^4+y^4+z^4) \,dz\,dy\,dx = 0$.
Now, all thats left is the $1+x^2z^2$ term.