Integral of determinant Good evening. I need help with this task
$$
\int\limits_{-\pi}^\pi\int\limits_{-\pi}^\pi\int\limits_{-\pi}^\pi{\det}^2\begin{Vmatrix}\sin \alpha x&\sin \alpha y&\sin \alpha z\\\sin \beta x&\sin \beta y&\sin \beta z\\\sin \gamma x&\sin \gamma y&\sin \gamma z\end{Vmatrix} \text{d}x\,\text{d}y\,\text{d}z
$$
where $\alpha,\beta,\gamma$ are integers.
Computations are horrible, I gave up.

Exact solution and Mathematica code to produce it (< 10 seconds computing time):
M = {{Sin[a x], Sin[a y], Sin[a z]},
     {Sin[b x], Sin[b y], Sin[b z]},
     {Sin[c x], Sin[c y], Sin[c z]}};
Expand[Det[M]^2] /. Plus -> List;
Total[Integrate[%, {x, -Pi, Pi}, {y, -Pi, Pi}, {z, -Pi, Pi}]]

It looks terrible but for a computer no problem:
$-\frac{8 \left(\pi -\frac{\sin (2 c \pi )}{2 c}\right) (b \cos (b \pi ) \sin (a \pi )-a \cos (a \pi ) \sin (b \pi ))^2}{\left(a^2-b^2\right)^2}-\frac{8 (2 c \pi -\sin (2 c \pi )) (b \cos (b \pi ) \sin (a \pi )-a \cos (a \pi ) \sin (b \pi ))^2}{\left(a^2-b^2\right)^2 c}+\frac{16 (a \cos (a \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (a \pi )) (2 c \cos (c \pi ) \sin (a \pi )-2 a \cos (a \pi ) \sin (c \pi )) (b \cos (b \pi ) \sin (c \pi )-c \cos (c \pi ) \sin (b \pi ))}{\left(a^2-b^2\right) \left(a^2-c^2\right) \left(b^2-c^2\right)}+\frac{16 (2 b \cos (b \pi ) \sin (a \pi )-2 a \cos (a \pi ) \sin (b \pi )) (a \cos (a \pi ) \sin (c \pi )-c \cos (c \pi ) \sin (a \pi )) (c \cos (c \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (c \pi ))}{\left(a^2-b^2\right) \left(a^2-c^2\right) \left(c^2-b^2\right)}+\frac{16 (a \cos (a \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (a \pi )) (a \cos (a \pi ) \sin (c \pi )-c \cos (c \pi ) \sin (a \pi )) (2 c \cos (c \pi ) \sin (b \pi )-2 b \cos (b \pi ) \sin (c \pi ))}{\left(a^2-b^2\right) \left(a^2-c^2\right) \left(b^2-c^2\right)}-\frac{(\sin (2 a \pi )-2 a \pi ) (2 b \pi -\sin (2 b \pi )) \left(\pi -\frac{\sin (2 c \pi )}{2 c}\right)}{2 a b}-\frac{(\sin (2 a \pi )-2 a \pi ) \left(\pi -\frac{\sin (2 b \pi )}{2 b}\right) (2 c \pi -\sin (2 c \pi ))}{2 a c}-\frac{\left(\pi -\frac{\sin (2 a \pi )}{2 a}\right) (\sin (2 b \pi )-2 b \pi ) (2 c \pi -\sin (2 c \pi ))}{2 b c}-\frac{8 \left(\pi -\frac{\sin (2 b \pi )}{2 b}\right) (c \cos (c \pi ) \sin (a \pi )-a \cos (a \pi ) \sin (c \pi ))^2}{\left(a^2-c^2\right)^2}-\frac{8 (2 b \pi -\sin (2 b \pi )) (c \cos (c \pi ) \sin (a \pi )-a \cos (a \pi ) \sin (c \pi ))^2}{b \left(a^2-c^2\right)^2}-\frac{8 \left(\pi -\frac{\sin (2 a \pi )}{2 a}\right) (c \cos (c \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (c \pi ))^2}{\left(b^2-c^2\right)^2}-\frac{8 (2 a \pi -\sin (2 a \pi )) (c \cos (c \pi ) \sin (b \pi )-b \cos (b \pi ) \sin (c \pi ))^2}{a \left(b^2-c^2\right)^2}$
Perhaps someone can simplify it from here?
Edit:
Confirmed
FullSimplify[%,
    a \[Element] Integers &&
    b \[Element] Integers &&
    c \[Element] Integers]

outputs $6\pi^3$.
