calculus and analysis - Is it possible to give the closed-form of the stiffness matrix of triangular prism element?

Suppose the vertices of triangular prism are 1, 2, 3 (bottom triangle), 4, 5, 6 (top triangle), the coordinates of the vertices are $$ (x_i, y_i, z_i), i=1,2,3,4,5,6. $$

We can write the stiffness matrix of FEM of the triangular prism as follows:

$$ K_{i,j} = \int_0^1\int_0^{1-\xi}\int_{-1}^1 \begin{pmatrix}\frac{\eth N_i}{\eth\xi} & \frac{\eth N_i}{\eth\eta} & \frac{\eth N_i}{\eth\zeta}\end{pmatrix} (\textbf{J}^{-1})^T\textbf{J}^{-1}\begin{pmatrix}\frac{\eth N_j}{\eth\xi} \\ \frac{\eth N_j}{\eth\eta} \\ \frac{\eth N_j}{\eth\zeta}\end{pmatrix}det(\mathbf{J})d\xi d\eta d\zeta\,\\ i,j=1,2,...,6. $$

where

$$ \textbf{J}=\begin{pmatrix}\frac{\eth x}{\eth\xi} & \frac{\eth y}{\eth\xi} & \frac{\eth z}{\eth\xi} \\ \frac{\eth x}{\eth\eta} & \frac{\eth y}{\eth\eta} & \frac{\eth z}{\eth\eta} \\ \frac{\eth x}{\eth\zeta} & \frac{\eth y}{\eth\zeta} & \frac{\eth z}{\eth\zeta} \end{pmatrix} $$

and $$ \textbf{N}=\begin{pmatrix}N_1 & N_2 & N_3 & N_4 & N_5 & N_6\end{pmatrix}^T\\ \left\{\begin{array}{} N_1=\frac{1}{2}(1-\zeta)(1-\xi-\eta)\\ N_2=\frac{1}{2}\xi(1-\zeta)\\ N_3=\frac{1}{‌​2}\eta(1-\zeta)\\ N_4=\frac{1}{2}(1+\zeta)(1-\xi-‌​‌​\eta)\\ N_5=\frac{1}{2}\xi(1+\zeta)\\ N_6=\frac{1}{2}\eta(1+\zeta) \end{array} \right.\ \\ \left\{\begin{array}{} x=\mathbf{N^T}\mathbf{x_e}\\ y=\mathbf{N^T}\mathbf{y_e}\\ z=\mathbf{N^T}\mathbf{z_e}\\ \end{array} \right.\ \\ \left\{\begin{array}{} \mathbf{x_e}=\begin{pmatrix}x_1 & x_2 & x_3 & x_4 & x_5 & x_6\end{pmatrix}^T\\ \mathbf{y_e}=\begin{pmatrix}y_1 & y_2 & y_3 & y_4 & y_5 & y_6\end{pmatrix}^T\\ \mathbf{z_e}=\begin{pmatrix}z_1 & z_2 & z_3 & z_4 & z_5 & z_6\end{pmatrix}^T \end{array} \right.\ $$

Is it possible to give the closed-form of the integration by the powerful mathematica?

Thanks, Tang Laoya