Background
The bivariate Gaussian copula function is
$$C_{\rho}(u,v)=∫_{-∞}^{Φ^{-1}(u)}∫_{-∞}^{Φ^{-1}(v)}\frac{1}{2π\sqrt{1-ρ^2}}×exp(-\frac{x^2+y^2-2ρxy}{2(1-ρ^2)})dxdy.$$
The multivariate Gaussian copula density (being the second mixed partial derivative of the copula function) is derived as follows:
Let $x=[\Phi^{-1}(u_1),\ldots,\Phi^{-1}(u_n)]^{\top}$. Then \begin{align} c(u_1,\ldots,u_n;\Sigma)&=\frac{\partial^nC(u_1,.\ldots,u_n;\Sigma)}{\partial u_1\cdots\partial u_n}=\frac{\Phi(x;0,\Sigma)}{\prod_{i=1}^n \phi(x_i)} \\ &=(2\pi)^{-\frac{n}{2}}|\Sigma|^{-\frac{1}{2}}\exp\!\left(-\frac{1}{2}x^{\top}\Sigma^{-1}x\right)\times\prod_{i=1}^n (2\pi)^{-\frac{1}{2}}\exp\!\left(\frac{1}{2}x_i^2\right) \\ &=|\Sigma|^{-\frac{1}{2}}\exp\!\left(-\frac{1}{2}x^{\top}\Sigma^{-1}x+\frac{1}{2}x^{\top}x\right) \\ &=|\Sigma|^{-\frac{1}{2}}\exp\!\left(-\frac{1}{2}x^{\top}(\Sigma^{-1}-I)x\right). \end{align}
This reduces nicely in the bivariate case, $n=2$, as $$ c\left(u_{1}, u_{2} ; \rho\right)=\frac{1}{\sqrt{1-\rho^{2}}} \exp \left\{-\frac{\rho^{2}\left(x_{1}^{2}+x_{2}^{2}\right)-2 \rho x_{1} x_{2}}{2\left(1-\rho^{2}\right)}\right\} $$
Question
How can we derive the entropy of the bivariate Gaussian copula density as a closed-form analytical solution? Here is my attempt based on helpful comments below: \begin{align} h(c(u,v)) &= \int c(u,v) \ln c(u,v) \mathrm{d} x_1 \mathrm{d} x_2\\ &= \mathbb{E}[ -\log c(u,v)] \\ &= \mathbb{E}\left[ -\log \frac{1}{\sqrt{1-\rho^{2}}} \exp \left\{- \frac{\rho^{2}\left(x_{1}^{2}+x_{2}^{2}\right)-2 \rho x_{1} x_{2}}{2\left(1-\rho^{2}\right)}\right\}\right] \\ &= \mathbb{E}\left[ \frac{1}{2}\log (1-\rho^2) + \frac{\rho^2( X_1^2 + X_2^2) - 2\rho X_1X_2}{2(1-\rho^2)} \right]\\ &= \frac{1}{2}\log (1-\rho^2) + \frac{\rho^2 \mathbb{E}( X_1^2 + X_2^2) - 2\rho \mathbb{E}(X_1 X_2)}{2(1-\rho^2)} \\ &= \frac{1}{2}\log (1-\rho^2) + \frac{\rho^2 \mathbb{E}( X_1^2) + \rho^2 \mathbb{E}( X_2^2) - 2\rho \mathbb{E}(X_1)\mathbb{E}(X_2)}{2(1-\rho^2)} \\ &= \log (1-\rho^2) + \frac{1}{(1-\rho^2)} \left[ \rho^2 \mathbb{E}( X_1^2) + \rho^2 \mathbb{E}( X_2^2) - 2\rho \mathbb{E}(X_1)\mathbb{E}(X_2) \right]\\ &= ? \end{align}