The circumference of a 2-dimensional ellipse and the surface area of a 3-dimensional ellipsoid can be given in close-forms using the Legendre’s elliptic integrals as $$C_2=4aE(\sqrt{1-b^2/a^2})$$ $$A_3=2\pi\left[c^2+\frac{bc^2}{\sqrt{a^2-c^2}}F\left(\arcsin\sqrt{1-c^2/a^2},\frac{a\sqrt{b^2-c^2}}{b\sqrt{a^2-c^2}}\right)+b\sqrt{a^2-c^2}E\left(\arcsin\sqrt{1-c^2/a^2},\frac{a\sqrt{b^2-c^2}}{b\sqrt{a^2-c^2}}\right)\right]$$ where $a$, $b$ and $c$ are positive real numbers representing the semi-axes (radii) and $E(k)$, $F(\phi,k)$ and $E(\phi,k)$ are the complete elliptic integral of the second kind and the incomplete elliptic integral of the first kind and the second kind respectively.
However, I have troubles when I attempt to calculate a formula for the surface volume of a 4-dimensional hyperellipsoid. Let $a$, $b$, $c$ and $d$ to be the semi-axes of the hyperellipsoid such that $$\frac{x_1^2}{a^2}+\frac{x_2^2}{b^2}+\frac{x_3^2}{c^2}+\frac{x_4^2}{d^2}=1$$ then the surface volume can be given by the integral $$V_4=16\int\limits_{t_1=0}^a\int\limits_{t_2=0}^{b\sqrt{1-t_1^2/a^2}}\int\limits_{t_3=0}^{c\sqrt{1-t_1^2/a^2-t_2^2/b^2}}\sqrt{\frac{1-(1-d^2/a^2)t_1^2/a^2-(1-d^2/b^2)t_2^2/b^2-(1-d^2/c^2)t_3^2/c^2}{1-t_1^2/a^2-t_2^2/b^2-t_3^2/c^2}}dt_3dt_2dt_1$$ that I cannot integrate. How can one integrate this integral to a close-form? Or, alternately, is there another way to calculate a formula for the surface volum of a 4-dimensional hyperellipsoid?
Edit: The integral can be reduced to $$V_4=\frac{16abc\pi}{3}\int\limits_0^1t^2\left(\frac{d^2/a^2}{1-(1-d^2/a^2)(1-t^2)^2}+\frac{d^2/b^2}{1-(1-d^2/b^2)(1-t^2)^2}+\frac{d^2/c^2}{1-(1-d^2/c^2)(1-t^2)^2}\right)\sqrt{\frac{2-t^2}{(1-(1-d^2/a^2)(1-t^2)^2)(1-(1-d^2/b^2)(1-t^2)^2)(1-(1-d^2/c^2)(1-t^2)^2)}}dt$$ but I still cannot solve it.