Excluding the elementary trivial cases $n=1,2$, I know that for $n=3,4,6,8,12,20$, this integral can be expressed in terms of elliptic integrals, as shown below.
$$ \begin{aligned} {\cal I}_3&=2\wp^{-1}(x;0,-4)\\ {\cal I}_4&=\wp^{-1}(x^2;-4,0)\\ {\cal I}_6&=-\wp^{-1}\left(\frac1{x^2};0,-4\right)\\ {\cal I}_8&=-\frac12\wp^{-1}\left(x^2+\frac1{x^2}+\frac23;\frac{40}3,\frac{224}{27}\right)+\frac12\wp^{-1}\left(x^2+\frac1{x^2}-\frac23;\frac{40}3,-\frac{224}{27}\right)\\ {\cal I}_{12}&=-\frac\zeta{\sqrt3}\wp^{-1}\left(\frac{x^2}\zeta+\frac\zeta{x^2};12,0\right)+\frac{\zeta^5}{\sqrt3}\wp^{-1}\left(\frac{x^2}{\zeta^5}+\frac{\zeta^5}{x^2};12,0\right),\quad\color{blue}{\zeta={\rm e}^{\frac{2\pi{\rm i}}6}}\\ {\cal I}_{20}&=\frac{\zeta^3}{\sqrt2\sqrt[8]{5^3}\sqrt{1+\sqrt5}}\wp^{-1}\left(\frac{\frac{x^2}{\zeta^4}+\frac{\zeta^4}{x^2}}{\sqrt[4]5}+\frac{\sqrt[4]5}{\frac{x^2}{\zeta^4}+\frac{\zeta^4}{x^2}}+\frac23;\frac{4(10+3\sqrt5)}3,\frac{16(14+9\sqrt5)}{27}\right)\\&\quad +\frac{\zeta^3}{\sqrt2\sqrt[8]{5^3}\sqrt{1+\sqrt5}}\wp^{-1}\left(\frac{\frac{x^2}{\zeta^4}+\frac{\zeta^4}{x^2}}{\sqrt[4]5}+\frac{\sqrt[4]5}{\frac{x^2}{\zeta^4}+\frac{\zeta^4}{x^2}}-\frac23;\frac{4(10+3\sqrt5)}3,\frac{16(14+9\sqrt5)}{27}\right)\\&\quad -\frac{\zeta^3}{\sqrt2\sqrt[8]{5^3}\sqrt{1+\sqrt5}}\wp^{-1}\left(\frac{\frac{x^2}{\zeta^{16}}+\frac{\zeta^{16}}{x^2}}{\sqrt[4]5}+\frac{\sqrt[4]5}{\frac{x^2}{\zeta^{16}}+\frac{\zeta^{16}}{x^2}}+\frac23;\frac{4(10+3\sqrt5)}3,-\frac{16(14+9\sqrt5)}{27}\right)\\&\quad -\frac{\zeta^3}{\sqrt2\sqrt[8]{5^3}\sqrt{1+\sqrt5}}\wp^{-1}\left(\frac{\frac{x^2}{\zeta^{16}}+\frac{\zeta^{16}}{x^2}}{\sqrt[4]5}+\frac{\sqrt[4]5}{\frac{x^2}{\zeta^{16}}+\frac{\zeta^{16}}{x^2}}-\frac23;\frac{4(10+3\sqrt5)}3,-\frac{16(14+9\sqrt5)}{27}\right),\\&\quad \quad\color{blue}{\zeta={\rm e}^{\frac{2\pi{\rm i}}{20}}} \end{aligned} $$
where $\wp^{-1}(z;g_2,g_3)$ denotes the inverse Weierstrass elliptic function.