In a youtube video titled "Generalizing the circumference" Prof. Michael Penn works out the integral
$$ L(n)=4 \int_{0}^{\pi/2}\sqrt{n^2 \cos^{2n-2}(\theta) \sin^2(\theta)+n^2 \sin^{2n-2}(\theta) \cos^{2}(\theta)} \, d\theta, $$ which gives the arclength of $(x^2)^{\frac1n}+(y^2)^{\frac1n}=1$ "supercircle" for various values of $n$. In particular the values $L(1)=2\pi$ is the circumference of the circle, $L(2)=4\sqrt{2}$ the circumference of a diamond shape, and $L(3)=6$. At the end of the video it is stated that the value when $n\geq 4$ has no closed form, but converges towards a known value. Surely, that's the case, but it seems that at least Wolfram alpha can work out some of the values for $n\geq 4$. I got for example $$ L(4) = 4 -\sqrt{2}\log(1-\frac{1}{\sqrt{2}})+\sqrt{2}\log(1+\frac{1}{\sqrt{2}})\approx 6.49 $$ Similarly, for $n=5$ we get an explicit form $$ L(5)=\frac{5}{6}\left(6+\sqrt{3}\log{\left(2+\sqrt{3}\right)}\right)\approx 6.90 $$ My follow up question is how do you derive the values for $n=4$ and $n=5$ and can you derive values for other values of $n$ as well? It seems integer values of $n\geq 4$ are possible?
We start with $$L(n)=4n\int_0^{\pi/2}\sqrt{\cos^{2n-2}\theta\sin^2\theta+\sin^{2n-2}\theta\cos^2\theta}\,d\theta$$ and then substitute $t=\sin\theta$, then $u=t^2$ to get $$L(n)=2n\int_0^1\sqrt{u^{n-2}+(1-u)^{n-2}}\,du$$ When $n=4$ or $n=5$ the polynomial under the square root is quadratic, which means the integral has an elementary solution – the ones you saw in Wolfram Alpha.
Below I will derive a solution for $n=6$, when said polynomial is quartic; an elliptic integral solution thus exists. $$L(6)=12\int_0^1\sqrt{u^4+(1-u)^4}\,du=7.2085721366\dots$$ The quartic can be linearly transformed into a biquadratic, at which point substituting $v=u^2$ yields a cubic under the square root: $$L(6)=\frac3{\sqrt2}\int_0^1\sqrt{\frac{(v-(-3+\sqrt8))(v-(-3-\sqrt8))}v}\,dv$$ Byrd & Friedman 237.08 states that this is equivalent to (after simplification) $$L(6)=3(2-\sqrt2)\int_0^{U=F(\varphi,m)}\operatorname{dc}^2(u,m)\operatorname{nc}^2(u,m)\,du$$ where the parameter $m=4(3\sqrt2-4)$ and $\sin\varphi=\frac{\sqrt{2+\sqrt2}}2$. B&F 361.13 transforms this into a sum of elliptic integrals: $$L(6)=3(2-\sqrt2)\frac1{3(1-m)}(2(1-m)U+(m-2)E(\varphi,m)+\operatorname{dn}(U,m)\operatorname{sc}(U,m)(2-m+(1-m)\operatorname{nc}^2(U,m)))$$ But the elliptic functions $\operatorname{xx}(U,m)$ are algebraic functions of $\sin\varphi$, so the whole thing massively simplifies to $$L(6)=2((2-\sqrt2)F(\varphi,m)-3(2+\sqrt2)E(\varphi,m)+8+3\sqrt2)$$ Now apply a descending Landen transformation (B&F 163.02) and the thing simplifies even further to its final form $$\boxed{L(6)=4(4-3E(1/2)+K(1/2))}$$ (The complete elliptic integrals here can themselves be reduced to gamma functions, but that makes things a bit more involved.)
By a similar process the $n=7$ case (the other one involving a quartic) can also be solved: $$L(7)=\frac76(9+\sqrt{2(5+\sqrt5)}(F(3\pi/5,1/\varphi^2)/\sqrt5-E(3\pi/5,1/\varphi^2)))$$ where $\varphi$ is the golden ratio.
For $n=8$, although the polynomial is now a sextic, a linear transformation removes all odd-power terms. Then the same $u=t^2$ substitution leaves a quartic under the square root, which means it can be solved by elliptic integrals again, this time involving all three types: $$L(8)=19+\frac{15}2\sqrt3+3^{1/4}((-75+44\sqrt3)F(\varphi,m)-15E(\varphi,m)+(55/8)(12-7\sqrt3)\Pi(n,\varphi,m))$$ where $m=\frac{2+\sqrt3}4$, $\tan\varphi=\sqrt{6+4\sqrt3}$, $n=\frac12+\frac7{8\sqrt3}$.
The same applies for $n=9$: let $[a,b,c]$ denote $a+bt+ct^2$ where $t=2\cos\frac{2\pi}7$, then with $$m=[14,-3,-6]\qquad\csc^2\varphi=\frac{[-5,5,4]}7\qquad n=\frac{[0,2,1]}4$$ we have $$L(9)=\frac{[117,45,45]}{16}+\frac9{16}\sqrt{\frac{[-76,41,16]}7}([52,6,-13]F(\varphi,m)-5[12,34,15]E(\varphi,m)+(13/4)[-15,1,5]\Pi(n,\varphi,m))$$
The original $L(n)$ algebraic integral can also be written as $$L(n)=2n\int_0^\infty\frac{\sqrt{1+u^{n-2}}}{(u+1)^{n/2+1}}\,du$$ which allows a representation for any $n$ using the Meijer $G$-function: $$L(m+2)=-\frac{m^{m/2+1}}{2^{(m-5)/2}m!!\pi^mf}G\left(\left\{\left\{-\frac12-\frac1m,-\frac12,\dots,\frac12-\frac2m,\color{blue}{1-\frac1m}\right\},\{\}\right\},\left\{\left\{\color{blue}{-\frac12-\frac1m},\frac0m,\frac1m,\dots,1-\frac1m\right\},\{\}\right\},1\right)$$ Here the argument order is as in Mathematica and mpmath and $f=1$ if $m$ is odd and $\sqrt{2/\pi}$ otherwise.
This Python script verifies both the exact expressions obtained above for $L(6)$ to $L(9)$ and the $G$-function "closed" form.