Arc length of the squircle The squircle is given by the equation $x^4+y^4=r^4$. Apparently, its circumference or arc length $c$ is given by 

$$c=-\frac{\sqrt[4]{3} r G_{5,5}^{5,5}\left(1\left|
\begin{array}{c}
 \frac{1}{3},\frac{2}{3},\frac{5}{6},1,\frac{4}{3} \\
 \frac{1}{12},\frac{5}{12},\frac{7}{12},\frac{3}{4},\frac{13}{12} \\
\end{array}
\right.\right)}{16 \sqrt{2} \pi ^{7/2} \Gamma \left(\frac{5}{4}\right)}$$

Where $G$ is the Meijer $G$ function. Where can I find the derivation of this result? Searching for any combination of squircle and arc length or circumference has led to nowhere.
You can do this using this empirical formula to find perimeter of general superellipse  

$L=a+b*(((2.5/(n+0.5))^{1/n})*b+a*(n-1)*0.566/n^2)/(b+a*(4.5/(0.5+n^2)))$
This mentioned in my research
By your definition, $\mathcal{C} = \{(x,y) \in \mathbb{R}^{2}: x^4 + y^4 = r^4\}$. Which can be parametrized as
\begin{align}
\mathcal{C} =
\begin{cases}
\left(+\sqrt{\cos (\theta )},+\sqrt{\sin (\theta )} \right)r\\
\left(+\sqrt{\cos (\theta )},-\sqrt{\sin (\theta )} \right)r\\
\left(-\sqrt{\cos (\theta )},+\sqrt{\sin (\theta )} \right)r\\
\left(-\sqrt{\cos (\theta )},-\sqrt{\sin (\theta )} \right)r
\end{cases}
, \qquad 0 \leq \theta \leq \frac{\pi}{2}, \, 0&lt;r
\end{align}



Now, look at this curve in $\mathbb{R}^{2}_{+}$ as $y = \sqrt[4]{r^4-x^4}$, then observe that symmetry with both axis. It yields the arc length is just: 
$$c = 4 \int_{0}^{r} \sqrt{1+\left(\dfrac{d}{dx}\sqrt[4]{r^4-x^4}\right)^2} \,dx = 4 \int_{0}^{r} \sqrt{1+\frac{x^6}{\left(r^4-x^4\right)^{3/2}}} \,dx$$