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. 

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<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: 
$$c = 4 \int_{0}^{r} \sqrt{1+\left(\dfrac{d}{dx}\sqrt[4]{r^4-x^4}\right)^2} \,dx = 4 \int_{0}^{r} \sqrt{1+\frac{x^6}{\left(r^4-x^4\right)^{3/2}}} \,dx$$
