What is the circumference (arc length) of $x^4 + x^2 + y^4 + y^2 = 2$? Consider
$$x^4 + x^2 + y^4 + y^2 = 2$$
It is a smooth non-intersecting circle like curve in the plane.
A bit like a Hyperellipse.
See https://en.m.wikipedia.org/wiki/Superellipse
What is the circumference (arc length) of $x^4 + x^2 + y^4 + y^2 = 2$ ?
Wolfram|Alpha plot
 A: In polar coordinates: $$r^4\cos^4t+r^2\cos^2t+r^4\sin^4t+r^2\sin^2t=2$$
$$r^4\left(\cos^4t+\sin^4t\right)+r^2-2=0$$
$$r^2=\frac{-1+\sqrt{1+8\left(\cos^4t+\sin^4t\right)}}{2\left(\cos^4t+\sin^4t\right)}$$
Now the arc length for polar coordinates is $$\int_0^{2\pi}\sqrt{r^2+\left(\frac{dr}{dt}\right)^2}dt$$ So it appears you have a challenging integral to compute, and you may need to resort to an approximation technique.

Note $$4r^3\left(\cos^4t+\sin^4t\right)\frac{dr}{dt}+r^4\cdot4\left(-\cos^3t\sin t+\sin^3t\cos t\right)+2r\frac{dr}{dt}=0$$
$$\frac{dr}{dt}=\frac{2r^3\sin t\cos t\left(\cos^2t-\sin^2t\right)}{2r^2\left(\cos^4t+\sin^4t\right)+1}=\frac{2r^3\sin t\cos t\left(\cos^2t-\sin^2t\right)}{\sqrt{1+8\left(\cos^4t+\sin^4t\right)}}$$
$$\left(\frac{dr}{dt}\right)^2=\frac{4r^6\sin^2 t\cos^2 t\left(\cos^2t-\sin^2t\right)^2}{1+8\left(\cos^4t+\sin^4t\right)}$$
You can either use this to try to conquer the integral, or you can use this to help with an approximation technique like say Simpson's Rule (where it would be wise to integrate from $0$ to $\frac{\pi}{4}$ and octuple the result.)
Using Simpson, with $n=2$ over $[0,\pi/4]$ and then octuple: $$8\cdot\frac{\pi}{24}\left(1+4\sqrt{\frac{-1+\sqrt{7}}{3/2}+\frac1{28}\left(\frac{-1+\sqrt7}{3/2}\right)^3}+\sqrt{-1+\sqrt{5}}\right)\approx6.6923\ldots$$ which is within $0.16\%$ of Robert's Maple decimal output.
A: Using the polar form, Maple evaluates the integral to 30 digits as
$6.68187645290337621429065046080$.  The Inverse Symbolic Calculator and Maple's identify function both come up with nothing. 
I'd guess that the probability of this having a "closed form" solution is rather small.
A: Easy estimation: $2\pi = 6.28\ldots \le s \le 8$.
$$
u = x^2 + 1/2 \Rightarrow u^2 = x^4 + x^2 + 1/4 \\
v = y^2 + 1/2 \Rightarrow v^2 = y^4 + y^2 + 1/4
$$
Then 
$$
r^2 = (\sqrt{2+1/2})^2 = u^2 + v^2 = x^4 + x^2 + y^4 + y^2 + 1/2 \iff \\
2 = x^4 + x^2 + y^4 + y^2 \quad (*)
$$
So the transformed curve is a circle in the $u$-$v$-plane with radius $r = \sqrt{5/2}$ and has the known arc length $S = \pi \sqrt{10}$.
I looked for a way to transform this arc length into the wanted arc length $s$, but had no success. 
For the upper half of the curve we get:
\begin{align}
v &= \sqrt{r^2 - u^2} \iff \\
y^2 + 1/2 &= \sqrt{9/4 - x^2 - x^4} \iff \\
y &= \sqrt{\sqrt{9/4 - x^2 -x^4} - 1/2} \quad (**) \\
\end{align}
This leads to the derivative
\begin{align}
y' &= \frac{1}{2\sqrt{\sqrt{9/4 - x^2 -x^4} - 1/2}} 
\frac{-2x-4x^3}{2\sqrt{9/4 - x^2 -x^4}} \\
&= - \frac{\sqrt{2}\,x\,(1 + 2x^2)}{\sqrt{\left(9 - 4x^2(1+x^2)\right)
\left(\sqrt{9 - 4x^2(1+x^2)} - 1\right)}}
\end{align}
The arc length is
\begin{align}
s &= 4 \int\limits_0^1\sqrt{1 + (y')^2} \, dx \\
  &= 4 \int\limits_0^1\sqrt{1 + \frac{2\,x^2\,(1 + 2x^2)^2}{\left(9 - 4x^2(1+x^2)\right)
\left(\sqrt{9 - 4x^2(1+x^2)} - 1\right)}} \, dx \\
&= 6.68188
\end{align}
where the rounded numerical value was obtained from WolframAlpha.
An alternative query using the simpler equation $(**)$ is this. The result has to be multiplied by $4$ however.
Addendum
Alex Jordan pointed out that $y' \to -\infty$ for $x \to 1$, see WA plot. So we go only for the first 1/8 of the arc, were there is a point $(\xi,\xi)$:
$$
2 = \xi^4 + \xi^2 + \xi^4 + \xi^2 = 2\xi^2(\xi^2+1) \iff \\
1 = \xi^4 + \xi^2 = (\xi^2 + 1/2) - 1/4 \Rightarrow \\
\xi = \sqrt{(\sqrt{5} - 1)/2} = \sqrt{\varphi - 1} = 0.78615\ldots
$$
where $\varphi = 1.618\ldots$ is the golden ratio.
This leads to
\begin{align}
s &= 8 \int\limits_0^\xi\sqrt{1 + (y')^2} \, dx \\
  &= 8 \int\limits_0^\xi\sqrt{1 + \frac{2\,x^2\,(1 + 2x^2)^2}{\left(9 - 4x^2(1+x^2)\right)
\left(\sqrt{9 - 4x^2(1+x^2)} - 1\right)}} \, dx \\
\end{align}
New query: link
A: Let us first rewrite the curve in the hope to perhaps parametrize it.
\begin{align*}
x^4+x^2+y^4+y^2=2\iff{}&\left(x^2+\frac12\right)^2+\left(y^2+\frac12\right)^2-\frac12=2\iff{} \\
{}\iff{}&\frac25\left(x^2+\frac12\right)^2+\frac25\left(y^2+\frac12\right)^2=1.
\end{align*}
So we must have $t$ such that:
$$\left\{\begin{array}{c}
\sqrt{\frac25}\left(x^2+\frac12\right)=\cos t \\
\sqrt{\frac25}\left(y^2+\frac12\right)=\sin t \\
\end{array}\right.
\iff
\left\{\begin{array}{c}
x^2=-\frac12+\sqrt{\frac52}\cos t \\
y^2=-\frac12+\sqrt{\frac52}\sin t \\
\end{array}\right.$$
Split a couple of cases depending on sign of $x,y$ (4 cases, I should think), then calculate the integrals of:
$$\gamma(t)=\left(\pm\sqrt{-\frac12+\sqrt{\frac52}\cos t},\pm\sqrt{-\frac12+\sqrt{\frac52}\sin t}\right)$$
with signs depending on the cases, sum them, and you're done.
