I am trying to compute arc length the area inside of closed curve $$\left(y-\sqrt[3]{x^2}\right)^2+x^2 = 1$$
It is known for the heart shape plotted at the end of the post.
I know how to compute the area using $$\int_{-1}^1 \left(\sqrt[3]{x^2} + \sqrt{1-x^2}\right) - \left(\sqrt[3]{x^2} - \sqrt{1-x^2}\right)dx = 2\int_{-1}^1\sqrt{1-x^2} = \pi$$
I found it superficially interesting because of the shape and the fact above that the area inside is just $\pi$. But I had a lot of trouble trying to compute the arc length of this curve:
$$\int_{-1}^1 \sqrt{1+\left(\frac{x}{\sqrt{1-x^2}}-\frac{2x}{3\sqrt[3]{x^4}}\right)^2}dx + \int_{-1}^1 \sqrt{1+\left(\frac{x}{\sqrt{1-x^2}}+\frac{2x}{3\sqrt[3]{x^4}}\right)^2}dx,$$ or $$\int_{-1}^1 \sqrt{1+\left(\frac{x}{\sqrt{1-x^2}}-\frac{2}{3\sqrt[3]{x}}\right)^2}dx + \int_{-1}^1 \sqrt{1+\left(\frac{x}{\sqrt{1-x^2}}+\frac{2}{3\sqrt[3]{x}}\right)^2}dx$$
How do I find the arc length of this curve?
