Solve $\int (y^4+by^2+c)^{-1/2}dy=x$ as $y$ is a real valued function of $x$.

Question

Solve $\int (y^4+by^2+c)^{-1/2}dy=x$ as y is a real valued function of x. Here $b$ and $c$ are real constants so that $c>b^2/4$.

Answer 1

Let $\Delta=b^2-4c<0$. Then $y^4+by^2+c$ never vanishes for $y\in\mathbb{R}$ and: $$\int_{-\infty}^{+\infty}\frac{dy}{\sqrt{y^4+by^2+c}}=\int_{0}^{+\infty}\frac{du}{\sqrt{u}\sqrt{u^2+bu+c}}=\int_{0}^{+\infty}\frac{du}{\sqrt{u}\sqrt{\left(u+\frac{b}{2}\right)^2-\frac{\Delta}{4}}}$$ is real and bounded by (assuming $b\neq 0$): $$\int_{0}^{|b|}\frac{2 du}{\sqrt{-\Delta u}}+\int_{|b|}^{+\infty}\frac{du}{\sqrt{u}\left(u-\frac{|b|}{2}\right)}=4\sqrt{\frac{|b|}{-\Delta}}+\log(1+\sqrt{2})\sqrt{\frac{8}{|b|}}.$$