$$f(x)=\int_1^{x} \frac{1}{\sqrt[m]{P(t)}}\;dt$$
$P(x)$ is polynomial with degree $n$.
$m$ is an positive integer and $m>1$
What is the algoritm to determine $f^{-1}(x)$ is periodic function via using $P(x)$ and $m$ without evaluting the integral ? Is there also a way to find period without evaluting the integral?
Example 1: $P(x)=x^2$ , $m=2$
$$f_1(x)=\int_1^{x} \frac{1}{t}\;dt=\ln x$$
$$f_1^{-1}(x)=e^x$$
$$f_1^{-1}(x+2k\pi i)=e^{x+2k\pi i}=e^x=f_1^{-1}(x)$$ $k$ is an integer $$f_1^{-1}(x+2k\pi i)=f_1^{-1}(x)$$ $f_1^{-1}(x)$ is a periodic function
Example 2: $P(x)=1-x^2$ , $m=2$ $$f_2(x)=\int_1^{x} \frac{1}{\sqrt{1-t^2}}\;dt=\arcsin x-\frac{\pi}{2}$$
$$f_2^{-1}(x)=\sin {(x+\frac{\pi}{2})}$$
$$f_2^{-1}(x+2k\pi )=\sin{(x+\frac{\pi}{2}+2k\pi )}=\sin{(x +\frac{\pi}{2})}=f_2^{-1}(x)$$ $k$ is an integer $$f_2^{-1}(x+2k\pi )=f_2^{-1}(x)$$ $f_2^{-1}(x)$ is a periodic function
Example 3: $P(x)=(1+x^2)^2=1+2x^2+x^4$ , $m=2$ $$f_3(x)=\int_1^{x} \frac{1}{1+t^2}\;dt=\arctan x -\frac{\pi}{4}$$
$$f_3^{-1}(x)=\tan (x+\frac{\pi}{4})$$
$$f_3^{-1}(x+k\pi )=\tan{(x+\frac{\pi}{4}+k\pi )}=\tan{(x+\frac{\pi}{4} )}=f_3^{-1}(x)$$ $k$ is an integer $$f_3^{-1}(x+k\pi )=f_3^{-1}(x)$$ $f_3^{-1}(x)$ is a periodic function
Example 4: $P(x)=x^4$ , $m=2$
$$f_4(x)=\int_1^{x} \frac{1}{t^2}\;dt=\frac{x-1}{x}$$
$$f_4^{-1}(x)=\frac{1}{1-x}$$
$f_4^{-1}(x)$ is not periodic function
Thanks a lot for answers