How do you handle the integral $\int_{0}^{\infty} \frac{d x}{\left(x^{m}+\frac{1}{x^{m}}\right)^{n}}$, where $n\in \mathbb{N}?$ Inspired by my post, I start to investigate a more general integral
$$
J_{m}=\int_{0}^{\infty} \frac{1}{\left(x^{m}+\frac{1}{x^{m}}\right)^{2}} d x \quad \text { where } m \in \mathbb{N} .
\\$$
Rearranging and integrate by parts yields
\begin{align*}
J_{m} &=\int_{0}^{\infty} \frac{x^{2 m}}{\left(x^{2 m}+1\right)^{2}} d x \\
&=-\frac{1}{2 m} \int_{0}^{\infty} x d\left(\frac{1}{x^{2 m}+1}\right) \\
&=-\left[\frac{x}{2 m\left(x^{2 m}+1\right)}\right]_{0}^{\infty}+\frac{1}{2 m} \int_{0}^{\infty} \frac{1}{x^{2 m}+1} d x \\
&=\frac{1}{2 m} \cdot\frac{\pi}{2 m} \csc \left(\frac{\pi}{2 m}\right)  \tag{$*$}\\
&=\frac{\pi}{4 m^{2}} \csc \left(\frac{\pi}{2 m}\right)
\end{align*}
where $(*)$ comes from the Theorem, $$
\int_{0}^{\infty} \frac{d x}{1+x^{n}}=\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right).
$$
In particular,
$$
\begin{array}{l} \displaystyle 
I_{3}=\frac{\pi}{36} \csc \left(\frac{\pi}{6}\right)=\frac{\pi}{18} \\
\displaystyle I_{4}=\frac{\pi}{64} \csc \left(\frac{\pi}{8}\right)=\frac{\pi}{32 \sqrt{2-\sqrt{2}}}
\end{array}
$$
My question:
Can we go further for
$$
I(m, n):=\int_{0}^{\infty} \frac{d x}{\left(x^{m}+\frac{1}{x^{m}}\right)^{n}}.
$$
where $m,n\in \mathbb{N}$ and $n\geq 2$?
Latest edit
I have found an answer for $I(m,n)$ using the result in one of my posts and share with you now. $$ I(m,n)=\frac{\pi}{2m(n-1) !} 
\csc \frac{(mn+1) \pi}{2m} \prod_{j=1}^{n-1}\left(j-\frac{mn+1}{2m}\right),\tag*{} $$
However, the proof is a bit complicated. Can you help give a simpler one?
 A: Too long for a comment
We can consider a general case:
$$I(a,b,c)=\int_0^\infty\frac{x^a}{(x^b+1)^c}dx$$
Making the substitution $x^b=t$
$$I(a,b,c)=\frac{1}{b}\int_0^\infty\frac{t^{\frac{a+1}{b}-1}}{(t+1)^c}dt$$
Making another substitution $x=\frac{1}{1+t}$
$$I(a,b,c)=\frac{1}{b}\int_0^1x^{c-\frac{a+1}{b}-1}(1-x)^{\frac{a+1}{b}-1}=\frac{1}{b}\frac{\Gamma\big(c-\frac{a+1}{b}\big)\Gamma\big(\frac{a+1}{b}\big)}{\Gamma(c)}$$
For our specific case $a=mn, b=2m, c=n$, and we get
$$I(m, n)=\frac{\Gamma\big(\frac{n}{2}-\frac{1}{2m}\big)\Gamma\big(\frac{n}{2}+\frac{1}{2m}\big)}{2m\,\Gamma(n)}$$
A: Still not elementary but quite fast.
$$I_{m,n}=\int\frac{d x}{\left(x^{m}+\frac{1}{x^{m}}\right)^{n}}=-\frac{ \, _2F_1\left(n,\frac{m n-1}{2 m};\frac{m (n+2)-1}{2 m};-x^{-2
   m}\right)}{(m n-1)\,x^{mn-1}}$$
$$J_{m,n}=\int_0^\infty\frac{d x}{\left(x^{m}+\frac{1}{x^{m}}\right)^{n}}=\frac{\Gamma \left(\frac{m n+1}{2 m}\right) \Gamma \left(\frac{m (n+2)-1}{2
   m}\right)}{(m n-1) \Gamma (n)}$$
A: I first convert the integral $$I(m,n)=\int_{0}^{\infty} \frac{d x}{\left(x^{m}+\frac{1}{x^{m}}\right)^{n}} $$into
$$
I(m, n)=\int_{0}^{\infty} \frac{x^{m n}}{\left(x^{2 m}+1\right)^{n}} d x .
$$
Using the result in my post,
$$ \int_{0}^{\infty} \frac{x^{r} d x}{\left(x^{m}+1\right)^{n}}=\frac{\pi}{m(n-1) !} 
\csc \frac{(r+1) \pi}{m} \prod_{j=1}^{n-1}\left(j-\frac{r+1}{m}\right),\tag*{} $$
we can conclude that
$$ I(m,n)=\int_{0}^{\infty} \frac{x^{mn} d x}{\left(x^{2m}+1\right)^{n}}=\frac{\pi}{2m(n-1) !} 
\csc \frac{(mn+1) \pi}{2m} \prod_{j=1}^{n-1}\left(j-\frac{mn+1}{2m}\right),\tag*{} $$
However, the solution is a bit complicated and non-elementary.
Is there any simpler or more elementary method?
