I have found a statement here equation 23 without explanation that $E\frac{X^k}{(\omega X^2+\sigma)^r}<\infty$
for $\omega,\sigma>0, k\in\{0,1,...,2r\}$, $r\in \mathbb{N}$, where we don't know if $EX^r$ exists. I have tried to understand it, obviously $E\frac{X^k}{(\omega X^2+\sigma)^r}<E\frac{X^k}{\omega^r X^{2r}}$, but maybe $\omega$ would be too small, so that the numerator is greater than the denominator, so we can not use 1 to bound it. Also I don't know if $E\frac{1}{X^{2r-k}}$ exists.
Thanks for any hints!
Let $k\in\{0,\dots,2r\}$. Replacing $X$ by $\lvert X\rvert$ if necessary, we assume that $X$ is non-negative.
Define $Y=\frac{X^k}{(\omega X^2+\sigma)^r}$. One has $$ Y\mathbf{1}_{\{X\leqslant 1\}}\leqslant \frac 1{\sigma^r} $$ hence $Y\mathbf{1}_{\{X\leqslant 1\}}$ is bounded.
Moreover, $$ Y\mathbf{1}_{\{X> 1\}}\leqslant \frac 1{\min\{w^r,\sigma^r\}}\frac{X^k}{(1+X^2)^r}\mathbf{1}_{\{X> 1\}} \leqslant \frac 1{\min\{w^r,\sigma^r\}}\left(\frac{X}{1+X^2}\right)^r\mathbf{1}_{\{X> 1\}}\leqslant \frac 1{\min\{2^rw^r,2^r\sigma^r\}} $$ hence $Y$ is bounded.