Is there a closed form or a better solution?

Question

This is another integral in the book "irresistible integral" I can find that: $$\int_0^\infty \frac{1}{\left( x^4 +2ax^2+1 \right)^{m+1}} \, \text{d}x =\frac{{{\left( -1 \right)}^m}\sqrt{2}\pi }{4\left( m! \right)}\left. \left( \frac{\text{d}^m}{\text{d}x^m} \sqrt{x}\sqrt{a+\sqrt{x}} \right) \right|_{x=1}$$ Do we have a better closed form? or better solution? My working: Let $$I(b)=\int_0^\infty \frac{1}{x^4+2ax^2+b}\,\text{d}x$$ then $$I^{(m)}(b)=m!(-1)^m\int_0^\infty \frac{1}{(x^4+2ax^2+b)^{m+1}} \, \text{d}x$$ Also $$I(b)=\int_0^\infty \frac{1}{(x^2+\alpha)(x^2+\beta)}\,\text{d}x$$ where $$\alpha=a+\sqrt{a^2-b}>0,\ \ \beta=a-\sqrt{a^2-b}>0$$ $$I(b)=\frac{1}{\beta-\alpha}\int_0^\infty \left(\frac{1}{x^2+\alpha}-\frac{1}{x^2+\beta}\right)$$ $$=\frac{\pi}{2(\beta-\alpha)}\left(\frac{1}{\sqrt{\alpha}}-\frac{1}{\sqrt{\beta}}\right)$$ $$=\frac{\pi}{4\sqrt{a^2-b}}\frac{\sqrt{2}\sqrt{a-\sqrt{b}}}{\sqrt{b}} = \frac{\sqrt{2}\pi}{4\sqrt{b}\sqrt{a+\sqrt{b}}}$$ Which follows that $$\int_0^\infty \frac{1}{( x^4+2ax^2+1 )^{m+1}}\text{d}x =\frac{(-1)^m\sqrt{2}\pi }{4m!} \left. \left( \frac{\text{d}^m}{\text{d}x^m}\sqrt{x}\sqrt{a+\sqrt{x}} \right) \right|_{x=1}$$

Answer 1

For every $a\gt-1$, consider $$ I_m(a)=\int_0^\infty \frac{\mathrm dx}{\left(x^4+2ax^2+1\right)^{m+1}},$$ then, for every $|t|$ small enough, $$ \sum_{m=0}^{+\infty}I_m(a)t^m=\frac{\pi}{2\sqrt2\sqrt{1-t}\sqrt{a+\sqrt{1-t}}}. $$ To deduce a closed form formula for each $I_m(a)$ from this expression does not seem obvious. However, one sees readily that the radius of convergence of the series in the LHS is $1-a^2$ if $-1\lt a\leqslant0$ and $1$ if $a\geqslant0$. Thus, when $m\to\infty$, $m^{-1}\log I_m(a)\to-\log(1-a^2)$ if $-1\lt a\leqslant0$ and $m^{-1}\log I_m(a)\to0$ if $a\geqslant0$.