Need help with $\int_{-\infty}^\infty \frac{x^2 \, dx}{x^4+2a^2x^2+b^4}$

Let we set $\lambda=\frac{b}{a}$ . We want: $$ I(a,b)=\int_{-\infty}^{+\infty}\frac{x^2\,dx}{(x^2+a^2)^2+(b^4-a^4)} = \frac{1}{a}\int_{-\infty}^{+\infty}\frac{x^2\,dx}{(x^2+1)^2+(\lambda^4-1)}$$ and assuming that $\zeta_1(\lambda),\zeta_2(\lambda)$ are the roots of $(x^2+1)^2=1-\lambda^4$ in the upper half-plane, the residue theorem gives: $$ I(a,b)=\frac{2\pi i}{a}\sum_{j=1}^{2}\text{Res}\left(\frac{x^2}{(x^2+1)^2+(\lambda^4-1)},x=\zeta_j\right)$$ that by De l'Hopital theorem simplifies to: $$ I(a,b)=\frac{\pi i}{2a}\sum_{j=1}^{2}\frac{\zeta_j}{1+\zeta_j^2}$$ where $1+\zeta_j^2$ is either $\sqrt{\lambda^4-1}$ or $-\sqrt{\lambda^4-1}$, $\zeta_1+\zeta_2+\overline{\zeta}_1+\overline{\zeta}_2=0$ by Viète's theorem and $\zeta_1-\zeta_2$ is deeply related with the discriminant of $(x^2+1)^2+(\lambda^4-1)$. Can you finish from here? I am getting:

$$ I(a,b) = \color{red}{\frac{\pi}{\sqrt{2(a^2+b^2)}}}. $$

We can write the integrand as

$$\begin{align} \frac{x^2}{x^4+2a^2x^2+b^4}&=\frac{x^2}{(x^2+a^2+\sqrt{a^4-b^4})(x^2+a^2-\sqrt{a^4-b^4})}\\\\ &=\frac{A}{x^2+a^2-\sqrt{a^4-b^4}}+\frac{B}{x^2+a^2+\sqrt{a^4-b^4}} \end{align}$$

where $A$ and $B$ are given respectively by

$$A=\frac{-a^2+\sqrt{a^4-b^4}}{2\sqrt{a^4-b^4}}$$

and

$$B=\frac{a^2+\sqrt{a^4-b^4}}{2\sqrt{a^4-b^4}}$$

Therefore, we find that for $a>b>0$

$$\begin{align} \int_{-\infty}^\infty \frac{x^2}{x^4+2a^2x^2+b^4}\,dx&=2\int_0^\infty \frac{x^2}{x^4+2a^2x^2+b^4}\,dx\\\\ &=2A\int_0^\infty \frac{1}{x^2+a^2-\sqrt{a^4-b^4}}\,dx+2B\int_0^\infty \frac{1}{x^2+a^2+\sqrt{a^4-b^4}}\,dx\\\\ &=\frac{\pi A}{\sqrt{a^2-\sqrt{a^4-b^4}}}+\frac{\pi B}{\sqrt{a^2+\sqrt{a^4-b^4}}}\\\\ &=\frac{\pi}{2\sqrt{a^4-b^4}}\left(\sqrt{a^2+\sqrt{a^4-b^4}}-\sqrt{a^2-\sqrt{a^4-b^4}}\right)\\\\ &=\frac{\pi}{2\sqrt{a^4-b^4}}\left(\sqrt{2(a^2-b^2)}\right)\\\\ &=\frac{\pi}{\sqrt{2(a^2+b^2)}} \end{align}$$

as expected!

Note that for $b>a>0$, result of the preceding development is unaltered. One needs only to proceed with caution when evaluating the integrals, $\int \frac{1}{x^2+a^2\pm \sqrt{a^4-b^4}}\,dx$ due to the complex constants $a^2\pm \sqrt{a^4-b^4}$.

For the case $a=b>0$, the result of the preceding developing is unaltered. The integral $\int_0^\infty \frac{x^2}{(x^2+a^2)^2}\,dx$ can easily be evaluated enforcing the substitution $x\to a\tan(x)$.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ With $\ds{a, b \in \mathbb{R}^{+}}$ and $\ds{p \equiv \pars{a/b}^{2}}$, \begin{align} \color{#f00}{% \int_{-\infty}^{\infty}{x^{2}\,\dd x \over x^{4} + 2a^{2}x^{2} + b^{4}}} & = {1 \over \verts{b}} \int_{-\infty}^{\infty}{x^{2}\,\dd x \over x^{4} + 2px^{2} + 1} = {1 \over \verts{b}} \int_{0}^{\infty}{x^{1/2}\,\dd x \over x^{2} + 2px + 1} \\[3mm] & = {1 \over \verts{b}} \int_{0}^{\infty}{x^{1/2}\,\dd x \over \pars{x - x_{+}}\pars{x - x_{-}}} \end{align}

where $\braces{x_{\pm}}$ are the roots of $x^{2} + 2px + 1 = 0$ which are given by $x_{\pm} = -p \pm \root{p^{2} - 1}$. The integration can be performed along a key-hole contour which takes into account the branch-cut of $z^{1/2} = \verts{z}^{1/2}\exp\pars{\ic\phi/2}\,,\ z \not= 0\,,\ 0 < \phi < 2\pi$. Then,

• ${\large p < 1}$
  The roots are given by $x_{\pm} = -p \pm \ic\root{1 - p^{2}}$ with $\verts{x_{\pm}} = 1$. \begin{align} 2\pi\ic\bracks{% {\verts{x_{+}}^{1/2}\expo{\ic\phi_{+}/2} \over x_{+} - x_{-}} + {\verts{x_{-}}^{1/2}\expo{\ic\phi_{-}/2} \over x_{-} - x_{+}}} & = \int_{0}^{\infty}{x^{1/2}\,\dd x \over x^{2} + 2px + 1} + \int_{\infty}^{0}{x^{1/2}\expo{\ic\pi}\,\dd x \over x^{2} + 2px + 1} \\[3mm] \int_{0}^{\infty}{x^{1/2}\,\dd x \over x^{2} + 2px + 1} & = {\pi \over 2\root{1 - p^{2}}}\pars{\expo{\ic\phi_{+}/2} - \expo{\ic\phi_{-}/2}} \\[3mm] \phi_{+} = {\pi \over 2} + \delta\phi\,,\quad \phi_{-} = {3\pi \over 2} - \delta\phi\,,\quad & \delta\phi \equiv \arctan\pars{{p \over \root{1 - p^{2}}}} \end{align} \begin{align} \int_{0}^{\infty}{x^{1/2}\,\dd x \over x^{2} + 2px + 1} & = {\pi \over 2\root{1 - p^{2}}}\pars{% \expo{\ic\pi/4}\expo{\ic\delta\phi/2} - \expo{3\ic\pi/4}\expo{-\ic\delta\phi/2}} \\[3mm] & = {\pi \over 2\root{1 - p^{2}}}\bracks{% {1 + \ic \over \root{2}}\,\expo{\ic\delta\phi/2} - {-1 + \ic \over \root{2}}\,\expo{-\ic\delta\phi/2}} \\[3mm] & = {\root{2}\pi \over 2\root{1 - p^{2}}}\bracks{% \cos\pars{\delta\phi \over 2} - \sin\pars{\delta\phi \over 2}} \\[3mm] & = {\root{2}\pi \over 2\root{1 - p^{2}}}\bracks{% \root{{1 + \cos\pars{\delta\phi} \over 2}} - \root{{1 - \cos\pars{\delta\phi}} \over 2}} \end{align} However, $\ds{\cos\pars{\delta\phi} = {1 \over \root{\tan^{2}\pars{\delta\phi} + 1}} = \root{1 - p^{2}} = {\root{b^{2} - a^{2}} \over \verts{b}}}$: \begin{align} \int_{0}^{\infty}{x^{1/2}\,\dd x \over x^{2} + 2px + 1} & = {\pi\verts{b}^{1/2} \over 2\root{b^{2} - a^{2}}}\bracks{% \root{\verts{b} + \root{b^{2} - a^{2}}} - \root{\verts{b} - \root{b^{2} - a^{2}}}} \end{align}
• ${\large p > 1}$. Both roots are negative: $x_{\pm} = -p \pm \root{p^{2} - 1}$. The calculation is similar to the previous one.