I want to show that $\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$

First note that because of

$$ \int_{\mathbb{R}}dx\frac{x^4}{(x^4-x^2+1)^2}=\int_{\mathbb{R}}dx\frac{x^4\overbrace{-x^2+1+(x^2-1)}^{=0}}{(x^4-x^2+1)^2}=\\ \int_{\mathbb{R}}dx\frac{1}{(x^4-x^2+1)}+\int_{\mathbb{R}}dx\frac{x^2-1}{(x^4-x^2+1)^2}=\\ \int_{\mathbb{R}}dx\frac{1}{(x^4-x^2+1)}+\int_{\mathbb{R}}dx\frac{-1}{(x^4-x^2+1)^2}+\color{blue}{\underbrace{\int_{\mathbb{R}}dx\frac{x^2}{(x^4-x^2+1)^2}}_{J}} $$

we only need to calulate $\color{\blue}{J}$ because everthing else is covered by your standard formulas.

Now let's define

$$ I(a)=\int_{\mathbb{R}}dx\frac{1}{(x^4-a x^2+1)} $$

from which it follows that

$$ \color{blue}{J}=I'(1) $$

so it can also derived using ur standard identities ($'$ denotes a derivative w.r.t. $a$)

Edit:

Note that the integrals with numerators containing odd powers of $x$ vanish due to symmetry!

Evaluating the integral of interest can be reduced to evaluating the integral

$$I(a,b)=\int_0^\infty \frac{1}{x^4+bx^2+a} \,dx\tag 1$$

To see this, we exploit first odd symmetry to write the integral of interest as

$$\begin{align} \int_{-\infty}^\infty \frac{(x^2-x+\pi)^2}{(x^4-x^2+1)^2}\,dx&=2\int_0^\infty \frac{x^4+(2\pi +1)x^2+\pi^2}{(x^4-x^2+1)^2}\,dx \tag 2 \end{align}$$

Enforcing the substitution $x\to 1/x$ in $(2)$ reveals $$\begin{align} \int_0^\infty \frac{x^4}{(x^4-x^2+1)^2}\,dx&=\int_0^\infty \frac{x^2}{(x^4-x^2+1)^2}\,dx\\\\ &=-\left.\frac{\partial I(a,b)}{\partial b}\right|_{(a,b)=(1,-1)} \tag 3 \end{align}$$

Moreover, we have

$$\int_0^\infty \frac{1}{(x^4-x^2+1)^2}\,dx=-\left.\frac{\partial I(a,b)}{\partial a}\right|_{(a,b)=(1,-1)} \tag 4$$

Using $(3)$ and $(4)$ in $(2)$ yields

$$\int_{-\infty}^\infty \frac{(x^2-x+\pi)^2}{(x^4-x^2+1)^2}\,dx=-4(\pi +1)\left.\frac{\partial I(a,b)}{\partial b}\right|_{(a,b)=(1,-1)}-2\pi^2\left.\frac{\partial I(a,b)}{\partial a}\right|_{(a,b)=(1,-1)}$$

where $I(a,b)$ as given by $(1)$ can be found using partial fraction expansion or contour integration, for examples.

Trick: the integral over $\mathbb{R}$ of a non-vanishing meromorphic function, $O\left(\frac{1}{\|z\|^2}\right)$ at infinity, is just $2\pi i$ times the sum of the residues for the poles in the upper half-plane. In our case such poles are located at $z=e^{\pi i/6}$ and $z=e^{5\pi i/6}$ (roots of $z^2-iz-1$) and the sum of the residues is $$ -\frac{i}{2}\left(1+\pi+\pi^2\right). $$ After that, it is simple math. A worked example of the same technique in a similar (but apparently much harder) problem can be found here.

Elaborating user @Dr. MV's answer, we have

\begin{equation} \int_0^\infty\frac{1}{a^2x^4+bx^2+c^2}\ dx=\frac{c\pi}{2a\sqrt{b+2ac}} \end{equation}

Putting $a=1$, $b=a$, and $c^2=b$, then

\begin{equation} I(a,b)=\int_0^\infty\frac{1}{x^4+ax^2+b}\ dx=\frac{\pi}{2}\sqrt{\frac{b}{a+2\sqrt{b}}} \end{equation}

Hence

\begin{equation} \frac{\partial}{\partial a}I(a,b)=\int_0^\infty\frac{x^2}{\left(x^4+ax^2+b\right)^2}\ dx=-\frac{\pi}{2b}\sqrt{\left(\!\frac{b}{a+2\sqrt{b}}\!\right)^3} \end{equation} and \begin{equation} \frac{\partial}{\partial b}I(a,b)=\int_0^\infty\frac{1}{\left(x^4+ax^2+b\right)^2}\ dx=\frac{\pi(a+\sqrt{b})}{2\sqrt{b\left(a+2\sqrt{b}\right)^3}} \end{equation}