A Generalisation of the Integral:
Given a fixed $n\in\mathbb{N}$, let $A$, $P$, $Q$ be polynomials satisfying the following conditions:
- $A(x)=Q(x)^{n+1}-2xP(x)Q(x)+P'(x)Q(x)-nP(x)Q'(x)$
- $\deg A=\deg Q$
- $P(0)=0, \ Q(0)\neq 0$
- $A(x)Q(x)^{-(n+1)}=A(-x)Q(-x)^{-(n+1)}$
Then the integral of $e^{-x^2}A(x)Q(x)^{-(n+1)}$ over $\mathbb{R}^+$ can be computed as such.
\begin{align}
\int^\infty_0 e^{-x^2}\frac{A(x)}{Q(x)^{n+1}}\ dx
&=\int^\infty_0 e^{-x^2}\left(1+\frac{-2xP(x)Q(x)+P'(x)Q(x)-nP(x)Q'(x)}{Q(x)^{n+1}}\right)\ dx\\
&=\frac{\sqrt{\pi}}{2}+\int^\infty_0\frac{((e^{-x^2})P(x))'Q(x)^n-e^{-x^2}P(x)(Q(x)^n)'}{Q(x)^{2n}}\ dx\\
&=\frac{\sqrt{\pi}}{2}+\left[e^{-x^2}\frac{P(x)}{Q(x)^n}\right]^\infty_0\\
&=\frac{\sqrt{\pi}}{2}
\end{align}
This means that as long as we can find polynomials $A$, $P$, $Q$ that satisfy all these conditions, we will be able to "construct" similar integrals to that which was posted in the question (at least in principle).
Two Useful Facts:
Before we proceed to determine suitable $A$, $P$ and $Q$, we first prove the following facts:
$\text{Fact 1}$: $$\deg P=n\deg Q-1\tag{*}$$
To deduce this fact, observe that the polynomials $Q(x)^{n+1}$, $-2xP(x)Q(x)$, $P'(x)Q(x)$ and $-nP(x)Q'(x)$ have degrees $(n+1)\deg Q$, $\deg P+\deg Q+1$, $\deg P+\deg Q-1$ and $\deg P+\deg Q-1$ respectively. In order for the sum of these polynomials to have degree $\deg Q$, for all $\deg Q < j\leq\max((n+1)\deg Q, \deg P+\deg Q+1)$, the coefficients of the terms $x^j$ in each of these four polynomials have to add up to equal $0$. This requires
$$\max((n+1)\deg Q, \deg P+\deg Q+1)=\min((n+1)\deg Q, \deg P+\deg Q+1)$$
and the desired result follows.
$\text{Fact 2}$: $$P(x)Q(x)^{-n}=-P(-x)Q(-x)^{-n}\tag{**}$$
This follows from Condition 4. Since $A(x)Q(x)^{-(n+1)}$ is even, $A(x)Q(x)^{-(n+1)}-1$ must also be even. But $A(x)Q(x)^{-(n+1)}-1=e^{x^2}(e^{-x^2}P(x)Q(x)^{-n})'$, so $P(x)Q(x)^{-n}$ must be odd.
Note that $(^{**})$ also implies Condition 3.
A Simple Example: $n=1$, $\deg Q=2$
Let $n=1$ and $Q(x)=x^2+c\ $ with $c\neq 0$. By $(^{*})$ and $(^{**})$, $P$ is an odd polynomial of degree $1$, i.e. $P(x)=kx$ for some constant $k$. By Condition 1,
\begin{align}
A(x)
&=(x^2+c)^2-2kx^2(x^2+c)+k(x^2+c)-2kx^2\\
&=(1-2k)x^4+(2c(1-k)-k)x^2+c(k+c)
\end{align}
Since $\deg A=2$, $k=\frac{1}{2}$ and $c\neq\frac{1}{2}$. Thus
$$A(x)=\left(c-\frac{1}{2}\right)x^2+c\left(c+\frac{1}{2}\right)$$
and we obtain the identity, for $c\in\mathbb{R}^+\setminus\{\frac{1}{2}\}$
$$\int^\infty_0e^{-x^2}\cdot\frac{x^2+\frac{c(2c+1)}{2c-1}}{(x^2+c)^2}\ dx=\frac{\sqrt{\pi}}{2c-1}$$
The Case in Question: $n=2$, $\deg Q=4$
We follow the exact same procedure outlined above. In this case $n=2$ and $Q(x)=x^4+px^2+q$. Then $P(x)=rx^7+sx^5+tx^3+ux$. Applying Condition 1 and noting that $\deg A=4$ (i.e. the coefficients of $x^{12}$, $x^{10}$, $x^8$, $x^6$ are all $0$) ,
\begin{align}
A(x)
&=(x^4+px^2+q)^3-2x(rx^7+sx^5+tx^3+ux)(x^4+px^2+q)\\
&\ \ \ \ \ +(7rx^6+5sx^4+3tx^2+u)(x^4+px^2+q)-2(rx^7+sx^5+tx^3+ux)(4x^3+2px)\\
&=(p(3pq-t-2u)+q(3q+5s-2t)-7u)x^4+(3p(q^2-u)+q(3t-2u))x^2+q(q^2+u)
\end{align}
where
\begin{align}
&\ \ \ \ \ 1-2r=3p-2s-r(1+2p)=p(3p-2s)+r(3p-2q)+3q-3s-2t\\
&=p(p^2+6q+s-2t)+q(7r-2s)-5t-2u=0
\end{align}
After some algebra, we may express $r,s,t,u$ in terms of the free variables $p,q$.
$$(r,s,t,u)=\left(\frac{1}{2},\frac{4p-1}{4},\frac{4p^2-4p+8q+3}{8},\frac{-4p^2+16pq+12p-8q-15}{16}\right)$$
Therefore
\begin{align}
A(x)
&=\tfrac{12p^2+16q^2-16pq-60p+24q+105}{16}x^4+\tfrac{12p^3-16p^2q+16pq^2-36p^2+64q^2-24pq+45p+48q}{16}x^2\\
&\ \ \ \ \ +\tfrac{16q^3-4p^2q+16pq^2-8q^2+12pq-15q}{16}
\end{align}
This yields, for $p,q,s,t,u\neq 0$,
$$\small{\int^\infty_0e^{-x^2}\tiny{\frac{(12p^2+16q^2-16pq-60p+24q+105)x^4+(12p^3-16p^2q+16pq^2-36p^2+64q^2-24pq+45p+48q)x^2+(16q^3-4p^2q+16pq^2-8q^2+12pq-15q)}{16(x^4+px^2+q)^3}}\ dx=\frac{\sqrt{\pi}}{2}}$$
If $p=1$ and $q=\frac34$, this integral reduces to the one posted in the question.
