How do we evaluate the closed form for $\int_{-\infty}^{+\infty}{(-1)^{n+1}x^{2n}+2n+1\over (1+x^2)^2}\cdot e^{-x^2}\mathrm dx?$ Proposed:

$$\int_{-\infty}^{+\infty}{(-1)^{n+1}x^{2n}+2n+1\over (1+x^2)^2}\cdot e^{-x^2}\mathrm dx={\sqrt{\pi}\over 2^{n-2}}\cdot F(n)\tag1$$
  Where is n integer, $n\ge1$

I am struggled to find the closed form for $(1)$
Where $F(1)=1, F(2)=3, F(3)=9, F(4)=21, F(5)=63, ...$
How can we find the closed form for $(1)?$
 A: 
Here is a way forward that provides a solution in terms of a finite double summation. 

Let $f(a)$ be defined by 
$$f(a)=\int_{-\infty}^\infty \frac{e^{-ax^2}}{(1+x^2)^2}\,dx=\sqrt{\pi a}-\frac{\pi}{2}(2a-1)e^a \text{erfc}(\sqrt {a}) \tag 1$$
Then, the $n$'th derivative of $f(a)$ can be written
$$f^{(n)}(a)=\int_{-\infty}^\infty \frac{(-1)^nx^{2n}e^{-ax^2}}{(1+x^2)^2}\,dx\tag 2$$
Using $(1)$ and $(2)$ reveals
$$\int_{-\infty}^\infty \frac{(-1)^{n+1}x^{2n}+(2n+1)}{(1+x^2)^2}e^{-x^2}\,dx=(2n+1)f(1)-f^{(n)}(1)\tag 3$$
We can then use $(1)$ and $(2)$ to generate evaluated at $1$ to evaluate the integral of interest in $(3)$.

First, it is trivial to see that 
$$(2n+1)f(1)=(2n+1)\sqrt{\pi}-(2n+1)\frac{e\pi}2 \,\text{erfc}(1)\tag 4$$
Second, it is straightforward to show that 
$$\left.\left(\frac{d^n }{da^n}\sqrt{\pi a}\right)\right|_{a=1}=\frac{(-1)^{n-1}\sqrt{\pi}(2n-3)!!}{2^n} \tag5$$
Third, using the General Leibniz's Rule for the n'th derivative of a product of two functions, we have 
$$\begin{align}
-\frac\pi2\left.\left(\frac{d^n }{da^n}\left( (2a-1)e^a\text{erfc}(\sqrt a)\right)\right)\right|_{a=1}&=-\frac\pi2\left.\left(\sum_{k=0}^n\binom{n}{k}\frac{d^k}{da^k}\left(\text{erfc}(\sqrt{a})\right)\,\frac{d^{n-k}}{da^{n-k}}\left((2a-1)e^a\right) \right)\right|_{a=1}\\\\
&=-\frac\pi2\sum_{k=0}^n\binom{n}{k}\left.\left(\frac{d^k}{da^k}\left(\text{erfc}(\sqrt{a})\right)\right)\right|_{a=1}\,\left(2(n-k)+1)e\right) \\\\
&=-(2n+1)\frac{e\pi}2 \,\text{erfc}(1)\\\\
&-\frac\pi2\sum_{k=1}^n\binom{n}{k}\left.\left(\frac{d^k}{da^k}\left(\text{erfc}(\sqrt{a})\right)\right)\right|_{a=1}\,\left(2(n-k)+1)e\right) \tag6
\end{align}$$
Using $(4)$, $(5)$, and $(6)$ in $(3)$ yields
$$\begin{align}
\int_{-\infty}^\infty \frac{(-1)^{n+1}x^{2n}+(2n+1)}{(1+x^2)^2}e^{-x^2}\,dx&=(2n+1)\sqrt{\pi}+\frac{(-1)^{n}\sqrt{\pi}}{2^n}(2n-3)!!\\\\
&+\frac\pi2\sum_{k=1}^n\binom{n}{k}\left.\left(\frac{d^k}{da^k}\left(\text{erfc}(\sqrt{a})\right)\right)\right|_{a=1}\,\left(2(n-k)+1)e\right)\tag 7
\end{align}$$

Evaluating that derivative(s) in $(7)$, $\frac{d^k}{da^k}\left(\text{erfc}(\sqrt{a})\right)$, we find 
$$\begin{align}
\frac{d^k}{da^k}\left(\text{erfc}(\sqrt{a})\right)&=-\frac{1}{\sqrt{\pi}}\frac{d^{k-1}}{da^{k-1}}\left(a^{-1/2}e^{-a}\right)\\\\
&=-\frac{1}{e\sqrt{\pi}}\sum_{m=0}^{k-1}\binom{k-1}{m}\left(\frac{(-1)^{k-1}(2m-1)!!}{2^m}\right)\tag 8
\end{align}$$ 
Finally, substituting $(8)$ into $(7)$ we have
$$\begin{align}
\int_{-\infty}^\infty \frac{(-1)^{n+1}x^{2n}+(2n+1)}{(1+x^2)^2}e^{-x^2}\,dx&=(2n+1)\sqrt{\pi}+\frac{(-1)^{n}\sqrt{\pi}}{2^n}(2n-3)!!\\\\
&-\frac{\sqrt {\pi}}{2}\sum_{k=1}^n\binom{n}{k}\,(2(n-k)+1)\sum_{m=0}^{k-1}\binom{k-1}{m}\left(\frac{(-1)^{k-1}(2m-1)!!}{2^m}\right)
\end{align}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\mc{J}_{n} \equiv \int_{-\infty}^{\infty}{\pars{-1}^{n + 1}x^{2n} + 2n + 1
\over \pars{1 + x^{2}}^{2}}\expo{-x^{2}}\,\dd x =
{\root{\pi} \over 2^{n - 2}}\,\mrm{F}\pars{n}\,,\qquad
n \in \mathbb{N}_{\ \geq\ 1}.\quad\mrm{F}\pars{n}:\ {\large ?}}$.

\begin{align}
\mc{J}_{n} & =
2\int_{0}^{\infty}{\pars{-1}^{n + 1}x^{2n} + 2n + 1
\over \pars{1 + x^{2}}^{2}}\expo{-x^{2}}\,\dd x
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
\int_{0}^{\infty}{\pars{-1}^{n + 1}x^{n -1/2} + \pars{2n + 1}x^{-1/2}
\over \pars{1 + x}^{2}}\expo{-x}\,\dd x
\\[5mm] & =
\int_{0}^{\infty}\bracks{%
\pars{-1}^{n + 1}x^{n -1/2} + \pars{2n + 1}x^{-1/2}}\expo{-x}
\bracks{\int_{0}^{\infty}t\expo{-\pars{1 + x}t}\,\dd t}\dd x
\\[5mm] & =
\int_{0}^{\infty}t\expo{-t}\int_{0}^{\infty}
\bracks{\pars{-1}^{n + 1}x^{n -1/2}\expo{-\pars{1 +t}x} + \pars{2n + 1}x^{-1/2}\expo{-\pars{1 + t}x}}
\dd x\,\dd t
\\[5mm] & =
\int_{0}^{\infty}t\expo{-t}\bracks{%
{\pars{-1}^{n + 1}\,\Gamma\pars{n + 1/2} \over \pars{1 + t}^{n + 1/2}} +
{\pars{2n + 1}\Gamma\pars{1/2} \over \pars{1 + t}^{1/2}}}\,\dd t
\\[5mm] & =
\pars{-1}^{n + 1}\,\Gamma\pars{n + {1 \over 2}}a_{n} +
\pars{2n + 1}\root{\pi}a_{0} 
\end{align}

where

\begin{align}
a_{n} & \equiv \int_{0}^{\infty}{t\expo{-t} \over \pars{t + 1}^{n + 1/2}}\,\dd t \,\,\,\stackrel{t + 1\ \mapsto\ t}{=}\,\,\,
\int_{1}^{\infty}{\pars{t - 1}\expo{-\pars{t - 1}} \over t^{n + 1/2}}\,\dd t
\\[5mm] & =
\expo{}\pars{\int_{1}^{\infty}{\expo{-t} \over t^{n - 1/2}}\,\dd t -
\int_{1}^{\infty}{\expo{-t} \over t^{n + 1/2}}}\,\dd t =
\expo{}\bracks{\mrm{E}_{n - 1/2}\pars{1} - \,\mrm{E}_{n + 1/2}\pars{1}}
\end{align}

$\ds{\,\mrm{E}_{p}}$ is the Generalized Exponential Integral.

\begin{align}
\mc{J}_{n} & \equiv
\int_{-\infty}^{\infty}{\pars{-1}^{n + 1}x^{2n} + 2n + 1
\over \pars{1 + x^{2}}^{2}}\expo{-x^{2}}\,\dd x
\\[5mm] & =
\expo{}\left\{\vphantom{\LARGE A}\pars{-1}^{n + 1}\,\Gamma\pars{n + {1 \over 2}}
\bracks{\mrm{E}_{n - 1/2}\pars{1} - \,\mrm{E}_{n + 1/2}\pars{1}}\right.
\\[5mm] & \left.\phantom{\expo{}\braces{}}+
\pars{2n + 1}\root{\pi}\bracks{\mrm{E}_{-1/2}\pars{1} - \,\mrm{E}_{1/2}\pars{1}}\vphantom{\LARGE A}\right\} =
{\root{\pi} \over 2^{n - 2}}\,\mrm{F}\pars{n}
\end{align}
