How to find the closed form of $\int_{-\infty}^{\infty} \frac{x^{2 n+1} \sin x}{\left(1+x^{2}\right)^{n+1}} d x, \textrm{ where }n=0,1,2,3,…?$ In my post, I had found the exact value of the integral
$$\displaystyle I:=\int_{-\infty}^{\infty} \frac{x^{3} \sin x}{\left(1+x^{2}\right)^{2}}dx= \frac{\pi}{2 e}\tag*{} $$
by differentiating $J(a)$w.r.t. $a$ once.
$\displaystyle J(a)=\int_{-\infty}^{\infty} \frac{x \sin x}{1+a x^{2}} dx =\frac{\pi e^{-\frac{1}{\sqrt{a}}}}{a}, \textrm{ where }a>0\tag*{} $
Then I generalize $I$ to $$
I_{n}=\int_{-\infty}^{\infty} \frac{x^{2 n+1} \sin x}{\left(1+x^{2}\right)^{n+1}} d x, \textrm{ where }n=0,1,2,3,…
$$
by differentiating $J(a)$ w.r.t. by $n$ times at $a=1$.
$$
\left. J^{(n)}(1) =\pi \frac{d^{n}}{d a^{n}}\left(a^{-1} e^{-\frac{1}{\sqrt{a}}}\right)\right|_{a=1}
$$
$$J^{(n)}(1) = \int_{-\infty}^{\infty} \frac{(-1)^{n} n ! x^{2n+1}\sin x}{\left(1+a x^{2}\right)^{n+1}} dx= (-1)^{n} n !I_n \tag*{}$$
Hence we can conclude that
$$ I_{n}=\left.\frac{(-1)^{n}}{n !} J^{(n)}(1) = \frac{(-1)^{n} \pi}{n !}\frac{d^{n}}{d a^{n}}\left(a^{-1} e^{-\frac{1}{\sqrt{a}}}\right)\right|_{a=1}
\tag*{} $$
Theoretically, we can find the exact value of $I_n$ by $J^{(n)}(1)$.
Urging to the closed form of $I_n$, I use  Leibniz’s Rule to find $J^{(n)}(1)$
\begin{aligned}
\left.\frac{d^{n}}{d a^{n}}\left(a^{-1} e^{-\frac{1}{\sqrt{a}}}\right) \right|_{a=1} &= \left.\sum_{k=0}^{n}{n\choose k} \left(a^{-1}\right)^{(n-k)}\left(e^{-\frac{1}{\sqrt{a}}}\right)^{(k)}\right|_{a=1} =
\left.\sum_{k=0}^{n} {n\choose k} (-1)^{n-k}(n-k)!( e^{-\frac{1}{\sqrt{a}}} )^{(k)} \right|_{a=1}
\end{aligned}
Hence  $$I_{n}=\frac{(-1)^{n}}{n !} J^{(n)}(1)= 
\left. \pi\sum_{k=0}^{n} \frac{(-1)^{k}}{k!}( e^{-\frac{1}{\sqrt{a}}} )^{(k)} \right|_{a=1}$$
By Wolframalpha, we have
$$
\frac{\partial^{k} e^{-1 / \sqrt{x}}}{\partial x^{k}}=e^{-1 / \sqrt{x}} x^{-k} \sum_{j=0}^{k} \sum_{i=0}^{j} \frac{(-1)^{i}\left(-\frac{1}{\sqrt{x}}\right)^{j}\left(\frac{1}{2}(2+i-j-2 k)\right)_{(k)}}{i !(j-i) !}
$$
Putting $x=1$ yields
$$
\left.\frac{\partial^{k} e^{-\frac{1}{\sqrt{x}}}}{\partial x^{k}}\right|_{x=1}=e^{-1} \sum_{j=0}^{k} \sum_{i=0}^{j} \frac{(-1)^{i}\left(\frac{1}{2}(2+i-j-2 k)\right) _{(k)}}{i !(j-i) !}
$$
Now we can conclude that
$$I_{n} = \pi\sum_{k=0}^{n} \frac{(-1)^{k}}{k!} \left.\frac{\partial^{k} e^{-\frac{1}{\sqrt{x}}}}{\partial x^{k}}\right|_{x=1}=\frac{\pi}{e} \sum_{k=0}^{n}   \sum_{j=0}^{k} \sum_{i=0}^{j} \frac{(-1)^{i+k}\left(\frac{1}{2}(2+i-j-2 k)\right)_{(k)}}{i !(j-i) !k!} $$
The closed form is rather complicated and ugly. Is there a simpler one?
 A: Only with Mathematica:
$$\int_{-\infty }^{\infty } \frac{x^{2 n+1} \sin (x)}{\left(1+x^2\right)^{n+1}} \, dx=\frac{1}{2} \pi  (-1)^{2 n} \, _1F_2\left(n+1;\frac{1}{2},1;\frac{1}{4}\right)+\frac{1}{2} \pi  \, _1F_2\left(n+1;\frac{1}{2},1;\frac{1}{4}\right)-\frac{\sqrt{\pi } (-1)^{2 n} \Gamma
   \left(n+\frac{3}{2}\right) \, _1F_2\left(n+\frac{3}{2};\frac{3}{2},\frac{3}{2};\frac{1}{4}\right)}{\Gamma (n+1)}-\frac{\sqrt{\pi } \Gamma \left(n+\frac{3}{2}\right) \,
   _1F_2\left(n+\frac{3}{2};\frac{3}{2},\frac{3}{2};\frac{1}{4}\right)}{\Gamma (n+1)}$$
MMA code:
Using:
Im[FourierTransform[x^(2 n + 1)/(1 + x^2)^(n + 1), x, -s,  FourierParameters -> {1, -1}]] /. s -> 1
We have:
(Pi*HypergeometricPFQ[{1 + n}, {1/2, 1}, 1/4])/2 + ((-1)^(2*n)*Pi*HypergeometricPFQ[{1 + n}, {1/2, 1}, 1/4])/2 - (Sqrt[Pi]*Gamma[3/2 + n]*HypergeometricPFQ[{3/2 + n}, {3/2, 3/2}, 1/4])/ Gamma[1 + n] - ((-1)^(2*n)*Sqrt[Pi]*Gamma[3/2 + n]*HypergeometricPFQ[{3/2 + n}, {3/2, 3/2}, 1/4])/Gamma[1 + n]
A: Using @Mariusz Iwaniuk's answer, a small simplification
$$I_n=\sqrt \pi \Bigg[\sqrt{\pi } \, _1F_2\left(n+1;\frac{1}{2},1;\frac{1}{4}\right) -2\frac{ \Gamma \left(n+\frac{3}{2}\right)}{\Gamma (n+1)}\, _1F_2\left(n+\frac{3}{2};\frac{3}{2},\frac{3}{2};\frac{1}{4}\right)\Bigg]$$ which is interesting to look at in details.
$$\, _1F_2\left(n+1;\frac{1}{2},1;\frac{1}{4}\right)=\frac 1 {a_n}\left(b_n\cosh(1)+c_n \sinh(1) \right)$$
$$\, _1F_2\left(n+\frac{3}{2};\frac{3}{2},\frac{3}{2};\frac{1}{4}\right)=\frac 1 {d_n}\left(c_n\cosh(1)+b_n \sinh(1) \right)$$ where
$$a_n=\frac 1 {2^n \,n!} \quad \text{and} \quad d_n=\frac 1 {(2n+1)!!} \implies I_n=\frac \pi {e}\,\,\frac{b_n-c_n}{2^n \,n! }$$ The $b_n$ form the sequence
$$\{1,2,9,63,592,6925,96451,1553762,28366983,578145897,13001344876,\cdots\}$$ and the $c_n$ form the sequence
$$\{0,1,7,58,587,7101,100332,1624099,29652373,602952562,13513681685,\cdots\}$$
Now, I better see why $I_n$ is positive for $0 \leq n \leq 4$ and for $n>44$.
A: This not an answer.
Computing
$$I_{n}=\left.\frac{(-1)^{n}}{n !} J^{(n)}(1) = \pi \frac{d^{n}}{d a^{n}}\left(a^{-1} e^{-\frac{1}{\sqrt{a}}}\right)\right|_{a=1}$$
$${I_n}= \frac \pi{2^n \,e }\, b_n$$ and the $b_n$ generate the sequence
$$\{-1,2,-5,5,176,-3881,70337,-1285390,24806665,-512336809,\cdots\}$$ which is not identified by $OEIS$.
Edit
For the generation of the sequence of $b_n$, let $a=b+1$ and use Taylor expansion around $b=0$ of
$$\frac 1a e^{-\frac{1}{\sqrt{a}}}=\frac 1{1+b} e^{-\frac{1}{\sqrt{1+b}}}$$ from inside to outside
$$\frac{1}{\sqrt{1+b}}=1-\frac{b}{2}+\frac{3 b^2}{8}-\frac{5 b^3}{16}+\frac{35 b^4}{128}-\frac{63
   b^5}{256}+\frac{231 b^6}{1024}-\frac{429 b^7}{2048}+O\left(b^{8}\right)$$
$$e^{-\frac{1}{\sqrt{1+b}}}=\frac{1}{e}+\frac{b}{2 e}-\frac{b^2}{4 e}+\frac{7 b^3}{48 e}-\frac{35 b^4}{384 e}+\frac{113
   b^5}{1920 e}-\frac{1769 b^6}{46080 e}+\frac{16003 b^7}{645120 e}+O\left(b^8\right)$$
$$\frac{e^{-\frac{1}{\sqrt{1+b}}}}{1+b}=\frac{1}{e}-\frac{b}{2 e}+\frac{b^2}{4 e}-\frac{5 b^3}{48 e}+\frac{5 b^4}{384 e}+\frac{11
   b^5}{240 e}-\frac{3881 b^6}{46080 e}+\frac{70337 b^7}{645120 e}+O\left(b^8\right)$$ Multiply by $e$ and notice that, multiplied by $2^n$ give whole numbers.
