Closed form of $ \int_0^\infty x e^{-b \pi x^2}(1+x^{-a})^{-n}\mathrm{d}x$ 
Is there an exact closed-form expression or a closed-form upper bound for the following integral?
  $$I = \int_{0}^{\infty}\frac{x e^{-b \pi x^2}}{(1+x^{-a})^{n}}\mathrm{d}x, \tag1$$
  where $a > 2$, $b, n > 0$.
  If yes, how can we obtain it?

My attempt: If we multiply $I$ in $(1)$ by $2\pi b$, we get the probability density function of Rayleigh random variable, i.e, $f_X(x) = 2\pi b xe^{-b \pi x^2}$. So we can write $I$ as
$$I = \mathbb{E}\left[\left(\frac{1}{1+X^{-a}}\right)^{n}\right].$$
I could not procced in this way.
 A: $$I = \int_{0}^{\infty}\frac{x e^{-b \pi x^2}}{(1+x^{-a})^{n}}\mathrm{d}x$$
$$I = \frac{1}{2}\int_{0}^{\infty}\frac{e^{-b \pi t}}{(1+t^{-a/2})^{n}}\mathrm{d}t\qquad\text{with}\quad t=x^2$$
$$I=\frac{1}{2}\mathscr{L}_t\left[\frac{1}{(1+t^{-a/2})^{n}}\right](b\pi)$$
$\mathscr{L}$ denotes the Laplace transform.
The hitch is that no closed form exists in the most extended tables of Laplace transform for the general form of this function. So, probably one cannot express the integral on a closed form in the general case (any $a$ and $n$).
Of course, closed form exist in particular cases. For example :
$n=1\:,\:a=2 \quad\to\quad I=\frac{1}{2b\pi}+\frac{1}{2}e^{b\pi}\mathrm{Ei}(-b\pi)\qquad \text{: Exponential integral}$
$n=1\:,\:a=4 \quad\to\quad I=\frac{1}{2b\pi}+\frac{1}{2}\cos(b\pi)\left(\mathrm{Si}(b\pi)-\frac{\pi}{2}\right)-\frac{1}{2}\sin(b\pi)\mathrm{Ci}(b\pi) \quad$
where Si and Ci are the sin and cos integrals.
A: WolframAlpha ist able to compute the anti-derivative of your integrand for $a=2$ and small fixed integer $n$. From these solutions I came up with a conjecture for $a=2$ and general integer $n$. 
Some notation:
$$ I_{a,n}(c)= \int_{0}^{\infty}\frac{x e^{-c x^2}}{(1+x^{-a})^{n}}\mathrm{d}x, \tag1$$
where $a > 2$, $c, n > 0$.
The conjecture for $a=2$:
$$ F_{2,n}(x,c)= \int\frac{x e^{-c x^2}}{(1+x^{-2})^{n}}\mathrm{d}x = p_1(c) ~ \mathrm{e}^c ~ \mathrm{Ei}(-c(x^2+1)) - \frac{\mathrm{e}^{-cx^2}~p_2(c,x)}{c(x^2+1)^{n-1}}$$
$$ I_{2,n}(c) = -F_{2,n}(0,c) = -p_1(c) ~ \mathrm{e}^c ~ \mathrm{Ei}(-c) + \frac{p_2(c,0)}{c}$$
Where $p_1(c)$ is a polynomial in $c$ of degree $n-1$ and $p_2(x,c)$ is a polynomial in $c$ of degree $n-1$ and in $x$ of degree $2(n-1)$ with only even terms in $x$.
I don't have a general form for $p_1$ and $p_2$, but using the above restrictions it's possible to solve for their coefficients given some fixed $n$ by differentiating $F_{2,n}$ and comparing with the integrand.
Some solutions:
$$I_{2,1} = \left(\frac{1}{2} \right)\mathrm{e}^c \mathrm{Ei}(-c) +
\frac{1}{c}\left( \frac{1}{2} \right) \text{(already given by JJacquelin)}$$
$$I_{2,2} = \left(\frac{1}{2} \, c + 1 \right)\mathrm{e}^c
\mathrm{Ei}(-c) + \frac{1}{c}\left( \frac{1}{2} \, c + \frac{1}{2}
\right)$$
$$I_{2,3} = \left(\frac{1}{4} \, c^{2} + \frac{3}{2} \, c + \frac{3}{2}
\right)\mathrm{e}^c \mathrm{Ei}(-c) + \frac{1}{c}\left( \frac{1}{4} \,
c^{2} + \frac{5}{4} \, c + \frac{1}{2} \right)$$
$$I_{2,4} = \left(\frac{1}{12} \, c^{3} + c^{2} + 3 \, c + 2
\right)\mathrm{e}^c \mathrm{Ei}(-c) + \frac{1}{c}\left( \frac{1}{12} \,
c^{3} + \frac{11}{12} \, c^{2} + \frac{13}{6} \, c + \frac{1}{2}
\right)$$
