Q:

How to evaluate the integral $\int_0^\infty \frac{x^{a-1}}{1+bx^a} e^{-x} dx$

How to evaluate this integral? 
\begin{equation}
  \int_0^\infty \frac{x^{a-1}}{1+bx^a} e^{-x} dx
\end{equation}
I think it will use a gamma function or a exponential integral. I really need an advice to continue.
$a$ and $b$ are real constants.

A:

I assume you are searching for the closed form. I doubt there is any nice one for the general $a$.
First, let's make $b$ more manageable. We make a substitution:
$$b^{1/a}x=t,~~~~~~~~b^{-1/a}=c$$
Then the integral becomes:
$$\frac{1}{b} I(a,c)= \frac{1}{b}\int_0^\infty \frac{t^{a-1}}{1+t^a} e^{-ct} dt$$
I was able to find the general closed form only for $a \in \mathbb{N}$ - positive integers. Of course, $c>0$.
The general expression was deduced using Mathematica:

$$I(a,c)=\int_0^\infty \frac{t^{a-1}}{1+t^a} e^{-ct} dt=\frac{1}{\sqrt{a} (2 \pi)^{(a-1)/2}} G_{1,a+1}^{a+1,1}\left( \frac{c^a}{a^a} \left|\begin{matrix}0\\
0,0,\frac{1}{a},\frac{2}{a},\dots,\frac{a-1}{a}\end{matrix}\right.\right)$$
$$a \in \mathbb{N},~~~~~c>0$$

Here $G$ - is the Meijer G Function. In the general case, there is no simpler expression.

There is a "duplication formula" for the G function here, first theorem on page 3, which might greatly simplify this formula, but I don't know how to apply it yet. I created a separate question here

For example:

$$I(3,1)=\int_0^\infty \frac{t^{2}}{1+t^3} e^{-t} dt=\frac{1}{2\sqrt{3} \pi} G_{1,4}^{4,1}\left( \frac{1}{27} \left|\begin{matrix}0\\
0,0,\frac{1}{3},\frac{2}{3} \end{matrix}\right.\right)$$

In some cases there is a more simple form:
$$I(1,1)= \int_0^\infty \frac{1}{1+t} e^{-t} dt=-e~\mathrm{Ei}(-1)$$
$$I(2,1)= \int_0^\infty \frac{t}{1+t^2} e^{-t} dt=\frac{\pi \sin (1)}{2}- \sin (1) \mathrm{Si} (1)-\cos (1) \mathrm{Ci} (1)$$
$$I(1/2,1)= \int_0^\infty \frac{1}{\sqrt{t}(1+\sqrt{t})} e^{-t} dt=\frac{1}{e} \left( \pi ~\mathrm{erfi}(1)- \mathrm{Ei}(1) \right)$$
$$I(1/3,1)= \int_0^\infty \frac{t^{-2/3}}{1+t^{1/3}} e^{-t} dt=-e \left(\text{Ei}(-1)+\Gamma \left(\frac{2}{3}\right) \Gamma \left(\frac{1}{3},1\right)-\Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{2}{3},1\right)\right)$$
In these cases exponential integral, sine integral, cosine integral, error function, gamma function and incomplete gamma functions are involved.
Thus, I would not expect a general closed form.

As for the series approach, I tried, but didn't find anything useful so far.

Numerically, this integral looks very boring:

