Approximating $\int_1^\infty x^a (x-1)^b e^{cx}dx$ After lengthy calculations, I arrived at
\begin{align*}
\int_1^\infty x^a (x-1)^b e^{cx}dx,
\end{align*}
which cannot be solved in closed-form. I thus seek to approximate the integral, potentially including special functions. The constants $a,b,c$ may be complex but are  chosen such that the integral exists.
Any ideas how to approximate the integral? I thought of rounding the real and imaginary part of $a$ and $b$ to integers. Then, simplifying the first two terms and plenty of partial integrations should give a solution. But that seems to be a very naive idea.
Different elements of solution.
I will consider only real values of $a,b$, with $c$ real and negative for evident convergence concerns.
If $a,b$ are positive integers, there is an easy answer because a function of the form $e^{cx}P_n(x)$ has a primitive function of the same form $e^{cx}Q_n(x)$ with degree$(P_n)  \ = \ $ degree$(Q_n)=n$ which is:
$$\int e^{ax}P_n(x)dx  = e^{ax}\sum\limits_{k = 0}^n (-1) ^k \frac{1} {a^{k+1}}\frac{d^k}{dx^k}\left(P_n(x)\right)  + C\tag{*}$$
Proof: First part of this answer (https://math.stackexchange.com/q/155808) by Robjohn.
See the remark at the bottom for a proof using a less &quot;symbolic&quot; approach.

In the general case, the different tracks I have obtained using the research tool https://approach0.xyz/ (especially efficient on formulas) were dealing with bounds either different from $1$ or $\infty$. Nevertheless, one of the main indications is the closed form expression of the indefinite integral (= primitive function), as found in the answer by Harry Peter here
$$\int e^{ax}x^b(1-x)^c~dx=\dfrac{x^{b+1}\Phi_1(b+1,-c,b+2;x,ax)}{b+1}+C$$
where $\Phi_1$ can be defined in the following way:
$$\text{For} \ \Re \,c&gt;\Re \,a&gt;0, \ \ \ \Phi _{1}(a,b,c;x,y):=$$
$${\frac {\Gamma (c)}{\Gamma (a)\Gamma (c-a)}}\int _{0}^{1}t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}\,\mathrm {d} t.$$
(see here).
With definite bounds, one finds, from the same Harry Peter here, a formula for
$$\int_0^\infty x^{a-1}(1-x)^{b-1}e^{t-cx}~dx=$$
$$=\dfrac{e^t\Gamma(a)\Gamma(b)M(a,a+b,-c)}{\Gamma(a+b)}-(-1)^be^{t-c}\Gamma(b)U(b,a+b,c)\tag{1}$$
(with an interesting splitting of the integral $\int_0^1+\int_1^{+\infty}$)
For $t=0$, formula (1) is nothing else than the Laplace Transform of $x^a (x-1)^b$, easy to find directly.
It remains to subtract to (1) the value of $\int_1^\infty x^a (x-1)^b e^{cx}dx$ as indicated here.
Remark: A more classical approach to formula (*):
If $I_n:= \int x^n e^{ax}dx$, we have: $$(-1)^na^n\frac{I_n}{n!}=e^{ax} \sum_{k=0}^n(-1)^ka^{k-1}\frac{x^k}{k!}$$
(proofs by Qiaochu Yuan and Olivier Oloa here). Please note that the summation above is a kind of truncated exponential.
$$I(a,b,c)=\int_1^\infty x^a(x-1)^b e^{cx}dx$$
$$I_c=I(a+1,b,c)$$
$$I_{ca}=(a+1)I\tag{1}$$
this is a pde so you could try and solve this, initial conditions with $c=0$ or $b=0$ should make this problem simple. However notice that this integral appear to diverge for $c&gt;0$