Evaluate or Simplify $\int_{a}^{+\infty} \frac{\exp(-bx)}{x+c} Ei(x) dx$ I am stuck trying to evaluate or simplify this integrale  :

$$I_{a,b,c} = \int_{a}^{+\infty} \frac{\exp(-bx)}{x+c} Ei(x) dx $$

with $a,b,c \in \mathbb{R}_+^*$.
and $ Ei(x) =\int_{-\infty}^{x} \frac{\exp(t)}{t}  \mathbb{d}t $ : The Exponential integral function
I already found this result:
$$\int\exp(-bx) \  Ei(x) = \frac{1}{b} \left[Ei((1-b)x)-\exp(-bx)\ Ei(x) \right] $$
but I can't see if it useful.
Any Hint ?
 A: By using the expression $\text{Ei}(x)=\gamma+\ln x+\int_0^x\dfrac{e^t-1}{t}dt$ mentioned in http://people.math.sfu.ca/~cbm/aands/page_230.htm,
$\int_a^\infty\dfrac{e^{-bx}\text{Ei}(x)}{x+c}dx$
$=\int_a^\infty\dfrac{e^{-bx}}{x+c}\left(\gamma+\ln x+\int_0^x\dfrac{e^t-1}{t}dt\right)~dx$
$=\gamma\int_a^\infty\dfrac{e^{-bx}}{x+c}dx+\int_a^\infty\dfrac{e^{-bx}\ln x}{x+c}dx+\int_a^\infty\dfrac{e^{-bx}}{x+c}\int_0^x\dfrac{e^t-1}{t}dt~dx$
$=\gamma\int_a^\infty\dfrac{e^{-bx}}{x+c}dx+\int_a^\infty\dfrac{e^{-bx}}{x+c}\int_1^x\dfrac{1}{t}dt~dx+\int_a^\infty\dfrac{e^{-bx}}{x+c}\int_0^x\dfrac{e^t-1}{t}dt~dx$
$=\gamma\int_a^\infty\dfrac{e^{-bx}}{x+c}dx+\int_a^\infty\int_1^x\dfrac{e^{-bx}}{t(x+c)}dt~dx+\int_a^\infty\int_0^x\dfrac{(e^t-1)e^{-bx}}{t(x+c)}dt~dx$
$=\gamma\int_a^\infty\dfrac{e^{-bx}}{x+c}dx+\int_1^a\int_a^\infty\dfrac{e^{-bx}}{t(x+c)}dx~dt+\int_a^\infty\int_t^\infty\dfrac{e^{-bx}}{t(x+c)}dx~dt+\int_0^a\int_a^\infty\dfrac{(e^t-1)e^{-bx}}{t(x+c)}dx~dt+\int_a^\infty\int_t^\infty\dfrac{(e^t-1)e^{-bx}}{t(x+c)}dx~dt$
$=\gamma\int_a^\infty\dfrac{e^{-bx}}{x+c}dx+\int_1^a\int_a^\infty\dfrac{e^{-bx}}{t(x+c)}dx~dt+\int_0^a\int_a^\infty\dfrac{(e^t-1)e^{-bx}}{t(x+c)}dx~dt+\int_a^\infty\int_t^\infty\dfrac{e^te^{-bx}}{t(x+c)}dx~dt$
$=\gamma\int_{a+c}^\infty\dfrac{e^{-b(x-c)}}{x}dx+\int_1^a\int_{a+c}^\infty\dfrac{e^{-b(x-c)}}{tx}dx~dt+\int_0^a\int_{a+c}^\infty\dfrac{(e^t-1)e^{-b(x-c)}}{tx}dx~dt+\int_a^\infty\int_{t+c}^\infty\dfrac{e^te^{-b(x-c)}}{tx}dx~dt$
$=\gamma e^{bc}E_1(b(a+c))+\int_1^a\dfrac{e^{bc}E_1(b(a+c))}{t}dt+\int_0^a\dfrac{(e^t-1)e^{bc}E_1(b(a+c))}{t}dt+\int_a^\infty\dfrac{e^{bc}e^tE_1(b(t+c))}{t}dt$
$=\gamma e^{bc}E_1(b(a+c))+e^{bc}E_1(b(a+c))\ln a+e^{bc}E_1(b(a+c))\int_0^a\dfrac{e^t-1}{t}dt+\int_a^\infty e^{bc}e^tE_1(b(t+c))~d(\ln t)$
$=e^{bc}E_1(b(a+c))\text{Ei}(a)+\int_{\ln a}^\infty e^{bc}e^{e^t}E_1(b(e^t+c))~dt$
