(New Solution needed) Explicit series expansion for inverse of $e^{-x}\left(\frac{x^2}2+x+1\right)$
Solution 1:
There is an efficient way to generate the coefficients
$$d_n = e^{-n}\frac{d^{n-1}}{dy^{n-1}}\Big( \frac{y-1}{e^{-y}(y^2/2+y+1) - 5/(2e)} \Big)^n\Big|_{y=1} $$ $$d_n=- \{2,4,32,384,6400,139536...\} \text{ for } n=1,2,...$$ Note the negative sign outside the leading bracket is meant to apply to each coefficient. Set $y \to y+1,$ and with algebra it can be shown that
$$d_n = \frac{d^{n-1}}{dy^{n-1}}\Big( \frac{y}{5/2(e^{-y}-1)+e^{-y}(2y+y^2/2)} \Big)^n\Big|_{y=0}$$
Interpret it in 'coefficient of' language. For a power series $A(y)=a_0+a_1y + ...$
$$ a_n= \frac{1}{n!}\frac{d^n}{dy^n} A(y)\Big|_{y=0} = [y^n]A(y)$$ Therefore $$d_n = (n-1)! \ [y^{n-1}] \Big( \frac{y}{5/2(e^{-y}-1)+e^{-y}(2y+y^2/2)} \Big)^n $$ $$ = (n-1)! \ [y^{-1}] \big(5/2(e^{-y}-1)+e^{-y}(2y+y^2/2)\big)^{-n} $$ $$ = (n-1)! \, [y^{n-1}] \big(\frac{5}{2y}(e^{-y}-1)+e^{-y}(2+y/2)\big)^{-n} $$ It has been written this way because the portion raised to the $-n$ power is now an ordinary power series, and everyone knows (or should know) how to calculate the coefficients recursively. That is, given
$$ \Big( \sum_{k=0}^\infty a_k \,x^{k} \Big)^{-n} = \sum_{m=0}^\infty c_m \, x^m, $$ with non-zero $a_0,$ the coefficients $c_m$ are found by
$$ c_0=a_0^{-n}$$ $$(*) \quad c_m=\frac{1}{m a_0} \, \sum_{k=1}^m \big( k(1-n)-m\big)a_k c_{m-k} $$
For this particular application, $$ \frac{5}{2y}(e^{-y}-1)+e^{-y}(2+y/2) = \sum_{k=0}^\infty \frac{(-y)^k}{k!}\big( 2 - \frac{k}{2} - \frac{5}{2(k+1)} \big) $$ so $$a_0=-1/2, \quad a_k=\frac{(-1)^k}{k!}\big(2 -\frac{k}{2} - \frac{5}{2(k+1)} \big) $$ Once the $c_n$ have been calculated, don't forget the multiplication by $(n-1)!$ to get the $d_n.$
Solution 2:
Remark. (Not closed form, just the first few terms.)
$e^{-x}\left(\frac{x^2}2+x+1\right)$
When $x=0$ we have $y=1$. There are three series solutions for $x$ in terms of $y$ near $y=1$.
For the real solution, Maple says
$$
x = -\sqrt [3]{6}(y-1)^{1/3}+\frac{{6}^{2/3}}{4} \left( y-1 \right) ^{2/3}-{
\frac {21\,(y-1)}{40}}+{\frac {101\,\sqrt [3]{6} \left( y-
1 \right) ^{4/3}}{480}}-{\frac {3049\,{6}^{2/3} \left( y-1 \right) ^{5
/3}}{33600}}+{\frac {11013\, \left( y-1 \right) ^{2}}{44800}}-{\frac {
733\,\sqrt [3]{6} \left( y-1 \right) ^{7/3}}{6400}}+{\frac {1760963\,{
6}^{2/3} \left( y-1 \right) ^{8/3}}{32256000}}-{\frac {87686441\,
\left( y-1 \right) ^{3}}{551936000}}+{\frac {34921261559\,\sqrt [3]{6
} \left( y-1 \right) ^{10/3}}{447068160000}}-{\frac {322388433709\,{6}
^{2/3} \left( y-1 \right) ^{11/3}}{8302694400000}}+{\frac {
670819407417\, \left( y-1 \right) ^{4}}{5740134400000}}-{\frac {
576727795744637\,\sqrt [3]{6} \left( y-1 \right) ^{13/3}}{
9763968614400000}}+{\frac {5867539686420271\,{6}^{2/3} \left( y-1
\right) ^{14/3}}{195279372288000000}}-{\frac {257105912707053\,
\left( y-1 \right) ^{5}}{2788065280000000}}+{\frac {
5035135228456419089\,\sqrt [3]{6} \left( y-1 \right) ^{16/3}}{
106231978524672000000}}-{\frac {8774289894655573392527\,{6}^{2/3}}{
358655502880604160000000} \left( y-1 \right) ^{{\frac{17}{3}}}}+O
\left( \left( y-1 \right) ^{6} \right)
$$