TITLE: Unknown Taylor expansion
QUESTION [5 upvotes]: I have come across a few apparently related Taylor expansions, as detailed below:
\begin{align}
&\dots\frac{a^7}{140}-\frac{a^6}{80}-\frac{3 a^5}{40}-\frac{a^4}{8}+\frac{a^2}{2}+a+1&&=\exp \left(a-\frac{a^3}{6}\right)&\tag{1}\\
\\
&\dots\frac{11 a^7 b}{1120}+\frac{a^6 b}{32}+\frac{a^5 b}{40}-\frac{a^4 b}{16}-\frac{a^3 b}{4}-\frac{a^2 b}{4}+\frac{b}{2}&&=?&\tag{2}\\
\end{align}
The first was fairly easy to guess the closed form on the RHS, but the bottom one I am stuck on. The extra $b$ is obviously doing something to the coefficients, but I don't know what. Are there any good strategies for finding a closed form for $(2$)?
Unfortunately, I don't have any background information on them, so much is pure guesswork.
Update
All known coefficients  for $(2)$:
$$\frac{1}{2},0,-\frac{1}{4},-\frac{1}{4},-\frac{1}{16},\frac{1}{40},\frac{1}{32},\frac{11}{1120},-\frac{1}{1280},-\frac{43}{20160}$$

REPLY [3 votes]: $$
\frac{1}{2} (1-a)\exp\left(a-\frac{a^3}{6}\right)=
\frac{1}{2}-\frac{a^2}{4}-\frac{a^3}{4}-\frac{a^4}{16}+\frac{a^5}{40}+\frac{a^6}{32}+\frac{11 a^7}{1120}-\frac{a^8}{1280}-\frac{43
   a^9}{20160}-\frac{69 a^{10}}{89600}+O\left(a^{11}\right)
$$