is there a closed algebraic solution to x(x+a)e^x=b, a,b positive reals?

I am looking at the following equation which is solvable in terms of the Lambert-W function when $a=0$ (but it is strictly positive in my case, i.e. $a>0$):

$x(x+a)e^x=b$ $(a,b>0)$

more generally, one can consider a generalization to the lambert function of the form : $(x-r_1)(x-r_2)e^x=b$ (in my case $r_1=0, r_2=-a$). Can the solution $x$ be expressed using the Lambert function or other special functions?

• arxiv.org/pdf/1501.00138v3.pdf could be of interest. – Claude Leibovici Mar 17 '16 at 12:52
• arxiv.org/pdf/1408.3999.pdf could be too – Renato Faraone Mar 17 '16 at 19:37
• Using the reference arxiv.org/pdf/1501.00138v3.pdf I was able to obtain the answer: $x=\sum_{n=1}^{\infty} \frac{(-\frac{-nb}{a})^n}{n n!}B_{n-1}(2a^{-1})$ for the equation $x(x+a)e^x=b$. – Gregor Samsa Mar 19 '16 at 7:25
• $B_{n-1}(x)$ in the above is the Bessel polynomial of order $n-1$. – Gregor Samsa Aug 11 '16 at 6:42

It is possible to solve $(x+a)e^x=b$ but $x(x+a)e^x=b$ is not solvable.
It would require you to get the exponent to become $x(x+a)$, which means you'd have to exponentiate everything by $x+a$ to get it into the exponent. And, that leads to problems.