How to evaluate a Lambert-type series in finite terms? Consider
$$\sum_{m = 1}^\infty \frac{mx^m}{1 - x^{2m}}.$$
Does anyone know how to evaluate these Lambert-type series in finite terms? The right-hand side is related to
$$\sum_{m = 1}^\infty \frac{x^m}{1 - x^{2m}} = L(x) - L(x^2),$$
where
$$L(x) = \sum_{m = 1}^\infty \frac{x^m}{1 - x^m}.$$
I know a related series
$$\sum_{n = 1}^\infty \frac{q^n}{1 + q^n}$$
can be expressed in terms of the Jacobi theta function $\vartheta_3(q)$ as
$$\sum_{n = 1}^\infty \frac{q^n}{1 + q^{2n}} = \frac{\vartheta_3(q)^2 - 1}{4},$$
but I cannot figure out how to relate the series in question to anything known to express it in finite terms so to speak. I am aware that $L(x)$ can be expressed in terms of the $q$-Polygamma function, but I do not think it can be expressed in finite terms.
 A: The Lambert series $$A(x) := \sum_{m = 1}^\infty \frac{mx^m}{1 - x^{2m}}$$ is the generating function of sequence A002131. See the OEIS link for more information. One formula similar to your $L(x)$ formula is
$$A(x) = (P(x^2) - P(x))/24$$ where $P(x)$ is a Ramanujan Eisenstein-Lambert series, the generating function of sequence A006352.
I know of several other expressions but this should be enough.
A: The function, say $f(x) $, of your question is connected to Ramanujan's $P$ function given by $$P(q) =1-24\sum_{i=1}^{\infty}\frac{iq^{i}} {1-q^{i}}\tag{1}$$ via $f(q) =(P(q^{2})-P(q))/24$. If $0<q<1$ then the above can be evaluated in terms of theta functions and elliptic integrals. Thus using Jacobi's theta functions $$\vartheta_{2}(q)=2q^{1/4}\sum_{i=0}^{\infty}q^{i(i+1)},\vartheta_{3}(q)=1+2\sum_{i=1}^{\infty}q^{i^{2}}\tag{2}$$ we get the modulus $k$ and the complete elliptic integrals $K(k) , E(k)$ as $$k=\frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)}, K(k) =\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^{2}\sin^{2}x}} = \frac{\pi} {2}\vartheta_{3}^{2}(q),\notag\\ E(k) =\int_{0}^{\pi/2}\sqrt{1-k^{2}\sin^{2}x}\,dx\tag{3}$$ and we have the following formulas:
\begin{align} 
P(q) & =\left(\frac{2K(k)}{\pi}\right)^{2}\left(\frac{6E(k)}{K(k)}+k^{2}-5\right)\tag{4}\\
P(-q) & =\left(\frac{2K(k)}{\pi}\right)^{2}\left(\frac{6E(k)}{K(k)}+4k^{2}-5\right)\tag{5}\\
P(q^{2})&=\left(\frac{2K(k)}{\pi}\right)^{2}\left(\frac{3E(k)}{K(k)}+k^{2}-2\right) \tag{6}
\end{align}
There are efficient algorithms to evaluate $K(k), E(k) $ from the value of $k$ using arithmetic-geometric mean. I don't think there are any closed forms which don't involve theta functions / elliptic integrals. 

Update: BTW Ramanujan proved that if $q=e^{-\pi\sqrt{n}} $ where $n$ is a positive rational number then $$P(q) +\frac{6}{\log q} =\left(\frac{2K(k)}{\pi}\right)^{2}\cdot A(n)\tag{7} $$ where $A(n) $ is an algebraic number dependent on rational number $n$. And further if $n$ is a perfect square then $K(k) $ can be expressed in terms of $\Gamma (1/4), \pi$ and certain algebraic numbers. Thus the function in your question has a finite closed form in terms of $\pi, \Gamma(1/4)$ and certain algebraic numbers for a range of special values of $q$. However, this is easier said than done and it is very difficult to evaluate the desired algebraic numbers. Ramanujan  was a master of such calculations and modern approach to such calculations is based on software like Maple, Mathematica and Macsyma.
Further Update: S. Chowla and A. Selberg proved that the elliptic integrals $K, E$ can be expressed in closed form using Gamma values and $\pi$ if $n$ is a positive rational. This result appears to be harder to establish than all the results mentioned above in my answer. Thus for all values of $q$ of the form $q=e^{-\pi\sqrt{n}}$ with $n$ a positive rational your sum has a closed form in terms of Gamma values, $\pi$ and certain algebraic numbers. 
