This is a cross-post from MSE since there wasn't having enough attention.
I need your help on proving the following identity
Theorem
Let $q=e^{-\pi \frac{K'}{K}}$ where $K$ denotes the complete elliptic integral of the first kind,
Then $$\left(\frac{2 K}{\pi}\right)^4(1-k^2+k^4)=1+240\sum_{n=1}^{\infty}\frac{n^3 q^{2n}}{1-q^{2n}}$$
This is stated on the book Pi and the AGM equation $(3.2.17)$. It is said that the identity follows from $$\theta^8_4(q)=\left(\sum_{n=-{\infty}}^{\infty}(-1)^n q^{n^2}\right)^8=1+16\sum_{n=1}^{\infty}\frac{(-1)^n n^3 q^n}{1-q^n}$$ equation $(3.2.25)$, Which I am unable to prove.
Possible approaches:
Famously, the theta functions have Lambert series like $$\theta^4_3(q)=1+8\sum_{n=1}^{\infty}\frac{nq^n}{1+(-q)^n}$$ $$\theta_2^4(q)=16 \sum_{n= 1}^{\infty} \frac{(2n-1)q^{2n-1}}{1-q^{4n-2}}$$ $$\theta_4^4(q)=1+8\sum_{n= 1}^{\infty} \frac{n(-1)^n q^n}{1+q^n}$$ Which could be proved using the identities $$\theta_3^4(q)=-4q\left(\frac{\theta'_4}{\theta_4}-\frac{\theta'_2}{\theta_2}\right)$$ $$\theta_4^4(q)=-4q\left(\frac{\theta'_3}{\theta_3}-\frac{\theta'_2}{\theta_2}\right)$$ $$\theta_2^4(q)=-4q\left(\frac{\theta'_4}{\theta_4}-\frac{\theta'_3}{\theta_3}\right)$$ See theorem 2.3, which it utilizes the fact that $\frac{dk}{dq}=\frac{2kk'^2K^2}{q\pi^2}$. This approach doesn't seem like that it would work since the summand has $n^3$ on the Lambert series of $\theta_4^8$ instead of $n$. Which suggests that it doesn't have a simple formula with $\log$ (I'd need to differentiate it three times). Paramanand's blog posts suggests that proving the identity algebraically feels quite tedious (Even for $\theta_3^4$). The only relevant formula I could think of using is $$\theta_3^4=\theta_2^4+\theta_4^4$$
