How can this equation be solved analytically? Is it possible, why? Is there a better alternative than the Newton approximation?
$k\tanh^2(ax+b) + l\tanh^2(cx+d) + m = 0 $ with $a,b,c,d,k,l,m \text{ constants } \in R $
Does an expression of $\tanh(ax+b)$ depending of $\tanh(x)$ exist?
Edit :
I found out that $\tanh(ax+b) = -\frac{(-(1 - \tanh(x))^a + \tanh(b) (1 - \tanh(x))^a + (1 + \tanh(x))^a + \tanh(b) (1 + \tanh(x))^a}{-(1 - \tanh(x))^a + \tanh(b) (1 - \tanh(x))^a - (1 + \tanh(x))^a - \tanh(b) (1 + \tanh(x))^a) }$
So if we pose $\tanh(x) = X$, the problem becomes :
$\frac{k(((1 - X)^a - e^{2 b} (1 + X)^a)^2} {((1 - X)^a + e^{2 b} (1 + X)^a)^2} + \frac{l( ((1 - X)^c - e^{2 d} (1 + X)^c)^2} {((1 - X)^c + e^{2 d} (1 + X)^c)^2} + m = 0$