One might ask a question whether it is possible to find all functions of the form $\frac{\cosh(\alpha x)}{\cosh x+c}$ that are self-reciprocal? There are some functions which are similar to these three, but actually are sums of several functions, e.g. $\frac{\cosh x-(1+\sqrt{3})/2}{\cosh(3x/2)}$ (see the unpublished paper "On the transformation of infinite series" by Bryden Cais). The three functions found by Ramanujan are composed of a single term, and we are interested only in such functions.
In view of the following integral evaluation from Gradshteyn and Ryzhik, (p. 539, formula 3.983, no. 6) $$ \int_0^\infty \frac{\cos ax\cosh \beta x}{\cosh \gamma x+\cos b}dx= \pi\frac{\cos\left[\frac{\beta}{\gamma}(\pi-b)\right]\cosh\left[\frac{a}{\gamma}(\pi+b)\right]- \cos\left[\frac{\beta}{\gamma}(\pi+b)\right]\cosh\left[\frac{a}{\gamma}(\pi-b)\right]} {\gamma\sin b\left(\cosh\frac{2\pi a}{\gamma}-\cos\frac{2\pi \beta}{\gamma}\right)},\\ |\text{Re} \beta|<\text{Re} \gamma,\quad 0< b <\pi,\quad a>0, $$ this question can be reduced to finding all $0< b <\pi, 0< \beta < 1$ such that $$ \frac{\cos\beta(\pi-b)\cdot \cosh a(\pi+b)-\cos\beta(\pi+b)\cdot \cosh a(\pi-b)}{\cosh 2\pi a-\cos 2\pi \beta}=\frac{\cosh\frac{\pi-b}{1-\beta}\beta a\cdot \cos\beta(\pi-b)}{\cosh\frac{\pi-b}{1-\beta}a+\cos b}\qquad (1) $$ for all $a\ge 0$.
(When $\beta=0$ the equation is $2\cdot\frac{\cosh a(\pi+b)-\cosh a(\pi-b)}{\cosh 2\pi a-1}=\frac{1}{\cosh(\pi-b)a+\cos b}$ which after substitution $a=0$ becomes $2b(1+\cos b)=\pi$ and has a solution $b=\frac{\pi}{3}$. This leads to Ramanujan's third function.)
Substitution $a=0$ in equation $(1)$ gives: $$ \sin\beta b\cdot(1+\cos b)=\cos\beta(\pi-b)\cdot\sin\pi\beta. \qquad (2) $$ Then we simplify equation $(1)$: \begin{align} &\cos\beta(\pi-b)\cosh\frac{2b-\pi\beta-b\beta}{1-\beta}a-\cos\beta(\pi+b)\cosh\frac{(\pi-b)(2-\beta)}{1-\beta}a+\\ &\cos(3\pi-b)\beta\cosh\frac{\pi-b}{1-\beta}\beta a+2\cos\beta(\pi-b)\cos b\cosh(\pi+b)a-\\ & 2\cos\beta(\pi+b)\cos b\cosh(\pi-b)a=\cos\beta(\pi-b)\cosh\frac{2\pi-3\pi\beta+b\beta}{1-\beta}a \end{align} Here we have 6 different exponents: $$ \frac{|2b-\pi\beta-b\beta|}{1-\beta},\quad\frac{(\pi-b)(2-\beta)}{1-\beta}, \quad\frac{\pi-b}{1-\beta}\beta,\quad\pi+b,\quad\pi-b,\quad\frac{|2\pi-3\pi\beta+b\beta|}{1-\beta}. $$ When $\beta<1/2$ the exponent $\frac{\pi-b}{1-\beta}\beta$ can be equal to only $\frac{|2b-\pi\beta-b\beta|}{1-\beta}$. When $\beta>1/2$ the exponent $\pi-b$ can equal either $\frac{|2b-\pi\beta-b\beta|}{1-\beta}$ or $\frac{|2\pi-3\pi\beta+b\beta|}{1-\beta}$. Together with $(2)$ we have two equations. All required $\beta$ and $b$ will be contained in the solutions of this system of equations. Solving these equations one can not only recover all of the Ramanujan's examples but find two more examples: $\beta=3/4,b=2\pi/3$ and $\beta=\sqrt{3}/2,b=(\sqrt{3}-1)\pi$ (note that in the last case $\cos\beta(\pi+b)=0$).
So there are five self-reciprocal functions of the form $\frac{\cosh(\alpha x)}{\cosh x+c}$: \begin{equation} \frac{1}{\cosh x}, \frac{\cosh x}{\cosh 2x},\frac{1}{1+2\cosh x},\frac{\cosh\frac{3x}{4}}{2\cosh x-1},\frac{\cosh\frac{\sqrt{3}x}{2}}{\cosh x-\cos(\sqrt{3}\pi)}. \end{equation}
