Let $\mathcal{S}$ denote the sum of the following (convergent) infinite series:
$$\mathcal{S}:=4\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}(2n+2)^{-2},\tag{1}$$
where here $n!!$ denotes the so-called double factorial of a number $n$.
(Note: My definition of $\mathcal{S}$ has an additional scalar factor of $4$ so as to simplify its expression in terms of the generalized hypergeometric function $_4F_3$.)
We'll make use of the following well-known integration formula for a subclass of Wallis' integrals (for proof see [wiki][1]):
$$\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\cos^{2n+1}{\left(\varphi\right)}=\frac{(2n)!!}{(2n+1)!!};~~~\small{n\in\mathbb{Z}_{\ge0}}.$$
Recall the definition of the [polylogarithm][2] as an infinite series. Given $s\in\mathbb{C}\land z\in\mathbb{C}\land|z|<1$, the polylogarithm $\operatorname{Li}_{s}{\left(z\right)}$ of order $s$ and argument $z$ is given by the (absolutely convergent) power series
$$\operatorname{Li}_{s}{\left(z\right)}=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{s}}.$$
For positive integer order, the polylogarithm can be defined iteratively by
$$\operatorname{Li}_{1}{\left(z\right)}:=-\ln{\left(1-z\right)};~~~\small{z\in\left(-\infty,1\right)},$$
$$\operatorname{Li}_{n+1}{\left(z\right)}:=\int_{0}^{z}\mathrm{d}t\,\frac{\operatorname{Li}_{n}{\left(t\right)}}{t};~~~\small{n\in\mathbb{N}\land z\in\left(-\infty,1\right]}.$$
Another useful integral representation for $\operatorname{Li}_{n+1}{\left(z\right)}$, which can be obtained from the previous one by repeated integration by parts, is
$$\operatorname{Li}_{n+1}{\left(z\right)}=\frac{(-1)^{n}}{n!}\int_{0}^{1}\mathrm{d}t\,\frac{z\ln^{n}{\left(t\right)}}{1-zt};~~~\small{n\in\mathbb{N}\land z\in\left(-\infty,1\right]}.$$
An important auxiliary function pertaining to the polylogarithm is the so-called Nielsen generalized polylogarithm, defined for positive integer parameters via the integral representation
$$S_{n,p}{\left(z\right)}:=\frac{(-1)^{n+p-1}}{(n-1)!\,p!}\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{n-1}{\left(t\right)}\ln^{p}{\left(1-zt\right)}}{t};~~~\small{\left(n,p\right)\in\mathbb{N}^{2}\land z\in\left(-\infty,1\right]}.$$
The following integration formula will be useful to have on hand later:
$$\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t\right)}\ln{\left(1-zt\right)}}{t}=\operatorname{Li}_{3}{\left(z\right)}+S_{1,2}{\left(z\right)};~~~\small{z\in\left(-\infty,1\right]}.$$
Proof:
$$\begin{align}
\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t\right)}\ln{\left(1-zt\right)}}{t}
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}+\ln^{2}{\left(1-zt\right)}-\left[\ln{\left(1-t\right)}-\ln{\left(1-zt\right)}\right]^{2}}{2t}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}}{2t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{2t}-\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(\frac{1-t}{1-zt}\right)}}{2t}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}}{2t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{2t}\\
&~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{\left(1-z\right)}{\left(1-zu\right)^{2}}\cdot\frac{\ln^{2}{\left(u\right)}}{2\left(\frac{1-u}{1-zu}\right)};~~~\small{\left[t=\frac{1-u}{1-zu}\right]}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}}{2t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{2t}\\
&~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{\left(1-z\right)\ln^{2}{\left(u\right)}}{2\left(1-u\right)\left(1-zu\right)}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-t\right)}}{2t}+\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{2t}\\
&~~~~~-\int_{0}^{1}\mathrm{d}u\,\frac{\ln^{2}{\left(u\right)}}{2\left(1-u\right)}+\int_{0}^{1}\mathrm{d}u\,\frac{z\ln^{2}{\left(u\right)}}{2\left(1-zu\right)}\\
&=\frac12\int_{0}^{1}\mathrm{d}t\,\frac{z\ln^{2}{\left(t\right)}}{1-zt}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1-zt\right)}}{t}\\
&=\operatorname{Li}_{3}{\left(z\right)}+S_{1,2}{\left(z\right)}.\\
\end{align}$$
Using the technique of interchanging the order of summation and integration, we obtain an expression for the power series $\mathcal{S}$ as a definite integral.
$$\begin{align}
\mathcal{S}
&=4\sum_{n=0}^{\infty}\frac{(2n)!!}{(2n+1)!!}(2n+2)^{-2}\\
&=\sum_{n=0}^{\infty}\frac{1}{(n+1)^{2}}\cdot\frac{(2n)!!}{(2n+1)!!}\\
&=\sum_{n=0}^{\infty}\frac{1}{(n+1)^{2}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\cos^{2n+1}{\left(\varphi\right)}\\
&=\sum_{n=0}^{\infty}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\cos^{2n+1}{\left(\varphi\right)}}{(n+1)^{2}}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sum_{n=0}^{\infty}\frac{\cos^{2n+1}{\left(\varphi\right)}}{(n+1)^{2}}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sum_{n=1}^{\infty}\frac{\cos^{2n-1}{\left(\varphi\right)}}{n^{2}}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{1}{\cos{\left(\varphi\right)}}\sum_{n=1}^{\infty}\frac{\left[\cos^{2}{\left(\varphi\right)}\right]^{n}}{n^{2}}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sec{\left(\varphi\right)}\operatorname{Li}_{2}{\left(\cos^{2}{\left(\varphi\right)}\right)}\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\cos{\left(\varphi\right)}\operatorname{Li}_{2}{\left(1-\sin^{2}{\left(\varphi\right)}\right)}}{1-\sin^{2}{\left(\varphi\right)}}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{2}{\left(1-x^{2}\right)}}{1-x^{2}};~~~\small{\left[\varphi=\arcsin{\left(x\right)}\right]}.\\
\end{align}$$
Then,
$$\begin{align}
\mathcal{S}
&=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{2}{\left(1-x^{2}\right)}}{1-x^{2}}\\
&=-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1+x}{1-x}\right)}}{2}\cdot\frac{2x\ln{\left(x^{2}\right)}}{1-x^{2}};~~~\small{I.B.P.s}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x\right)}\ln{\left(\frac{1-x}{1+x}\right)}}{1-x^{2}}\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x\right)}\ln{\left(1-x^{2}\right)}}{1-x^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x\right)}\ln{\left((1+x)^2\right)}}{1-x^{2}}\\
&=\frac12\int_{0}^{1}\mathrm{d}x\,\frac{2x\ln{\left(x^{2}\right)}\ln{\left(1-x^{2}\right)}}{1-x^{2}}-\int_{0}^{1}\mathrm{d}x\,\frac{4x\ln{\left(x\right)}\ln{\left(1+x\right)}}{1-x^{2}}\\
&=\frac12\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(y\right)}\ln{\left(1-y\right)}}{1-y};~~~\small{\left[x^{2}=y\right]}\\
&~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{2\ln{\left(x\right)}\ln{\left(1+x\right)}}{1+x}-\int_{0}^{1}\mathrm{d}x\,\frac{2\ln{\left(x\right)}\ln{\left(1+x\right)}}{1-x}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}\ln{\left(1-t\right)}}{2t};~~~\small{\left[y=1-t\right]}\\
&~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{\ln^{2}{\left(1+x\right)}}{x};~~~\small{I.B.P.s}\\
&~~~~~-\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1-t\right)}\ln{\left(2-t\right)}}{t};~~~\small{\left[x=1-t\right]}\\
&=\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(t\right)}\ln{\left(1-t\right)}}{2t}-\int_{0}^{1}\mathrm{d}t\,\frac{\ln^{2}{\left(1+t\right)}}{t}\\
&~~~~~-\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1-t\right)}\ln{\left(2\right)}}{t}-\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(1-t\right)}\ln{\left(1-\frac12t\right)}}{t}\\
&=\frac12\,S_{2,1}{\left(1\right)}-2S_{1,2}{\left(-1\right)}\\
&~~~~~+2\ln{\left(2\right)}\operatorname{Li}_{2}{\left(1\right)}-2\left[\operatorname{Li}_{3}{\left(\frac12\right)}+S_{1,2}{\left(\frac12\right)}\right]\\
&=3\ln{\left(2\right)}\,\zeta{\left(2\right)}-\frac74\,\zeta{\left(3\right)}.\blacksquare\\
\end{align}$$
Best Answer
Since the integrand is even, we have $$\mathcal{I} \stackrel{def}{=}\int_{-\infty}^\infty \frac{dx}{\cosh^2(x-\frac{a}{x})} dx = 2\int_0^\infty \frac{dx}{\cosh^2(x-\frac{a}{x})}dx$$ In one copy of the integral on RHS, change variables to $y = \frac{a}{x}$, one get $$\int_0^\infty \frac{dx}{\cosh^2(x-\frac{a}{x})} = \int_0^\infty \frac{1}{\cosh^2(\frac{a}{y} - y)} \frac{a}{y^2}dy$$ Renaming $y$ back to $x$ and add back to another copy of integral on RHS, we get $$\mathcal{I} = \int_0^\infty \frac{1}{\cosh^2(x - \frac{a}{x})}\left(1 + \frac{a}{x^2}\right)dx = \int_0^\infty \frac{1}{\cosh^2(x - \frac{a}{x})}\frac{d}{dx}\left(x - \frac{a}{x}\right) dx$$ Change variable to $z = x - \frac{a}{x}$, this becomes $$\mathcal{I} = \int_{-\infty}^\infty \frac{dz}{\cosh^2(z)}$$ i.e. transform your integral $(3)$ to integral $(2)$, the one you already know.
Let's look at the more general identity $(1)$.
As you increases $x$ from $-\infty$ to $0$, $x - \frac{a}{x}$ increases from $-\infty$ to $\infty$ once. If one further increases $x$ from $0$ to $\infty$, $x - \frac{a}{x}$ increases from $-\infty$ to $\infty$ the second time. For any $t \in \mathbb{R}$, let $x_1(t) < 0$, $x_2(t) > 0$ be the two roots of $$x - \frac{a}{x} = t \equiv x^2 - tx - a = 0$$ This is a quadratic equation in $x$. By Vieta's formula, we have
$$x_1(t) + x_2(t) = t \implies x_1'(t) + x_2'(t) = 1$$
If you change variable to $t = x - \frac{a}{x}$ for both $(-\infty,0)$ and $(0,\infty)$, we obtain:
$$\begin{align} \int_{-\infty}^\infty f(x - \frac{a}{x}) dx &= \left(\int_{-\infty}^0 + \int_0^\infty\right) f(x - \frac{a}{x})dx\\ &= \int_{-\infty}^{\infty} f(t) x'_1(t) dt + \int_{-\infty}^{\infty} f(t) x'_2(t) dt\\ &= \int_{-\infty}^\infty f(t) (x'_1(t) + x'_2(t))dt\\ &= \int_{-\infty}^\infty f(t)dt \end{align}$$
This is the identity $(1)$ we seek.