Computing the Integral $\int\tanh[b(x-a)]\cos\beta x\,dx$
If we consider the series expansion of the hypergeometric functions about $x=0$ $$ _2F_1(a,b;c;z) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!} $$ we also have $$ \frac{\left(-\frac{i \beta}{2b}\right)_n}{\left(1-\frac{i \beta}{2 b}\right)_n} = \frac{\beta}{\beta+2inb}\\ \frac{\left(\frac{i \beta}{2b}\right)_n}{\left(1+\frac{i \beta}{2 b}\right)_n} = \frac{\beta}{\beta-2inb} $$ with $(\cdot)_n$ the rising Pochhammer function, otherwise the first argument $(1)_n$ will cancel with the $n!$ in the series definition. So we can write the result as $$ _2F_1\left(1,-\frac{i \beta}{2b},1-\frac{i \beta}{2 b},-e^{2b(x-a)}\right)=\sum_{n=0}^\infty \frac{(-1)^n\beta e^{2nb(x-a)}}{\beta+2inb}=\beta\sum_{n=0}^\infty \frac{(-1)^n(\beta-2inb)e^{2nb(x-a)}}{\beta^2+4b^2n^2}\\ _2F_1\left(1,\frac{i \beta}{2b},1+\frac{i \beta}{2 b},-e^{2b(x-a)}\right)=\sum_{n=0}^\infty \frac{(-1)^n\beta e^{2nb(x-a)}}{\beta-2inb}=\beta\sum_{n=0}^\infty \frac{(-1)^n(\beta+2inb)e^{2nb(x-a)}}{\beta^2+4b^2n^2} $$ Then your integral is written $$ I=\frac{\sin(\beta x)}{\beta} - ie^{-i\beta x}\sum_{n=0}^\infty \frac{(-1)^n(\beta-2inb)e^{2nb(x-a)}}{\beta^2+4b^2n^2}+ie^{i\beta x}\sum_{n=0}^\infty \frac{(-1)^n(\beta+2inb)e^{2nb(x-a)}}{\beta^2+4b^2n^2} $$ we can have that $$ \sum_{n=0}^\infty \frac{\beta(-1)^n}{\beta^2+4 b^2 n^2} = \frac{2b + \beta\pi\text{csch}\left(\frac{\beta \pi}{2b}\right)}{4b\beta} $$ so we can reduce $I$ further to $$ I=\frac{\sin(\beta x)}{\beta}-\left(2b+\beta \pi \text{csch}\left(\frac{\beta \pi}{2 b}\right)\right)\frac{\sin(\beta x)}{2 b \beta} -2be^{-i\beta x}\sum_{n=0}^\infty \frac{(-1)^nne^{2nb(x-a)}}{\beta^2+4b^2n^2}-2be^{i\beta x}\sum_{n=0}^\infty \frac{(-1)^nne^{2nb(x-a)}}{\beta^2+4b^2n^2} $$ which is $$ I=-\beta \pi \text{csch}\left(\frac{\beta \pi}{2 b}\right)\frac{\sin(\beta x)}{2 b \beta} -4b\cos(\beta x)\sum_{n=0}^\infty \frac{(-1)^nne^{2nb(x-a)}}{\beta^2+4b^2n^2} $$ I'm not sure how to evaluate the last sum, but there are no more $i$'s in the expression. I may have made some mistakes somewhere but hopefully this is a tiny bit useful.
Edit: To see the Pochhammer identity use this relation which is just stating $$ \frac{a(a+1)(a+2)\cdots(a+n-1)}{(a+1)(a+2)\cdots(a+n)}=\frac{a}{(a+n)} $$ substitute $a=-\frac{i \beta}{2b}$ and $a=\frac{i \beta}{2b}$ to get the two identities used above.