The trigonometric integral
$$I(a,b)= \int_0^{\sinh^{-1} b}\frac{\tan^{-1}(\frac ab \cosh t)}{a\cosh^2t+\frac1a\sinh^2t}\ dt
$$
exhibits the symmetry between the parameters $a$ and $b$,
i.e.
$$I(a,b)=I(b,a)$$
I established the relationship based on heuristics as well as with enough trials and fails. Besides, I have verified it numerically with a large number of parameter pairs.
However, I was unable to prove it. To be clear, I am not interested in a closed-form result for the integral, which may be more challenging. For the question, I am merely interested in showing that the symmetry can be derived analytically, perhaps via some conventional yet clever integration techniques.
Best Answer
$$I(a,b)=\int_0^{\operatorname{arcsinh} b} \frac{1}{\sinh^2 t} \frac{a\coth t}{1+a^2\coth^2 t} \frac{\arctan\left(\frac ab \coth t \sinh t\right)}{\coth t}dt$$
$$=\int_0^{\operatorname{arcsinh} b} \frac{1}{\sinh^2 t} \frac{a\coth t}{1+a^2\coth^2 t} \left(\int_0^{\sinh t} \frac{\frac{a}{b}}{1+\frac{a^2}{b^2}x^2\coth^2 t} dx\right)dt$$
$$=\int_0^{b} \int_{\operatorname{arcsinh} x}^{\operatorname{arcsinh} b} \frac{1}{\sinh^2 t} \frac{a\coth t}{1+a^2\coth^2 t} \frac{\frac{a}{b}}{1+\frac{a^2}{b^2}x^2\coth^2 t} \ dtdx$$
$$\overset{\large \coth t\to t}=\int_0^{b} \int_{\large \frac{\sqrt{1+b^2}}{b}}^{\large \frac{\sqrt{1+x^2}}{x}} \frac{at}{1+a^2t^2} \frac{\frac{a}{b}}{1+\frac{a^2}{b^2}x^2t^2} dtdx$$
$$=\frac{b}{2}\int_0^b\ln\left(\frac{(a^2+a^2x^2+x^2)(b^4+a^2x^2+a^2b^2x^2)}{x^2(a^2+a^2+a^2b^2)(a^2+b^2+a^2x^2)}\right) \frac{dx}{b^2-x^2}$$
$$\overset{x\to bx}=\frac12\int_0^1 \ln\left(\frac{(a^2+b^2x^2+a^2b^2x^2)(b^2+a^2x^2+a^2b^2x^2)}{x^2(a^2+b^2+a^2b^2)(a^2+b^2+a^2b^2x^2)}\right)\frac{dx}{1-x^2}$$
$$\Rightarrow I(a,b)=I(b,a)$$