I cannot find the following integral in an integral table.

Question

In the appendix A of this paper there is an integral that the author says can be solved using any good integral table. However I cannot seem to find it on any integral table (ex: gradshteyn and ryzhik) - although it could be that I am not using the table properly as it is the first time I am doing it. I have solved it using Wolfram but it does not give the same result as in the paper.

The integral is:

$$\int{\frac{-1}{1-F \cdot\text{sech}^2(x)}dx}$$

Note that the author gives the result for F < 1 which is what I want. Maybe that is why the wolfram solution is different, I am not sure.

Answer 1

Factor $-\dfrac1F$ outside the integral sign. Then, since $F<1$, it follows that the $\dfrac1F$ in the integrand's

denominator can be written as $1+a^2$. At the same time, similarly to $\sec^2x=1+\tan^2x$ which

we know to hold for its trigonometric counterpart, we also have $\text{sech}^2x=1-\tanh^2x$, making

our integral $\displaystyle-\frac1F\int\frac{dx}{a^2+\tanh^2x}$ . At this point we recognize $\arctan't=\dfrac1{1+t^2}$ . Substituting

$u=\dfrac{\tanh x}a$, we finally arrive at the desired result, $-I=x+\dfrac1a\arctan\dfrac{\tanh x}a$.

Answer 2

If we put $x=\log t$ we have: $$\int \frac{dx}{1-\frac{F}{\cosh^2 x}} = \int\frac{(t^2+1)^2 dt}{t^5+(2-4F)t^3+t},$$ and by putting $t=\sqrt{u}$ we get: $$\int \frac{dx}{1-\frac{F}{\cosh^2 x}} = \frac{1}{2}\int\frac{(u+1)^2 du}{u^3+(2-4F)u^2+u}.$$ Decomposing the last integrand function in simple fractions leads to the answer.