Problem with some specific indefinite integral solved with Wolfram Alpha online integrator

Question

Looking for solution to integral below I have found on Wolfram Alpha that result \begin{equation} \int{\sqrt{\frac{1+2x}{(1+x)^3 (1-ax)}}} dx=-i~2\sqrt{2}\sqrt{\frac{1}{1+a}}~E\Bigg(\arcsin{\sqrt{\frac{1+a}{a (1+x)}}}, \frac{a}{2+2 a}\Bigg). \tag{1} \end{equation} However, because my argument in $arcsin$ function is always greater as $1$ the result of integration (right side of equation 1) is a complex number what cannot be true. The variable $x$ and parameter $a$ are positive and less or equal $1$, i.e. the integrand is a positive real function. How to solve that problem?

Answer 1

The solution to the problem described above is to realize that an indefinite integral can be determined only up to some arbitrary constant. As it happens the expression on the right side of equation $(1)$ is the sum of imaginary constant and a real function. Thus, an appropriate choice of integration constant can make the result a real function, as showed below \begin{equation} \int{\sqrt{\frac{1+2x}{(1+x)^3 (1-ax)}}} dx=-i~2\sqrt{2}\sqrt{\frac{1}{1+a}}~E\Bigg(\arcsin{\sqrt{\frac{1+a}{a (1+x)}}}, \frac{a}{2+2 a}\Bigg)+~i~2\sqrt{2}\sqrt{\frac{1}{1+a}}~E\Bigg(\arcsin{\sqrt{\frac{1+a}{a (1+x_0)}}}, \frac{a}{2+2 a}\Bigg) \tag{2} \end{equation} where $x_0$ is an arbitrary chosen value of the integration variable.