Consider an indefinite "elliptic integral”, $$ J_n(r):=\int^r_{r_0}\frac{x^n}{\sqrt{f(x)}}, \quad f(x):=(x-x_1)(x-x_2)(x-x_3). $$ where $r_0>0, n=1,0,−1$, and $x_{1,2,3}$ are roots of the cubic equation $()=0$.
I am interested in a case when $()=0$ has one real root $_1>0$ and two complex roots $_2$ and $_3$ (since it's a cubic equation, they are also complex conjugate, $_3=\bar{x_2}$).
I want to know if $_{−1}$ can be written without imaginary number $$, because this integral appears in a problem of physics, $_$ should be real. But, so far, I failed to make it. I write down what I faced:
I have tried an integration by substitution. In the present case, let $_2=+$ and thus $_3=\bar{x_2}=−$. Then we have $()=(−_1)((−)^2+^2)$.
Now, transform the variable from $$ to $$ with $$ =_1+ \tan^2, :=\sqrt{(_1−)^2+^2}. $$ After some manipulation, $_$ reduces to $$ \frac{\sqrt{p}}{2}_()=\int\frac{(x_1+\tan^2y)^n {\rm d}y}{\sqrt{1−\sin^2\cos^2}}, $$ where $:=2(−(_1−))/$ and thus $0<<2$ by definition.
For $=0$ and $n=+1$, there is no problem: For $=0$, a software "Mathematica" returns $$ \frac{\sqrt{p}}{2}_0()=\frac{1}{2}(2,\frac{A}{4}), $$ where $(\phi,)$ is Elliptic integral of the first kind. For $=1$, Mathematica returns a combination of elliptic functions and elementary functions (I do not express them here, though).
Note that there is no imaginary number $i$ so far.
For $=−1$, however, integration returns a messy form with $i$:
where $\Pi(n,\phi,m)$ is incomplete Elliptic integral of the third kind.
I guess there should be a simpler form of $_{−1}$ without $i$. Maybe the above transformation is not good for $_{−1}$? Is there some identities that eliminate $i$? I am not comfortable that in the messy form, because $\sqrt{A-4}$ appears, although $0<<2$.
I am beginner of elliptic integral, any comments are welcome. Thank you.