I am trying to find the solution to:
$$I=\int_0^1\int_b^c \frac{u}{\sqrt{1-u^2t^2}\sqrt{1-(u-a)^2}}dudt,$$ this definitely has the form of an elliptic integral, but for the life of me I can't figure out how to reduce this.
It's similar in form to $$F(\phi,t)=\int^{\sin\phi} \frac{du}{\sqrt{1-u^2t^2}\sqrt{1-u^2}},$$ but the factor of $u$ and the shifted argument make this difficult.
I have tried integration by parts ($[fg]_b^c-\int_b^cgdf=\int_b^cfdg$) where $$f=u$$ and $$dg=\frac{1}{\sqrt{1-u^2t^2}\sqrt{1-(u-a)^2}}.$$ Letting $$g=\mathcal{G}(u,t):=\int^u\frac{dx}{\sqrt{1-x^2t^2}\sqrt{1-(x-a)^2}},$$ and $df=du$ we have $$\int_b^c \frac{u}{\sqrt{1-u^2t^2}\sqrt{1-(u-a)^2}}du=\left[u\mathcal{G}(u,t)\right]^c_b-\int_b^c\mathcal{G}(u,t)du,$$ thus $$I=\int_0^1\left[u\mathcal{G}(u,t)\right]^c_bdt-\int_0^1\int_b^c\mathcal{G}(u,t)dudt.$$ The closest I get to an expression for $\mathcal{G}(u,t)$ is $$\mathcal{G}(u,t)=-\int^u\frac{d\phi}{\sqrt{1-(\cos\phi+a)^2t^2}}.$$ Mathematica gives a very complicated expression but I was wondering if there was anything simpler?
