I have the elliptic integral below, and currently I am stuck on how I can solve it, could someone please help me out:
$$\int^{\frac{L}{15}}_{0}2\pi\sqrt{\frac{a^{2}+b^{2}}{2}}dx$$
where: $a = W-\frac{0.608W^{2}}{L^{2}}x^{2}$ and $b = H-\frac{1.68H^{2}}{L^{2}}x^{2}$ ($H$, $W$ and $L$ are all positive constants).
Substituting into the initial equation:
$$\int^{\frac{L}{15}}_{0}2\pi\sqrt{\frac{(W-\frac{0.608W^{2}}{L^{2}}x^{2})^{2}+(H-\frac{1.68H^{2}}{L^{2}}x^{2})^{2}}{2}}dx$$
How can I solve the equation above?
Using whole numbers $$\frac{a^2+b^2}2=A-B x^2+C x^4$$ $$A=\frac{H^2+W^2}{2} \qquad B=\frac{2 \left(21 H^3+76 W\right)}{25 L^2} \qquad C=\frac{2 \left(441 H^4+5776\right)}{625 L^4}$$ and the antiderivative $$I=\int \sqrt{\frac{a^2+b^2}2}\,dx$$ contains, as you expected , the elliptic integrals of the first and second kinds with very nasty arguments.
We can complete the square and write $$A-B x^2+C x^4=C\left( \left(x^2-\frac{B}{2 C}\right)^2-\frac{B^2-4 A C}{4 C^2}\right)$$ Now, let $$D=\frac{B}{2 C} \qquad \qquad E=\frac{4 A C-B^2}{4 C^2}$$ to make $$I=\sqrt C \int \sqrt{(x^2- D)^2+E}\,dx$$
Have a look to the monster.