<s>[INST] Help with solving an elliptic integral 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?
 [/INST] 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.
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