Maple recognizes the complete elliptic integrals

     ${\int }_0^{\frac{\mathrm{\π}}{2}}\frac{1}{\sqrt{1-k^2\sin {\left(t\right)}^2}} \ⅆt$  (First Kind)

     ${\int }_0^{\frac{\mathrm{\π}}{2}}\sqrt{1-k^2\sin {\left(t\right)}^2} \ⅆt$  (Second Kind)

     ${\int }_0^{\frac{\mathrm{\π}}{2}}\frac{1}{\left(1-a \sin {\left(t\right)}^2\right) \sqrt{1-k^2\sin {\left(t\right)}^2}} \ⅆt$   (Third Kind)

where $0<k<1$.  We see this as follows:

First, we tell Maple that the usual parameter k lies between 0 and 1.

Now we evaluate the above integrals:

The tilde (~) following each k reminds us of the assumption of restricted range.

Maple can integrate the same type of integrals by using a hyperbolic trig argument. For instance, the integral

     ${\int }_{\mathrm{arcsinh}\left(\frac{1}{2}\right)}^{\mathrm{arcsinh}\left(1\right)}\frac{1}{\sqrt{1\&plus;k^2\sinh {\left(t\right)}^2}} \&DifferentialD;t$ evaluates to:

As another example, to compute the value of the integral ${\int }_{\mathrm{arcsinh}\left(\frac{1}{2}\right)}^{\mathrm{arcsinh}\left(1\right)}\sqrt{2\&plus;k^2\sinh {\left(t\right)}^2} \&DifferentialD;t$,  we evaluate:

Because elliptic integration works for variable arguments that take advantage of the Maple assume facility, you can investigate formulae that cross branches of the square root function. The assumptions are the following:

Consider the following integral, ${\int }_0^z\frac{1}{\sqrt{\left(1-x^2\right)\left(1-k^2 x^2\right)}} \&DifferentialD;x$. Evaluate and assign to answer1:

The next integral, ${\int }_0^z\sqrt{\left(1-x^2\right)\left(1-k^2 x^2\right)} \&DifferentialD;x$,  assigned to answer2, evaluates to

We can do a simple check for the integral directly above by trying some values. For example, try $k\&equals;\frac{1}{2}$ and $z\&equals;2$.

Compute check - check2. This should evaluate to zero.

Integrals can be reduced to normal form in terms of the three Legendre elliptic functions: EllipticF, EllipticE, and EllipticPi. We begin by declaring some assumptions.

Firstly, the EllipticF function is given by ${\int }_0^z\frac{1}{\sqrt{\left(1-x^2\right) \left(1-k^2 x^2\right)}} \&DifferentialD;x$.

Secondly, the EllipticE function has the form ${\int }_0^z\frac{\sqrt{1-k^2{x}^2}}{\sqrt{1-x^2}} \&DifferentialD;x$.

Thirdly, the EllipticPi function is of the form ${\int }_0^z\frac{1}{\left(1-a x^2\right) \sqrt{\left(1-k^2 x^2\right) \left(1-x^2\right)}} \&DifferentialD;x$.

For all of the above functions, the variable k must lie between 0 and 1. The Maple integrator facility reduces

to a normal form expression. This can then be evaluated numerically to 30 (or more) digits.

Compare this to the Maple numerical integrator, and we see that the answers are the same (at least up to round errors).

In Maple, the integrator also recognizes the trigonometric form of these integrals:

${\int }_a^{b}R\left(\sin \left(\mathrm{\&theta;}\right)\&comma;\cos \left(\mathrm{\&theta;}\right)\right) \sqrt{y\left(\sin \left(\mathrm{\&theta;}\right)\&comma; \cos \left(\mathrm{\&theta;}\right)\right)} \&DifferentialD;\mathrm{\&theta;}$, where $R$ is a rational function of sin and cos, and $y$ is a quadratic polynomial in sin and cos.

assume, int