oriol.-reducing elliptic integrals to binomials

The projective transformation (one to one, unless at point of double root) y=(az+b)/(cz+d) (1) has been used for doing integrable a lot of functions. The integrands having the squared root of polynomials de 3d or 4th degree if have a double root they are not elliptics any more. Elemental integration for a second degree factor under radical. The solution for elliptic integrals, is to get that they were no elliptic any more after a variable change. The transformation of Möbius have two freedom degrees and can give one or two double roots to any polynomial. For calculation simplicity, we apply (1) to the 3d degree reduced form (*) Besides we shall see that all trinomials can be reduced to integrable binomials of Chebishev

After reduction (*), the integrand is 2ª espècie √(y+1)√(y-h)/(y-1)5/2 or better, the form of 2 roots:

I=∫dy(y2-1)1/2(y-h)1/2/(y-1)3 (2) the transformation is (1) y=(z+r)/(z+c)

y2=(z2+r2+2rz)/(z2+c2+2cz) x>1 y2-1=[ 2z(r-c)+r2-

c2]/(z+c)2=(z+p)/(z+c)2 p=(r+c)/2

y-h=[z(1-h)+r-ch]/(z+c) dy=dz/(z+c)2 y-h=(z+q)/(z+c) q=(r-hc)/(1-h)

y-1=(r-c)/(z+c) (y-1)3=k/(z+c)3

I=∫ dz√(z+q)√[z+(r+p)]/(z+c)1/2 we do p=-q 2(r-hc)=(c+r)(h-1)=1+r-

h-hr

3r-2hc=1-h(1+r) r(1+h)=1+(2c-1)h r(3-h)+(1-3h)c=0 r/c

=(3h-1)/(3-h) we do c=3-h r=3h-1 p=2(h+1) and doing z+c=t2

dz=tdt I=∫(z2-p2)1/2dt integrable, monomial and now the change

will be z=pchw I=∫shw·shw·dw=shwchw-w being

w=argch(z/p) The transformation will be (1) y=(z+r)/(rz-1)

1/r=c

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The 1ª espècie I(1ª)=∫dy/√(y2-1)√(y-h) de (1) dy=dz/(x-c)2

y2=(z2+2rz+r2)/(z2-2cz+c2) y2-1=[2z(r-c)+(r2-c2)]/(z-c)2=(z+p)/(z-c)2 p=(r+c)/2

y-h=[z(1-rh)+r-h]/(z-c)=(z+q)/(z-c) q=(r-h)/(1-rh) con r=1 q=1 I=∫dz/(z-1) 1/2(z+1) (1+h)/(1+h)=r=1=c p=1 I(1ª)=∫dz/(z+1)√(z-1)

Con z=1+t2 I=∫tdt/(t2+2)t integrable, t=√2·shw I=∫dw·chw/ch2w I=arctg(ew)

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All from biquadrate trinomial (All elliptic) are integrable

(*)The reducction can be seen at the paper nº3