Solving reduced elliptic integrands

Reducing an integrand to third degree, generalizing the lineal Abelian unicursal method.

We do’nt care constants affecting the whole integrand: we work for possibility We sall cut the curve integrand with a bunch of curves. The parameter of the bunch ρ will be the new variable. Bunch is y=ρx2(b+x2) P2=(b+x2)(a+x2)=y1=ρx2(b+x2) x2+a=ρx2 x=√a /√(ρ-1) dx=dρ /√(ρ-1)3

x2=a/(ρ-1)

2c+1=h=2r-1 r=a/b ratio of semi-axes

With this change from x to ρ, the polynomial P under radical is of 2nd degree, (and with dx, it will be 3 factors under root) x2+a=(a+aρ-a)/(ρ-1) = ρ/(ρ-1) ρ>1

x2+b=(a+bρ-b)/(ρ-1)

1ª espècie: polynomial at denominator ρ=t2-c dρ=tdt

P2=ρ(ρ+c)(ρ-1)-2

I=∫dρ(ρ-1)/√ρ√(ρ+c)

c=a/b-1

now the variable change is

I=∫tdt[t2-(c+1)]/t√(t2-c) I=∫dt[t2-(c+1)]/√(t2-c) we do t=√c·chw I=∫dwshw[ch2w(c+1)/c]/shw I=∫dwch2w-(1+1/c)·chw from wikipedia tables

∫dwch2w=(1/2)shwchw+(1/2)w

being w=arg ch[√(1+ρ/c)] ρ=1+a/x2

***************************** 2ª espècie

polynomial at numerator

doρ=t-c/2

P2=ρ(ρ+c)/(ρ-1)2

I=∫dρ·√(ρ+c)√ρ/(ρ-1)

we

N=(1+c/2)

I=∫dt√(t2-c2/4)/(t-N)

t=(c/2)·chw dt=dwchw

I=∫d(chw)√(ch2w-1)/(chw-2N/c) s=y+2N/c=y+M root (y+M)2-1 we do y+M=chw

M=(2+c)/c

we call chw=s

I=∫sh2wdw/(chw-2N/c) I=∫ds√(s2-1)/(s-2N/c) we do

I=∫dy√(y2+2yM+M2-1)/y

y=-M+/-√[M2-M2+1]

I=∫sh2wdw/(chw-M)=∫ewdw(e4w+1-2e2w)/e3w(ew+e-w+2M)=

under

I=∫d(ew)(e4w+1-2e2w)/(e4w+e2w+2Me3w)=∫ds(s4+12s2)/s2(s2+1+2Ms)=∫s4ds/R+∫s2ds/R+2∫ds·s3/R R=s2(s2+2Ms+1) s=Ag-M

s=-M+/-√(M2-1) R=s2[(s+M)2-(M2-1)] we do s+M=√(M2-1)chq

we call g=chq A=√(M2-1) ds=shq·dq 1)=H(g2-1) H=s2 R=(M2g+A2g2-2M√Ag)(g2-1) degree 4 second

R=s2(ch2q-1)= (M2+A2g2-2M√A·g)(g2s4 is 4th degree in g, and s3 of third an s2 of

Polynomials: elemental. As we wanted to show. No squared root g=chq I(1)=∫HAsh2qdq/(g2-1)=AM2∫dq+A2∫dq-A2∫dq·sh2q-2M√A∫chq

1)= ∫dq I(3)= ∫dq·shq·(Ag-M)/(g2-1)=A∫dq·chq/shq-M∫dq/shq

I(2)= ∫shqdqshq/(g2-