The binomials have the form I=∫dx·xm(1+xn)p only were integrable if p (m+1)/n or ( m+1)/n +p were integers. Abel cut the conicals with bunches of straight lines, for getting them in parametric form. Generalizing the abelian method with bunches of curves, the difficult integrands can also be put in parametric form. We shall have several freedom degrees. I=∫dx·xm(1+xn)p a bunch is y=rxm r=(1+xn)p
x=(r1/p-1)1/n dx=dr·r(1-p)/p(r1/p-1)(1-n)/n
All binomials are this one Are M=1/p-1
(the parameter r is the new variable)
N=1/p
∫dr·r1/p (r1/p-1)(m+1-n)/n and now the news m, n and p
P=(m+1)/n-1
(M+1)/N=1 all integrable with the
Chebichev criterion! Near all elliptic integrands, if binomials (they can be)(*) are integrable, **************************** (*) Transforming to binomials all biquadrate polynomials Applying the projective transformation (1) on reduced biquadrate (2). I do not care constants affecting the whole integrand
x>1 e is a data in every case.
(1) x=(z+r)/(z+e) r y c to be determinate (2) I= ∫dx(x2-1)1/2/(x-e)1/2
x2=(z2+r2+2rz)/(z2+e2+2ez) x-e=[z(1-e2)+r2-e2]/(z+e)
x2-1=[r2-e2+2z(r-e)]/(z+e)2 =(z+q)/(z+e)2 dx=dz[(z+e)-(z+r)]/(z+e)2dx=(e-r)dz/(z+e)2
2ª especies I=∫dρ·√(ρ+c)√ρ/(ρ-1)
ρ=y-c/2 I=∫dy√(y2-c/4) /[y-(c/2+1)]
y=(√c/2)chw I=∫dwsh2w/(ch-(c/2+1)/√c/2)] I=∫dwsh2w/(ch-A)=∫dw(e2w+e-2w-2)/[ew+e-w+2(1-A)]
I=∫ewdw·e2w(e2w+e-2w-2)/e3w(ew+e-w+2(1-A) I=∫ds(s4+1-2s2)/(s4+s2+Ns3) N=2(1-A) I=∫dss2/R+∫ds/s2R-2∫ds·s/R R=(s2+1+Ns) elemental With the other reduced
I=∫ dz(z+q)1/2/(z+p)1/2(z+e)5/2
p=(r2-e2)/(1-e2) q=(r+e)/2 We do p=q 2r2-2e2=(1-e2)(r+e)=r+e-re2-e3 r/e=u 2e2(u-1)(u+1)=e(u+1)-e3(u+1) e2+2e (u-1)-1 u=1+(1-e2)/2e=(1+2e-e2)/2=r with this r I=∫dz/(z+e)5/2 z=t2-e good binomial m=0 n=1 p=-5/2
(m+1)/n=1
I=∫tdt/t5=t-3=(z+e)-3/2
******************+ 1ª especie
I=∫dρ(ρ-1)/√ρ√(ρ+c)
ρ=y-c/2
I=∫dy(y-(1+c/2)/√(y2-c2/4) (2)* c=2f
I=∫dy(y-(1+f)/√(y2-f2) y=fchw I=∫dwshw(chw-(1+f)/f)/shw=shw-(2+c)w/c * we separate
I=∫dy(y)/(y2-c2/4)1/2-(1+c/2)∫dy/√(y2-c2/4) m=1 p=-1/2 n=2 (m+1)/n=1
All good binomial [(It was immediate: d(y2)]
With this, all biquadrate (2ª especies) become The binomial I=∫ dz·z-1/2(z+c)-2 Which is converted, with the change dz=dt·ta-1 in m=-a-1
n=a
z=ta-c
I=∫ dt t-(a+1)(ta-c)-1/2 for any a (m+1)/n=-1 p=-1/2
all biquadrates ( 2ªespecies)
integrate. I=∫ dt (t3a+2-ct2a+2)-1/2 con a=-1 I=∫ dtt1/2 (1-ct)-1/2 t=s2 I=∫ s2ds /(1/c-s2)1/2
can be
s=c-1/2sinw I=∫sin2wdw =½(w-sinwcosw)
siendo w=arsin(t/c)1/2