
Re: Rubi 4.5 released
Posted:
Jun 22, 2014 8:15 AM


Albert Rich schrieb: > > On Friday, June 20, 2014 4:56:06 PM UTC10, Nasser M. Abbasi wrote: > > > > > Int[ArcSin[Sqrt[x + 1]  Sqrt[x]], x] > > > > Out[68]= (x)*ArcSin[Sqrt[x]  Sqrt[1 + x]] + > > Subst[Int[Sqrt[1  x^2 + x*Sqrt[1 + x^2]], x], x, Sqrt[1 + x]]/Sqrt[2] > > As the above output shows, using integration by parts Rubi is able to > find one term of the antiderivative (ie (x)*ArcSin[Sqrt[x]). However > it is unable to integrate the resulting algebraic integrand > > Sqrt[x + Sqrt[x]*Sqrt[1 + x]]/Sqrt[1 + x] > > to get the antiderivative > > 1/2*(Sqrt[x] + 3*Sqrt[1 + x])*Sqrt[x + Sqrt[x]*Sqrt[1 + x]]  > 3*ArcSin[Sqrt[x]  Sqrt[1 + x]]/(2*Sqrt[2]) > > Mathematica is able to integrate it but gets an antiderivative 3 times > the above size. I do not know how to integrate this algebraic > integrand. Tell me the integration steps you would use to integrate > it, and I will be happy to teach Rubi how to do it. >
Perhaps the best way is to manipulate the complex roots, as we did when Rubi 2 was taught to handle Bondarenko's 'An exact 1D integration challenge  58  (sqrt)' in 2010. For arbitrary complex x we have:
SQRT(x + 1) = SQRT(1  #i*SQRT(x))*SQRT(1 + #i*SQRT(x))
SQRT(SQRT(x)*(SQRT(x + 1)  SQRT(x))) = SQRT(SQRT(x))*SQRT(SQRT(x + 1)  SQRT(x))
SQRT(2)*SQRT(SQRT(x + 1)  SQRT(x)) = SQRT(#i)*SQRT(1 + #i*SQRT(x)) + SQRT(#i)*SQRT(1  #i*SQRT(x))
The first identity is well known, the second holds since PHASE(SQRT(x)) <= pi/2 and PHASE(SQRT(x+1)  SQRT(x)) < pi/2, and the third was derived in 2010. So we can write:
INT(SQRT(SQRT(x)*(SQRT(x + 1)  SQRT(x)))/SQRT(x + 1), x) = INT(SQRT(SQRT(x))*SQRT(SQRT(x + 1)  SQRT(x))/SQRT(x + 1), x) = INT(SQRT(SQRT(x))*(SQRT(#i)*SQRT(1 + #i*SQRT(x)) + SQRT(#i)*SQRT(1  #i*SQRT(x))) / (SQRT(2)*SQRT(1  #i*SQRT(x))*SQRT(1 + #i*SQRT(x))), x) = INT(SQRT(#i)*x^(1/4)/(SQRT(2)*SQRT(1  #i*SQRT(x))), x) + INT(SQRT(#i)*x^(1/4)/(SQRT(2)*SQRT(1 + #i*SQRT(x))), x)
where the final integrals should be no problem for Rubi. The alternative would be to use variable substitution; one way is described by Charlwood, another is to apply the substitution t = SQRT(x + 1)  SQRT(x), SQRT(x) = (1  t^2)/(2*t), dx = (t^4  1)/ (2*t^3)*dt to the integral:
INT(SQRT(SQRT(x))*SQRT(SQRT(x + 1)  SQRT(x))/SQRT(x + 1), x)
Martin.

