Albert Rich schrieb:   
   >   
   > On Friday, June 20, 2014 4:56:06 PM UTC-10, 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 1-D 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.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)