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|    sci.math.symbolic    |    Symbolic algebra discussion    |    10,432 messages    |
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|    Message 9,146 of 10,432    |
|    Albert Rich to Nasser M. Abbasi    |
|    Re: Rubi 4.9.2 do not integrate Timofeev    |
|    23 Aug 16 19:24:08    |
      From: Albert_Rich@msn.com              On Tuesday, August 16, 2016 at 8:15:34 PM UTC-10, Nasser M. Abbasi wrote:       > I am updating CAS integration for Mathematica 11 and Rubi 4.9.2       > (before it was Mathematica 10.4 and Rubi 4.9). (it has been compiling       > for last 7 days, should be done in few more days).       >        > I noticed that using Rubi 4.9.2 (downloaded August 10, 2016),       > Rubi do not integrate Timofeev 319 any more. But it       > did when using 4.9 downloaded March 14 2016 from Albert Rich web site.              >        > 4.9 (March 14 2016 downloaded)       > ================================       > ShowSteps = False;       > Int[1/((3*x + 3*x^2 + x^3)*(3 + 3*x + 3*x^2 + x^3)^(1/3)), x]       >        >        > -(((1 + x)*       > Hypergeometric2F1[1/3, 1,       > 4/3, (3*(1 + x)^3)/(2 + (1 + x)^3)])/(2 + (1 + x)^3)^(1/3))       >        > 4.9.2 (August 10, 2016 downloaded)       > ===================================       > The above integral do not evaluate any more. i.e Rubi fails       > to integrate it, it gives (notice it has Int remaining inside)       >        > ((1 + 2^(1/3) + x)^(1/3)*(2 - 2^(1/3)*(1 + I*Sqrt[3]) + 2*x)^(1/3)*       > (2 + I*2^(1/3)*(I + Sqrt[3]) + 2*x)^(1/3)*       > Int[1/(x*(3 + 3*2^(1/3) + 3*x)^(1/3)*(3 -       > (3*(1 - I*Sqrt[3]))/2^(2/3) + 3*x)^(1/3)*(3 - (3*(1 + I*Sqrt[3]))/2^(2/3) +       3*x)^(1/3)), x])/       > (2^(2/3)*(3 + 3*x + 3*x^2 + x^3)^(1/3)) -       > ((1 - I*Sqrt[3])*(1 + 2^(1/3) + x)^(1/3)*(2 - 2^(1/3)*       > (1 + I*Sqrt[3]) + 2*x)^(1/3)*       > (2 + I*2^(1/3)*(I + Sqrt[3]) + 2*x)^(1/3)*Int[1/((3 - I*Sqrt[3] + 2*x)       > *(3 + 3*2^(1/3) + 3*x)^(1/3)*(3 - (3*(1 - I*Sqrt[3]))/2^(2/3) + 3*x)^(1/3)*       > (3 - (3*(1 + I*Sqrt[3]))/2^(2/3) + 3*x)^(1/3)), x])/(2^(2/3)*       > (3 + 3*x + 3*x^2 + x^3)^(1/3)) -       > ((1 + I*Sqrt[3])*(1 + 2^(1/3) + x)^(1/3)*(2 - 2^(1/3)*(1 + I*Sqrt[3]) +       2*x)^(1/3)       > *(2 + I*2^(1/3)*(I + Sqrt[3]) + 2*x)^(1/3)*       > Int[1/((3 + I*Sqrt[3] + 2*x)*(3 + 3*2^(1/3) + 3*x)^(1/3)*       > (3 - (3*(1 - I*Sqrt[3]))/2^(2/3) + 3*x)^(1/3)*       > (3 - (3*(1 + I*Sqrt[3]))/2^(2/3) + 3*x)^(1/3)),       > x])/(2^(2/3)*(3 + 3*x + 3*x^2 + x^3)^(1/3))       >        > I see the rules have changed. Is there a problem here?       >        > So now Rubi fails 3 Timofeev integrals, while before it only failed       > on two integrals.       >        > ps. on Version numbers of Rubi, it will help if there is a command       > to check Rubi version number as the Rubi notebook title says       > 4.9 for both, but I know they are not the same, one is 4.9 and one is       > I assume 4.9.2.       >        > I keep Rubi package in different folders, with time-stamp when I       > downloaded it so I know. But it will be better to put       > the current version number in the notebook title at least, so       > it is clear or provide a command to use to get the exact version.       >        > thanks,       > --Nasser              Thank you for pointing out this deficiency in Rubi. Now available for       downloading on Rubi's website is version 4.9.8 (you missed a few sub-versions       :). It returns a near optimal, 3 term, elementary antiderivative for Timofeev       problem #319 (i.e. problem        #112 in Chapter 4) as well as for problem #314. As you note above, Rubi use       to return the hypergeometric F1 function for these problems, and so in       hindsight were nonoptimal.              Although no where near Martin's ideal of always returning pseudo-elliptic       integrals in elementary form and elliptic integrals in Legendre's canonical       form, Rubi 4.9.8 is able to integrate more classes of such integrals. For       example, an elementary or        elliptic result is returned for integrals of the form               x^m*(a+b*x^2)^(n/3)              when m,n are integers, and for               (a+b*x^2)^(1/3+n)*(c+d*x^2)^m              when b*c+3*a*d=0 and m,n are integers. Formerly hypergeometric or Appell F1       functions were returned.              I took your advice and will include the sub-version number in the file       Rubi4.9.nb.              Thanks and aloha,       Albert              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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