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|    sci.math.symbolic    |    Symbolic algebra discussion    |    10,432 messages    |
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|    Message 9,782 of 10,432    |
|    clicliclic@freenet.de to Albert Rich    |
|    Re: More teething help    |
|    25 Jan 18 08:36:05    |
      Albert Rich schrieb:       >       > On Monday, January 22, 2018 at 3:29:26 AM UTC-10, clicl...@freenet.de wrote:       > >       > > I don't think the example on which your rule 482 is based was provided       > > by me (no such example was in my files until very recently), but perhaps       > > you transformed an example containing (1 - x^2)^(1/3) by making this       > > radical the new integration variable. Anyway, the following result       > > should be much simpler than the elementary antiderivative returned by       > > FriCAS:       > >       > > INT(18*x/((x^3 + 8)*SQRT(x^3 - 1)), x)       > > = SQRT(3)*ATANH(SQRT(3)*(x - 1)/SQRT(x^3 - 1))       > > - ATAN(3/SQRT(x^3 - 1))       > > + ATAN((x - 1)^2/(3*SQRT(x^3 - 1)))       > >       > > where (if you prefer):       > >       > > - ATAN((x - 1)^2/(3*SQRT(x^3 - 1)))       > > = ATAN((x - 1)/SQRT(x^3 - 1)) + ATAN((2*x + 1)/SQRT(x^3 - 1))       > >       >       > Thanks! I generalized the above result to integrate expressions of the       > form x/((a+b*x^3)*sqrt(c+d*x^3)) when 8*b*c+a*d=0 and c is positive       > and when c is negative. The forthcoming Rubi 4.14.5 will incorporate       > these two new rules, as well as two corresponding rules for       > expressions of the form 1/((a+b*x^2)^(1/3)*(c+d*x^2)) when       > b*c-9*a*d=0.              However, the integrand 1/((3 + x^2)*(1 + 3*x^2)^(1/3)) has appeared on       sci.math.symbolic, e.g. on 25th August 2017 in the thread "multiple       non-linear factors & trace zero". Euler's paper is entitled "Integratio       Succincta Formulae Integralis Maxime Memorabilis ..." and appeared in       the Nova Acta Acad. Imp. Scient. Tom. X (1792) pp. 20-26.              >       > However, for your first example integrand x/((4-x^3)*sqrt(1-x^3)), I       > have tried in vain to discover a simple antiderivative involving just       > 3 or 4 arctan(h) and/or log terms. If you or someone derives an       > antiderivative simpler than the one Rubi currently returns, please let       > me know. Rubi will appreciate it.       >              In my files I have:              INT(x/((4 - x^3)*SQRT(1 - x^3)), x)        = 2^(1/3)/18*(ATANH(SQRT(1 - x^3))        - 3*ATANH((1 + 2^(1/3)*x)/SQRT(1 - x^3))        - SQRT(3)*ATAN((2^(1/3) - 2^(2/3)*x - x^2)        /(SQRT(3)*2^(1/3)*SQRT(1 - x^3))))              where              ATAN((2^(1/3) - 2^(2/3)*x - x^2)/(SQRT(3)*2^(1/3)*SQRT(1 - x^3)))        = ATAN(1/SQRT(3) - 2^(2/3)*(1 - SQRT(1 - x^3))/(SQRT(3)*x))        - ATAN(1/SQRT(3) - 2^(2/3)*(1 + SQRT(1 - x^3))/(SQRT(3)*x))        = ATAN(SQRT(3)/SQRT(1 - x^3))        + ATAN(SQRT(3)*(1 - 2^(1/3)*x)/SQRT(1 - x^3))              Unfortunately, the ATANH combination is discontinuous at x = 0; I could       not improve this defect so far.              Martin.              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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