Forums before death by AOL, social media and spammers... "We can't have nice things"
|    sci.math.symbolic    |    Symbolic algebra discussion    |    10,432 messages    |
[   << oldest   |   < older   |   list   |   newer >   |   newest >>   ]
|    Message 9,181 of 10,432    |
|    Nasser M. Abbasi to Nasser M. Abbasi    |
|    Re: FriCAS web interface failure    |
|    30 Oct 16 02:36:10    |
      From: nma@12000.org              On 10/4/2016 11:17 AM, Nasser M. Abbasi wrote:              >       > btw, I tried the above on the new Mathematica 11.0.1       > and on Maple 2016.1 and Rubi 4.9.2, Mathematica gave       > the simplist answer. Maple did not do it. Rubi answer       > contained complex numbers. Here is the result       >              .....              >       > Rubi: 4.9.2       > ============       > In[62]:= ShowSteps = False;       > In[63]:= Int[(a + b*x)/((3 + x^2)*(1 - x^2)^(1/3)), x]       >       > Out[63]= ((I*Sqrt[3]*a -       > 3*b)*((1 - x)/(I*Sqrt[3] - x))^(1/       > 3)*(-((1 + x)/(I*Sqrt[3] - x)))^(1/3)*       > AppellF1[2/3, 1/3, 1/3, 5/3, -((1 - I*Sqrt[3])/(I*Sqrt[3] - x)),       > (1 + I*Sqrt[3])/(I*Sqrt[3] - x)])/(4*(1 - x^2)^(1/3)) - ((I*       > Sqrt[3]*a +       > 3*b)*(-((1 - x)/(I*Sqrt[3] + x)))^(1/       > 3)*((1 + x)/(I*Sqrt[3] + x))^(1/3)*       > AppellF1[2/3, 1/3, 1/3,       > 5/3, (1 + I*Sqrt[3])/(I*Sqrt[3] +       > x), -((1 - I*Sqrt[3])/(I*Sqrt[3] + x))])/(4*(1 - x^2)^(1/3))       >       > In[64]:= Int[(a + b*x)/((3 - x^2)*(1 + x^2)^(1/3)), x]       >       > Out[64]= ((Sqrt[3]*a +       > 3*b)*((I - x)/(Sqrt[3] - x))^(1/       > 3)*(-((I + x)/(Sqrt[3] - x)))^(1/3)*       > AppellF1[2/3, 1/3, 1/3, 5/3, (I + Sqrt[3])/(Sqrt[3] - x),       > -((I - Sqrt[3])/(Sqrt[3] - x))])/(4*(1 + x^2)^(1/       > 3)) - ((Sqrt[3]*a -       > 3*b)*(-((I - x)/(Sqrt[3] + x)))^(1/       > 3)*((I + x)/(Sqrt[3] + x))^(1/3)*       > AppellF1[2/3, 1/3, 1/3,       > 5/3, -((I - Sqrt[3])/(Sqrt[3] + x)), (I + Sqrt[3])/(Sqrt[3] +       > x)])/(4*(1 + x^2)^(1/3))       >              fyi, Rubi 4.10.1 now gives better answer in terms of only       elementary functions and no complex numbers to the above       two problems:              In[6]:= ShowSteps = False;       In[7]:= Int[(a + b*x)/((3 + x^2)*(1 - x^2)^(1/3)), x]              Out[7]= (a*        ArcTan[1/        Sqrt[3] - (2^(2/3)*(1 - x)^(2/3))/(Sqrt[        3]*(1 + x)^(1/3))])/(2*2^(2/3)*Sqrt[3]) - (a*        ArcTan[1/        Sqrt[3] - (2^(2/3)*(1 + x)^(2/3))/(Sqrt[        3]*(1 - x)^(1/3))])/(2*2^(2/3)*Sqrt[3]) +        (Sqrt[3]*b*        ArcTan[(1 + 2^(1/3)*(1 - x^2)^(1/3))/Sqrt[3]])/(2*2^(2/3)) - (b*        Log[3 + x^2])/(4*2^(2/3)) + (a*        Log[(2 - 2*x)^(1/3) + (1 + x)^(2/3)])/(4*2^(2/3)) - (a*        Log[(1 - x)^(2/3) + (2 + 2*x)^(1/3)])/(4*2^(2/3)) +        (3*b*Log[2^(2/3) - (1 - x^2)^(1/3)])/(4*2^(2/3))              In[8]:= Int[(a + b*x)/((3 - x^2)*(1 + x^2)^(1/3)), x]              Out[8]= -((a*        ArcTan[1/        Sqrt[3] + (2^(2/3)*(Sqrt[3] -        x))/(3*(1 + x^2)^(1/3))])/(6*2^(2/3))) + (a*        ArcTan[1/        Sqrt[3] + (2^(2/3)*(Sqrt[3] +        x))/(3*(1 + x^2)^(1/3))])/(6*2^(2/3)) -        (Sqrt[3]*b*        ArcTan[(1 + 2^(1/3)*(1 + x^2)^(1/3))/Sqrt[3]])/(2*2^(2/3)) + (b*        Log[3 - x^2])/(4*2^(2/3)) - (3*b*        Log[2^(2/3) - (1 + x^2)^(1/3)])/(4*2^(2/3)) -        (a*Log[-1 - x/Sqrt[3] + 2^(1/3)*(1 + x^2)^(1/3)])/(4*2^(2/3)*        Sqrt[3]) + (a*        Log[-1 + x/Sqrt[3] + 2^(1/3)*(1 + x^2)^(1/3)])/(4*2^(2/3)*        Sqrt[3]) -        (a*Log[-((Sqrt[3] - x)^3/(3*Sqrt[3])) + 2*(1 + x^2)])/(12*2^(2/3)*        Sqrt[3]) + (a*        Log[-((Sqrt[3] + x)^3/(3*Sqrt[3])) + 2*(1 + x^2)])/(12*2^(2/3)*        Sqrt[3])                     --Nasser              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
[   << oldest   |   < older   |   list   |   newer >   |   newest >>   ]
(c) 1994, bbs@darkrealms.ca