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   sci.math.symbolic      Symbolic algebra discussion      10,432 messages   

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   Message 10,299 of 10,432   
   Sam Blake to Richard Fateman   
   Re: A challenging nested radical integra   
   13 Apr 23 16:29:33   
   
   From: samuel.thomas.blake@gmail.com   
      
   On Monday, April 10, 2023 at 6:46:48 AM UTC+10, Richard Fateman wrote:   
   > Maxima 5.45.1 returns the (unevaluated) integral in about 0.00 seconds.    
   > Using quadpack numerical integration, quad_qag() it is possible to integrate    
   > this function between any two points, (essentially instantaneously) and   
   therefore    
   > plot it, if you wish.    
   > Since the integrand does not have any symbolic constants, there is no need   
   to have    
   > a surface or higher-dimensional plot.    
   >    
   > The usefulness of the symbolic result you quote, from    
   > Mathematica, is somewhat difficult to grasp; perhaps plotting it would be   
   easier to comprehend?    
   >    
   > If I just use Mathematica's Integrate[] , it fails to find a closed form. It   
   returns, after 10.21 seconds    
   > {10.2188, \[Integral](-2 + 2^(1/3) (x + Sqrt[4 + x^2])^(2/3))^(    
   > 1/3)/(x + Sqrt[4 + x^2])^(1/9) \[DifferentialD]x}    
   >    
   > My system, version 13.0.0 does not have IntegrateAlgebraic[]    
   > so that's it.    
   >    
   > While doing symbolic indefinite integration is an interesting test for   
   computer algebra systems,    
   > and it is perhaps useful for helping with calculus homework, it has not, for   
   the most part,    
   > entered the mainstream of applied scientific computing.    
   >    
   > RJF   
   > On Sunday, March 5, 2023 at 7:14:42 PM UTC-8, Sam Blake wrote:    
   > > On Thursday, February 23, 2023 at 11:57:47 AM UTC+10, Sam Blake wrote:    
   > > > Greetings,    
   > > >    
   > > > I was hoping the Risch-Trager-Bronstein algorithm of FriCAS and/or AXIOM   
   would compute the following integral in a reasonable amount of time    
   > > >    
   > > > -> integrate((x+(4+x^2)^(1/2))^(-1/9)*(-2+2^(1/3)*(x+(4+x^   
   )^(1/2))^(2/3))^(1/3),x)    
   > > >    
   > > > Was it wishful thinking? My computer has been running FriCAS 1.3.6 for   
   over an hour. The answer to this integral does not require any logarithms, so   
   it should be computable with the algebraic extensions of Hermite reduction,   
   right?    
   > > >    
   > > > Sam    
   > > I guess not. This does make me question the practicality of the   
   Trager/Bronstein algorithm. Here's the solution from IntegrateAlgebraic in   
   Mathematica:    
   > >    
   > > In[5117]:= IntegrateAlgebraic[(-2 + 2^(1/3) (x + Sqrt[4 + x^   
   ])^(2/3))^(1/3)/(x + Sqrt[4 + x^2])^(1/9), x] // Timing    
   > >    
   > > Out[5117]= {0.241908, (x (-2 + 2^(1/3) (x + Sqrt[4 + x^2])^(   
   /3))^(1/3))/(x + Sqrt[4 + x^2])^(1/9) + ((-2 + 2^(1/3) (x + Sqrt[4 +   
   x^2])^(2/3))^(4/3) (-(1/10) - (11 (x + Sqrt[4 + x^2])^(2/3))/(20 2^(2/3)) - (x   
   + Sqrt[4 + x^2])^(4/3)/(20 2^(1/3))))/(x +   
    Sqrt[4 + x^2])^(10/9)}    
   > >    
   > > Sam   
      
   FYI IntegrateAlgebraic is an open source package I have been working on in my   
   spare time. It has recently been incorporated into Mathematica. When the   
   latest version is included in mathematica then Integrate will compute this   
   integral.    
      
   Sam    
      
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