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    Nasser M. Abbasi
    @nasser1

    How to find number of Rules in Rubi? I am using 14.6.1.

    Is there a command that returns this number? my loop now goes up to 6874 with no problem. But this is trial and error. I'd like to know before the loop starts exactly the number of rules.

    Thanks.

    Nasser M. Abbasi
    @nasser1
    opps typo above, I meant using Rubi version 4.16.1 and not 14.6.1
    Albert D. Rich
    @AlbertRich
    @nasser1 The Mathematica command Length[DownValues[Int]] returns 6881 Int rules for Rubi 4.16.1.
    Nasser M. Abbasi
    @nasser1

    @AlbertRich Thanks. That works. Do you know why when I print the rules using the following, it shows always as Removed[Int] in there and not just Int? Here is an example

    $LoadShowSteps = False;
     << Rubi`
     BeginPackage["Rubi`"];
     Begin["`Private`"];
     Do[
         Print[InputForm[DownValues[Int][[n]]]]
        ,
         {n, 1, 1}(*Length[DownValues[Int]]}*)
      ]

    Which gives

    HoldPattern[Removed["Int"][(u_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol]] :> Removed["Int"][u*(b*x^n)^p, x] /; FreeQ[{a, b, n, p}, x] && EqQ[a, 0]

    Now I remove this as follows

      BeginPackage["Rubi`"];
      Begin["`Private`"];
      Do[
          res = InputForm[DownValues[Int][[n]]];
          res = ToString[ReleaseHold[res]];
          res = StringReplace[res, "Defer[Removed[\"Int\"]]" -> "Int"];
          res = StringReplace[res, "Removed[\"Int\"]" -> "Int"];
          Print[res]
           ,
            {n, 1, 1}(*Length[DownValues[Int]]}*)
           ]

    Which gives

    Int[(u_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(b*x^n)^p, x] /; FreeQ[{a, b, n, p}, x] && EqQ[a, 0]

    Starting from clean Kernel did not help. The Removed[Int] is always there. It is not big deal as I can remove it, this is for just printing the rules, but I wondered why it happens.

    tringocao
    @tringocao
    image.png
    Why WildCards have option S(1)? Can you help me to explain that?
    Anixx
    @Anixx
    I would be glad if anyone commented on my work, particularly, the code that allows to multiply divergent integrals. Here is my post with a Mathematica code, it is like a calculator that allows to multiply two divergent integrals: https://mathoverflow.net/q/421354/10059
    edmontz
    @edmontz
    Int[Sin[Sin[a*x]],x] appears to be beyond any CAS I've encountered thus far. Is this type of function beyond integration. Maybe related to a
    Bessel function.
    Nasser M. Abbasi
    @nasser1
    AAA.png

    You can't get closed form antiderivative. You could find approximate series solution.

       (a x^2)/2 - (a^3 x^4)/12 + (a^5 x^6)/60 - (a^7 x^8)/315 + (
       13 a^9 x^10)/25200 - (47 a^11 x^12)/598752 + (
      15481 a^13 x^14)/1362160800 - (3947 a^15 x^16)/2554051500 + (
      451939 a^17 x^18)/2273570208000 - (
      23252857 a^19 x^20)/950352346944000 + (
      186846623 a^21 x^22)/64568056512960000 - (
      831520891 a^23 x^24)/2524611009656736000 + (
      1108990801 a^25 x^26)/30644204599008000000 - (
      143356511198507 a^27 x^28)/37217815504359602112000000 + (
      920716137922619 a^29 x^30)/2312821392056632416960000000 + 
      C[1]

    Here is a plot comparing the derivative of the above antiderivative above with the integrand showing good agreement. More terms gives better agreemeent.

       ode = y'[x] == Sin[Sin[a*x]];
       antiderivative = AsymptoticDSolveValue[ode, y[x], {x, 0, 30}]
       Plot[Evaluate[{Sin[Sin[a*x]], D[antiderivative, x]} /. a -> 1], {x, -3, 3}]
    edmontz
    @edmontz
    I tried something different, I attempted an adaptive method to find the principle components from a sample that I applied an fft to. Initially just two peaks appeared to be relevant, more iterations showed otherwise.
    So time to investigate your approach.
    edmontz
    @edmontz
    I did manage to run all of the RUBI test suites, this took about a day; I did wonder if this might be sped up with a GPU. It required about a days worth of CPU time, I just let run in the background while I continued with all of the usual computer related tasks.
    edmontz
    @edmontz
    Then, with the RUBI package loaded I attempted to solve some demanding nonlinear, second order differential equations; thinking that with RUBI the process is much more proficient . This didn't appear to be the instance. Perhaps RUBI isn't completely incorporated into Mathematica calculus routines yet.
    edmontz
    @edmontz
    Okay, I just tried your method, for a particular domain this gives the appearance of a good fit; outside of that it becomes asymptotically inaccurate. The formula for a geometric sequence, as found in math tables, sometimes returns a closed form representation.
    Albert D. Rich
    @AlbertRich

    The following Issue was recently posted on the Rubi's GitHub repository:

    • With the last public update to this repo being over a year ago, it would be great to hear if this is still actively being worked on ­čÖé.

    I posted the following response:

    • Yes, Rubi 4 is currently undergoing a major redesign that will significantly expand the class of mathematical expressions it can integrate, produce simpler antiderivatives, and provide more elegant derivations. Also it is necessary to perfect Rubi 4 before compiling its pattern matching rules into an if-then-else decision tree for the next release of Rubi.

    • I'm working fulltime on what has turned out to be a massive project. It's driven by my figuring out the math required to integrate expressions symbolically. I will release a new version of Rubi 4 as-soon-as it's perfected to my satisfaction. But it's impossible to predict when that will be, given the creative nature of this open-ended work.

    Albert

    Miguel Raz Guzmán Macedo
    @miguelraz
    Hello @AlbertRich - how goes the Rubi revamp?
    Wish you all the best an dlooking forward to it.
    Albert D. Rich
    @AlbertRich
    @miguelraz Hang tight. Hopefully a new release of Rubi will be available by the end of the year. Sorry I can't be more definitive. It's hard to predict where the FTOC will take me... Albert
    Miguel Raz Guzmán Macedo
    @miguelraz
    All the support for you @AlbertRich! Keep up the great work!
    toshiki
    @toshiki:matrix.org
    [m]
    Hi, just started using Rubi. When I use Integrate[], Mathematica often returns a result with a condition, for example, a>0 or something, I'm wondering if Rubi has this functionality. Thank you.
    Also, when I try to access the documentation using ?Int, my Mathematica freezes. I don't know if it's only a problem for my computer or there's a bug though...
    Albert D. Rich
    @AlbertRich
    @toshiki:matrix.org The antiderivatives produced by Rubi are valid for all real and complex values of their parameters. Therefore, there is no need place conditions on them. That's a good thing.
    Tci Gravifer Fang
    @Gravifer
    Hello guys! I have been through an academic hiatus for almost a year now, and only coming back to coding and stuff recently. I noticed that the rubi repo is not currently very active, and wonder what's the situation of the project at this point. Can somebody catch me up?
    Albert D. Rich
    @AlbertRich

    @Gravifer Thanks for your interest in Rubi. As previously posted:

    Rubi 4 is currently undergoing a major redesign that will significantly expand the class of mathematical expressions it can integrate, produce simpler antiderivatives, and provide more elegant derivations. Also it is necessary to perfect Rubi 4 before compiling its pattern matching rules into an if-then-else decision tree for the next release of Rubi.

    I'm working fulltime on what has turned out to be a massive project. It's driven by my figuring out the math required to integrate expressions symbolically. I will release a new version of Rubi 4 as-soon-as it's perfected to my satisfaction. But it's impossible to predict when that will be, given the creative nature of this open-ended work.

    Miguel Raz Guzmán Macedo
    @miguelraz
    @AlbertRich Happy Holidays!
    Wish you the best this year and thank you for all your hard work <3