oldk1331@gmail.com schrieb:   
   >   
   > Off topic a little:   
   >   
   > First, I appreciate your effort to compile this   
   > "Computer Algebra Independent Integration Tests".   
   >   
   > However, I have a small advice for you:   
   >   
   > Could you mark a test which returns non-elementary result   
   > as "half-solved" if there exists elementary results?   
   >   
   > Take this #319 as an example:   
   >   
   > There are elementary results for this #319 test, Timofeev's   
   > book page 194 has one elementary result.   
   >   
   > Rubi's result used to contain F1 function, I consider such   
   > result as not "fully solved", just "half-solved".   
   >   
   > So I'd like to see there's a third state other than "passed"   
   > and "failed", that'll reflect each system's strength more   
   > accurately.   
   >   
   > It should be simple: write a function to scan its argument   
   > for non-elementary function, compare the difference   
   > between optimal antiderivative and CAS's result.   
   >   
   > FriCAS recently can solve this integral, the result is long,   
   > elementary and correct.   
   >   
   > You can verify (in MMA) by:   
   >   
   > y=(2*3^(1/2)*Log[((3*x+3)*(3^(1/3))^2*((x^3+3*x^2+3*x+3)^(1/3))^2+((-9)*   
   > x^2+(-18)*x+(-9))*(x^3+3*x^2+3*x+3)^(1/3)+(2*x^3+6*x^2+6*x)*3^(1/3))/(   
   > x^3+3*x^2+3*x)]+((-1)*3^(1/2)*Log[((63*x^4+252*x^3+378*x^2+270*x+81)*(   
   > (x^3+3*x^2+3*x+3)^(1/3))^2+(45*x^5+225*x^4+450*x^3+486*x^2+297*x+81)*3   
   > ^(1/3)*(x^3+3*x^2+3*x+3)^(1/3)+(31*x^6+186*x^5+465*x^4+666*x^3+603*x^2   
   > +324*x+81)*(3^(1/3))^2)/(x^6+6*x^5+15*x^4+18*x^3+9*x^2)]+(6*ArcTan[((63   
   > *x^4+252*x^3+378*x^2+270*x+81)*((x^3+3*x^2+3*x+3)^(1/3))^2+(63*x^5+315*   
   > x^4+630*x^3+648*x^2+351*x+81)*3^(1/3)*(x^3+3*x^2+3*x+3)^(1/3)+(38*x^6+228   
   > *x^5+570*x^4+801*x^3+693*x^2+351*x+81)*(3^(1/3))^2)/((7*x^6+42*x^5+105*x^   
   > 4+135*x^3+90*x^2+27*x)*3^(1/2)*(3^(1/3))^2)]+6*ArcTan[((18*x^2+36*x+18)*3   
   > ^(1/2)*(x^3+3*x^2+3*x+3)^(1/3)+(5*x^3+15*x^2+15*x+9)*3^(1/2)*3^(1/3))/((27   
   > *x^3+81*x^2+81*x+27)*3^(1/3))])))/(18*3^(1/2)*3^(1/3))   
   >   
   > Simplify[D[y,x]-1/((3*x + 3*x^2 + x^3)*(3 + 3*x + 3*x^2 + x^3)^(1/3))]   
   >   
      
   This and other suggestions were initially made in Albert's post "Who is   
   fastest AND optimal?" of Wed, 27 Apr 2016 15:45:46 -0700 (PDT):   
      
   > [...] I have included an "I" when an integrator's result involves the   
   > imaginary unit but the optimal antiderivative does not. Similarly, an   
   > "R" following a leaf count indicates the result involves the RootOf   
   > function and is thus technically not a closed-form antiderivative.   
   >   
   > This extra information in the table makes clear that although   
   > Mathematicas's results for integrals #1 and #2 are smaller than   
   > Rubi's, they involve the imaginary unit.   
   >   
   > Therefore I recommend in test result summaries, an "I", "S" and/or "H"   
   > follow leaf counts when a result contains the imaginary unit, a   
   > special function, or a hypergeometric function, respectively when the   
   > optimal antiderivative does NOT involve them.   
      
   I enthusiastically support the pledge for counting such "I", "R" and "N"   
   (needlessly nonelementary) test results as "half done" only.   
      
   And FriCAS has evaluated a moderately difficult Goursat pseudo-elliptic   
   cube-root integral correctly! While the antiderivative involving two   
   LOG's and two ATAN's jumps at x = -1 - 98^(1/3)/7 and x = -1, this seems   
   to introduce the first Risch integrator not helpless in the face of cube   
   roots! Is it hardcoded knowledge about the splitting field of special   
   Trager resultants - as mentioned by Waldek on Mon, 29 Feb 2016 02:28:58   
   +0000 (UTC) - that enables this feat?   
      
   Martin.   
      
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