"Nasser M. Abbasi" schrieb:   
   >   
   > On 1/21/2018 1:50 AM, clicliclic@freenet.de wrote:   
   >   
   > >>   
   > >> Rubi 4.14.4 uses Appell for the last 3 only   
   > >> ===========   
   > >>   
   > >> In[33]:= Int[x/((4-x^3)*Sqrt[1-x^3]),x]   
   > >> Out[33]= -(ArcTan[1/Sqrt[3]-(2^(2/3) (1-Sqrt[1-x^3]))/(Sqrt[3] x)]/(3   
   2^(2/3) Sqrt[3]))+ArcTan[1/Sqrt[3]-(2^(2/3) (1+Sqrt[1-x^3]))/(Sqrt[3] x)]/(3   
   2^(2/3) Sqrt[3])+ArcTanh[Sqrt[1-x^3]]/(9 2^(2/3))+Log[1+2^(1/3)    
   -Sqrt[1-x^3]]/(6 2^(2/3))-Log[1+2^(1/   
   3) x+Sqrt[1-x^3]]/(6 2^(2/3))   
   > >>   
   > >   
   > > Wow! But what makes this first integral special?   
   > >   
   >   
   > Looks like Rule 482 kicks in. For integral of the form   
   >   
   >             x/( (a + b x^3) sqrt( c + d x^3) )   
   >   
   > This "good" rule activates when   
   >   
   >            (b c- a d ) /= 0 AND  (4 b c - a d ) = 0   
   >   
   > In this case, Int[x/((4-x^3)*Sqrt[1-x^3]),x] we see that   
   > a=4,b=-1,c=1,d=-1. Hence both conditions above are met   
   > and this leads to solution with elementary functions.   
   >   
   > For the other integrals, such as Int[x/((x^3+8)*Sqrt[x^3-1]),x],   
   > here a=8,b=1,c=-1,d=1, therefore  (4 b c- a d )=4*(-1)-(8)=-12   
   > which is not zero. Hence it uses other rule which ends up   
   > using AppellF1.   
   >   
      
   Thank you for analyzing this. To me it makes no sense to have Rubi   
   equipped with a rule for INT(x/((4 - x^3)*SQRT(1 - x^3)), x) but no   
   rule for the very similar INT(x/((x^3 + 8)*SQRT(x^3 - 1)), x). This   
   should be remedied, while Rubi's handling of the third and fourth   
   integrals need not be changed in my view.   
      
   Martin.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)