clicliclic@freenet.de  wrote:   
   >   
   > antispam@math.uni.wroc.pl schrieb:   
   > >   
   > > clicliclic@freenet.de  wrote:   
   > > >   
   > > > What does this FriCAS failure on Timofeev Chapter 5, Example 81 (p.   
   > > > 276 in the book, #417 in the test suite) mean:   
   > > >   
   > > >   integrate((-2*sin(2*x)+(cos(x)*sin(x)^3)^(1/2))   
   > > >   /(-(cos(x)^3*sin(x))^(1/2)+tan(x)^(1/2)), x)   
   > > >   
   > > >   Error detected within library code: not invertible   
   > >   
   > > This is rather excessive example of dependent roots:   
   > >   
   > > (cos(x)*sin(x)^3)^(1/2) = +-sin(x)*(cos(x)*sin(x))^(1/2)   
   > > (cos(x)^3*sin(x))^(1/2) = +-cos(x)*(cos(x)*sin(x))^(1/2)   
   > > tan(x)^(1/2) = +-(cos(x)*sin(x))^(1/2)/cos(x)   
   > >   
   > > So there are 4 essentially different combinations of branches   
   > > (8 together, but they are equvalent in pairs).   
   > >   
   > > As I wrote, dependent roots may lead to division by zero,   
   > > which happens in this example.   
   > >   
   >   
   > If this error message is specific to the detection of dependent roots   
   > that FriCAS cannot handle, it can serve as an indication for the user   
   > to rewrite his integrand in terms of independent roots.   
      
   Unfortunately, no.  This is really generic "division by zero"   
   error which is possible for several different reasons.   
      
   > The problem of dependent roots would thereby be addressed adequately in   
   > my view even though the test cases are currently counted as failed.   
      
   Well, it is hard to know what user wants as an answer.  Apparently   
   Timofeev completely disregared any branch issues.  This makes   
   sense for integration as branch issues are almost independent   
   of finding correct formula.  But there are also subtelties,   
   because correct formula may be different for different   
   branches.  And it may happen that one branch has elementary   
   integral while other branch does not.   
      
   Anyway, I plan to add detection of dependent roots to FriCAS,   
   but ATM there is no such routine and results are mixed:   
   - FriCAS may signal division by zero   
   - FriCAS may produce result for some choice of branches   
   - various ad-hoc choices may contradict themself leading   
     to wrong result   
      
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