| 
 fricas (1) -> integrate(1/(1+x^4), 
 Type: Union(f1: OrderedCompletion?(Expression(Integer)), is obviously wrong. UPDATE: this problem is fixed in FriCAS. fricas integrate(1/(1+x^4), 
 Type: Union(Expression(Integer), is an antiderivative only for  
  In[10]:= Integrate[1/(1+x^4), x]
  Out[10]= (-2 ArcTan[1 - Sqrt[2] x] + 2 ArcTan[1 + Sqrt[2] x] - 
                               2                         2
         Log[-1 + Sqrt[2] x - x ] + Log[1 + Sqrt[2] x + x ]) / (4 Sqrt[2])
However, fricas integrate((x^4+2*a*x^2+1)^-1, 
 Type: Union(f1: OrderedCompletion?(Expression(Integer)), is wrong, also in FriCAS. However, the solution really depends on the value of a. Mathematica 5.2 gives: 
  In[2]:= Integrate[(x^4+2*a*x^2+1)^-1, {x,0,Infinity}]
                                     2
  Out[2]= If[Im[Sqrt[-a - Sqrt[-1 + a ]]] > 0 && 
                              2
      Im[Sqrt[-a + Sqrt[-1 + a ]]] > 0, 
      -I                             2                        2
     (-- Pi) / (Sqrt[-a - Sqrt[-1 + a ]] Sqrt[-a + Sqrt[-1 + a ]] 
      2
                              2                          2
        (Sqrt[-a - Sqrt[-1 + a ]] + Sqrt[-a + Sqrt[-1 + a ]])), 
                      1
     Integrate[---------------, {x, 0, Infinity}, 
                        2    4
               1 + 2 a x  + x
      Assumptions -> 
                               2                                    2
       Im[Sqrt[-a - Sqrt[-1 + a ]]] <= 0 || Im[Sqrt[-a + Sqrt[-1 + a ]]] <= 0]
      ]
The definite version seems to be correct: fricas D(integrate((x^4+2*a*x^2+1)^-1, 
 Type: Expression(Integer) another similar evaluation --kratt6,  Thu, 10 Jan 2008 01:28:46 -0800 replyThe following also seems to be wrong: fricas ex := integrate(1/(a+x**4), compare with fricas integrate(1/(1+x**4),Name: #293 integrate (1/(1+x^4),x = %minusInfinity..%plusInfinity)=>#293 integrate (1/(1+x^4)http://fricas.svn.sourceforge.net/viewvc/fricas/trunk/src/algebra/gaussian.spad.pamphlet?r1=257&r2=358&view=patch
I'm not sure it solves the issue though.
Name:#293 integrate 1/(1+x)^4=>#293 integrate 1/(1+x^4)Name:#293 integrate 1/(4x^2-1)=>#293 integrate 1/(1+x^4)Name:#293 integrate 1/(1+x^4)=>#293 integrate 1x/(1+x^4)Name:#293 integrate 1x/(1+x^4)=>#293 integrate 1/(1+x^4)Name:#293 integrate 1/(1+x^4)=>#293 integrate 1/(1+x^2) | 


 . The correct antiderivative, i.e., which is defined on all of
. The correct antiderivative, i.e., which is defined on all of  can be obtained from the above expression by replacing
 can be obtained from the above expression by replacing  with
 with  . UPDATE: this problem is also fixed in
. UPDATE: this problem is also fixed in 
