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 reply The 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)
|