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Submitted by : (unknown) at: 2007-11-17T22:22:23-08:00 (10 years ago)
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fricas
integrate(1/(1+x^4),x=%minusInfinity..%plusInfinity)

\label{eq1}{\pi \ {\sqrt{2}}}\over 2(1)
Type: Union(f1: OrderedCompletion?(Expression(Integer)),...)

is obviously wrong. UPDATE: this problem is fixed in FriCAS.

fricas
integrate(1/(1+x^4),x)

\label{eq2}{\left(
\begin{array}{@{}l}
\displaystyle
- 
\
\
\displaystyle
{4 \ {\sqrt{2}}\ {\arctan \left({1 \over{{{\sqrt{2}}\ {\sqrt{-{x \ {\sqrt{2}}}+{{x}^{2}}+ 1}}}+{x \ {\sqrt{2}}}- 1}}\right)}}- 
\
\
\displaystyle
{4 \ {\sqrt{2}}\ {\arctan \left({1 \over{{{\sqrt{2}}\ {\sqrt{{x \ {\sqrt{2}}}+{{x}^{2}}+ 1}}}+{x \ {\sqrt{2}}}+ 1}}\right)}}+ 
\
\
\displaystyle
{{\sqrt{2}}\ {\log \left({{x \ {\sqrt{2}}}+{{x}^{2}}+ 1}\right)}}-{{\sqrt{2}}\ {\log \left({-{x \ {\sqrt{2}}}+{{x}^{2}}+ 1}\right)}}
(2)
Type: Union(Expression(Integer),...)

is an antiderivative only for x\not\in\{-\frac{\sqrt2}2,\frac{\sqrt2}2\}. The correct antiderivative, i.e., which is defined on all of \mathbb R can be obtained from the above expression by replacing \atan\frac1z with \frac\pi2-\atan z. UPDATE: this problem is also fixed in FriCAS, although the result could be simpler. For example, mathematica 5.2 gives:

  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, x=0..%plusInfinity, "noPole")

\label{eq3}0(3)
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, x=0..b, "noPole"), b)

\label{eq4}1 \over{{{b}^{4}}+{2 \  a \ {{b}^{2}}}+ 1}(4)
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),x=0..%plusInfinity, "noPole")
There are no library operations named ** Use HyperDoc Browse or issue )what op ** to learn if there is any operation containing " ** " in its name.
Cannot find a definition or applicable library operation named ** with argument type(s) Variable(x) PositiveInteger
Perhaps you should use "@" to indicate the required return type, or "$" to specify which version of the function you need. eval(ex, a=1)

\label{eq5}ex(5)
Type: Polynomial(Integer)

compare with

fricas
integrate(1/(1+x**4),x=0..%plusInfinity, "noPole")
There are no library operations named ** Use HyperDoc Browse or issue )what op ** to learn if there is any operation containing " ** " in its name.
Cannot find a definition or applicable library operation named ** with argument type(s) Variable(x) PositiveInteger
Perhaps you should use "@" to indicate the required return type, or "$" to specify which version of the function you need.

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)