When I type: fricas (1) -> ex1:=(x-x/(x+exp(x)/(x*log(x+x))))*exp(2)
Type: Expression(Integer)
fricas ex2:=D(ex1,
Type: Expression(Integer)
fricas ex3:=simplify(ex2)
Type: Expression(Integer)
fricas integrate(ex3,
Type: Union(Expression(Integer),
then Axiom 3.9 returns unevaluated integral. Note, that if I remove the exp(2) factor, then Axiom is able to compute the integral. fricas ex4:=(x-x/(x+exp(x)/(x*log(x+x))))
Type: Expression(Integer)
fricas ex5:=D(ex4,
Type: Expression(Integer)
fricas ex6:=simplify(ex5)
Type: Expression(Integer)
fricas integrate(ex6,
Type: Union(Expression(Integer),
We claim that Axiom is good at integration, therefore critical --kratt6, Wed, 10 May 2006 01:28:56 -0500 reply Severity: normal => critical
We claim that Axiom is good at integration, therefore critical --kratt6, Wed, 10 May 2006 01:29:14 -0500 reply Category: Axiom Mathematics => Axiom Library
If I try:
fricas integrate(simplify(((x^4*log(2*x)^2+(3*x^2-2*x)*exp(x)*log(2*x)+exp(2*x)-x*exp(x))/(x^4*log(2*x)^2+2*x^2*exp(x)*log(2*x)+exp(2*x))-1)*exp(2)),
Type: Union(Expression(Integer),
I see that Axiom failed to recognize that e^4 - (e^2)^2 is zero. AFAICS similar problem appears in original example: when I trace rischNormalize I see that both e^2 and e^4 appear in "normalized" function. But this may cause trouble later, since Axiom assumes that different kernels are algebraically independent. I do not know if rischNormalize is supposed to simplify constants (maybe this shoud be done by a separtate pass) but failing to recognize "transcendental" zero may give wrong results and otherwise Risch algorithm may declare an integrable function as non integrable. Status: open => fixed somewhere Status: fixed somewhere => closed |