Maxima has the tools for calculating these integrals, but sadly doesn't
handle these cases directly:
ex: 1/ ( (x-a)*(x-b)*sqrt(1-x^2) )
... integrate(ex,x) => nounform
but
... integrate(partfrac(ex,x),x) => reasonable solution (with a couple of
questions)
ex: 1/ ( (x-1)*(x+1)*sqrt(1-x^2) )
... integrate(ex,x) => nounform
... integrate(radcan(ex),x) => nounform
... integrate(factor(radcan(ex)),x) => nounform
but
... integrate(partfrac(ex,x),x) => reasonable solution
... integrate(ratsimp(ex),x) => reasonable solution
On Thu, Oct 17, 2013 at 7:00 PM, Michel Talon <talon at lpthe.jussieu.fr>wrote:
> Raymond Toy wrote:
> >>>>>> "maryam" == maryam shamssolary <shamssolary at gmail.com> writes:
> >
> > maryam> Dear Master
> >
> >
> > maryam> I am a young researcher in numerical linear algebra. For some
> calculations, I used your "Maxima"
> > maryam> software. But I have some problems with some commands such
> as:
> > maryam> integrate(1/(r+s*x+t*x^2)*1/sqrt(1-x^2), x, -1, 1);
> >
> >
> > maryam> give me this message:
> > maryam> "Is "4*r*t-s^2" positive, negative, or zero?"
> >
> >
> > maryam> I tried different answers but did not?! What do you suggest?
> >
> > Maxima cannot do all integrals of this form. It appears that maxima
> > can only do this if t*x^2+s*x+r has real roots. The case for complex
> > roots is not handled. This is why maxima is asking for the sign of
> > 4*r*t-s^2. The roots are real if 4*r*t-s^2 is zero or negative.
> >
> > After this maxima may ask many more questions as it breaks the
> > original integral into two new integrals. I find the questions hard to
> > answer. :-(
> >
> > You might get a feel for what is going on by using some numerical
> > values for r, s, and t.
> >
> > Sorry that maxima can't easily produce an answer for you.
> >
> > Ray
>
>
> By the way i have tried with maple and mathematica which all give very
> complicated answers. So i have tried to find something simpler by hand.
>
> Suppose you know the roots of r+s*x+t*x^2. Then you can as well replace
> that by (x-a)*(x-b).
>
> The inverse of which can be written as
> 1/(a-b)*(1/(x-a)-1/(x-b)).
>
> This reduces the problem to one that maxima easily handles:
>
> (%i3) integrate(1/(x-a)*1/sqrt(1-x^2), x, -1, 1);
> Is a - 1 positive, negative, or zero?
>
> p;
> 2
> 2 %pi sqrt(a - 1)
> (%o3) - ------------------
> 2
> 2 a - 2
> If you answer n; you have some more sign questions, for example:
>
> (%i2) integrate(1/(x-a)*1/sqrt(1-x^2), x, -1, 1);
> Is a - 1 positive, negative, or zero?
>
> n;
> Is a + 1 positive, negative, or zero?
>
> p;
> Principal Value
> 2 2
> log(2) sqrt(1 - a ) log(2 - 4 a) sqrt(1 - a )
> (%o2) ------------------- - -------------------------
> 2 2
> a - 1 a - 1
>
>
> As you can see these formulas require appropriate signs for the arguments
> of the logs to be "correct" in an elementary sense. Are they true? I
> don't know. For exemple in the second case one has -1<a<1 so for
> 1/2<a<1 one has 2-4*a <0 which renders the formula highly suspect.
>
>
> --
> Michel Talon
>
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