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Re: Integrate
*To*: mathgroup at smc.vnet.net
*Subject*: [mg74464] Re: Integrate
*From*: "dimitris" <dimmechan at yahoo.com>
*Date*: Thu, 22 Mar 2007 01:12:38 -0500 (EST)
*References*: <esls78$q0v$1@smc.vnet.net><et5o76$jor$1@smc.vnet.net>
> Anyway, all indefinite integrals in Mathematica are evaluated by
> application of the Risch Algorithm (R.A.)
> Exceptions are elliptic integrals, antiderivatives that require
> special functions and antiderivatives of
> special functions.
I should have added...
Exceptions are elliptic integrals, antiderivatives that require
special functions and antiderivatives of special functions. For these
exceptions table lookup takes place.
=CF/=C7 dimitris =DD=E3=F1=E1=F8=E5:
> Evaluating your integral was a really tough procedure for Mathematica.
>
> I want to know how you actually evaluated the integral by hand
> "easily"...
>
> Anyway, all indefinite integrals in Mathematica are evaluated by
> application of the Risch Algorithm (R.A.)
> Exceptions are elliptic integrals, antiderivatives that require
> special functions and antiderivatives of
> special functions.
>
> (see http://library.wolfram.com/infocenter/Conferences/5832/
> http://library.wolfram.com/infocenter/Conferences/4684/
> http://72.14.203.104/search?q=cache:_LK2UsI5d_UJ:www-sop.inria.fr/cafe/=
Ma=
> nuel.Bronstein/publications/issac98.pdf+symbolic+integration+tutorial&hl=
==
> en&ct=clnk&cd=1
> and references therein.)
>
> Hence, since you integrand contain only elementary functions Integrate
> will use R.A.
> As a critical concept required in the application of the R.A. is the
> notion of the partial
> fraction decomposition (I believe that) the problem must be the
> considerable difficult partial fraction
> decomposition of your integrand.
>
> However helping a little Integrate you get the desired results very
> quickly!
> Note that helping Mathematica does not say anything for her
> efficiencies.
> Mathematica and as well the other CAS are here to help you; but some
> times
> they also require help by you. It's an interactive relation!
> So, here goes...
>
>
> In[104]:=
> Quit
>
> In[1]:=
> $Version
> Out[1]=
> "5.2 for Microsoft Windows (June 20, 2005)"
>
> In[2]:=
> SessionTime[]
> rep = {b1 -> a1 + t, b2 -> a2 + t, b3 -> a3 + t, b4 -> a4 + t}
> test = -((1/((b1*b3 - t13^2)^3*((-b2)*b4 + t24^2)^2))*(b3*b4*t13^2*((-
> b2)*b4 + t24^2) -
> b1^2*b3*(b2*b4^2 - b4*t24^2 + b3*(b4^2 + t24^2)) + b1*((-
> b2)*b4^2*(b3^2 + t13^2) + b3^2*b4*t24^2 + b4*t13^2*t24^2 +
> b3*t13^2*(b4^2 + t24^2)))) /. rep
> test2 = FullSimplify[test /. t -> y - a4 /. {a3 - a4 -> c, a2 - a4 ->
> d, a1 - a4 -> e}](*helping Mathematica*)
> Integrate[test2, y]
> Reverse /@ {a3 - a4 -> c, a2 - a4 -> d, a1 - a4 -> e}
> ff=%% /. y -> t + a4 /. {c -> a3 - a4, d -> a2 - a4, e -> a1 - a4}
> (*returning to original variable and parameters*)
> D[ff,t]==test//Simplify (*check*)
> SessionTime[]
>
> Out[2]=
> 0=2E375`7.0255762612236925
>
> Out[3]=
> {b1 -> a1 + t, b2 -> a2 + t, b3 -> a3 + t, b4 -> a4 + t}
>
> Out[4]=
> -(((a3 + t)*(a4 + t)*t13^2*((-(a2 + t))*(a4 + t) + t24^2) - (a1 +
> t)^2*(a3 + t)*((a2 + t)*(a4 + t)^2 - (a4 + t)*t24^2 +
> (a3 + t)*((a4 + t)^2 + t24^2)) + (a1 + t)*((-(a2 + t))*(a4 +
> t)^2*((a3 + t)^2 + t13^2) + (a3 + t)^2*(a4 + t)*t24^2 +
> (a4 + t)*t13^2*t24^2 + (a3 + t)*t13^2*((a4 + t)^2 + t24^2)))/
> (((a1 + t)*(a3 + t) - t13^2)^3*
> ((-(a2 + t))*(a4 + t) + t24^2)^2))
>
> Out[5]=
> -(((-(c + y))*(e + y)^2*(c*t24^2 + (c + d)*y^2 + 2*y^3) + t13^2*y*(c +
> y)*(t24^2 - y*(d + y)) +
> (e + y)*(t13^2*t24^2*y + t24^2*y*(c + y)^2 + t13^2*(c + y)*(t24^2
> + y^2) - y^2*(d + y)*(t13^2 + (c + y)^2)))/
> ((t24^2 - y*(d + y))^2*(-t13^2 + (c + y)*(e + y))^3))
>
> Out[6]=
> (y*(c + y)*(e + y))/((t24^2 - y*(d + y))*(-t13^2 + c*(e + y) + y*(e +
> y))^2)
>
> Out[7]=
> {c -> a3 - a4, d -> a2 - a4, e -> a1 - a4}
>
> Out[8]=
> ((a1 + t)*(a3 + t)*(a4 + t))/(((a3 - a4)*(a1 + t) + (a1 + t)*(a4 + t)
> - t13^2)^2*((-(a2 + t))*(a4 + t) + t24^2))
>
> Out[9]=
> True
>
> Out[10]=
> 17.28125`8.689120146480764
>
>
> I think almost 17 seconds is more than just reasonable timing!
>
>
> "Stay tuned" with this forum for other Integrate queries!!!
>
>
> Kind Regards
> Dimitris
>
> =CF/=C7 Michael Weyrauch =DD=E3=F1=E1=F8=E5:
> > Hello,
> >
> > I seem to have a problem with Integrate using Mathematica 5.2 on a
> > Windows PC.
> >
> > In fact, I would like to integrate the following somewhat lengthy
> > rational function of t. (I use a replacement rule to save some typing)
> >
> > rep={b1\[Rule]a1+t,b2\[Rule]a2+t,b3\[Rule]a3+t,b4\[Rule]a4+t}
> >
> > \!\(test = \(-\(\((b3\ b4\ t13\^2\ \((\(-b2\)\ b4 + t24\^2)\) -
> > b1\^2\ b3\ \((b2\ b4\^2 - b4\ t24\^2 +
> > b3\ \((b4\^2 + t24\^2)\))\) +
> > b1\ \((\(-b2\)\ b4\^2\ \((b3\^2 + t13\^2)\) +
> > b3\^2\ b4\ t24\^2 + b4\ t13\^2\ t24\^2 +
> > b3\ t13\^2\ \((b4\^2 +
> > t24\^2)\))\))\)\/\(\((b1\ b3 - t13\^2)\)\^3=
\ \
> > \((\(-b2\)\ b4 + t24\^2)\)\^2\)\)\) /. rep\)
> >
> > If I try to determine the indefinite integral using
> >
> > Integrate[test,t]
> >
> > Mathematica never returns a result on my computer. It works hard, consu=
me=
> s eventually all memory of my computer ( I have up to 32 GB
> > on a Linux workstation) and eventually shuts down the kernel. But that =
ta=
> kes very long!
> >
> > However, the integral is actually relatively easily determined "by hand=
"=2E=
> The result is
> >
> > \!\(test3 = \(\((a1 + t)\)\ \((a3 + t)\)\ \((a4 + t)\)\)\/\(\((\(-t1=
3\=
> ^2\) + \
> > \((a1 + t)\)\ \((a3 + t)\))\)\^2\ \((t24\^2 - \((a2 + t)\)\ \((a4 + t)\=
))=
> \)\)\
> > \)
> >
> > as can be verified by differentiation.
> >
> > Since I have to do more of such integrals, it would be nice if I could =
ge=
> t Mathematica to do the work. Does anyone have an idea what
> > could be done about that? Is there any trick or option in order to do s=
uc=
> h integrals using Mathematica?
> >
> > Moreover, it would be nice to know what Mathematica is actually doing w=
hi=
> le it suffers through that calculation, it also does not
> > give me an error message.
> >
> > Thanks for any help.
> >
> > Michael Weyrauch
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