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MathGroup Archive 2007

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Integrate

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74034] Integrate
  • From: "Michael Weyrauch" <michael.weyrauch at gmx.de>
  • Date: Wed, 7 Mar 2007 03:08:42 -0500 (EST)

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, consumes eventually all memory of my computer ( I have up to 32 GB 
on a Linux workstation) and eventually shuts down the kernel. But that takes very long!

However, the integral is actually relatively easily determined "by hand". The result is

 \!\(test3 = \(\((a1 + t)\)\ \((a3 + t)\)\ \((a4 + t)\)\)\/\(\((\(-t13\^2\) + \
\((a1 + t)\)\ \((a3 + t)\))\)\^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 get 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 such integrals using Mathematica?

Moreover, it would be nice to know what Mathematica is actually doing while it suffers through that calculation, it also does not 
give me an error message.

Thanks for any help.

Michael Weyrauch



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