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Two small problems compute OK, but not their sum.


Hi,

I needed to intergate a linear function times Log of something, and
Mathematica would give me a too-compicated answer. So I broke it up
into two smaller programs and it worked much better as you can see
below. Although the answer is probably correct in both cases, I'm just
trying to understand why the difference in appearance is quite so
significant.

Thanks!

Aaron Fude

In[31]:=
Zero=A*Log[c^2+x^2]/2;
One=(B-A)/a*x*Log[c^2+x^2]/2;
Assuming[a>0&&c>0, Integrate[Zero, {x, 0, a}]]
Assuming[a>0&&c>0, Integrate[One, {x, 0, a}]]
Assuming[a>0&&c>0, Integrate[Zero+One, {x, 0, a}]]

Out[33]=
\!\(1\/2\ A\ \((2\ c\ ArcTan[a\/c] + a\ \((\(-2\) + Log[a\^2 +
c\^2])\))\)\)

Out[34]=
\!\(\(-\(\(\((A -
      B)\)\ \((\(-a\^2\) - 2\ c\^2\ Log[
            c] + \((a\^2 + c\^2)\)\ Log[a\^2 + c\^2])\)\)\/\(4\
a\)\)\)\)

Out[35]=
\!\(\(\(1\/\(4\ a\ c\)\)\((\(-\[ImaginaryI]\)\ a\
    A\ \((a\^2 + c\^2)\)\ HypergeometricPFQ[{1\/2, 1, 1}, {2,
          2}, 1 + a\^2\/c\^2] + c\ \((\((
          a + \[ImaginaryI]\ c)\)\ \((a\ \((A + B)\) + \[ImaginaryI]\
\((A - \
B)\)\ c)\)\ Log[1 + a\^2\/c\^2] + a\ \((a\ \((A -
      B)\) - 2\ \[ImaginaryI]\ A\ c\ \((\(-2\) + Log[4])\) + 2\ a\ \((A
+ 
                B)\)\ Log[c])\))\))\)\)\)


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