On Tuesday, August 29, 2017 at 2:06:56 AM UTC+10,
clicl...@freenet.de wrote:
... as promised last year (on Tue, 15 Nov 2016 23:04:20 +0100 in the
thread "FriCAS web interface failure") here are some pseudo-elliptic integration exercises involving fourth roots. Does your favorite
integrator arrive at the elementary results?
INT((a + b*x)/((2 - x^2)*(x^2 - 1)^(1/4)), x)
INT((a + b*x)/((2 + x^2)*(1 + x^2)^(1/4)), x)
INT((3 + x^2)/((1 + x^2)*(1 + 6*x^2 + x^4)^(1/4)), x)
INT((3 - x^2)/((1 - x^2)*(1 - 6*x^2 + x^4)^(1/4)), x)
The integrals are taken from or based on papers by Euler dated to 26th
March 1777, which can be found as items E688 and E690 at the Euler
Archive <http://eulerarchive.maa.org>, or on pp. 98-117 and 127-131 of
the Nova Acta Academiae Scientiarum Imperialis Petropolitanae IX (1795
for 1791).
Martin.
5 years later... if you split the first two integrals into a sum of two integrals, then Mathematica 13.2 can compute these:
In[7391]:= Integrate[a/((2 - x^2) (x^2 - 1)^(1/4)), x] +
Integrate[(b x)/((2 - x^2) (x^2 - 1)^(1/4)), x]
Out[7391]= (a (-ArcTan[(Sqrt[2] (-1 + x^2)^(1/4))/x] +
ArcTanh[x/(Sqrt[2] (-1 + x^2)^(1/4))]))/(2 Sqrt[2]) +
b (-ArcTan[(-1 + x^2)^(1/4)] + ArcTanh[(-1 + x^2)^(1/4)])
In[7392]:= Integrate[a/((2 + x^2) (1 + x^2)^(1/4)), x] +
Integrate[(b x)/((2 + x^2) (1 + x^2)^(1/4)), x]
Out[7392]= (b (ArcTan[(-1 + Sqrt[1 + x^2])/(Sqrt[2] (1 + x^2)^(1/4))] -
ArcTanh[(Sqrt[2] (1 + x^2)^(1/4))/(1 + Sqrt[1 + x^2])]))/Sqrt[2] +
a (-(1/4) ArcTan[(-x + (1 + x^2)^(1/4))/(1 + x^2)^(1/4)] +
1/4 ArcTan[(x + (1 + x^2)^(1/4))/(1 + x^2)^(1/4)] +
1/4 ArcTanh[(2 x (1 + x^2)^(1/4))/(x^2 + 2 Sqrt[1 + x^2])])
and Mathematica gets the last two integrals in around 8 seconds each:
In[7393]:= Integrate[(3 + x^2)/((1 + x^2) (1 + 6 x^2 + x^4)^(1/4)), x] // Timing
Out[7393]= {8.00801,
ArcTan[(-1 + x)/(1 + 6 x^2 + x^4)^(1/4)] +
ArcTan[(1 + x)/(1 + 6 x^2 + x^4)^(1/4)] +
ArcTanh[(-1 + x)/(1 + 6 x^2 + x^4)^(1/4)] +
ArcTanh[(1 + x)/(1 + 6 x^2 + x^4)^(1/4)]}
In[7394]:= Integrate[(3 - x^2)/((1 - x^2) (1 - 6 x^2 + x^4)^(1/4)), x] // Timing
Out[7394]= {8.30879,
ArcTan[(I + x)/(1 - 6 x^2 + x^4)^(1/4)] -
ArcTan[(1 - 6 x^2 + x^4)^(1/4)/(-I + x)] +
ArcTanh[(I + x)/(1 - 6 x^2 + x^4)^(1/4)] +
ArcTanh[(1 - 6 x^2 + x^4)^(1/4)/(-I + x)]}
the general form of these integrals takes a bit longer and times-out (10 seconds) before IntegrateAlgebraic can compute it:
In[7397]:=
Integrate`IntegrateAlgebraic[((b^2 - 2 a b x + a^2 x^2) (b^2 +
2 a b x + a^2 x^2))/((b^2 + a^2 x^2) (b^4 + 6 a^2 b^2 x^2 +
a^4 x^4)^(3/4)), x] // Timing
Out[7397]= {14.2513,
ArcTan[(b - a x)/(b^4 + 6 a^2 b^2 x^2 + a^4 x^4)^(1/4)]/a +
ArcTan[(b^4 + 6 a^2 b^2 x^2 + a^4 x^4)^(1/4)/(b + a x)]/a -
ArcTanh[(b - a x)/(b^4 + 6 a^2 b^2 x^2 + a^4 x^4)^(1/4)]/a +
ArcTanh[(b^4 + 6 a^2 b^2 x^2 + a^4 x^4)^(1/4)/(b + a x)]/a}
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