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Topic: The Charlwood Fifty
Replies: 52   Last Post: Jun 24, 2013 10:24 PM

 Messages: [ Previous | Next ]
 Nasser Abbasi Posts: 6,677 Registered: 2/7/05
Re: The Charlwood Fifty
Posted: Jun 16, 2013 9:46 PM

On 6/16/2013 11:30 AM, clicliclic@freenet.de wrote:
>

>
> Oops, this should have read:
>
> INT(COS(x)^2/SQRT(COS(x)^4 + COS(x)^2 + 1), x) =
> = x/3 + 1/3*ATAN(SIN(x)*COS(x)*(COS(x)^2 + 1)
> /(COS(x)^2*SQRT(COS(x)^4 + COS(x)^2 + 1) + 1))
>

fyi, for the above, which #5, the LeafCount is 45

---------------------------------------

f1= x/3+1/3*ArcTan[Sin[x]*Cos[x]*(Cos[x]^2+1)/(Cos[x]^2*Sqrt[Cos[x]^4+Cos[x]^2+1]+1)];

LeafCount[f1]
-----------------------------------------
45

For the one listed in the table, it is 37:

------------------------------------------

f0 =(-ArcSin[Cos[x]^3]*Sqrt[1-Cos[x]^6]*Csc[x])/(3*Sqrt[1+Cos[x]^2+Cos[x]^4]);

LeafCount[f0]

Out[13]= 37
---------------------------------------

I was wondering: assuming both antiderivates contain
just elemetary functions, can one then use the leaf count as a
measure of which is most optimal answer? Or will there
be other considerations one should look at or better
way to measure which one is more "optimal" than the
other?

reference:
http://reference.wolfram.com/mathematica/ref/LeafCount.html

thanks,
--Nasser