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Topic: first anniversary of the IITS
Replies: 21   Last Post: Jan 10, 2015 4:01 PM

 Messages: [ Previous | Next ]
 clicliclic@freenet.de Posts: 1,245 Registered: 4/26/08
Re: first anniversary of the IITS
Posted: Mar 2, 2014 5:29 AM

Albert Rich schrieb:
>
> On Monday, February 24, 2014 7:45:35 AM UTC-10, clicl...@freenet.de wrote:
>

> > To celebrate the 1st anniversary of the Independent Integration Test
> > Suite, I am offering streamlined versions of some evaluations given on
> > the Rubi website <http://www.apmaths.uwo.ca/~arich/> for the example
> > integrals from Chapter 5 of Timofeev's book.
> >
> > [...]

>
> I updated the Timofeev integration problems and test results on the
> Rubi website as per your suggestions, with the exception of examples
> 62, 64, 66 and 118.
>
> Happy Anniversary!

I have done some extra work and arrived at the following alternative
evaluations for the Examples 62, 64, 66 (p. 268) and 118 (p. 309) from
Chapter 5:

INT(SQRT(TAN(x)),x)=-1/SQRT(2)*ATANH((1+TAN(x))/(SQRT(2)*SQRT(TA~
N(x))))+1/SQRT(2)*ATAN(1+SQRT(2)*SQRT(TAN(x)))-1/SQRT(2)*ATAN(1-~
SQRT(2)*SQRT(TAN(x)))=-1/SQRT(2)*LN((1+TAN(x)+SQRT(2)*SQRT(TAN(x~
)))/SEC(x))+1/SQRT(2)*ATAN(1+SQRT(2)*SQRT(TAN(x)))-1/SQRT(2)*ATA~
N(1-SQRT(2)*SQRT(TAN(x)))

INT(1/(4+3*TAN(2*x))^(3/2),x)=-3/(25*SQRT(4+3*TAN(2*x)))+13*SQRT~
(2)/500*ACOTH((3+TAN(2*x))/(SQRT(2)*SQRT(4+3*TAN(2*x))))+9*SQRT(~
2)/500*ATAN(3+SQRT(2)*SQRT(4+3*TAN(2*x)))-9*SQRT(2)/500*ATAN(3-S~
QRT(2)*SQRT(4+3*TAN(2*x)))=-3/(25*SQRT(4+3*TAN(2*x)))+13*SQRT(2)~
/500*LN((3+TAN(2*x)+SQRT(2)*SQRT(4+3*TAN(2*x)))/SEC(2*x))+9*SQRT~
(2)/500*ATAN(3+SQRT(2)*SQRT(4+3*TAN(2*x)))-9*SQRT(2)/500*ATAN(3-~
SQRT(2)*SQRT(4+3*TAN(2*x)))

INT(TAN(x)/(SQRT(TAN(x))-1)^2,x)=-x/2+1/(1-SQRT(TAN(x)))+1/2*LN(~
COS(x))+LN(1-SQRT(TAN(x)))+1/SQRT(2)*ATANH((1+TAN(x))/(SQRT(2)*S~
QRT(TAN(x))))-1/SQRT(2)*ATAN(1+SQRT(2)*SQRT(TAN(x)))+1/SQRT(2)*A~
TAN(1-SQRT(2)*SQRT(TAN(x)))=-x/2+1/(1-SQRT(TAN(x)))+1/2*LN(COS(x~
))+LN(1-SQRT(TAN(x)))+1/SQRT(2)*LN((1+TAN(x)+SQRT(2)*SQRT(TAN(x)~
))/SEC(x))-1/SQRT(2)*ATAN(1+SQRT(2)*SQRT(TAN(x)))+1/SQRT(2)*ATAN~
(1-SQRT(2)*SQRT(TAN(x)))

INT(SIN(x)^6*TAN(x)/COS(2*x)^(3/4),x)=(71+13*SIN(x)^2+5*SIN(x)^4~
)*COS(2*x)^(1/4)/45-1/SQRT(2)*ATANH(SQRT(2)*COS(2*x)^(1/4)/(1+SQ~
RT(COS(2*x))))-1/SQRT(2)*ATAN(1+SQRT(2)*COS(2*x)^(1/4))+1/SQRT(2~
)*ATAN(1-SQRT(2)*COS(2*x)^(1/4))=(71+13*SIN(x)^2+5*SIN(x)^4)*COS~
(2*x)^(1/4)/45+1/SQRT(2)*LN((1+SQRT(COS(2*x))-SQRT(2)*COS(2*x)^(~
1/4))/(SQRT(2)*COS(x)))-1/SQRT(2)*ATAN(1+SQRT(2)*COS(2*x)^(1/4))~
+1/SQRT(2)*ATAN(1-SQRT(2)*COS(2*x)^(1/4))

Martin.