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

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 clicliclic@freenet.de Posts: 1,220 Registered: 4/26/08
Re: first anniversary of the IITS
Posted: Mar 26, 2014 4:42 PM

Albert Rich schrieb:
>
> On Tuesday, March 25, 2014 7:03:13 AM UTC-10, clicl...@freenet.de wrote:

> >
> > This must be because you are keeping everybody busy struggling along
> > behind Rubi, trying to catch up ...

>
> Thanks for stroking my ego, but I can't believe the legions of PhDs
> and grad students behind what are advertised as the world's most
> sophisticated computer algebra systems can't keep up with a beekeeper
> in Germany and a beach bum in Hawaii...
>

Now you are hiding your light under a bushel. To cheer you up I took
another look at the hypergeometric evaluations for integrals from
Chapter 8 of Timofeev's book.

- Example 3.n (p. 346) avoiding the 2F1 branch cut:

INT((#e^x-#e^(-x))^n,x)=1/(2*(n+1))*#e^x*(#e^x-#e^(-x))^(n+1)*F2~
1(1,1+n/2,2+n,1-#e^(2*x))

- Example 5b.n (p. 346) avoiding the 2F1 branch cut:

INT((a^(k*x)-a^(l*x))^n,x)=1/((k-l)*(n+1)*LN(a))*a^(-l*x)*(a^(k*~
x)-a^(l*x))^(n+1)*F21(1,1+k*n/(k-l),2+n,1-a^((k-l)*x))

- Example 6a.n (p. 346) avoiding the 2F1 branch cut (F21(-a^(m*x))
cannot be used here, compare example 5a.n for l=0):

INT((1+a^(m*x))^n,x)=1/(m*n*LN(a))*a^(-m*x)*(1+a^(m*x))^(n+1)*F2~
1(1,1,1-n,-a^(-m*x))

- Example 14 (p. 347) alternatives avoiding the 2F1 branch cut in
dependence of SIGN(a/b):

INT((a+b*#e^(n*x))^(r/s),x)=s/(n*r*b)*#e^(-n*x)*(a+b*#e^(n*x))^(~
r/s+1)*F21(1,1,1-r/s,-a/b*#e^(-n*x))=-s/(n*(r+s)*a)*(a+b*#e^(n*x~
))^(r/s+1)*F21(1,1+r/s,2+r/s,1+b/a*#e^(n*x))

- Example 17 (p. 348) avoiding the 2F1 branch cut:

INT((#e^(7*x)-3)^(2/3)/#e^(2*x),x)=1/(35*3^(2/7))*(#e^(7*x)-3)^(~
5/3)*F21(9/7,5/3,8/3,1-1/3*#e^(7*x))

- Example 35 (p. 355) made more compact:

INT(#e^(m*x)/COS(x)^3,x)=8*(#e^((m+3*#i)*x)/(m+3*#i))*F21(3,(3-#~
i*m)/2,(5-#i*m)/2,-#e^(2*#i*x))

- Example 38 (p. 356) made more compact (compare example 36b):

INT(#e^x*(1+SIN(x))/(1-COS(x)),x)=#e^x*SIN(x)/(1-COS(x))+2*INT(#~
e^x/(1-COS(x)),x)=#e^x*SIN(x)/(1-COS(x))-2*(1-#i)*#e^((1+#i)*x)*~
F21(2,1-#i,2-#i,#e^(#i*x))

- Example 40 (p. 356) made more compact (compare example 36a):

INT(#e^x*(1-SIN(x))/(1+COS(x)),x)=-#e^x*SIN(x)/(1+COS(x))+2*INT(~
#e^x/(1+COS(x)),x)=-#e^x*SIN(x)/(1+COS(x))+2*(1-#i)*#e^((1+#i)*x~
)*F21(2,1-#i,2-#i,-#e^(#i*x))

- Example 41 (p. 356) made more compact (compare example 36d):

INT(#e^x*(1-COS(x))/(1-SIN(x)),x)=-#e^x*COS(x)/(1-SIN(x))+2*INT(~
#e^x/(1-SIN(x)),x)=-#e^x*COS(x)/(1-SIN(x))+2*(1+#i)*#e^((1+#i)*x~
)*F21(2,1-#i,2-#i,-#i*#e^(#i*x))

- Example 43 (p. 356) made more compact (compare example 36c):

INT(#e^x*(1+COS(x))/(1+SIN(x)),x)=#e^x*COS(x)/(1+SIN(x))+2*INT(#~
e^x/(1+SIN(x)),x)=#e^x*COS(x)/(1+SIN(x))-2*(1+#i)*#e^((1+#i)*x)*~
F21(2,1-#i,2-#i,#i*#e^(#i*x))

I didn't look at the elementary evaluations in Chapter 8.

Martin.