Artifact 4c7b3ccc6e8a8ab1b43060b1ddcb9def1cf455da90dbb7b326fcccd4d0b8d412:
- File
r30/int.tst
— part of check-in
[eb17ceb7f6]
at
2020-04-21 19:40:01
on branch master
— Add Reduce 3.0 to the historical section of the archive, and some more
files relating to version sof PSL from the early 1980s. Thanks are due to
Paul McJones and Nelson Beebe for these, as well as to all the original
authors.git-svn-id: https://svn.code.sf.net/p/reduce-algebra/code/historical@5328 2bfe0521-f11c-4a00-b80e-6202646ff360 (user: arthurcnorman@users.sourceforge.net, size: 10109) [annotate] [blame] [check-ins using] [more...]
COMMENT THE REDUCE INTEGRATION TEST PACKAGE Edited By Anthony C. Hearn The Rand Corporation This file is designed to provide a set of representative tests of the Reduce integration package. Not all examples go through, even when an integral exists, since some of the arguments are outside the domain of applicability of the current package. However, future improvements to the package will result in more closed-form evaluations in later releases. We would appreciate any additional contributions to this test file either because they illustrate some feature (good or bad) of the current package, or suggest domains which future versions should handle. Any suggestions for improved organization of this test file (e.g., in a way which corresponds more directly to the organization of a standard integration table book such as Gradshteyn and Ryznik) are welcome. Acknowledgments: The examples in this file have been contributed by the following. Any omissions to this list should be reported to the Editor. David M. Dahm John P. Fitch Steven Harrington Anthony C. Hearn K. Siegfried Koelbig Ernst Krupnikov Arthur C. Norman Herbert Stoyan ; Comment we first set up a suitable testing function; SYMBOLIC OPERATOR TIME; PROCEDURE TESTINT(A,B); BEGIN SCALAR DIFFCE,RES,TT; TT:=TIME(); RES:=INT(A,B); WRITE "Time for Integral: ",TIME()-TT," ms"; DIFFCE := DF(RES,B)-A; IF DIFFCE NEQ 0 THEN BEGIN FOR ALL X LET TAN X=SIN(2*X)/(1+COS(2*X)), SIN X**2=1-COS X**2, TANH X= (E**(X)-E**(-X))/(E**X+E**(-X)); DIFFCE := DIFFCE; FOR ALL X CLEAR TAN X,SIN X**2,TANH X END; %hopefully, difference appeared non-zero due to absence of %above transformations; IF DIFFCE NEQ 0 THEN WRITE "DERIVATIVE OF INTEGRAL NOT EQUAL TO INTEGRAND"; RETURN RES END; % REFERENCES ARE TO GRADSHTEYN & RYZHIK; testint(1/x,x); % 2.01 #2; testint((x+1)**3/(x-1)**4,x); testint(log x,x); testint(x*log x,x); testint(x**2*log x,x); testint(x**p*log x,x); testint((log x)**2,x); testint(x**9*log x**11,x); testint(log x**2/x,x); testint(1/log x,x); testint(1/(x*log x),x); testint(sin log x,x); testint(cos log x,x); testint((log x)**p/x,x); testint(log x *(a*x+b),x); testint((a*x+b)**2*log x,x); testint(log x/(a*x+b)**2,x); testint(log x/sqrt(a*x+b),x); testint(x*log (a*x+b),x); testint(x**2*log(a*x+b),x); testint(log(x**2+a**2),x); testint(x*log(x**2+a**2),x); testint(x**2*log(x**2+a**2),x); testint(x**4*log(x**2+a**2),x); testint(log(x**2-a**2),x); testint(log(log(log(log(x)))),x); testint(sin x,x); % 2.01 #5; testint(cos x,x); % #6; testint(tan x,x); % #11; testint(1/tan(x),x); % 2.01 #12; testint(1/cos x,x); testint(1/sin x,x); testint(sin x**2,x); testint(x**3*sin(x**2),x); testint(sin x**3,x); testint(sin x**p,x); testint((sin x**2+1)**2*cos x,x); testint(cos x**2,x); testint(cos x**3,x); testint(sin(a*x+b),x); testint(1/cos x**2,x); testint(1/(1+cos x),x); testint(1/(1-cos x),x); testint(sqrt(1-cos x),x); testint(sin x* sin (2*x),x); testint(x*sin x,x); testint(x**2*sin x,x); testint(x*sin x**2,x); testint(x**2*sin x**2,x); testint(x*sin x**3,x); testint(x*cos x,x); testint(x**2*cos x,x); testint(x*cos x**2,x); testint(x**2*cos x**2,x); testint(x*cos x**3,x); testint(sin x/x,x); testint(cos x/x,x); testint(sin x/x**2,x); testint(sin x**2/x,x); testint(tan x**3,x); testint(e**x,x); % 2.01 #3; testint(a**x,x); % 2.01 #4; testint(e**(a*x),x); testint(e**(a*x)/x,x); testint(1/(a+b*e**(m*x)),x); testint(e**(2*x)/(1+e**x),x); testint(1/(a*e**(m*x)+b*e**(-m*x)),x); testint(x*e**(a*x),x); testint(x**20*e**x,x); testint(a**x/b**x,x); testint(a**x*b**x,x); testint(a**x/x**2,x); testint(x*a**x/(1+b*x)**2,x); testint(x*e**(a*x)/(1+a*x)**2,x); testint(x*k**(x**2),x); testint(e**(x**2),x); testint(x*e**(x**2),x); testint((2*x**3+x)*(e**(x**2))**2*e**(1-x*e**(x**2))/(1-x*e**(x**2))**2, x); testint(e**(e**(e**(e**x))),x); testint(e**x*log x,x); testint(x*e**x*log x,x); testint(e**(2*x)*log(e**x),x); z:=a+b*x; testint(z**p,x); testint(x*z**p,x); testint(x**2*z**p,x); testint(1/z,x); testint(1/z**2,x); testint(x/z,x); testint(x**2/z,x); testint(1/(x*z),x); testint(1/(x**2*z),x); testint(1/(x*z)**2,x); testint(1/(c**2+x**2),x); testint(1/(c**2-x**2),x); u:=sqrt(a+b*x); v:=sqrt(c+d*x); testint(u*v,x); testint(u,x); testint(x*u,x); testint(x**2*u,x); testint(u/x,x); testint(u/x**2,x); testint(1/u,x); testint(x/u,x); testint(x**2/u,x); testint(1/(x*u),x); testint(1/(x**2*u),x); testint(u**p,x); testint(x*u**p,x); testint(sin z,x); testint(cos z,x); testint(tan z,x); testint(1/tan z,x); testint(1/sin z,x); testint(1/cos z,x); testint(sin z**2,x); testint(sin z**3,x); testint(cos z**2,x); testint(cos z**3,x); testint(1/cos z**2,x); testint(1/(1+sin x),x); testint(1/(1-sin x),x); testint(x**2*sin z**2,x); testint(cos x*cos(2*x),x); testint(x**2*cos z**2,x); testint(1/tan x**3,x); testint(x**3*tan(x)**4,x); testint(x*tan(x)**2,x); testint(sin(2*x)*cos(3*x),x); testint(sin x**2*cos x**2,x); testint(1/(sin x**2*cos x**2),x); testint(d**x*sin x,x); testint(x*d**x*sin x,x); testint(x**2*d**x*sin x,x); testint(d**x*cos x,x); testint(x*d**x*cos x,x); testint(x**2*d**x*cos x,x); testint(x**3*d**x*sin x,x); testint(x**3*d**x*cos x,x); testint(sin x*sin(2*x)*sin(3*x),x); testint(cos x*cos(2*x)*cos(3*x),x); testint(x*cos(xi/sin(x))*cos(x)/sin(x)**2,x); Comment this integral has given trouble at various times; testint(atan((-sqrt(2)+2*x)/sqrt(2)),x); Comment many of these integrals used to require Steve Harrington's code to evaluate. They originated in Novosibirsk as examples of using Analytik. There are still a few examples which could be evaluated using better heuristics; testint(a*sin(3*x+5)**2*cos(3*x+5),x); testint(log(x**2)/x**3,x); testint(x*sin(x+a),x); testint((log(x)*(1-x)-1)/(e**x*log(x)**2),x); testint(x**3*(a*x**2+b)**(-1),x); testint(x**(1/2)*(x+1)**(-7/2),x); testint(x**(-1)*(x+1)**(-1),x); testint(x**(-1/2)*(2*x-1)**(-1),x); testint((x**2+1)*x**(1/2),x); testint(x**(-1)*(x-a)**(1/3),x); testint(x*sinh(x),x); testint(x*cosh(x),x); testint(x**2*(2*x**2+x)**2,x); testint(x*(x**2+2*x+1),x); testint(sinh(2*x)/cosh(2*x),x); testint(sin(2*x+3)*cos(x)**2,x); testint(x*atan(x),x); testint(x*acot(x),x); testint(x*log(x**2+a),x); testint(sin(x+a)*cos(x),x); testint(cos(x+a)*sin(x),x); testint((2+2*sin(x))**(1/2),x); testint((2-2*sin(x))**(1/2),x); testint((2+2*cos(x))**(1/2),x); testint((2-2*cos(x))**(1/2),x); testint(1/(x**(1/2)-(x-1)**(1/2)),x); testint(1/(1-(x+1)**(1/2)),x); testint(x/(x**4+36)**(1/2),x); int(1/(x**(1/3)+x**(1/2)),x); testint(log(2+3*x**2),x); testint(cot(x),x); int(cot x**4,x); testint(tanh(x),x); testint(coth(x),x); testint(b**x,x); testint((x**4+x**(-4)+2)**(1/2),x); testint((2*x+1)/(3*x+2),x); testint(x*log(x+(x**2+1)**(1/2)),x); testint(x*(e**x*sin(x)+1)**2,x); testint(x*e**x*cos(x),x); Comment the following set came from Herbert Stoyan who used to be in Dresden; testint(1/(x-3)**4,x); testint(x/(x**3-1),x); testint(x/(x**4-1),x); testint(log(x)*(x**3+1)/(x**4+2),x); testint(log(x)+log(x+1)+log(x+2),x); testint(1/(x**3+5),x); testint(sqrt(x**2+3),x); testint(x/(x+1)**2,x); COMMENT The following integrals were contributed by David M. Dahm. He also developed the code to make most of them integrable; testint(1/(2*x**3-1),x); testint(1/(x**3-2),x); testint(1/(a*x**3-b),x); testint(1/(x**4-2),x); testint(1/(5*x**4-1),x); testint(1/(3*x**4+7),x); testint(1/(x**4+3*x**2-1),x); testint(1/(x**4-3*x**2-1),x); testint(1/(x**4-3*x**2+1),x); testint(1/(x**4-4*x**2+1),x); testint(1/(x**4+4*x**2+1),x); testint(1/(x**4+x**2+2),x); testint(1/(x**4-x**2+2),x); testint(1/(x**6-2),x); testint(1/(x**6+2),x); testint(1/(x**8+1),x); testint(1/(x**8-x**4+1),x); COMMENT The following integrals were used among others as a test of Moses' SIN program; testint(asin x,x); testint(x**2*asin x,x); testint(sec x**2/(1+sec x**2-3*tan x),x); testint(1/sec x**2,x); testint((5*x**2-3*x-2)/(x**2*(x-2)),x); testint(1/(4*x**2+9)**(1/2),x); testint((x**2+4)**(-1/2),x); testint(1/(9*x**2-12*x+10),x); testint(1/(x**8-2*x**7+2*x**6-2*x**5+x**4),x); testint((a*x**3+b*x**2+c*x+d)/((x+1)*x*(x-3)),x); testint(1/(2-log(x**2+1))**5,x); testint((2*x**3+x)*e**(x**2)**2*e**(1-x*e**(x**2))/(1-x*e**(x**2))**2 ,x); testint(2*x*e**(x**2)*log(x)+e**(x**2)/x+(log(x)-2)/(log(x)**2+x)**2+ ((2/x)*log(x)+(1/x)+1)/(log(x)**2+x),x); Comment here is an example of using the integrator with pattern matching; for all m,n let int(k1**m*log(k1)**n/(p**2-k1**2),k1)=foo(m,n), int(k1*log(k1)**n/(p**2-k1**2),k1)=foo(1,n), int(k1**m*log(k1)/(p**2-k1**2),k1)=foo(m,1), int(k1*log(k1)/(p**2-k1**2),k1)=foo(1,1), int(log(k1)**n/(k1*(p**2-k1**2)),k1)=foo(-1,n); int(k1**2*log(k1)/(p**2-k1**2),k1); COMMENT It is interesting to see how much of this one can be done; let f1s= (12*log(s/mc**2)*s**2*pi**2*mc**3*(-8*s-12*mc**2+3*mc) + pi**2*(12*s**4*mc+3*s**4+176*s**3*mc**3-24*s**3*mc**2 -144*s**2*mc**5-48*s*mc**7+24*s*mc**6+4*mc**9-3*mc**8)) /(384*e**(s/y)*s**2); int(f1s,s); factor int; ws; Comment Some definite integrals; algebraic procedure dint(f,x,x1,x2); begin scalar y; y := int(f,x); return sub(x=x2,y) - sub(x=x1,y) end; dint(sin x,x,0,pi/2); dint(x/(x+2),x,2,6); dint(log(x),x,1,5); dint((1+x**2/p**2)**(1/2),x,0,p); dint(x**9+y+y**x+x,x,0,2); Comment the following integrals reveal deficiencies in the current integrator; %this one seems to run forever; %testint(x**7/(x**12+1),x); %high degree denominator; %testint(1/(2-log(x**2+1))**5,x); %the next two integrals should return a closed-form solution; testint(1/(a+b*sin x),x); testint(1/(a+b*sin x+cos x),x); %this example should evaluate; testint(sin(2*x)/cos(x),x); %this example, which appeared in Tobey's thesis, needs factorization %over algebraic fields. It currently gives an ugly answer; int((7*x**13+10*x**8+4*x**7-7*x**6-4*x**3-4*x**2+3*x+3)/ (x**14-2*x**8-2*x**7-2*x**4-4*x**3-x**2+2*x+1),x); end;