Can you write down exactly what you expected the answer to be? Then I can tell you how to get that answer.

@Ahmadomari My code is just intended to replace the one procedure Chisqr1. Your code outside of that procedure was fine.

@Ahmadomari My code is just intended to replace the one procedure Chisqr1. Your code outside of that procedure was fine.

@Alejandro Jakubi It's like you say. In Maple 17.00/32/Windows 8, I killed it after 90 seconds. In Maple 16.02/64/Windows 8, it finished in 9-10 seconds (obtaining essentially the same answer).

@Alejandro Jakubi It's like you say. In Maple 17.00/32/Windows 8, I killed it after 90 seconds. In Maple 16.02/64/Windows 8, it finished in 9-10 seconds (obtaining essentially the same answer).

seq(evalf(1/(1+50*(x[0]+i*h)^2)), i= 1..5);

seq(evalf(1/(1+50*(x[0]+i*h)^2)), i= 1..5);

@Alejandro Jakubi wrote:

On the other hand the computation of this integral in Maple 17.01 by method elliptic, or equivalently leaving the default, does not finish here (on Linux 32-bit) in any reasonable time. What version have you used?

I got 39 seconds on 17.01 / 64 / Windows 8:

restart:
J:= Int(sin(x)/cos(x)/sqrt(1+sin(x)^3), x= 0..Pi/4):

IntegrationTools:-Change(J, t= sin(x)):
CodeTools:-Usage(value(%)):
memory used=3.72GiB, alloc change=390.01MiB, cpu time=38.88s, real time=37.92s
kernelopts(version);
   Maple 17.01, X86 64 WINDOWS, Jun 25 2013, Build ID 849430

I've done it multiple times without failure.

@Alejandro Jakubi wrote:

On the other hand the computation of this integral in Maple 17.01 by method elliptic, or equivalently leaving the default, does not finish here (on Linux 32-bit) in any reasonable time. What version have you used?

I got 39 seconds on 17.01 / 64 / Windows 8:

restart:
J:= Int(sin(x)/cos(x)/sqrt(1+sin(x)^3), x= 0..Pi/4):

IntegrationTools:-Change(J, t= sin(x)):
CodeTools:-Usage(value(%)):
memory used=3.72GiB, alloc change=390.01MiB, cpu time=38.88s, real time=37.92s
kernelopts(version);
   Maple 17.01, X86 64 WINDOWS, Jun 25 2013, Build ID 849430

I've done it multiple times without failure.

@Markiyan Hirnyk Addressing a minor point from earlier in this thread: Function application does indeed distribute over the list-building operator [] and the set-building operator {}. So, in particular,

{seq}(expr, k= 1..n) = {seq(expr, k= 1..n)},

and there was no problem with Mehdi's syntax on this point.

The undistributed syntax is especially useful when there are multiple functions in the list or set:

[f,g](x,y) = [f(x,y), g(x,y)]

@Markiyan Hirnyk Addressing a minor point from earlier in this thread: Function application does indeed distribute over the list-building operator [] and the set-building operator {}. So, in particular,

{seq}(expr, k= 1..n) = {seq(expr, k= 1..n)},

and there was no problem with Mehdi's syntax on this point.

The undistributed syntax is especially useful when there are multiple functions in the list or set:

[f,g](x,y) = [f(x,y), g(x,y)]

@spradlig asked: Can you please tell us more about who you are and why you are asking?

The Poster is an employee of Maplesoft, as is indicated by the blue background on the name and maple leaf icon in the lower right of the gravatar.

@Mathematix

Can you describe in words what you were intending by subs(exp= 3, Y)? That does not do anything mathematically valid; rather it changes the exponential function into a constant function, the constant being 3. This works:

plot(Y, x= -1..1);

@Mathematix

Can you describe in words what you were intending by subs(exp= 3, Y)? That does not do anything mathematically valid; rather it changes the exponential function into a constant function, the constant being 3. This works:

plot(Y, x= -1..1);

@Mathematix I don't know. Those don't look like difference equations or differential equations. There's no independent variable and no derivatives. Also, it is obvious that the system does not satisfy its own initial conditions.