@Bendesarts 

plots[sparsematrixplot](Phi, matrixview, tickmarks=[[seq(i=q||i, i=1..3)], [seq(-i=Eq||i, i=1..3)]], scaling=constrained);

@Earl  I do not have Maple 15, but in the standard interface Maple 16 this is working properly.

@Bendesarts   I think that there is no single simplifying commands for all occasions. Your last expression can easily be simplified in a similar manner, applied to the right side:

-cos(gamma0(t))*cos(beta0(t)) = -(sin(psi[1](t))*cos(theta[1](t))*sin(gamma[1](t))-sin(psi[1](t))*sin(theta[1](t))*cos(gamma[1](t))+cos(theta[1](t))*cos(psi[1](t))*cos(gamma[1](t))+cos(psi[1](t))*sin(theta[1](t))*sin(gamma[1](t)))*cos(beta[1](t));

lhs(%)=(trigsubs(-2*combine(rhs(%)))/2)[];

@Leinad  What you asked about:

k:=4+f*I:

evalc(abs(k)-2=0) assuming f::real;

                                            (f^2+16)^(1/2)-2 = 0

@vv  Thank you, I didn't know before about  trigsubs  command.

@Bendesarts  Automatically you can do so:

S:=indets(eq_liaison);

S1:=indets(select(t->has(t, sin) or has(t, cos), S));

S minus S1 ;

@Bendesarts  Substitution  T  replaces more complex names with  simple names. I do not know why without these changes  the simplification is not fully implemented.

@acer  Thank you for your attention! Of course should be  s^(11/2)   instead.

@nMaple  Here is an animation with front moving part filled with yellow color (the initial yellow region bounded by y=x^2+1, x=0, and y=5 and rotation about x=-1):

restart;

with(plots): with(plottools):

ax := spacecurve([-1, t, 0], t = -1 .. 6, color = red, thickness = 2, linestyle = dash):

s := plot3d([x, y, 0], x = 0 .. 2, y = x^2+1 .. 5, style = surface, color = yellow):

F := q->rotate(q, alpha, [[-1, 0, 0], [-1, 1, 0]]):

s0 := animate(display, [F(s)], alpha = 0 .. 2*Pi, frames = 60):

s1 := animate(plot3d, [[(x+1)*cos(t)-1, x^2+1, -(x+1)*sin(t)], x = 0 .. 2, t = 0 .. alpha, style = surface, color = blue], alpha = 0 .. 2*Pi, frames = 60):

s2 := animate(plot3d, [[(x+1)*cos(t)-1, 5, -(x+1)*sin(t)], x = 0 .. 2, t = 0 .. alpha, style = surface, color = blue], alpha = 0 .. 2*Pi, frames = 60):

s3 := animate(plot3d, [[cos(t)-1, y, -sin(t)], y = 1 .. 5, t = 0 .. alpha, style = surface, color = blue], alpha = 0 .. 2*Pi, frames = 60):

display([s0, ax, seq(s || i, i = 1 .. 3), s], scaling = constrained, lightmodel = light3, axes = normal, orientation = [60, 70], orientation = [70, 75], labels = [x, y, z]);

@nMaple 

restart;

with(plots):  with(plottools):

ax := spacecurve([-1, t, 0], t = -1 .. 6, color = red, thickness = 2, linestyle = dash):

s1 := animate(plot3d, [[(x+1)*cos(t)-1, x^2+1, -(x+1)*sin(t)], x = 0 .. 2, t = 0 .. alpha, style = surface, color = blue], alpha = 0 .. 2*Pi, frames = 100):

s2 := animate(plot3d, [[(x+1)*cos(t)-1, 5, -(x+1)*sin(t)], x = 0 .. 2, t = 0 .. alpha, style = surface, color = blue], alpha = 0 .. 2*Pi, frames = 100):

s3 := animate(plot3d, [[cos(t)-1, y, -sin(t)], y = 1 .. 5, t = 0 .. alpha, style = surface, color = blue], alpha = 0 .. 2*Pi, frames = 100):

display(ax, seq(s||i, i = 1 .. 3), scaling = constrained, lightmodel = light3, axes = normal, orientation = [60, 70], labels = [x, y, z]);

See my answer in this thread

@Markiyan Hirnyk  I didn't see your comment before posting mine. In any case, a visualization is always helpful as it gives confidence that there are no other solutions.

@patient   of course. It is not necessary to write  method=rkf45, since this method is a method by default. 

restart;

Eq:=diff(u1(z),z)=9/4*(84*(-z)^(11/2)-8*z^7+540*(-z)^(5/2)+324*z^4-324*z)*u1(z)/((z^2+3*sqrt(-z))*(2*(-z)^(3/2)+3)*(8*z^6+66*(-z)^(9/2)-189*z^3+216*(-z)^(3/2)+81)):

Sol:=dsolve({Eq, u1(-2)=1}, numeric, output=listprocedure);

u1:=rhs(Sol[2]);

u0(z):=z^2/3+sqrt(-z):

U:=unapply(eval(u0(z)+t*u1(z), z=x/t),x,t);

plot(U(x,1), x=-2..0, color=red, thickness=2); 

@Markiyan Hirnyk  

In your example, a significant difference between the volume and the number of integer points can be explained by the small  radius of the ball  (R=5). If you increase the radius, the relative error of this difference decreases. But IntegerPoints2  procedure is ill-suited for counting the number of points if the number of points is very large (the procedure seeking the points themselves).

This is a simple code, which finds the exact number of integer points inside the 6-dimensional ball of radius 20:

N:=0:

for i from -19 to 19 do

for j from -isqrt(400-i^2) to isqrt(400-i^2) do

for k from -isqrt(400-i^2-j^2) to isqrt(400-i^2-j^2) do

for l from -isqrt(400-i^2-j^2-k^2) to isqrt(400-i^2-j^2-k^2) do

for m from -isqrt(400-i^2-j^2-k^2-l^2) to isqrt(400-i^2-j^2-k^2-l^2) do

for n from -isqrt(400-i^2-j^2-k^2-l^2-m^2) to isqrt(400-i^2-j^2-k^2-l^2-m^2) do

if i^2+j^2+k^2+l^2+m^2+n^2<400 then N:=N+1 fi;

od: od: od: od: od: od:

N;  # The exact number of integer points in 6-D ball of radius 20

V:=(1/6)*Pi^3*20^6:  # The exact volume of 6-D ball of radius 20

evalf((V-N)/V*100);   # The relative error is 0.3 percent

                                                      329715877

                                                     0.3077223011

Another variant of Preben's example:

u:=4^(1/2)+sqrt(x^2)/(1+x);

map(`^`, u, 2);

u^~2;  # Maple just ignores  ~  operator

We see that  map  operator has a wide range of applications than  ~ operator.