@gaurav_rs  Maple often does not take into account restrictions on variables imposed by  assume  command.

An example:

assume(x>Pi/2, x<Pi):
solve(sin(x)=1/2);
            Warning, solve may be ignoring assumptions on the input variables.
                              1   
                              - Pi
                              6   
 

@Magma  See help on  ?remember . There everything is explained in detail with examples.

@vv   ``( ... )  construction prevents the automatic opening of the parentheses.

@mehdi jafari   Probably no.

@digerdiga 

f:=ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2)) + ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2)) + ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2)) + ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2))*ln((1-x)^2*(x+1)^2/((-I*x-I+sqrt(-x^2+1))^2*(I*x+I+sqrt(-x^2+1))^2));
subsindets(f, `function`, p->applyop(t->numer(t)/expand(denom(t)),1, p));
simplify(%);
expand(%);

@waseem 

with(plots):
A:=plot([sin(x), cos(x)], x=-Pi..2*Pi, linestyle=[dash,dot], color=[red,blue], thickness=2):
B:=textplot([[3.6,0.8,sin(x)], [-1.8,0.5,cos(x)]], font=[times,bold,
15]):
C:=plot([[t,0.8,t=2.6..3.3],[t,0.5,t=-2.6..-2.2]], color=[red,blue], linestyle=[dash,dot], thickness=2):
display(A, B, C, size=[950,200], scaling=constrained);
 

@waseem   Add  evalf  command:

evalf(1-cos(0.1*Pi)^2);
                                             0.0954915028

I can not confirm it (Maple 2018.2  64 bit on Windows 10 without Physics version 272):

restart;
pde := diff(u(x, t), t) + diff(u(x, t),x) =0;
sol:=pdsolve(pde,u(x,t));
pdetest(sol,pde);
  
bc:=u(0,t)=0;
ic:=u(x,0)=sin(x);
sol:=pdsolve([pde,ic,bc],u(x,t)) assuming x>0;
pdetest(sol,pde);
                                

@bliengme 

Replace the line  
proc GCD3:=proc(n1,n2)  
with the line  
GCD3:=proc(n1,n2)

@Earl  If a fraction makes sense (the denominator is not 0) and this fraction is 0, then this means that its numerator is 0:

a/b-c/d=0;
# It is the same as
numer(lhs(%))=0;
 

@vanzzy  "The new version" above.

This is the new version of my answer that was deleted a few minutes ago.

Try the following:

plotmp_new1.mw

@greenworld   Use  continuous  option for this. In this case, Maple calculates the integral using the Newton – Leibniz formula, without checking the continuity of the integrand for continuity in the interval of integration:

Expr:=sin(x)/((a*b+a^2*sin(x)^2-d^2*cos(x)^2)*(c+cos(x)));
int(Expr, x=0..x, continuous);
 

@digerdiga  You wrote  "Apparently it can be transformed into a polylog(3,I)."

But

evalf([polylog(3,1+I) + polylog(3,1-I), polylog(3,I)]);
                   [1.742317767+0.*I, -.1126928347+.9689461463*I]

@vanzzy  Here are the plotting of 3 points for B=1, k=2  and 3 values of  V=[0.5, 1.5, 2] :

restart;
gm:=V->1/sqrt(1-V^2): T:=w-k*V: S:=w*V-k:
f:=unapply(I*B*gm(V)^3*T^3+gm(V)^2*T^2-I*gm(V)^3*T*S^2-1/3*I*B*gm(V)^3*S^3*T-1/3*gm(V)^2*S^2, w,B,V,k);
F:=(B,V,k)->fsolve(f(w,B,V,k),w, complex);
B:=1: k:=2: V:=[0.5, 1.5, 2]:
plots:-pointplot([seq([Re,Im](F(B,v,k)[1]), v=V)], color=red, symbol=solidcircle, symbolsize=17);