Probably you want:

with(Fractals:-LSystem); with(LSystemExamples):

PlotExample(DragonCurve, 15);

Lindenmayer System Plot Generator:

states := "FX";

rules := ["Y" = "FX-Y", "X" = "X+YF"];

cons := ["F" = "draw:1", "+" = "turn:-90", "-" = "turn:90"];

newstate1 := Iterate(states, rules, 10);

LSystemPlot(newstate1, cons);

EDITED:

Code from book Geometry of Curves and Surfaces with MAPLE. https://books.google.pl/books/about/Geometry_of_Curves_and_Surfaces_with_MAP.html?id=78w0CseXgvMC&redir_esc=y

dragon := proc (k::algebraic, N::integer) local t, i, q1, q2, q3, q4, d; global p; q2 := [k, 0]; q3 := [1-k, 0]; d := evalm(q3-q2); p[0] := plot([[q2[1], q2[2]], [q3[1], q3[2]]]); for i to N do if `mod`(i, 2) = 0 then t[i] := t[(1/2)*i] else t[i] := (`mod`(i, 4))-2 end if; q4 := evalm(q3+k*d/(1-2*k)); d := evalm(t[i]*[d[2], -d[1]]); q1 := evalm(q3); q2 := evalm(q4+k*d/(1-2*k)); q3 := evalm(q2+d); p[i] := plot([[q1[1], q1[2]], [q2[1], q2[2]], [q3[1], q3[2]]]) end do; return plots:-display([seq(p[i], i = 0 .. N)]) end proc;

dragon(0.1, 500);

f := x^2*y^2+3*x*y^2;

fdiff(f, [x, y], {x = 1, y = 2});

See in Help for more information,exectute code below:

?fdiff 

restart;

de := diff(y(t), t, t) = -1; ic := y(0) = 1, (D(y))(0) = 0;

Events := [y(t), diff(y(t), t) = -.5*(diff(y(t), t))];

Events2 := [y(t) = -2*(1/1000), halt];

dsol := dsolve({de, ic}, numeric, events = [Events, Events2], range = 0 .. 5);

plots[odeplot](dsol, thickness = 3, color = red)

You can do it numerically:

c[1]:=1;
c[2]:=1;
int(12.*x^3*c[2]+6.*x^2*c[1]+x^2*exp(x^3*c[2])*exp(x^2*c[1]), x = 0. .. 1.,numeric);

#6.155281446

If you looking a analyticaly solution it's probably not possible.

See :https://en.wikipedia.org/wiki/Integral#Symbolic

PDE := diff(f(x, y), x, x)+diff(f(x, y), y, y) = 0;
sol := PDEtools:-SimilaritySolutions(PDE);
sol[1];

OR:

pdsolve(PDE, HINT = `+`, build);
pdsolve(PDE, HINT = `*`, build);

I doubt there's a closed form for the integral for general variables.

Only for a=0,1 and k=-1,0,1,2,3. can be found:

By Maple 2018.

sol := x*a*k*((x-g)/b)^(a-1)*(1+((x-g)/b)^a)^(k+1)/b;

`assuming`([[seq([int(eval(sol, a = j), x = g .. t)], j = 0 .. 1)]], [a > 0, b > 0, k in integer, k > 0, g > 0, t > 0]);

#[[0], [k*(-g*k+b^(-k)*(b-g+t)^k*k*t-2*b^(-k-1)*(b-g+t)^k*g*k*t+2*b^(-k-1)*(b-g+t)^k*t^2*k+b^(-k-2)*(b-g+t)^k*g^2*k*t-2*b^(-k-2)*(b-g+t)^k*t^2*k*g+b^(-k-2)*(b-g+t)^k*t^3*k+b-3*g-b^(-k+1)*(b-g+t)^k+3*b^(-k)*(b-g+t)^k*g-3*b^(-k-1)*(b-g+t)^k*g^2+3*b^(-k-1)*(b-g+t)^k*t^2+b^(-k-2)*(b-g+t)^k*g^3-3*b^(-k-2)*(b-g+t)^k*t^2*g+2*b^(-k-2)*(b-g+t)^k*t^3)/(k^2+5*k+6)]]


`assuming`([[seq([int(eval(sol, k = j), x = g .. t)], j = -1 .. 3)]], [a > 0, b > 0, g > 0, t > 0, a > 0]);

#[[-b^(-a)*(t-g)^a*(a*t+g)/(a+1)], [0], [(6*b^(-3*a)*(t-g)^(3*a)*a^3*t+2*b^(-3*a)*(t-g)^(3*a)*a^2*g+9*b^(-3*a)*(t-g)^(3*a)*a^2*t+18*b^(-2*a)*(t-g)^(2*a)*a^3*t+3*b^(-3*a)*(t-g)^(3*a)*a*g+3*b^(-3*a)*(t-g)^(3*a)*a*t+9*b^(-2*a)*(t-g)^(2*a)*a^2*g+24*b^(-2*a)*(t-g)^(2*a)*a^2*t+18*b^(-a)*(t-g)^a*a^3*t+b^(-3*a)*(t-g)^(3*a)*g+12*b^(-2*a)*(t-g)^(2*a)*a*g+6*b^(-2*a)*(t-g)^(2*a)*a*t+18*b^(-a)*(t-g)^a*a^2*g+15*b^(-a)*(t-g)^a*a^2*t+3*g*b^(-2*a)*(t-g)^(2*a)+15*b^(-a)*(t-g)^a*a*g+3*b^(-a)*(t-g)^a*a*t+3*g*b^(-a)*(t-g)^a)/(3*(6*a^3+11*a^2+6*a+1))], [(168*b^(-3*a)*(t-g)^(3*a)*a^3*t+56*b^(-3*a)*(t-g)^(3*a)*a^2*g+84*b^(-3*a)*(t-g)^(3*a)*a^2*t+228*b^(-2*a)*(t-g)^(2*a)*a^3*t+28*b^(-3*a)*(t-g)^(3*a)*a*g+12*b^(-3*a)*(t-g)^(3*a)*a*t+114*b^(-2*a)*(t-g)^(2*a)*a^2*g+96*b^(-2*a)*(t-g)^(2*a)*a^2*t+104*b^(-a)*(t-g)^a*a^3*t+48*b^(-2*a)*(t-g)^(2*a)*a*g+12*b^(-2*a)*(t-g)^(2*a)*a*t+104*b^(-a)*(t-g)^a*a^2*g+36*b^(-a)*(t-g)^a*a^2*t+36*b^(-a)*(t-g)^a*a*g+4*b^(-a)*(t-g)^a*a*t+4*b^(-3*a)*(t-g)^(3*a)*g+6*g*b^(-2*a)*(t-g)^(2*a)+4*g*b^(-a)*(t-g)^a+11*b^(-4*a)*(t-g)^(4*a)*a^2*g+24*b^(-4*a)*(t-g)^(4*a)*a^2*t+6*b^(-4*a)*(t-g)^(4*a)*a*g+4*b^(-4*a)*(t-g)^(4*a)*a*t+96*b^(-a)*(t-g)^a*a^3*g+24*b^(-4*a)*(t-g)^(4*a)*a^4*t+6*b^(-4*a)*(t-g)^(4*a)*a^3*g+44*b^(-4*a)*(t-g)^(4*a)*a^3*t+96*b^(-3*a)*(t-g)^(3*a)*a^4*t+32*b^(-3*a)*(t-g)^(3*a)*a^3*g+144*b^(-2*a)*(t-g)^(2*a)*a^4*t+72*b^(-2*a)*(t-g)^(2*a)*a^3*g+96*b^(-a)*(t-g)^a*a^4*t+b^(-4*a)*(t-g)^(4*a)*g)/(2*(24*a^4+50*a^3+35*a^2+10*a+1))], [(3*(b^(-5*a)*(t-g)^(5*a)*g+600*b^(-a)*(t-g)^a*a^4*g+600*b^(-a)*(t-g)^a*a^5*t+120*b^(-5*a)*(t-g)^(5*a)*a^5*t+1470*b^(-3*a)*(t-g)^(3*a)*a^3*t+490*b^(-3*a)*(t-g)^(3*a)*a^2*g+360*b^(-3*a)*(t-g)^(3*a)*a^2*t+1180*b^(-2*a)*(t-g)^(2*a)*a^3*t+120*b^(-3*a)*(t-g)^(3*a)*a*g+30*b^(-3*a)*(t-g)^(3*a)*a*t+590*b^(-2*a)*(t-g)^(2*a)*a^2*g+260*b^(-2*a)*(t-g)^(2*a)*a^2*t+355*b^(-a)*(t-g)^a*a^3*t+130*b^(-2*a)*(t-g)^(2*a)*a*g+20*b^(-2*a)*(t-g)^(2*a)*a*t+355*b^(-a)*(t-g)^a*a^2*g+70*b^(-a)*(t-g)^a*a^2*t+70*b^(-a)*(t-g)^a*a*g+5*b^(-a)*(t-g)^a*a*t+10*b^(-3*a)*(t-g)^(3*a)*g+10*g*b^(-2*a)*(t-g)^(2*a)+5*g*b^(-a)*(t-g)^a+205*b^(-4*a)*(t-g)^(4*a)*a^2*g+220*b^(-4*a)*(t-g)^(4*a)*a^2*t+55*b^(-4*a)*(t-g)^(4*a)*a*g+20*b^(-4*a)*(t-g)^(4*a)*a*t+770*b^(-a)*(t-g)^a*a^3*g+1220*b^(-4*a)*(t-g)^(4*a)*a^4*t+305*b^(-4*a)*(t-g)^(4*a)*a^3*g+820*b^(-4*a)*(t-g)^(4*a)*a^3*t+2340*b^(-3*a)*(t-g)^(3*a)*a^4*t+780*b^(-3*a)*(t-g)^(3*a)*a^3*g+2140*b^(-2*a)*(t-g)^(2*a)*a^4*t+1070*b^(-2*a)*(t-g)^(2*a)*a^3*g+770*b^(-a)*(t-g)^a*a^4*t+5*b^(-4*a)*(t-g)^(4*a)*g+24*b^(-5*a)*(t-g)^(5*a)*a^4*g+250*b^(-5*a)*(t-g)^(5*a)*a^4*t+50*b^(-5*a)*(t-g)^(5*a)*a^3*g+175*b^(-5*a)*(t-g)^(5*a)*a^3*t+35*b^(-5*a)*(t-g)^(5*a)*a^2*g+50*b^(-5*a)*(t-g)^(5*a)*a^2*t+10*b^(-5*a)*(t-g)^(5*a)*a*g+5*b^(-5*a)*(t-g)^(5*a)*a*t+600*b^(-4*a)*(t-g)^(4*a)*a^5*t+150*b^(-4*a)*(t-g)^(4*a)*a^4*g+1200*b^(-3*a)*(t-g)^(3*a)*a^5*t+400*b^(-3*a)*(t-g)^(3*a)*a^4*g+1200*b^(-2*a)*(t-g)^(2*a)*a^5*t+600*b^(-2*a)*(t-g)^(2*a)*a^4*g))/(5*(120*a^5+274*a^4+225*a^3+85*a^2+15*a+1))]]

Mathematica 11.3 can find solution for: - infinity > a >= 2 and -1<= k < infinity

Download The_Hard_Spring_by_Maple_2018.mw

sol := `assuming`([dsolve(Eq2 = 0)], [phi(xi) > 0]);
with(DEtools);
odeadvisor(op([2, 2], sol)[1, 1]);

dsolve can't solve symbolical Abel's equation. Try numerically.

 I invert that laplace transform numerically,but answer is different than :(section 4.1 page 13629, data used for FIG. 3 )

Download FracPDE_vers_2.mw

You can restrict T  using assumptions.

plot([Re(sqrt(sin(x))), Im(sqrt(sin(x)))], x = 0 .. 4*Pi, legend = [typeset("Curve: ", Re(sqrt(sin(x)))), typeset("Curve: ", Im(sqrt(sin(x))))]);

for: 0 < T < Pi

simplify(int(sqrt(sin(x)), x = 0 .. T)) assuming 0 < T < Pi

#-sqrt(2)*(EllipticF(sqrt(2)*sqrt(sin(T))/sqrt(sin(T)+1), (1/2)*sqrt(2))-EllipticPi(sqrt(2)*sqrt(sin(T))/sqrt(sin(T)+1), 1/2, (1/2)*sqrt(2)))

and for: Pi < T < 2*Pi

simplify(int(sqrt(sin(x)), x = 0 .. T)) assuming Pi < T < 2*Pi

#sqrt(2)*(I*EllipticF(2*sqrt(sin((1/2)*T))*sqrt(-cos((1/2)*T))/(cos((1/2)*T)-sin((1/2)*T)), (1/2)*sqrt(2))-I*EllipticPi(2*sqrt(sin((1/2)*T))*sqrt(-cos((1/2)*T))/(cos((1/2)*T)-sin((1/2)*T)), 1/2, (1/2)*sqrt(2))+4*EllipticE((1/2)*sqrt(2))-2*EllipticK((1/2)*sqrt(2)))

EDITED:

Generally my method works if Maple does not have errors(Bugs).

Workaround:

f := proc (T) options operator, arrow; int(sqrt(sin(x)), x = 0 .. T) end proc;
[f((1/4)*Pi), f((1/2)*Pi), f(3*Pi*(1/4)), f(Pi)];

#[-sqrt(2)*EllipticF(sqrt(2)*sqrt(sqrt(2)/(2+sqrt(2))), (1/2)*sqrt(2))+sqrt(2)*EllipticPi(sqrt(2)*sqrt(sqrt(2)/(2+sqrt(2))), 1/2, (1/2)*sqrt(2)), -sqrt(2)*EllipticK((1/2)*sqrt(2))+2*sqrt(2)*EllipticE((1/2)*sqrt(2)), -sqrt(2)*EllipticK((1/2)*sqrt(2))+2*sqrt(2)*EllipticE((1/2)*sqrt(2))-sqrt(2)*EllipticF(cos(3*Pi*(1/8))*sqrt(2), (1/2)*sqrt(2))+2*sqrt(2)*EllipticE(cos(3*Pi*(1/8))*sqrt(2), (1/2)*sqrt(2)), -2*sqrt(2)*EllipticK((1/2)*sqrt(2))+4*sqrt(2)*EllipticE((1/2)*sqrt(2))]

Regards,MI

Well Maple is very weak for solving symbolic sum and can't solve your example.

`assuming`([sum((d*k+a)^r, k = 1 .. n)], [n::posint, r::posint, r > 0, a > 0, d > 0]);

# returns unevaluated

Mathematica 11.2 solution:

Translated code form MMA to Maple:

Sum((d*k+a)^r, k = 1 .. n) = d^r*(Zeta(0, -r, (a+d)/d)-Zeta(0, -r, (d*n+a+d)/d));

Regards,MI

sol:=identify(evalf(2*ln(3)-3*ln(2)));

# arcsinh(17/144)

simplify(convert(sol, arccoth), symbolic);

# 2*arctanh(1/17)

Only for inverse hyperbolic function.

with(gfun);
l := [8, 32, 128, 512, 2048];
rec := listtorec(l, u(n), [ogf]);

# rec := [{u(1+n)-4*u(n), u(0) = 8}, ogf]

You must only change to:  u(1) = 8

sol:=rsolve({u(1+n)-4*u(n), u(1) = 8}, u(n));

#Answer is: 2*4^n
#or:
sol:=rsolve({eval(op([1, 2], rec), 0 = 1), op([1, 1], rec) = 0}, u(n))

[seq(sol, n = 1 .. 6)]; #Check:

#[8, 32, 128, 512, 2048, 8192]

I don't know how to change: 2*4^n = 2*2^(2 n) = 2^(2n+1) in Maple!

simplify(2*4^n, power);# Dosen't work!

For first example:

restart: 
Digits := 30:
convert((k-2)*(k^2+5)*(k^3-k^2+7*k+8)/(6*k*(k^2-3*k+8)), parfrac, k);

and factor expression k^2-3k+8  by:

(1/6)*k^3+k-1/2-5/(3*k)+convert((-3*k-6)/identify(factor(denom(op(4, sol)), complex)), parfrac, k);

 and last 2 expression:

identify(factor(k^3+3*k^2+11*k-3, complex));
identify(factor(k^2+2*k+9, complex))

f := proc (x, z) options operator, arrow; ln(x^z) end proc;
simplify(f(x, 2), symbolic);
collect(expand(simplify(f(x, 1/2+I*y), symbolic)), ln(x));

Note: When the symbolic option is specified, any branch of a multi-valued function can be chosen during the simplification process. The result of such an operation is in general not valid over the whole complex plane and can lead to incorrect results if you assume the expressions represent analytical functions.