Added June 2, 2019.
From example 3.5.11, page 219 nonlinear pde’s by Lokenath Debnath, 3rd edition.
Solve for \(u(x,y)\) \[ u_t+u u_x=x \] with \(u(x,0)=f(x)\)
Mathematica ✗
ClearAll["Global`*"]; pde = D[u[x, t], t] +u[x,t]*D[u[x, t], x] ==x; ic = u[x,0]==f[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde,ic}, u[x, t], {x, t}], 60*10]];
Failed
Maple ✓
restart; pde := diff(u(x,t),t)+u(x,t)*diff(u(x,t),x)=x; ic := u(x,0)=f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,ic],u(x,t))),output='realtime'));
\[u \left ( x,t \right ) ={{\rm e}^{t}}f \left ( \RootOf \left ( \left ( f \left ( {\it \_Z} \right ) +{\it \_Z} \right ) \left ( f \left ( {\it \_Z} \right ) {{\rm e}^{2\,t}}+{{\rm e}^{2\,t}}{\it \_Z}-2\,x{{\rm e}^{t}}-f \left ( {\it \_Z} \right ) +{\it \_Z} \right ) \right ) \right ) +{{\rm e}^{t}}\RootOf \left ( \left ( f \left ( {\it \_Z} \right ) +{\it \_Z} \right ) \left ( f \left ( {\it \_Z} \right ) {{\rm e}^{2\,t}}+{{\rm e}^{2\,t}}{\it \_Z}-2\,x{{\rm e}^{t}}-f \left ( {\it \_Z} \right ) +{\it \_Z} \right ) \right ) -x\]
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