Added December 27, 2018.
Taken from https://en.wikipedia.org/wiki/List_of_nonlinear_partial_differential_equations
Fisher’s equation. Solve for \(u(x,t)\) \[ u_t = u(1-u)+u_{xx} \]
Mathematica ✓
ClearAll["Global`*"]; pde = D[u[x, t], t] == u[x, t]*(1 - u[x, t]) + D[u[x, t], {x, 2}]; sol = AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, t], {x, t}], 60*10]];
\begin {align*} & \left \{u(x,t)\to \frac {1}{4} \left (\tanh \left (\frac {1}{12} \left (-12 c_3+5 t-\sqrt {6} x\right )\right )+1\right ){}^2\right \}\\& \left \{u(x,t)\to -\frac {1}{4} \left (-3+\tanh \left (\frac {1}{12} \left (-12 c_3+5 t-i \sqrt {6} x\right )\right )\right ) \left (1+\tanh \left (\frac {1}{12} \left (-12 c_3+5 t-i \sqrt {6} x\right )\right )\right )\right \}\\& \left \{u(x,t)\to -\frac {1}{4} \left (-3+\tanh \left (\frac {1}{12} \left (-12 c_3+5 t+i \sqrt {6} x\right )\right )\right ) \left (1+\tanh \left (\frac {1}{12} \left (-12 c_3+5 t+i \sqrt {6} x\right )\right )\right )\right \}\\& \left \{u(x,t)\to \frac {1}{4} \left (\tanh \left (\frac {1}{12} \left (-12 c_3+5 t+\sqrt {6} x\right )\right )+1\right ){}^2\right \}\\& \left \{u(x,t)\to \frac {1}{4} \left (\tanh \left (c_3+\frac {5 t}{12}-\frac {x}{2 \sqrt {6}}\right )+1\right ){}^2\right \}\\& \left \{u(x,t)\to -\frac {1}{4} \left (-3+\tanh \left (c_3+\frac {5 t}{12}-\frac {i x}{2 \sqrt {6}}\right )\right ) \left (1+\tanh \left (c_3+\frac {5 t}{12}-\frac {i x}{2 \sqrt {6}}\right )\right )\right \}\\& \left \{u(x,t)\to -\frac {1}{4} \left (-3+\tanh \left (c_3+\frac {5 t}{12}+\frac {i x}{2 \sqrt {6}}\right )\right ) \left (1+\tanh \left (c_3+\frac {5 t}{12}+\frac {i x}{2 \sqrt {6}}\right )\right )\right \}\\& \left \{u(x,t)\to \frac {1}{4} \left (\tanh \left (c_3+\frac {5 t}{12}+\frac {x}{2 \sqrt {6}}\right )+1\right ){}^2\right \}\\ \end {align*}
Maple ✓
restart; pde := diff(u(x,t),t)= u(x,t)*(1-u(x,t))+ diff(u(x,t),x$2); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',PDEtools:-TWSolutions(pde,u(x,t))),output='realtime'));
\begin {align*} & \left \{ u \left ( x,t \right ) =1 \right \} \\& \left \{ u \left ( x,t \right ) =1/4\, \left ( \tanh \left ( -{\frac {5\,t}{12}}+1/12\,\sqrt {6}x+{\it \_C1} \right ) \right ) ^{2}-1/2\,\tanh \left ( -{\frac {5\,t}{12}}+1/12\,\sqrt {6}x+{\it \_C1} \right ) +1/4 \right \} \\& \left \{ u \left ( x,t \right ) =1/4\, \left ( \tanh \left ( {\frac {5\,t}{12}}-1/12\,\sqrt {6}x+{\it \_C1} \right ) \right ) ^{2}+1/2\,\tanh \left ( {\frac {5\,t}{12}}-1/12\,\sqrt {6}x+{\it \_C1} \right ) +1/4 \right \} \\& \left \{ u \left ( x,t \right ) =1/4\, \left ( \tanh \left ( -{\frac {5\,t}{12}}-1/12\,\sqrt {6}x+{\it \_C1} \right ) \right ) ^{2}-1/2\,\tanh \left ( -{\frac {5\,t}{12}}-1/12\,\sqrt {6}x+{\it \_C1} \right ) +1/4 \right \} \\& \left \{ u \left ( x,t \right ) =1/4\, \left ( \tanh \left ( {\frac {5\,t}{12}}+1/12\,\sqrt {6}x+{\it \_C1} \right ) \right ) ^{2}+1/2\,\tanh \left ( {\frac {5\,t}{12}}+1/12\,\sqrt {6}x+{\it \_C1} \right ) +1/4 \right \} \\& \left \{ u \left ( x,t \right ) =-1/4\, \left ( \tanh \left ( -{\frac {5\,t}{12}}+i/12\sqrt {6}x+{\it \_C1} \right ) \right ) ^{2}-1/2\,\tanh \left ( -{\frac {5\,t}{12}}+i/12\sqrt {6}x+{\it \_C1} \right ) +3/4 \right \} \\& \left \{ u \left ( x,t \right ) =-1/4\, \left ( \tanh \left ( {\frac {5\,t}{12}}-i/12\sqrt {6}x+{\it \_C1} \right ) \right ) ^{2}+1/2\,\tanh \left ( {\frac {5\,t}{12}}-i/12\sqrt {6}x+{\it \_C1} \right ) +3/4 \right \} \\& \left \{ u \left ( x,t \right ) =-1/4\, \left ( \tanh \left ( -{\frac {5\,t}{12}}-i/12\sqrt {6}x+{\it \_C1} \right ) \right ) ^{2}-1/2\,\tanh \left ( -{\frac {5\,t}{12}}-i/12\sqrt {6}x+{\it \_C1} \right ) +3/4 \right \} \\& \left \{ u \left ( x,t \right ) =-1/4\, \left ( \tanh \left ( {\frac {5\,t}{12}}+i/12\sqrt {6}x+{\it \_C1} \right ) \right ) ^{2}+1/2\,\tanh \left ( {\frac {5\,t}{12}}+i/12\sqrt {6}x+{\it \_C1} \right ) +3/4 \right \} \\ \end {align*}
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