ODE
\[ f\left (y(x) y'(x)+x\right )=y(x)^2 \left (y'(x)^2+1\right ) \] ODE Classification
[`x=_G(y,y')`]
Book solution method
Change of variable
Mathematica ✓
cpu = 0.0989407 (sec), leaf count = 42
\[\text {Solve}\left [\left \{\text {K$\$$278007} y(\text {K$\$$278007})+x=f^{(-1)}\left (\left (\text {K$\$$278007}^2+1\right ) y(\text {K$\$$278007})^2\right ),y(x)=\frac {c_1}{\sqrt {\text {K$\$$278007}^2+1}}\right \},\{y(x),\text {K$\$$278007}\}\right ]\]
Maple ✓
cpu = 0.151 (sec), leaf count = 47
\[ \left \{ [x \left ( {\it \_T} \right ) ={1 \left ( {\it RootOf} \left ( {{\it \_C1}}^{2}-f \left ( {\it \_Z} \right ) \right ) \sqrt {{{\it \_T}}^{2}+1}-{\it \_C1}\,{\it \_T} \right ) {\frac {1}{\sqrt {{{\it \_T}}^{2}+1}}}},y \left ( {\it \_T} \right ) ={{\it \_C1}{\frac {1}{\sqrt {{{\it \_T}}^{2}+1}}}}] \right \} \] Mathematica raw input
DSolve[f[x + y[x]*y'[x]] == y[x]^2*(1 + y'[x]^2),y[x],x]
Mathematica raw output
Solve[{x + K$278007*y[K$278007] == InverseFunction[f, 1, 1][(1 + K$278007^2)*y[K$278007]^2], y[x] == C[1]/Sqrt[1 + K$278007^2]}, {y[x], K$278007}]
Maple raw input
dsolve(f(y(x)*diff(y(x),x)+x) = (1+diff(y(x),x)^2)*y(x)^2, y(x),'implicit')
Maple raw output
[x(_T) = (RootOf(_C1^2-f(_Z))*(_T^2+1)^(1/2)-_C1*_T)/(_T^2+1)^(1/2), y(_T) = _C1/(_T^2+1)^(1/2)]