4.9.6 \(\sqrt {x^4+x^2+1} y'(x)=\sqrt {y(x)^4+y(x)^2+1}\)

ODE
\[ \sqrt {x^4+x^2+1} y'(x)=\sqrt {y(x)^4+y(x)^2+1} \] ODE Classification

[_separable]

Book solution method
Separable ODE, Neither variable missing

Mathematica
cpu = 0.231231 (sec), leaf count = 149

\[\left \{\left \{y(x)\to \text {InverseFunction}\left [\frac {(-1)^{2/3} \sqrt {\sqrt [3]{-1} \text {$\#$1}^2+1} \sqrt {1-(-1)^{2/3} \text {$\#$1}^2} F\left (i \sinh ^{-1}\left ((-1)^{5/6} \text {$\#$1}\right )|(-1)^{2/3}\right )}{\sqrt {\text {$\#$1}^4+\text {$\#$1}^2+1}}\& \right ]\left [c_1+\frac {(-1)^{2/3} \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} F\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right )|(-1)^{2/3}\right )}{\sqrt {x^4+x^2+1}}\right ]\right \}\right \}\]

Maple
cpu = 0.016 (sec), leaf count = 34

\[ \left \{ \int \!{\frac {1}{\sqrt {{x}^{4}+{x}^{2}+1}}}\,{\rm d}x-\int ^{y \relax (x ) }\!{\frac {1}{\sqrt {{{\it \_a}}^{4}+{{\it \_a}}^{2}+1}}}{d{\it \_a}}+{\it \_C1}=0 \right \} \] Mathematica raw input

DSolve[Sqrt[1 + x^2 + x^4]*y'[x] == Sqrt[1 + y[x]^2 + y[x]^4],y[x],x]

Mathematica raw output

{{y[x] -> InverseFunction[((-1)^(2/3)*EllipticF[I*ArcSinh[(-1)^(5/6)*#1], (-1)^(
2/3)]*Sqrt[1 + (-1)^(1/3)*#1^2]*Sqrt[1 - (-1)^(2/3)*#1^2])/Sqrt[1 + #1^2 + #1^4]
 & ][C[1] + ((-1)^(2/3)*Sqrt[1 + (-1)^(1/3)*x^2]*Sqrt[1 - (-1)^(2/3)*x^2]*Ellipt
icF[I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)])/Sqrt[1 + x^2 + x^4]]}}

Maple raw input

dsolve(diff(y(x),x)*(x^4+x^2+1)^(1/2) = (1+y(x)^2+y(x)^4)^(1/2), y(x),'implicit')

Maple raw output

Int(1/(x^4+x^2+1)^(1/2),x)-Intat(1/(_a^4+_a^2+1)^(1/2),_a = y(x))+_C1 = 0