##### 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 = 0.443052 (sec), leaf count = 149

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

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

$\left [\int \frac {1}{\sqrt {x^{4}+x^{2}+1}}d x -\left (\int _{}^{y \left (x \right )}\frac {1}{\sqrt {\textit {\_a}^{4}+\textit {\_a}^{2}+1}}d \textit {\_a} \right )+\textit {\_C1} = 0\right ]$ 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))

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]