4.15.38 $$\left (x \sqrt {x^2+y(x)^2+1}-y(x) \left (x^2+y(x)^2\right )\right ) y'(x)=\sqrt {x^2+y(x)^2+1} y(x)+x \left (x^2+y(x)^2\right )$$

ODE
$\left (x \sqrt {x^2+y(x)^2+1}-y(x) \left (x^2+y(x)^2\right )\right ) y'(x)=\sqrt {x^2+y(x)^2+1} y(x)+x \left (x^2+y(x)^2\right )$ ODE Classiﬁcation

[[_1st_order, _with_linear_symmetries]]

Book solution method
Exact equation, integrating factor

Mathematica
cpu = 0.588689 (sec), leaf count = 27

$\text {Solve}\left [\sqrt {x^2+y(x)^2+1}+\tan ^{-1}\left (\frac {x}{y(x)}\right )=c_1,y(x)\right ]$

Maple
cpu = 0.234 (sec), leaf count = 27

$\left [\arctan \left (\frac {y \left (x \right )}{x}\right )-\sqrt {1+x^{2}+y \left (x \right )^{2}}-\textit {\_C1} = 0\right ]$ Mathematica raw input

DSolve[(-(y[x]*(x^2 + y[x]^2)) + x*Sqrt[1 + x^2 + y[x]^2])*y'[x] == x*(x^2 + y[x]^2) + y[x]*Sqrt[1 + x^2 + y[x]^2],y[x],x]

Mathematica raw output

Solve[ArcTan[x/y[x]] + Sqrt[1 + x^2 + y[x]^2] == C[1], y[x]]

Maple raw input

dsolve((x*(1+x^2+y(x)^2)^(1/2)-y(x)*(x^2+y(x)^2))*diff(y(x),x) = x*(x^2+y(x)^2)+y(x)*(1+x^2+y(x)^2)^(1/2), y(x))

Maple raw output

[arctan(y(x)/x)-(1+x^2+y(x)^2)^(1/2)-_C1 = 0]