\[ x^2 y^{(3)}(x)+x (y(x)-1) y''(x)+x y'(x)^2+(1-y(x)) y'(x)=0 \] ✓ Mathematica : cpu = 0.166297 (sec), leaf count = 286
\[\left \{\left \{y(x)\to \frac {2 x \left (c_3 \left (J_{\frac {\sqrt {c_2+2}}{\sqrt {2}}}\left (-\frac {1}{2} i x \sqrt {c_1}\right )-\frac {1}{4} i \sqrt {c_1} x \left (J_{\frac {\sqrt {c_2+2}}{\sqrt {2}}-1}\left (-\frac {1}{2} i x \sqrt {c_1}\right )-J_{\frac {\sqrt {c_2+2}}{\sqrt {2}}+1}\left (-\frac {1}{2} i x \sqrt {c_1}\right )\right )\right )+Y_{\frac {\sqrt {c_2+2}}{\sqrt {2}}}\left (-\frac {1}{2} i x \sqrt {c_1}\right )-\frac {1}{4} i \sqrt {c_1} x \left (Y_{\frac {\sqrt {c_2+2}}{\sqrt {2}}-1}\left (-\frac {1}{2} i x \sqrt {c_1}\right )-Y_{\frac {\sqrt {c_2+2}}{\sqrt {2}}+1}\left (-\frac {1}{2} i x \sqrt {c_1}\right )\right )\right )}{c_3 x J_{\frac {\sqrt {c_2+2}}{\sqrt {2}}}\left (-\frac {1}{2} i x \sqrt {c_1}\right )+x Y_{\frac {\sqrt {c_2+2}}{\sqrt {2}}}\left (-\frac {1}{2} i x \sqrt {c_1}\right )}\right \}\right \}\] ✓ Maple : cpu = 0.721 (sec), leaf count = 190
\[\ln \relax (x )+2 \left (\int _{}^{y \relax (x )}\frac {1}{2 \RootOf \left (-2 \sqrt {4+c_{1}}\, \BesselY \left (\frac {\sqrt {4+c_{1}}}{2}, \frac {\sqrt {2}\, \textit {\_Z}}{2}\right ) c_{2}+2 \BesselY \left (\frac {\sqrt {4+c_{1}}}{2}, \frac {\sqrt {2}\, \textit {\_Z}}{2}\right ) c_{2} \textit {\_h} -4 \BesselY \left (\frac {\sqrt {4+c_{1}}}{2}, \frac {\sqrt {2}\, \textit {\_Z}}{2}\right ) c_{2}+2 \BesselY \left (\frac {\sqrt {4+c_{1}}}{2}+1, \frac {\sqrt {2}\, \textit {\_Z}}{2}\right ) \sqrt {2}\, c_{2} \textit {\_Z} +2 \BesselJ \left (\frac {\sqrt {4+c_{1}}}{2}+1, \frac {\sqrt {2}\, \textit {\_Z}}{2}\right ) \sqrt {2}\, \textit {\_Z} -2 \BesselJ \left (\frac {\sqrt {4+c_{1}}}{2}, \frac {\sqrt {2}\, \textit {\_Z}}{2}\right ) \sqrt {4+c_{1}}+2 \BesselJ \left (\frac {\sqrt {4+c_{1}}}{2}, \frac {\sqrt {2}\, \textit {\_Z}}{2}\right ) \textit {\_h} -4 \BesselJ \left (\frac {\sqrt {4+c_{1}}}{2}, \frac {\sqrt {2}\, \textit {\_Z}}{2}\right )\right )^{2}+\textit {\_h}^{2}-c_{1}-4 \textit {\_h}}d \textit {\_h} \right )-c_{3} = 0\]