ODE
\[ y'(x)^3=f(x) \left (a+b y(x)+c y(x)^2\right ) \] ODE Classification
[[_1st_order, `_with_symmetry_[F(x),G(x)*y+H(x)]`]]
Book solution method
No Missing Variables ODE, Solve for \(y'\)
Mathematica ✓
cpu = 0.893304 (sec), leaf count = 353
\[\left \{\left \{y(x)\to \text {InverseFunction}\left [\frac {(2 \text {$\#$1} c+b) \sqrt [3]{\frac {c (\text {$\#$1} (\text {$\#$1} c+b)+a)}{4 a c-b^2}} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {3}{2};\frac {(b+2 c \text {$\#$1})^2}{b^2-4 a c}\right )}{\sqrt [3]{2} c \sqrt [3]{\text {$\#$1} (\text {$\#$1} c+b)+a}}\& \right ]\left [\int _1^x\sqrt [3]{f(K[1])}dK[1]+c_1\right ]\right \},\left \{y(x)\to \text {InverseFunction}\left [\frac {(2 \text {$\#$1} c+b) \sqrt [3]{\frac {c (\text {$\#$1} (\text {$\#$1} c+b)+a)}{4 a c-b^2}} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {3}{2};\frac {(b+2 c \text {$\#$1})^2}{b^2-4 a c}\right )}{\sqrt [3]{2} c \sqrt [3]{\text {$\#$1} (\text {$\#$1} c+b)+a}}\& \right ]\left [\int _1^x-\sqrt [3]{-1} \sqrt [3]{f(K[2])}dK[2]+c_1\right ]\right \},\left \{y(x)\to \text {InverseFunction}\left [\frac {(2 \text {$\#$1} c+b) \sqrt [3]{\frac {c (\text {$\#$1} (\text {$\#$1} c+b)+a)}{4 a c-b^2}} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {3}{2};\frac {(b+2 c \text {$\#$1})^2}{b^2-4 a c}\right )}{\sqrt [3]{2} c \sqrt [3]{\text {$\#$1} (\text {$\#$1} c+b)+a}}\& \right ]\left [\int _1^x(-1)^{2/3} \sqrt [3]{f(K[3])}dK[3]+c_1\right ]\right \}\right \}\]
Maple ✓
cpu = 2.103 (sec), leaf count = 191
\[\left [\int _{}^{y \left (x \right )}\frac {1}{\left (\textit {\_a}^{2} c +b \textit {\_a} +a \right )^{\frac {1}{3}}}d \textit {\_a} +\int _{}^{x}-\frac {\left (\left (a +b y \left (x \right )+c y \left (x \right )^{2}\right ) f \left (\textit {\_a} \right )\right )^{\frac {1}{3}}}{\left (a +b y \left (x \right )+c y \left (x \right )^{2}\right )^{\frac {1}{3}}}d \textit {\_a} +\textit {\_C1} = 0, \int _{}^{y \left (x \right )}\frac {1}{\left (\textit {\_a}^{2} c +b \textit {\_a} +a \right )^{\frac {1}{3}}}d \textit {\_a} +\int _{}^{x}\frac {\left (\left (a +b y \left (x \right )+c y \left (x \right )^{2}\right ) f \left (\textit {\_a} \right )\right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}{2 \left (a +b y \left (x \right )+c y \left (x \right )^{2}\right )^{\frac {1}{3}}}d \textit {\_a} +\textit {\_C1} = 0, \int _{}^{y \left (x \right )}\frac {1}{\left (\textit {\_a}^{2} c +b \textit {\_a} +a \right )^{\frac {1}{3}}}d \textit {\_a} +\int _{}^{x}-\frac {\left (\left (a +b y \left (x \right )+c y \left (x \right )^{2}\right ) f \left (\textit {\_a} \right )\right )^{\frac {1}{3}} \left (-1+i \sqrt {3}\right )}{2 \left (a +b y \left (x \right )+c y \left (x \right )^{2}\right )^{\frac {1}{3}}}d \textit {\_a} +\textit {\_C1} = 0\right ]\] Mathematica raw input
DSolve[y'[x]^3 == f[x]*(a + b*y[x] + c*y[x]^2),y[x],x]
Mathematica raw output
{{y[x] -> InverseFunction[(Hypergeometric2F1[1/3, 1/2, 3/2, (b + 2*c*#1)^2/(b^2 - 4*a*c)]*(b + 2*c*#1)*((c*(a + #1*(b + c*#1)))/(-b^2 + 4*a*c))^(1/3))/(2^(1/3)*c*(a + #1*(b + c*#1))^(1/3)) & ][C[1] + Inactive[Integrate][f[K[1]]^(1/3), {K[1], 1, x}]]}, {y[x] -> InverseFunction[(Hypergeometric2F1[1/3, 1/2, 3/2, (b + 2*c*#1)^2/(b^2 - 4*a*c)]*(b + 2*c*#1)*((c*(a + #1*(b + c*#1)))/(-b^2 + 4*a*c))^(1/3))/(2^(1/3)*c*(a + #1*(b + c*#1))^(1/3)) & ][C[1] + Inactive[Integrate][-((-1)^(1/3)*f[K[2]]^(1/3)), {K[2], 1, x}]]}, {y[x] -> InverseFunction[(Hypergeometric2F1[1/3, 1/2, 3/2, (b + 2*c*#1)^2/(b^2 - 4*a*c)]*(b + 2*c*#1)*((c*(a + #1*(b + c*#1)))/(-b^2 + 4*a*c))^(1/3))/(2^(1/3)*c*(a + #1*(b + c*#1))^(1/3)) & ][C[1] + Inactive[Integrate][(-1)^(2/3)*f[K[3]]^(1/3), {K[3], 1, x}]]}}
Maple raw input
dsolve(diff(y(x),x)^3 = (a+b*y(x)+c*y(x)^2)*f(x), y(x))
Maple raw output
[Intat(1/(_a^2*c+_a*b+a)^(1/3),_a = y(x))+Intat(-((a+b*y(x)+c*y(x)^2)*f(_a))^(1/3)/(a+b*y(x)+c*y(x)^2)^(1/3),_a = x)+_C1 = 0, Intat(1/(_a^2*c+_a*b+a)^(1/3),_a = y(x))+Intat(1/2*((a+b*y(x)+c*y(x)^2)*f(_a))^(1/3)*(1+I*3^(1/2))/(a+b*y(x)+c*y(x)^2)^(1/3),_a = x)+_C1 = 0, Intat(1/(_a^2*c+_a*b+a)^(1/3),_a = y(x))+Intat(-1/2*((a+b*y(x)+c*y(x)^2)*f(_a))^(1/3)*(-1+I*3^(1/2))/(a+b*y(x)+c*y(x)^2)^(1/3),_a = x)+_C1 = 0]