\(\int \sqrt [3]{b x+c x^2} \, dx\) [31]

Optimal result

Integrand size = 13, antiderivative size = 387 \[ \int \sqrt [3]{b x+c x^2} \, dx=\frac {3 \sqrt [3]{-\frac {c x (b+c x)}{b^2}} (b+2 c x) \sqrt [3]{b x+c x^2}}{10 c \sqrt [3]{-\frac {c \left (b x+c x^2\right )}{b^2}}}+\frac {3^{3/4} \sqrt {2-\sqrt {3}} b^2 \sqrt [3]{b x+c x^2} \left (1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right ) \sqrt {\frac {1+2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}+2 \sqrt [3]{2} \left (-\frac {c x (b+c x)}{b^2}\right )^{2/3}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {1+\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}\right ),-7+4 \sqrt {3}\right )}{5\ 2^{2/3} c (b+2 c x) \sqrt [3]{-\frac {c \left (b x+c x^2\right )}{b^2}} \sqrt {-\frac {1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right )^2}}} \]

[Out]

3/10*(-c*x*(c*x+b)/b^2)^(1/3)*(2*c*x+b)*(c*x^2+b*x)^(1/3)/c/(-c*(c*x^2+b*x)/b^2)^(1/3)+1/10*3^(3/4)*b^2*(c*x^2
+b*x)^(1/3)*(1-2^(2/3)*(-c*x*(c*x+b)/b^2)^(1/3))*EllipticF((1-2^(2/3)*(-c*x*(c*x+b)/b^2)^(1/3)+3^(1/2))/(1-2^(
2/3)*(-c*x*(c*x+b)/b^2)^(1/3)-3^(1/2)),2*I-I*3^(1/2))*(1/2*6^(1/2)-1/2*2^(1/2))*((1+2^(2/3)*(-c*x*(c*x+b)/b^2)
^(1/3)+2*2^(1/3)*(-c*x*(c*x+b)/b^2)^(2/3))/(1-2^(2/3)*(-c*x*(c*x+b)/b^2)^(1/3)-3^(1/2))^2)^(1/2)*2^(1/3)/c/(2*
c*x+b)/(-c*(c*x^2+b*x)/b^2)^(1/3)/((-1+2^(2/3)*(-c*x*(c*x+b)/b^2)^(1/3))/(1-2^(2/3)*(-c*x*(c*x+b)/b^2)^(1/3)-3
^(1/2))^2)^(1/2)

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 387, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {636, 633, 201, 242, 225} \[ \int \sqrt [3]{b x+c x^2} \, dx=\frac {3^{3/4} \sqrt {2-\sqrt {3}} b^2 \sqrt [3]{b x+c x^2} \left (1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right ) \sqrt {\frac {2 \sqrt [3]{2} \left (-\frac {c x (b+c x)}{b^2}\right )^{2/3}+2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}+1}{\left (-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}-\sqrt {3}+1\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}+\sqrt {3}+1}{-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}-\sqrt {3}+1}\right ),-7+4 \sqrt {3}\right )}{5\ 2^{2/3} c (b+2 c x) \sqrt [3]{-\frac {c \left (b x+c x^2\right )}{b^2}} \sqrt {-\frac {1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{\left (-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}-\sqrt {3}+1\right )^2}}}+\frac {3 \sqrt [3]{-\frac {c x (b+c x)}{b^2}} (b+2 c x) \sqrt [3]{b x+c x^2}}{10 c \sqrt [3]{-\frac {c \left (b x+c x^2\right )}{b^2}}} \]

[In]

Int[(b*x + c*x^2)^(1/3),x]

[Out]

(3*(-((c*x*(b + c*x))/b^2))^(1/3)*(b + 2*c*x)*(b*x + c*x^2)^(1/3))/(10*c*(-((c*(b*x + c*x^2))/b^2))^(1/3)) + (
3^(3/4)*Sqrt[2 - Sqrt[3]]*b^2*(b*x + c*x^2)^(1/3)*(1 - 2^(2/3)*(-((c*x*(b + c*x))/b^2))^(1/3))*Sqrt[(1 + 2^(2/
3)*(-((c*x*(b + c*x))/b^2))^(1/3) + 2*2^(1/3)*(-((c*x*(b + c*x))/b^2))^(2/3))/(1 - Sqrt[3] - 2^(2/3)*(-((c*x*(
b + c*x))/b^2))^(1/3))^2]*EllipticF[ArcSin[(1 + Sqrt[3] - 2^(2/3)*(-((c*x*(b + c*x))/b^2))^(1/3))/(1 - Sqrt[3]
- 2^(2/3)*(-((c*x*(b + c*x))/b^2))^(1/3))], -7 + 4*Sqrt[3]])/(5*2^(2/3)*c*(b + 2*c*x)*(-((c*(b*x + c*x^2))/b^
2))^(1/3)*Sqrt[-((1 - 2^(2/3)*(-((c*x*(b + c*x))/b^2))^(1/3))/(1 - Sqrt[3] - 2^(2/3)*(-((c*x*(b + c*x))/b^2))^
(1/3))^2)])

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt
[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sq
rt[(-s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3])*s + r*x)/((1 - Sqrt[3])*s + r
*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 242

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Dist[3*(Sqrt[b*x^2]/(2*b*x)), Subst[Int[1/Sqrt[-a + x^3], x], x
, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b}, x]

Rule 633

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*(-4*(c/(b^2 - 4*a*c)))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 636

Int[((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(b*x + c*x^2)^p/((-c)*((b*x + c*x^2)/b^2))^p, Int[((-c
)*(x/b) - c^2*(x^2/b^2))^p, x], x] /; FreeQ[{b, c}, x] && RationalQ[p] && 3 <= Denominator[p] <= 4

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{b x+c x^2} \int \sqrt [3]{-\frac {c x}{b}-\frac {c^2 x^2}{b^2}} \, dx}{\sqrt [3]{-\frac {c \left (b x+c x^2\right )}{b^2}}} \\ & = -\frac {\left (b^2 \sqrt [3]{b x+c x^2}\right ) \text {Subst}\left (\int \sqrt [3]{1-\frac {b^2 x^2}{c^2}} \, dx,x,-\frac {c}{b}-\frac {2 c^2 x}{b^2}\right )}{2\ 2^{2/3} c^2 \sqrt [3]{-\frac {c \left (b x+c x^2\right )}{b^2}}} \\ & = \frac {3 \sqrt [3]{-\frac {c x (b+c x)}{b^2}} (b+2 c x) \sqrt [3]{b x+c x^2}}{10 c \sqrt [3]{-\frac {c \left (b x+c x^2\right )}{b^2}}}-\frac {\left (b^2 \sqrt [3]{b x+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\left (1-\frac {b^2 x^2}{c^2}\right )^{2/3}} \, dx,x,-\frac {c}{b}-\frac {2 c^2 x}{b^2}\right )}{5\ 2^{2/3} c^2 \sqrt [3]{-\frac {c \left (b x+c x^2\right )}{b^2}}} \\ & = \frac {3 \sqrt [3]{-\frac {c x (b+c x)}{b^2}} (b+2 c x) \sqrt [3]{b x+c x^2}}{10 c \sqrt [3]{-\frac {c \left (b x+c x^2\right )}{b^2}}}+\frac {\left (3 \sqrt [3]{b x+c x^2} \sqrt {-1-\frac {4 c x}{b}-\frac {4 c^2 x^2}{b^2}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-1+x^3}} \, dx,x,2^{2/3} \sqrt [3]{-\frac {c x \left (1+\frac {c x}{b}\right )}{b}}\right )}{10\ 2^{2/3} \left (-\frac {c}{b}-\frac {2 c^2 x}{b^2}\right ) \sqrt [3]{-\frac {c \left (b x+c x^2\right )}{b^2}}} \\ & = \frac {3 \sqrt [3]{-\frac {c x (b+c x)}{b^2}} (b+2 c x) \sqrt [3]{b x+c x^2}}{10 c \sqrt [3]{-\frac {c \left (b x+c x^2\right )}{b^2}}}+\frac {3^{3/4} \sqrt {2-\sqrt {3}} b^2 \sqrt [3]{b x+c x^2} \sqrt {-1-\frac {4 c x}{b}-\frac {4 c^2 x^2}{b^2}} \left (1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right ) \sqrt {\frac {1+2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}+2 \sqrt [3]{2} \left (-\frac {c x (b+c x)}{b^2}\right )^{2/3}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right )^2}} F\left (\sin ^{-1}\left (\frac {1+\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}\right )|-7+4 \sqrt {3}\right )}{5\ 2^{2/3} c (b+2 c x) \sqrt [3]{-\frac {c \left (b x+c x^2\right )}{b^2}} \sqrt {-1-\frac {4 c x (b+c x)}{b^2}} \sqrt {-\frac {1-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}}{\left (1-\sqrt {3}-2^{2/3} \sqrt [3]{-\frac {c x (b+c x)}{b^2}}\right )^2}}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.12 \[ \int \sqrt [3]{b x+c x^2} \, dx=\frac {3 x \sqrt [3]{x (b+c x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {4}{3},\frac {7}{3},-\frac {c x}{b}\right )}{4 \sqrt [3]{1+\frac {c x}{b}}} \]

[In]

Integrate[(b*x + c*x^2)^(1/3),x]

[Out]

(3*x*(x*(b + c*x))^(1/3)*Hypergeometric2F1[-1/3, 4/3, 7/3, -((c*x)/b)])/(4*(1 + (c*x)/b)^(1/3))

Maple [F]

\[\int \left (c \,x^{2}+b x \right )^{\frac {1}{3}}d x\]

[In]

int((c*x^2+b*x)^(1/3),x)

[Out]

int((c*x^2+b*x)^(1/3),x)

Fricas [F]

\[ \int \sqrt [3]{b x+c x^2} \, dx=\int { {\left (c x^{2} + b x\right )}^{\frac {1}{3}} \,d x } \]

[In]

integrate((c*x^2+b*x)^(1/3),x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x)^(1/3), x)

Sympy [F]

\[ \int \sqrt [3]{b x+c x^2} \, dx=\int \sqrt [3]{b x + c x^{2}}\, dx \]

[In]

integrate((c*x**2+b*x)**(1/3),x)

[Out]

Integral((b*x + c*x**2)**(1/3), x)

Maxima [F]

\[ \int \sqrt [3]{b x+c x^2} \, dx=\int { {\left (c x^{2} + b x\right )}^{\frac {1}{3}} \,d x } \]

[In]

integrate((c*x^2+b*x)^(1/3),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x)^(1/3), x)

Giac [F]

\[ \int \sqrt [3]{b x+c x^2} \, dx=\int { {\left (c x^{2} + b x\right )}^{\frac {1}{3}} \,d x } \]

[In]

integrate((c*x^2+b*x)^(1/3),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x)^(1/3), x)

Mupad [B] (verification not implemented)

Time = 9.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.09 \[ \int \sqrt [3]{b x+c x^2} \, dx=\frac {3\,x\,{\left (c\,x^2+b\,x\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {4}{3};\ \frac {7}{3};\ -\frac {c\,x}{b}\right )}{4\,{\left (\frac {c\,x}{b}+1\right )}^{1/3}} \]

[In]

int((b*x + c*x^2)^(1/3),x)

[Out]

(3*x*(b*x + c*x^2)^(1/3)*hypergeom([-1/3, 4/3], 7/3, -(c*x)/b))/(4*((c*x)/b + 1)^(1/3))