3.13.97 \(\int \sqrt [3]{x+x^3} \, dx\)

Optimal. Leaf size=94 \[ \frac {1}{2} \sqrt [3]{x^3+x} x-\frac {1}{6} \log \left (\sqrt [3]{x^3+x}-x\right )-\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+x}+x}\right )}{2 \sqrt {3}}+\frac {1}{12} \log \left (\sqrt [3]{x^3+x} x+\left (x^3+x\right )^{2/3}+x^2\right ) \]

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Rubi [A]  time = 0.13, antiderivative size = 178, normalized size of antiderivative = 1.89, number of steps used = 11, number of rules used = 11, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.222, Rules used = {2004, 2032, 329, 275, 331, 292, 31, 634, 618, 204, 628} \begin {gather*} \frac {1}{2} \sqrt [3]{x^3+x} x-\frac {\left (x^2+1\right )^{2/3} x^{2/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{x^2+1}}\right )}{6 \left (x^3+x\right )^{2/3}}+\frac {\left (x^2+1\right )^{2/3} x^{2/3} \log \left (\frac {x^{4/3}}{\left (x^2+1\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{x^2+1}}+1\right )}{12 \left (x^3+x\right )^{2/3}}-\frac {\left (x^2+1\right )^{2/3} x^{2/3} \tan ^{-1}\left (\frac {\frac {2 x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{2 \sqrt {3} \left (x^3+x\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + x^3)^(1/3),x]

[Out]

(x*(x + x^3)^(1/3))/2 - (x^(2/3)*(1 + x^2)^(2/3)*ArcTan[(1 + (2*x^(2/3))/(1 + x^2)^(1/3))/Sqrt[3]])/(2*Sqrt[3]
*(x + x^3)^(2/3)) - (x^(2/3)*(1 + x^2)^(2/3)*Log[1 - x^(2/3)/(1 + x^2)^(1/3)])/(6*(x + x^3)^(2/3)) + (x^(2/3)*
(1 + x^2)^(2/3)*Log[1 + x^(4/3)/(1 + x^2)^(2/3) + x^(2/3)/(1 + x^2)^(1/3)])/(12*(x + x^3)^(2/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 292

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 618

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 2004

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(x*(a*x^j + b*x^n)^p)/(n*p + 1), x] + Dist[(
a*(n - j)*p)/(n*p + 1), Int[x^j*(a*x^j + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] &&  !IntegerQ[p] && LtQ[0,
j, n] && GtQ[p, 0] && NeQ[n*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \int \sqrt [3]{x+x^3} \, dx &=\frac {1}{2} x \sqrt [3]{x+x^3}+\frac {1}{3} \int \frac {x}{\left (x+x^3\right )^{2/3}} \, dx\\ &=\frac {1}{2} x \sqrt [3]{x+x^3}+\frac {\left (x^{2/3} \left (1+x^2\right )^{2/3}\right ) \int \frac {\sqrt [3]{x}}{\left (1+x^2\right )^{2/3}} \, dx}{3 \left (x+x^3\right )^{2/3}}\\ &=\frac {1}{2} x \sqrt [3]{x+x^3}+\frac {\left (x^{2/3} \left (1+x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (1+x^6\right )^{2/3}} \, dx,x,\sqrt [3]{x}\right )}{\left (x+x^3\right )^{2/3}}\\ &=\frac {1}{2} x \sqrt [3]{x+x^3}+\frac {\left (x^{2/3} \left (1+x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+x^3\right )^{2/3}} \, dx,x,x^{2/3}\right )}{2 \left (x+x^3\right )^{2/3}}\\ &=\frac {1}{2} x \sqrt [3]{x+x^3}+\frac {\left (x^{2/3} \left (1+x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x}{1-x^3} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 \left (x+x^3\right )^{2/3}}\\ &=\frac {1}{2} x \sqrt [3]{x+x^3}+\frac {\left (x^{2/3} \left (1+x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 \left (x+x^3\right )^{2/3}}-\frac {\left (x^{2/3} \left (1+x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1-x}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 \left (x+x^3\right )^{2/3}}\\ &=\frac {1}{2} x \sqrt [3]{x+x^3}-\frac {x^{2/3} \left (1+x^2\right )^{2/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 \left (x+x^3\right )^{2/3}}+\frac {\left (x^{2/3} \left (1+x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{12 \left (x+x^3\right )^{2/3}}-\frac {\left (x^{2/3} \left (1+x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 \left (x+x^3\right )^{2/3}}\\ &=\frac {1}{2} x \sqrt [3]{x+x^3}-\frac {x^{2/3} \left (1+x^2\right )^{2/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 \left (x+x^3\right )^{2/3}}+\frac {x^{2/3} \left (1+x^2\right )^{2/3} \log \left (1+\frac {x^{4/3}}{\left (1+x^2\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{12 \left (x+x^3\right )^{2/3}}+\frac {\left (x^{2/3} \left (1+x^2\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 \left (x+x^3\right )^{2/3}}\\ &=\frac {1}{2} x \sqrt [3]{x+x^3}-\frac {x^{2/3} \left (1+x^2\right )^{2/3} \tan ^{-1}\left (\frac {1+\frac {2 x^{2/3}}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{2 \sqrt {3} \left (x+x^3\right )^{2/3}}-\frac {x^{2/3} \left (1+x^2\right )^{2/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{6 \left (x+x^3\right )^{2/3}}+\frac {x^{2/3} \left (1+x^2\right )^{2/3} \log \left (1+\frac {x^{4/3}}{\left (1+x^2\right )^{2/3}}+\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{12 \left (x+x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 38, normalized size = 0.40 \begin {gather*} \frac {3 x \sqrt [3]{x^3+x} \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};-x^2\right )}{4 \sqrt [3]{x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + x^3)^(1/3),x]

[Out]

(3*x*(x + x^3)^(1/3)*Hypergeometric2F1[-1/3, 2/3, 5/3, -x^2])/(4*(1 + x^2)^(1/3))

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IntegrateAlgebraic [A]  time = 0.17, size = 94, normalized size = 1.00 \begin {gather*} \frac {1}{2} x \sqrt [3]{x+x^3}-\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )}{2 \sqrt {3}}-\frac {1}{6} \log \left (-x+\sqrt [3]{x+x^3}\right )+\frac {1}{12} \log \left (x^2+x \sqrt [3]{x+x^3}+\left (x+x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x + x^3)^(1/3),x]

[Out]

(x*(x + x^3)^(1/3))/2 - ArcTan[(Sqrt[3]*x)/(x + 2*(x + x^3)^(1/3))]/(2*Sqrt[3]) - Log[-x + (x + x^3)^(1/3)]/6
+ Log[x^2 + x*(x + x^3)^(1/3) + (x + x^3)^(2/3)]/12

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fricas [A]  time = 0.63, size = 90, normalized size = 0.96 \begin {gather*} -\frac {1}{6} \, \sqrt {3} \arctan \left (-\frac {196 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \sqrt {3} {\left (539 \, x^{2} + 507\right )} - 1274 \, \sqrt {3} {\left (x^{3} + x\right )}^{\frac {2}{3}}}{2205 \, x^{2} + 2197}\right ) + \frac {1}{2} \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - \frac {1}{12} \, \log \left (3 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 3 \, {\left (x^{3} + x\right )}^{\frac {2}{3}} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+x)^(1/3),x, algorithm="fricas")

[Out]

-1/6*sqrt(3)*arctan(-(196*sqrt(3)*(x^3 + x)^(1/3)*x - sqrt(3)*(539*x^2 + 507) - 1274*sqrt(3)*(x^3 + x)^(2/3))/
(2205*x^2 + 2197)) + 1/2*(x^3 + x)^(1/3)*x - 1/12*log(3*(x^3 + x)^(1/3)*x - 3*(x^3 + x)^(2/3) + 1)

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giac [A]  time = 0.36, size = 67, normalized size = 0.71 \begin {gather*} \frac {1}{2} \, x^{2} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {1}{12} \, \log \left ({\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + 1\right ) - \frac {1}{6} \, \log \left ({\left | {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+x)^(1/3),x, algorithm="giac")

[Out]

1/2*x^2*(1/x^2 + 1)^(1/3) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*(1/x^2 + 1)^(1/3) + 1)) + 1/12*log((1/x^2 + 1)^(
2/3) + (1/x^2 + 1)^(1/3) + 1) - 1/6*log(abs((1/x^2 + 1)^(1/3) - 1))

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maple [C]  time = 4.24, size = 17, normalized size = 0.18

method	result	size
meijerg	\(\frac {3 x^{\frac {4}{3}} \hypergeom \left (\left [-\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -x^{2}\right )}{4}\)	\(17\)
trager	\(\frac {x \left (x^{3}+x \right )^{\frac {1}{3}}}{2}-\frac {\ln \left (-36 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}-72 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+27 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x +57 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}+36 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}-24 x \left (x^{3}+x \right )^{\frac {1}{3}}+5 x^{2}+27 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+2\right )}{6}+\frac {\ln \left (-9 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}+72 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-45 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x -24 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}+9 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}+24 x \left (x^{3}+x \right )^{\frac {1}{3}}-16 x^{2}+3 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )-12\right )}{6}-\frac {\ln \left (-9 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}+72 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-45 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x -24 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}+9 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}+24 x \left (x^{3}+x \right )^{\frac {1}{3}}-16 x^{2}+3 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )-12\right ) \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )}{2}\)	\(419\)
risch	\(\frac {x \left (x \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{2}+\frac {\left (\frac {\RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \ln \left (-\frac {2 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{4}+11 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{4}+15 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}-40 x^{4}+15 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}-48 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} x^{2}+28 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}-48 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}-2 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}-70 x^{2}-48 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}+17 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )-30}{x^{2}+1}\right )}{12}-\frac {\ln \left (\frac {-2 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{4}+19 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{4}+15 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}+10 x^{4}+15 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+18 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} x^{2}+28 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}+18 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}+2 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}+14 x^{2}+18 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}+9 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )+4}{x^{2}+1}\right ) \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )}{12}+\frac {\ln \left (\frac {-2 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{4}+19 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{4}+15 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}+10 x^{4}+15 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+18 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} x^{2}+28 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}+18 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}+2 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}+14 x^{2}+18 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}+9 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )+4}{x^{2}+1}\right )}{6}\right ) \left (x \left (x^{2}+1\right )\right )^{\frac {1}{3}} \left (x^{2} \left (x^{2}+1\right )^{2}\right )^{\frac {1}{3}}}{x \left (x^{2}+1\right )}\)	\(727\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+x)^(1/3),x,method=_RETURNVERBOSE)

[Out]

3/4*x^(4/3)*hypergeom([-1/3,2/3],[5/3],-x^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (x^{3} + x\right )}^{\frac {1}{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+x)^(1/3),x, algorithm="maxima")

[Out]

integrate((x^3 + x)^(1/3), x)

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mupad [B]  time = 0.80, size = 27, normalized size = 0.29 \begin {gather*} \frac {3\,x\,{\left (x^3+x\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},\frac {2}{3};\ \frac {5}{3};\ -x^2\right )}{4\,{\left (x^2+1\right )}^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + x^3)^(1/3),x)

[Out]

(3*x*(x + x^3)^(1/3)*hypergeom([-1/3, 2/3], 5/3, -x^2))/(4*(x^2 + 1)^(1/3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt [3]{x^{3} + x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+x)**(1/3),x)

[Out]

Integral((x**3 + x)**(1/3), x)

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