3.16.50 \(\int x^4 \sqrt [3]{x+x^3} \, dx\)

Optimal. Leaf size=107 \[ -\frac {5}{162} \log \left (\sqrt [3]{x^3+x}-x\right )-\frac {5 \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3+x}+x}\right )}{54 \sqrt {3}}+\frac {1}{108} \sqrt [3]{x^3+x} \left (18 x^5+3 x^3-5 x\right )+\frac {5}{324} \log \left (\sqrt [3]{x^3+x} x+\left (x^3+x\right )^{2/3}+x^2\right ) \]

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Rubi [A]  time = 0.23, antiderivative size = 210, normalized size of antiderivative = 1.96, number of steps used = 13, number of rules used = 12, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {2021, 2024, 2032, 329, 275, 331, 292, 31, 634, 618, 204, 628} \begin {gather*} \frac {1}{36} \sqrt [3]{x^3+x} x^3-\frac {5}{108} \sqrt [3]{x^3+x} x+\frac {1}{6} \sqrt [3]{x^3+x} x^5-\frac {5 \left (x^2+1\right )^{2/3} x^{2/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{x^2+1}}\right )}{162 \left (x^3+x\right )^{2/3}}+\frac {5 \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 )}{324 \left (x^3+x\right )^{2/3}}-\frac {5 \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 )}{54 \sqrt {3} \left (x^3+x\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*(x + x^3)^(1/3),x]

[Out]

(-5*x*(x + x^3)^(1/3))/108 + (x^3*(x + x^3)^(1/3))/36 + (x^5*(x + x^3)^(1/3))/6 - (5*x^(2/3)*(1 + x^2)^(2/3)*A
rcTan[(1 + (2*x^(2/3))/(1 + x^2)^(1/3))/Sqrt[3]])/(54*Sqrt[3]*(x + x^3)^(2/3)) - (5*x^(2/3)*(1 + x^2)^(2/3)*Lo
g[1 - x^(2/3)/(1 + x^2)^(1/3)])/(162*(x + x^3)^(2/3)) + (5*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)])/(324*(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 2021

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

Rule 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + 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 x^4 \sqrt [3]{x+x^3} \, dx &=\frac {1}{6} x^5 \sqrt [3]{x+x^3}+\frac {1}{9} \int \frac {x^5}{\left (x+x^3\right )^{2/3}} \, dx\\ &=\frac {1}{36} x^3 \sqrt [3]{x+x^3}+\frac {1}{6} x^5 \sqrt [3]{x+x^3}-\frac {5}{54} \int \frac {x^3}{\left (x+x^3\right )^{2/3}} \, dx\\ &=-\frac {5}{108} x \sqrt [3]{x+x^3}+\frac {1}{36} x^3 \sqrt [3]{x+x^3}+\frac {1}{6} x^5 \sqrt [3]{x+x^3}+\frac {5}{81} \int \frac {x}{\left (x+x^3\right )^{2/3}} \, dx\\ &=-\frac {5}{108} x \sqrt [3]{x+x^3}+\frac {1}{36} x^3 \sqrt [3]{x+x^3}+\frac {1}{6} x^5 \sqrt [3]{x+x^3}+\frac {\left (5 x^{2/3} \left (1+x^2\right )^{2/3}\right ) \int \frac {\sqrt [3]{x}}{\left (1+x^2\right )^{2/3}} \, dx}{81 \left (x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{x+x^3}+\frac {1}{36} x^3 \sqrt [3]{x+x^3}+\frac {1}{6} x^5 \sqrt [3]{x+x^3}+\frac {\left (5 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 )}{27 \left (x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{x+x^3}+\frac {1}{36} x^3 \sqrt [3]{x+x^3}+\frac {1}{6} x^5 \sqrt [3]{x+x^3}+\frac {\left (5 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 )}{54 \left (x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{x+x^3}+\frac {1}{36} x^3 \sqrt [3]{x+x^3}+\frac {1}{6} x^5 \sqrt [3]{x+x^3}+\frac {\left (5 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 )}{54 \left (x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{x+x^3}+\frac {1}{36} x^3 \sqrt [3]{x+x^3}+\frac {1}{6} x^5 \sqrt [3]{x+x^3}+\frac {\left (5 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 )}{162 \left (x+x^3\right )^{2/3}}-\frac {\left (5 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 )}{162 \left (x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{x+x^3}+\frac {1}{36} x^3 \sqrt [3]{x+x^3}+\frac {1}{6} x^5 \sqrt [3]{x+x^3}-\frac {5 x^{2/3} \left (1+x^2\right )^{2/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{162 \left (x+x^3\right )^{2/3}}+\frac {\left (5 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 )}{324 \left (x+x^3\right )^{2/3}}-\frac {\left (5 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 )}{108 \left (x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{x+x^3}+\frac {1}{36} x^3 \sqrt [3]{x+x^3}+\frac {1}{6} x^5 \sqrt [3]{x+x^3}-\frac {5 x^{2/3} \left (1+x^2\right )^{2/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{162 \left (x+x^3\right )^{2/3}}+\frac {5 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 )}{324 \left (x+x^3\right )^{2/3}}+\frac {\left (5 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 )}{54 \left (x+x^3\right )^{2/3}}\\ &=-\frac {5}{108} x \sqrt [3]{x+x^3}+\frac {1}{36} x^3 \sqrt [3]{x+x^3}+\frac {1}{6} x^5 \sqrt [3]{x+x^3}-\frac {5 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 )}{54 \sqrt {3} \left (x+x^3\right )^{2/3}}-\frac {5 x^{2/3} \left (1+x^2\right )^{2/3} \log \left (1-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{162 \left (x+x^3\right )^{2/3}}+\frac {5 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 )}{324 \left (x+x^3\right )^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 61, normalized size = 0.57 \begin {gather*} \frac {x \sqrt [3]{x^3+x} \left (5 \, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};-x^2\right )+\sqrt [3]{x^2+1} \left (6 x^4+x^2-5\right )\right )}{36 \sqrt [3]{x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*(x + x^3)^(1/3),x]

[Out]

(x*(x + x^3)^(1/3)*((1 + x^2)^(1/3)*(-5 + x^2 + 6*x^4) + 5*Hypergeometric2F1[-1/3, 2/3, 5/3, -x^2]))/(36*(1 +
x^2)^(1/3))

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IntegrateAlgebraic [A]  time = 0.31, size = 107, normalized size = 1.00 \begin {gather*} \frac {1}{108} \sqrt [3]{x+x^3} \left (-5 x+3 x^3+18 x^5\right )-\frac {5 \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x+x^3}}\right )}{54 \sqrt {3}}-\frac {5}{162} \log \left (-x+\sqrt [3]{x+x^3}\right )+\frac {5}{324} \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^4*(x + x^3)^(1/3),x]

[Out]

((x + x^3)^(1/3)*(-5*x + 3*x^3 + 18*x^5))/108 - (5*ArcTan[(Sqrt[3]*x)/(x + 2*(x + x^3)^(1/3))])/(54*Sqrt[3]) -
(5*Log[-x + (x + x^3)^(1/3)])/162 + (5*Log[x^2 + x*(x + x^3)^(1/3) + (x + x^3)^(2/3)])/324

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fricas [A]  time = 0.93, size = 103, normalized size = 0.96 \begin {gather*} -\frac {5}{162} \, \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}{108} \, {\left (18 \, x^{5} + 3 \, x^{3} - 5 \, x\right )} {\left (x^{3} + x\right )}^{\frac {1}{3}} - \frac {5}{324} \, \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^4*(x^3+x)^(1/3),x, algorithm="fricas")

[Out]

-5/162*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)) + 1/108*(18*x^5 + 3*x^3 - 5*x)*(x^3 + x)^(1/3) - 5/324*log(3*(x^3 + x)^(1/3)*x - 3*(x^3 +
x)^(2/3) + 1)

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giac [A]  time = 0.41, size = 88, normalized size = 0.82 \begin {gather*} -\frac {1}{108} \, {\left (5 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {7}{3}} - 13 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {4}{3}} - 10 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )} x^{6} + \frac {5}{162} \, \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 {5}{324} \, \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 {5}{162} \, \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^4*(x^3+x)^(1/3),x, algorithm="giac")

[Out]

-1/108*(5*(1/x^2 + 1)^(7/3) - 13*(1/x^2 + 1)^(4/3) - 10*(1/x^2 + 1)^(1/3))*x^6 + 5/162*sqrt(3)*arctan(1/3*sqrt
(3)*(2*(1/x^2 + 1)^(1/3) + 1)) + 5/324*log((1/x^2 + 1)^(2/3) + (1/x^2 + 1)^(1/3) + 1) - 5/162*log(abs((1/x^2 +
1)^(1/3) - 1))

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maple [C]  time = 3.81, size = 17, normalized size = 0.16

method	result	size
meijerg	\(\frac {3 x^{\frac {16}{3}} \hypergeom \left (\left [-\frac {1}{3}, \frac {8}{3}\right ], \left [\frac {11}{3}\right ], -x^{2}\right )}{16}\)	\(17\)
trager	\(\frac {x \left (18 x^{4}+3 x^{2}-5\right ) \left (x^{3}+x \right )^{\frac {1}{3}}}{108}-\frac {5 \ln \left (-36 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}+72 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}-27 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x -33 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}+36 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}-9 \left (x^{3}+x \right )^{\frac {2}{3}}-15 x \left (x^{3}+x \right )^{\frac {1}{3}}+20 x^{2}-51 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )+15\right )}{162}+\frac {5 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \ln \left (-9 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2} x^{2}-72 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {2}{3}}+45 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) \left (x^{3}+x \right )^{\frac {1}{3}} x +30 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right ) x^{2}+9 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )^{2}+15 \left (x^{3}+x \right )^{\frac {2}{3}}+9 x \left (x^{3}+x \right )^{\frac {1}{3}}-25 x^{2}-9 \RootOf \left (9 \textit {\_Z}^{2}-3 \textit {\_Z} +1\right )-10\right )}{54}\)	\(299\)
risch	\(\frac {x \left (18 x^{4}+3 x^{2}-5\right ) \left (x \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{108}+\frac {\left (-\frac {5 \ln \left (-\frac {-5 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{4}-38 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{4}+18 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}+16 x^{4}+30 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+60 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} x^{2}-70 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}-96 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+18 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}+5 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}+28 x^{2}+60 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}-32 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )+12}{x^{2}+1}\right )}{162}+\frac {5 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2} x^{4}+20 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{4}-48 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}-100 x^{4}+30 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}+60 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}} x^{2}+14 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) x^{2}+36 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {2}{3}}-48 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right ) \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}+\RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )^{2}-140 x^{2}+60 \left (x^{6}+2 x^{4}+x^{2}\right )^{\frac {1}{3}}-6 \RootOf \left (\textit {\_Z}^{2}-2 \textit {\_Z} +4\right )-40}{x^{2}+1}\right )}{324}\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 )}\)	\(508\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(x^3+x)^(1/3),x,method=_RETURNVERBOSE)

[Out]

3/16*x^(16/3)*hypergeom([-1/3,8/3],[11/3],-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}} x^{4}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(x^3+x)^(1/3),x, algorithm="maxima")

[Out]

integrate((x^3 + x)^(1/3)*x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^4\,{\left (x^3+x\right )}^{1/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(x + x^3)^(1/3),x)

[Out]

int(x^4*(x + x^3)^(1/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{4} \sqrt [3]{x \left (x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(x**3+x)**(1/3),x)

[Out]

Integral(x**4*(x*(x**2 + 1))**(1/3), x)

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