3.17.12 \(\int x^9 \sqrt [3]{1+x^6} \, dx\)

Optimal. Leaf size=110 \[ \frac {1}{54} \log \left (\sqrt [3]{x^6+1}-x^2\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x^2}{2 \sqrt [3]{x^6+1}+x^2}\right )}{18 \sqrt {3}}+\frac {1}{36} \sqrt [3]{x^6+1} \left (3 x^{10}+x^4\right )-\frac {1}{108} \log \left (\left (x^6+1\right )^{2/3}+x^4+\sqrt [3]{x^6+1} x^2\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 119, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.769, Rules used = {275, 279, 321, 331, 292, 31, 634, 618, 204, 628} \begin {gather*} \frac {1}{12} \sqrt [3]{x^6+1} x^{10}+\frac {1}{36} \sqrt [3]{x^6+1} x^4+\frac {1}{54} \log \left (1-\frac {x^2}{\sqrt [3]{x^6+1}}\right )+\frac {\tan ^{-1}\left (\frac {\frac {2 x^2}{\sqrt [3]{x^6+1}}+1}{\sqrt {3}}\right )}{18 \sqrt {3}}-\frac {1}{108} \log \left (\frac {x^4}{\left (x^6+1\right )^{2/3}}+\frac {x^2}{\sqrt [3]{x^6+1}}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9*(1 + x^6)^(1/3),x]

[Out]

(x^4*(1 + x^6)^(1/3))/36 + (x^10*(1 + x^6)^(1/3))/12 + ArcTan[(1 + (2*x^2)/(1 + x^6)^(1/3))/Sqrt[3]]/(18*Sqrt[
3]) + Log[1 - x^2/(1 + x^6)^(1/3)]/54 - Log[1 + x^4/(1 + x^6)^(2/3) + x^2/(1 + x^6)^(1/3)]/108

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 279

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

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 321

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

Rubi steps

\begin {align*} \int x^9 \sqrt [3]{1+x^6} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int x^4 \sqrt [3]{1+x^3} \, dx,x,x^2\right )\\ &=\frac {1}{12} x^{10} \sqrt [3]{1+x^6}+\frac {1}{12} \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^3\right )^{2/3}} \, dx,x,x^2\right )\\ &=\frac {1}{36} x^4 \sqrt [3]{1+x^6}+\frac {1}{12} x^{10} \sqrt [3]{1+x^6}-\frac {1}{18} \operatorname {Subst}\left (\int \frac {x}{\left (1+x^3\right )^{2/3}} \, dx,x,x^2\right )\\ &=\frac {1}{36} x^4 \sqrt [3]{1+x^6}+\frac {1}{12} x^{10} \sqrt [3]{1+x^6}-\frac {1}{18} \operatorname {Subst}\left (\int \frac {x}{1-x^3} \, dx,x,\frac {x^2}{\sqrt [3]{1+x^6}}\right )\\ &=\frac {1}{36} x^4 \sqrt [3]{1+x^6}+\frac {1}{12} x^{10} \sqrt [3]{1+x^6}-\frac {1}{54} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\frac {x^2}{\sqrt [3]{1+x^6}}\right )+\frac {1}{54} \operatorname {Subst}\left (\int \frac {1-x}{1+x+x^2} \, dx,x,\frac {x^2}{\sqrt [3]{1+x^6}}\right )\\ &=\frac {1}{36} x^4 \sqrt [3]{1+x^6}+\frac {1}{12} x^{10} \sqrt [3]{1+x^6}+\frac {1}{54} \log \left (1-\frac {x^2}{\sqrt [3]{1+x^6}}\right )-\frac {1}{108} \operatorname {Subst}\left (\int \frac {1+2 x}{1+x+x^2} \, dx,x,\frac {x^2}{\sqrt [3]{1+x^6}}\right )+\frac {1}{36} \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\frac {x^2}{\sqrt [3]{1+x^6}}\right )\\ &=\frac {1}{36} x^4 \sqrt [3]{1+x^6}+\frac {1}{12} x^{10} \sqrt [3]{1+x^6}+\frac {1}{54} \log \left (1-\frac {x^2}{\sqrt [3]{1+x^6}}\right )-\frac {1}{108} \log \left (1+\frac {x^4}{\left (1+x^6\right )^{2/3}}+\frac {x^2}{\sqrt [3]{1+x^6}}\right )-\frac {1}{18} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 x^2}{\sqrt [3]{1+x^6}}\right )\\ &=\frac {1}{36} x^4 \sqrt [3]{1+x^6}+\frac {1}{12} x^{10} \sqrt [3]{1+x^6}+\frac {\tan ^{-1}\left (\frac {1+\frac {2 x^2}{\sqrt [3]{1+x^6}}}{\sqrt {3}}\right )}{18 \sqrt {3}}+\frac {1}{54} \log \left (1-\frac {x^2}{\sqrt [3]{1+x^6}}\right )-\frac {1}{108} \log \left (1+\frac {x^4}{\left (1+x^6\right )^{2/3}}+\frac {x^2}{\sqrt [3]{1+x^6}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 34, normalized size = 0.31 \begin {gather*} \frac {1}{12} x^4 \left (\left (x^6+1\right )^{4/3}-\, _2F_1\left (-\frac {1}{3},\frac {2}{3};\frac {5}{3};-x^6\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9*(1 + x^6)^(1/3),x]

[Out]

(x^4*((1 + x^6)^(4/3) - Hypergeometric2F1[-1/3, 2/3, 5/3, -x^6]))/12

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IntegrateAlgebraic [A]  time = 0.82, size = 110, normalized size = 1.00 \begin {gather*} \frac {1}{36} \sqrt [3]{1+x^6} \left (x^4+3 x^{10}\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x^2}{x^2+2 \sqrt [3]{1+x^6}}\right )}{18 \sqrt {3}}+\frac {1}{54} \log \left (-x^2+\sqrt [3]{1+x^6}\right )-\frac {1}{108} \log \left (x^4+x^2 \sqrt [3]{1+x^6}+\left (1+x^6\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^9*(1 + x^6)^(1/3),x]

[Out]

((1 + x^6)^(1/3)*(x^4 + 3*x^10))/36 + ArcTan[(Sqrt[3]*x^2)/(x^2 + 2*(1 + x^6)^(1/3))]/(18*Sqrt[3]) + Log[-x^2
+ (1 + x^6)^(1/3)]/54 - Log[x^4 + x^2*(1 + x^6)^(1/3) + (1 + x^6)^(2/3)]/108

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fricas [A]  time = 0.52, size = 100, normalized size = 0.91 \begin {gather*} -\frac {1}{54} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2} + 2 \, \sqrt {3} {\left (x^{6} + 1\right )}^{\frac {1}{3}}}{3 \, x^{2}}\right ) + \frac {1}{36} \, {\left (3 \, x^{10} + x^{4}\right )} {\left (x^{6} + 1\right )}^{\frac {1}{3}} + \frac {1}{54} \, \log \left (-\frac {x^{2} - {\left (x^{6} + 1\right )}^{\frac {1}{3}}}{x^{2}}\right ) - \frac {1}{108} \, \log \left (\frac {x^{4} + {\left (x^{6} + 1\right )}^{\frac {1}{3}} x^{2} + {\left (x^{6} + 1\right )}^{\frac {2}{3}}}{x^{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(x^6+1)^(1/3),x, algorithm="fricas")

[Out]

-1/54*sqrt(3)*arctan(1/3*(sqrt(3)*x^2 + 2*sqrt(3)*(x^6 + 1)^(1/3))/x^2) + 1/36*(3*x^10 + x^4)*(x^6 + 1)^(1/3)
+ 1/54*log(-(x^2 - (x^6 + 1)^(1/3))/x^2) - 1/108*log((x^4 + (x^6 + 1)^(1/3)*x^2 + (x^6 + 1)^(2/3))/x^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (x^{6} + 1\right )}^{\frac {1}{3}} x^{9}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(x^6+1)^(1/3),x, algorithm="giac")

[Out]

integrate((x^6 + 1)^(1/3)*x^9, x)

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maple [C]  time = 3.53, size = 17, normalized size = 0.15

method	result	size
meijerg	\(\frac {x^{10} \hypergeom \left (\left [-\frac {1}{3}, \frac {5}{3}\right ], \left [\frac {8}{3}\right ], -x^{6}\right )}{10}\)	\(17\)
risch	\(\frac {x^{4} \left (3 x^{6}+1\right ) \left (x^{6}+1\right )^{\frac {1}{3}}}{36}-\frac {x^{4} \hypergeom \left (\left [\frac {2}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], -x^{6}\right )}{36}\)	\(37\)
trager	\(\frac {x^{4} \left (3 x^{6}+1\right ) \left (x^{6}+1\right )^{\frac {1}{3}}}{36}+\frac {\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{6}-2 x^{6} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {1}{3}} x^{4}+x^{6}-3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {2}{3}} x^{2}-\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+1\right )}{54}-\frac {\ln \left (\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{6}+4 x^{6} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {1}{3}} x^{4}+4 x^{6}+3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {2}{3}} x^{2}+3 x^{4} \left (x^{6}+1\right )^{\frac {1}{3}}+3 x^{2} \left (x^{6}+1\right )^{\frac {2}{3}}+\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+2\right ) \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{54}-\frac {\ln \left (\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{6}+4 x^{6} \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {1}{3}} x^{4}+4 x^{6}+3 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{6}+1\right )^{\frac {2}{3}} x^{2}+3 x^{4} \left (x^{6}+1\right )^{\frac {1}{3}}+3 x^{2} \left (x^{6}+1\right )^{\frac {2}{3}}+\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+2\right )}{54}\)	\(323\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9*(x^6+1)^(1/3),x,method=_RETURNVERBOSE)

[Out]

1/10*x^10*hypergeom([-1/3,5/3],[8/3],-x^6)

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maxima [A]  time = 0.42, size = 121, normalized size = 1.10 \begin {gather*} -\frac {1}{54} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (x^{6} + 1\right )}^{\frac {1}{3}}}{x^{2}} + 1\right )}\right ) - \frac {\frac {2 \, {\left (x^{6} + 1\right )}^{\frac {1}{3}}}{x^{2}} + \frac {{\left (x^{6} + 1\right )}^{\frac {4}{3}}}{x^{8}}}{36 \, {\left (\frac {2 \, {\left (x^{6} + 1\right )}}{x^{6}} - \frac {{\left (x^{6} + 1\right )}^{2}}{x^{12}} - 1\right )}} - \frac {1}{108} \, \log \left (\frac {{\left (x^{6} + 1\right )}^{\frac {1}{3}}}{x^{2}} + \frac {{\left (x^{6} + 1\right )}^{\frac {2}{3}}}{x^{4}} + 1\right ) + \frac {1}{54} \, \log \left (\frac {{\left (x^{6} + 1\right )}^{\frac {1}{3}}}{x^{2}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(x^6+1)^(1/3),x, algorithm="maxima")

[Out]

-1/54*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^6 + 1)^(1/3)/x^2 + 1)) - 1/36*(2*(x^6 + 1)^(1/3)/x^2 + (x^6 + 1)^(4/3)/
x^8)/(2*(x^6 + 1)/x^6 - (x^6 + 1)^2/x^12 - 1) - 1/108*log((x^6 + 1)^(1/3)/x^2 + (x^6 + 1)^(2/3)/x^4 + 1) + 1/5
4*log((x^6 + 1)^(1/3)/x^2 - 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^9\,{\left (x^6+1\right )}^{1/3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9*(x^6 + 1)^(1/3),x)

[Out]

int(x^9*(x^6 + 1)^(1/3), x)

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sympy [C]  time = 1.18, size = 31, normalized size = 0.28 \begin {gather*} \frac {x^{10} \Gamma \left (\frac {5}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {5}{3} \\ \frac {8}{3} \end {matrix}\middle | {x^{6} e^{i \pi }} \right )}}{6 \Gamma \left (\frac {8}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9*(x**6+1)**(1/3),x)

[Out]

x**10*gamma(5/3)*hyper((-1/3, 5/3), (8/3,), x**6*exp_polar(I*pi))/(6*gamma(8/3))

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