3.13.28 \(\int \frac {\sqrt [3]{1+x^3}}{x^4} \, dx\)

Optimal. Leaf size=90 \[ -\frac {\sqrt [3]{x^3+1}}{3 x^3}+\frac {1}{9} \log \left (\sqrt [3]{x^3+1}-1\right )-\frac {1}{18} \log \left (\left (x^3+1\right )^{2/3}+\sqrt [3]{x^3+1}+1\right )-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{x^3+1}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

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Rubi [A]  time = 0.04, antiderivative size = 70, normalized size of antiderivative = 0.78, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {266, 47, 57, 618, 204, 31} \begin {gather*} -\frac {\sqrt [3]{x^3+1}}{3 x^3}+\frac {1}{6} \log \left (1-\sqrt [3]{x^3+1}\right )-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{x^3+1}+1}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {\log (x)}{6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^3)^(1/3)/x^4,x]

[Out]

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

Rule 31

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

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 266

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

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 26, normalized size = 0.29 \begin {gather*} \frac {1}{4} \left (x^3+1\right )^{4/3} \, _2F_1\left (\frac {4}{3},2;\frac {7}{3};x^3+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^3)^(1/3)/x^4,x]

[Out]

((1 + x^3)^(4/3)*Hypergeometric2F1[4/3, 2, 7/3, 1 + x^3])/4

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IntegrateAlgebraic [A]  time = 0.08, size = 90, normalized size = 1.00 \begin {gather*} -\frac {\sqrt [3]{1+x^3}}{3 x^3}-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^3}}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{9} \log \left (-1+\sqrt [3]{1+x^3}\right )-\frac {1}{18} \log \left (1+\sqrt [3]{1+x^3}+\left (1+x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x^3)^(1/3)/x^4,x]

[Out]

-1/3*(1 + x^3)^(1/3)/x^3 - ArcTan[1/Sqrt[3] + (2*(1 + x^3)^(1/3))/Sqrt[3]]/(3*Sqrt[3]) + Log[-1 + (1 + x^3)^(1
/3)]/9 - Log[1 + (1 + x^3)^(1/3) + (1 + x^3)^(2/3)]/18

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fricas [A]  time = 0.46, size = 78, normalized size = 0.87 \begin {gather*} -\frac {2 \, \sqrt {3} x^{3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + x^{3} \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) - 2 \, x^{3} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) + 6 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{18 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(2*sqrt(3)*x^3*arctan(2/3*sqrt(3)*(x^3 + 1)^(1/3) + 1/3*sqrt(3)) + x^3*log((x^3 + 1)^(2/3) + (x^3 + 1)^(
1/3) + 1) - 2*x^3*log((x^3 + 1)^(1/3) - 1) + 6*(x^3 + 1)^(1/3))/x^3

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giac [A]  time = 0.16, size = 67, normalized size = 0.74 \begin {gather*} -\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x^{3}} - \frac {1}{18} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{9} \, \log \left ({\left | {\left (x^{3} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 1)^(1/3) + 1)) - 1/3*(x^3 + 1)^(1/3)/x^3 - 1/18*log((x^3 + 1)^(2/3)
+ (x^3 + 1)^(1/3) + 1) + 1/9*log(abs((x^3 + 1)^(1/3) - 1))

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maple [C]  time = 2.23, size = 55, normalized size = 0.61

method	result	size
meijerg	\(-\frac {\frac {\Gamma \left (\frac {2}{3}\right ) x^{3} \hypergeom \left (\left [1, 1, \frac {5}{3}\right ], \left [2, 3\right ], -x^{3}\right )}{3}-\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}-1+3 \ln \relax (x )\right ) \Gamma \left (\frac {2}{3}\right )+\frac {3 \Gamma \left (\frac {2}{3}\right )}{x^{3}}}{9 \Gamma \left (\frac {2}{3}\right )}\)	\(55\)
risch	\(-\frac {\left (x^{3}+1\right )^{\frac {1}{3}}}{3 x^{3}}+\frac {-\frac {2 \Gamma \left (\frac {2}{3}\right ) x^{3} \hypergeom \left (\left [1, 1, \frac {5}{3}\right ], \left [2, 2\right ], -x^{3}\right )}{3}+\left (\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+3 \ln \relax (x )\right ) \Gamma \left (\frac {2}{3}\right )}{9 \Gamma \left (\frac {2}{3}\right )}\)	\(59\)
trager	\(-\frac {\left (x^{3}+1\right )^{\frac {1}{3}}}{3 x^{3}}+\frac {\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (-\frac {4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+17 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}+15 x^{3}+24 \left (x^{3}+1\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}-4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+24 \left (x^{3}+1\right )^{\frac {1}{3}}+11 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+20}{x^{3}}\right )}{9}-\frac {\ln \left (\frac {-4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+9 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}-2 x^{3}-9 \left (x^{3}+1\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}+4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-9 \left (x^{3}+1\right )^{\frac {1}{3}}+19 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-5}{x^{3}}\right ) \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{9}-\frac {\ln \left (\frac {-4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+9 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}-2 x^{3}-9 \left (x^{3}+1\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}+4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-9 \left (x^{3}+1\right )^{\frac {1}{3}}+19 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-5}{x^{3}}\right )}{9}\)	\(359\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9/GAMMA(2/3)*(1/3*GAMMA(2/3)*x^3*hypergeom([1,1,5/3],[2,3],-x^3)-(1/6*Pi*3^(1/2)-3/2*ln(3)-1+3*ln(x))*GAMMA
(2/3)+3*GAMMA(2/3)/x^3)

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maxima [A]  time = 0.44, size = 66, normalized size = 0.73 \begin {gather*} -\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {{\left (x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x^{3}} - \frac {1}{18} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{9} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 1)^(1/3) + 1)) - 1/3*(x^3 + 1)^(1/3)/x^3 - 1/18*log((x^3 + 1)^(2/3)
+ (x^3 + 1)^(1/3) + 1) + 1/9*log((x^3 + 1)^(1/3) - 1)

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mupad [B]  time = 0.90, size = 78, normalized size = 0.87 \begin {gather*} \frac {\ln \left (\frac {{\left (x^3+1\right )}^{1/3}}{9}-\frac {1}{9}\right )}{9}+\ln \left ({\left (x^3+1\right )}^{1/3}+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )-\ln \left ({\left (x^3+1\right )}^{1/3}+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )-\frac {{\left (x^3+1\right )}^{1/3}}{3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 + 1)^(1/3)/x^4,x)

[Out]

log((x^3 + 1)^(1/3)/9 - 1/9)/9 + log((x^3 + 1)^(1/3) - (3^(1/2)*1i)/2 + 1/2)*((3^(1/2)*1i)/18 - 1/18) - log((3
^(1/2)*1i)/2 + (x^3 + 1)^(1/3) + 1/2)*((3^(1/2)*1i)/18 + 1/18) - (x^3 + 1)^(1/3)/(3*x^3)

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sympy [C]  time = 0.88, size = 32, normalized size = 0.36 \begin {gather*} - \frac {\Gamma \left (\frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, \frac {2}{3} \\ \frac {5}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{2} \Gamma \left (\frac {5}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(2/3)*hyper((-1/3, 2/3), (5/3,), exp_polar(I*pi)/x**3)/(3*x**2*gamma(5/3))

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