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

Optimal. Leaf size=81 \[ \sqrt [3]{x^3+1}+\frac {1}{3} \log \left (\sqrt [3]{x^3+1}-1\right )-\frac {1}{6} \log \left (\left (x^3+1\right )^{2/3}+\sqrt [3]{x^3+1}+1\right )-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{x^3+1}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3}} \]

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Rubi [A]  time = 0.05, antiderivative size = 61, normalized size of antiderivative = 0.75, 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, 50, 57, 618, 204, 31} \begin {gather*} \sqrt [3]{x^3+1}+\frac {1}{2} \log \left (1-\sqrt [3]{x^3+1}\right )-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{x^3+1}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\log (x)}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^3)^(1/3)/x,x]

[Out]

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

Rule 31

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt [3]{1+x}}{x} \, dx,x,x^3\right )\\ &=\sqrt [3]{1+x^3}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x (1+x)^{2/3}} \, dx,x,x^3\right )\\ &=\sqrt [3]{1+x^3}-\frac {\log (x)}{2}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1+x^3}\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1+x^3}\right )\\ &=\sqrt [3]{1+x^3}-\frac {\log (x)}{2}+\frac {1}{2} \log \left (1-\sqrt [3]{1+x^3}\right )+\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1+x^3}\right )\\ &=\sqrt [3]{1+x^3}-\frac {\tan ^{-1}\left (\frac {1+2 \sqrt [3]{1+x^3}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\log (x)}{2}+\frac {1}{2} \log \left (1-\sqrt [3]{1+x^3}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 80, normalized size = 0.99 \begin {gather*} \sqrt [3]{x^3+1}+\frac {1}{3} \log \left (1-\sqrt [3]{x^3+1}\right )-\frac {1}{6} \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}+1}{\sqrt {3}}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^3)^(1/3)/x,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.06, size = 81, normalized size = 1.00 \begin {gather*} \sqrt [3]{1+x^3}-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^3}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (-1+\sqrt [3]{1+x^3}\right )-\frac {1}{6} \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,x]

[Out]

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

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fricas [A]  time = 0.53, size = 63, normalized size = 0.78 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + {\left (x^{3} + 1\right )}^{\frac {1}{3}} - \frac {1}{6} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \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,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 62, normalized size = 0.77 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + {\left (x^{3} + 1\right )}^{\frac {1}{3}} - \frac {1}{6} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \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,x, algorithm="giac")

[Out]

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

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maple [C]  time = 1.93, size = 48, normalized size = 0.59

method	result	size
meijerg	\(-\frac {-\Gamma \left (\frac {2}{3}\right ) x^{3} \hypergeom \left (\left [\frac {2}{3}, 1, 1\right ], \left [2, 2\right ], -x^{3}\right )-3 \left (3+\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+3 \ln \relax (x )\right ) \Gamma \left (\frac {2}{3}\right )}{9 \Gamma \left (\frac {2}{3}\right )}\)	\(48\)
trager	\(\left (x^{3}+1\right )^{\frac {1}{3}}+\frac {\RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \ln \left (-\frac {4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+17 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}+15 x^{3}+24 \left (x^{3}+1\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}-4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}+24 \left (x^{3}+1\right )^{\frac {1}{3}}+11 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+20}{x^{3}}\right )}{3}-\frac {\ln \left (\frac {-4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+9 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}-2 x^{3}-9 \left (x^{3}+1\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}+4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-9 \left (x^{3}+1\right )^{\frac {1}{3}}+19 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-5}{x^{3}}\right ) \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{3}-\frac {\ln \left (\frac {-4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{3}+9 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{3}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}-2 x^{3}-9 \left (x^{3}+1\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}+4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2}-9 \left (x^{3}+1\right )^{\frac {1}{3}}+19 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-5}{x^{3}}\right )}{3}\)	\(354\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 61, normalized size = 0.75 \begin {gather*} -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + {\left (x^{3} + 1\right )}^{\frac {1}{3}} - \frac {1}{6} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {2}{3}} + {\left (x^{3} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{3} \, \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,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.83, size = 75, normalized size = 0.93 \begin {gather*} \frac {\ln \left ({\left (x^3+1\right )}^{1/3}-1\right )}{3}+{\left (x^3+1\right )}^{1/3}+\ln \left (3\,{\left (x^3+1\right )}^{1/3}+\frac {3}{2}-\frac {\sqrt {3}\,3{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (3\,{\left (x^3+1\right )}^{1/3}+\frac {3}{2}+\frac {\sqrt {3}\,3{}\mathrm {i}}{2}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 + 1)^(1/3)/x,x)

[Out]

log((x^3 + 1)^(1/3) - 1)/3 + (x^3 + 1)^(1/3) + log(3*(x^3 + 1)^(1/3) - (3^(1/2)*3i)/2 + 3/2)*((3^(1/2)*1i)/6 -
1/6) - log((3^(1/2)*3i)/2 + 3*(x^3 + 1)^(1/3) + 3/2)*((3^(1/2)*1i)/6 + 1/6)

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sympy [C]  time = 0.94, size = 34, normalized size = 0.42 \begin {gather*} - \frac {x \Gamma \left (- \frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{3}, - \frac {1}{3} \\ \frac {2}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 \Gamma \left (\frac {2}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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