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

Optimal. Leaf size=95 \[ \frac {\left (x^3+1\right )^{2/3} (2-3 x)}{6 x^3}+\frac {1}{9} \log \left (\sqrt [3]{x^3+1}-1\right )-\frac {1}{18} \log \left (\left (x^3+1\right )^{2/3}+\sqrt [3]{x^3+1}+1\right )+\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{x^3+1}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

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Rubi [A]  time = 0.08, antiderivative size = 86, normalized size of antiderivative = 0.91, number of steps used = 9, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1844, 266, 51, 55, 618, 204, 31, 264} \begin {gather*} \frac {\left (x^3+1\right )^{2/3}}{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 {\left (x^3+1\right )^{2/3}}{2 x^2}-\frac {\log (x)}{6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 + x)/(x^4*(1 + x^3)^(1/3)),x]

[Out]

(1 + x^3)^(2/3)/(3*x^3) - (1 + x^3)^(2/3)/(2*x^2) + 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 51

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

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), 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 264

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

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]

Rule 1844

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {-1+x}{x^4 \sqrt [3]{1+x^3}} \, dx &=\int \left (-\frac {1}{x^4 \sqrt [3]{1+x^3}}+\frac {1}{x^3 \sqrt [3]{1+x^3}}\right ) \, dx\\ &=-\int \frac {1}{x^4 \sqrt [3]{1+x^3}} \, dx+\int \frac {1}{x^3 \sqrt [3]{1+x^3}} \, dx\\ &=-\frac {\left (1+x^3\right )^{2/3}}{2 x^2}-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt [3]{1+x}} \, dx,x,x^3\right )\\ &=\frac {\left (1+x^3\right )^{2/3}}{3 x^3}-\frac {\left (1+x^3\right )^{2/3}}{2 x^2}+\frac {1}{9} \operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{1+x}} \, dx,x,x^3\right )\\ &=\frac {\left (1+x^3\right )^{2/3}}{3 x^3}-\frac {\left (1+x^3\right )^{2/3}}{2 x^2}-\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 {\left (1+x^3\right )^{2/3}}{3 x^3}-\frac {\left (1+x^3\right )^{2/3}}{2 x^2}-\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 {\left (1+x^3\right )^{2/3}}{3 x^3}-\frac {\left (1+x^3\right )^{2/3}}{2 x^2}+\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 = 0.02, size = 35, normalized size = 0.37 \begin {gather*} -\frac {\left (x^3+1\right )^{2/3} \left (x^2 \, _2F_1\left (\frac {2}{3},2;\frac {5}{3};x^3+1\right )+1\right )}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x)/(x^4*(1 + x^3)^(1/3)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 8.02, size = 95, normalized size = 1.00 \begin {gather*} \frac {(2-3 x) \left (1+x^3\right )^{2/3}}{6 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)/(x^4*(1 + x^3)^(1/3)),x]

[Out]

((2 - 3*x)*(1 + x^3)^(2/3))/(6*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 = 0.73, size = 104, normalized size = 1.09 \begin {gather*} -\frac {2 \, \sqrt {3} x^{3} \arctan \left (-\frac {\sqrt {3} {\left (x^{3} + 1\right )} - 2 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {2}{3}} + 4 \, \sqrt {3} {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x^{3} + 9}\right ) - x^{3} \log \left (\frac {x^{3} - 3 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} + 3 \, {\left (x^{3} + 1\right )}^{\frac {1}{3}}}{x^{3}}\right ) + 3 \, {\left (x^{3} + 1\right )}^{\frac {2}{3}} {\left (3 \, x - 2\right )}}{18 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*(2*sqrt(3)*x^3*arctan(-(sqrt(3)*(x^3 + 1) - 2*sqrt(3)*(x^3 + 1)^(2/3) + 4*sqrt(3)*(x^3 + 1)^(1/3))/(x^3
+ 9)) - x^3*log((x^3 - 3*(x^3 + 1)^(2/3) + 3*(x^3 + 1)^(1/3))/x^3) + 3*(x^3 + 1)^(2/3)*(3*x - 2))/x^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x - 1}{{\left (x^{3} + 1\right )}^{\frac {1}{3}} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 3.35, size = 90, normalized size = 0.95

method	result	size
meijerg	\(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (\frac {4 \pi \sqrt {3}\, x^{3} \hypergeom \left (\left [1, 1, \frac {7}{3}\right ], \left [2, 3\right ], -x^{3}\right )}{27 \Gamma \left (\frac {2}{3}\right )}-\frac {2 \left (2-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+3 \ln \relax (x )\right ) \pi \sqrt {3}}{9 \Gamma \left (\frac {2}{3}\right )}-\frac {2 \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right ) x^{3}}\right )}{6 \pi }-\frac {\left (x^{3}+1\right )^{\frac {2}{3}}}{2 x^{2}}\)	\(90\)
risch	\(-\frac {3 x^{4}-2 x^{3}+3 x -2}{6 x^{3} \left (x^{3}+1\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (-\frac {2 \pi \sqrt {3}\, x^{3} \hypergeom \left (\left [1, 1, \frac {4}{3}\right ], \left [2, 2\right ], -x^{3}\right )}{9 \Gamma \left (\frac {2}{3}\right )}+\frac {2 \left (-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \relax (3)}{2}+3 \ln \relax (x )\right ) \pi \sqrt {3}}{3 \Gamma \left (\frac {2}{3}\right )}\right )}{18 \pi }\)	\(91\)
trager	\(-\frac {\left (-2+3 x \right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{6 x^{3}}-\frac {\ln \left (-\frac {16 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{3}+34 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x^{3}+30 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}+15 x^{3}-16 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2}+30 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}+24 \left (x^{3}+1\right )^{\frac {2}{3}}+22 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )+24 \left (x^{3}+1\right )^{\frac {1}{3}}+20}{x^{3}}\right )}{9}-\frac {2 \ln \left (-\frac {16 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{3}+34 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x^{3}+30 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}+15 x^{3}-16 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2}+30 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}+24 \left (x^{3}+1\right )^{\frac {2}{3}}+22 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )+24 \left (x^{3}+1\right )^{\frac {1}{3}}+20}{x^{3}}\right ) \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )}{9}+\frac {2 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \ln \left (-\frac {16 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{3}-18 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x^{3}-30 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}+2 x^{3}-16 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2}-30 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}+1\right )^{\frac {1}{3}}+9 \left (x^{3}+1\right )^{\frac {2}{3}}-38 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )+9 \left (x^{3}+1\right )^{\frac {1}{3}}+5}{x^{3}}\right )}{9}\)	\(446\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/Pi*3^(1/2)*GAMMA(2/3)*(4/27*Pi*3^(1/2)/GAMMA(2/3)*x^3*hypergeom([1,1,7/3],[2,3],-x^3)-2/9*(2-1/6*Pi*3^(1/
2)-3/2*ln(3)+3*ln(x))*Pi*3^(1/2)/GAMMA(2/3)-2/3*Pi*3^(1/2)/GAMMA(2/3)/x^3)-1/2*(x^3+1)^(2/3)/x^2

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maxima [A]  time = 0.47, size = 78, normalized size = 0.82 \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 {2}{3}}}{2 \, x^{2}} + \frac {{\left (x^{3} + 1\right )}^{\frac {2}{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((-1+x)/x^4/(x^3+1)^(1/3),x, algorithm="maxima")

[Out]

1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(x^3 + 1)^(1/3) + 1)) - 1/2*(x^3 + 1)^(2/3)/x^2 + 1/3*(x^3 + 1)^(2/3)/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 = 1.34, size = 104, normalized size = 1.09 \begin {gather*} \frac {\ln \left (\frac {{\left (x^3+1\right )}^{1/3}}{9}-\frac {1}{9}\right )}{9}+\ln \left (\frac {{\left (x^3+1\right )}^{1/3}}{9}-9\,{\left (-\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )}^2\right )\,\left (-\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )-\ln \left (\frac {{\left (x^3+1\right )}^{1/3}}{9}-9\,{\left (\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )}^2\right )\,\left (\frac {1}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )-\frac {{\left (x^3+1\right )}^{2/3}}{2\,x^2}+\frac {{\left (x^3+1\right )}^{2/3}}{3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 1)/(x^4*(x^3 + 1)^(1/3)),x)

[Out]

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

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sympy [C]  time = 1.81, size = 53, normalized size = 0.56 \begin {gather*} \frac {\left (1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {2}{3}\right )}{3 \Gamma \left (\frac {1}{3}\right )} + \frac {\Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {4}{3} \\ \frac {7}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{4} \Gamma \left (\frac {7}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1 + x**(-3))**(2/3)*gamma(-2/3)/(3*gamma(1/3)) + gamma(4/3)*hyper((1/3, 4/3), (7/3,), exp_polar(I*pi)/x**3)/(
3*x**4*gamma(7/3))

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