3.16.92 \(\int \frac {-1+x}{x \sqrt [3]{1+2 x+2 x^2+x^3}} \, dx\)

Optimal. Leaf size=132 \[ -\log \left (\sqrt [3]{x^3+2 x^2+2 x+1}-x-1\right )+\frac {1}{2} \log \left (x^2+\left (x^3+2 x^2+2 x+1\right )^{2/3}+(x+1) \sqrt [3]{x^3+2 x^2+2 x+1}+2 x+1\right )-\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x^3+2 x^2+2 x+1}}{\sqrt [3]{x^3+2 x^2+2 x+1}+2 x+2}\right ) \]

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Rubi [F]  time = 0.52, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1+x}{x \sqrt [3]{1+2 x+2 x^2+x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(3*(1 + x)*(1 - (2*(1 + x))/(1 - I*Sqrt[3]))^(1/3)*(1 - (2*(1 + x))/(1 + I*Sqrt[3]))^(1/3)*AppellF1[2/3, 1/3,
1/3, 5/3, (2*(1 + x))/(1 - I*Sqrt[3]), (2*(1 + x))/(1 + I*Sqrt[3])])/(2*(1 + 2*x + 2*x^2 + x^3)^(1/3)) - ((1 +
x)^(1/3)*(1 + x + x^2)^(1/3)*Defer[Subst][Defer[Int][1/((-2/3 + x)*(1/3 + x)^(1/3)*(7/9 - x/3 + x^2)^(1/3)),
x], x, 2/3 + x])/(1 + 2*x + 2*x^2 + x^3)^(1/3)

Rubi steps

\begin {align*} \int \frac {-1+x}{x \sqrt [3]{1+2 x+2 x^2+x^3}} \, dx &=\int \left (\frac {1}{\sqrt [3]{1+2 x+2 x^2+x^3}}-\frac {1}{x \sqrt [3]{1+2 x+2 x^2+x^3}}\right ) \, dx\\ &=\int \frac {1}{\sqrt [3]{1+2 x+2 x^2+x^3}} \, dx-\int \frac {1}{x \sqrt [3]{1+2 x+2 x^2+x^3}} \, dx\\ &=\operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{\frac {7}{27}+\frac {2 x}{3}+x^3}} \, dx,x,\frac {2}{3}+x\right )-\operatorname {Subst}\left (\int \frac {1}{\left (-\frac {2}{3}+x\right ) \sqrt [3]{\frac {7}{27}+\frac {2 x}{3}+x^3}} \, dx,x,\frac {2}{3}+x\right )\\ &=\frac {\left (\sqrt [3]{1+x} \sqrt [3]{1+x+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{\frac {1}{3}+x} \sqrt [3]{\frac {7}{9}-\frac {x}{3}+x^2}} \, dx,x,\frac {2}{3}+x\right )}{\sqrt [3]{1+2 x+2 x^2+x^3}}-\frac {\left (\sqrt [3]{1+x} \sqrt [3]{1+x+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-\frac {2}{3}+x\right ) \sqrt [3]{\frac {1}{3}+x} \sqrt [3]{\frac {7}{9}-\frac {x}{3}+x^2}} \, dx,x,\frac {2}{3}+x\right )}{\sqrt [3]{1+2 x+2 x^2+x^3}}\\ &=\frac {\left (\sqrt [3]{\frac {-i+\sqrt {3}-2 i x}{i+\sqrt {3}}} \sqrt [3]{\frac {i+\sqrt {3}+2 i x}{-i+\sqrt {3}}} \sqrt [3]{1+x}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{x} \sqrt [3]{1-\frac {2 x}{1-i \sqrt {3}}} \sqrt [3]{1-\frac {2 x}{1+i \sqrt {3}}}} \, dx,x,1+x\right )}{\sqrt [3]{1+2 x+2 x^2+x^3}}-\frac {\left (\sqrt [3]{1+x} \sqrt [3]{1+x+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-\frac {2}{3}+x\right ) \sqrt [3]{\frac {1}{3}+x} \sqrt [3]{\frac {7}{9}-\frac {x}{3}+x^2}} \, dx,x,\frac {2}{3}+x\right )}{\sqrt [3]{1+2 x+2 x^2+x^3}}\\ &=\frac {3 \sqrt [3]{-\frac {i-\sqrt {3}+2 i x}{i+\sqrt {3}}} \sqrt [3]{-\frac {i+\sqrt {3}+2 i x}{i-\sqrt {3}}} (1+x) F_1\left (\frac {2}{3};\frac {1}{3},\frac {1}{3};\frac {5}{3};\frac {2 (1+x)}{1-i \sqrt {3}},\frac {2 (1+x)}{1+i \sqrt {3}}\right )}{2 \sqrt [3]{1+2 x+2 x^2+x^3}}-\frac {\left (\sqrt [3]{1+x} \sqrt [3]{1+x+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-\frac {2}{3}+x\right ) \sqrt [3]{\frac {1}{3}+x} \sqrt [3]{\frac {7}{9}-\frac {x}{3}+x^2}} \, dx,x,\frac {2}{3}+x\right )}{\sqrt [3]{1+2 x+2 x^2+x^3}}\\ \end {align*}

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Mathematica [F]  time = 0.10, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+x}{x \sqrt [3]{1+2 x+2 x^2+x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.28, size = 132, normalized size = 1.00 \begin {gather*} -\log \left (\sqrt [3]{x^3+2 x^2+2 x+1}-x-1\right )+\frac {1}{2} \log \left (x^2+\left (x^3+2 x^2+2 x+1\right )^{2/3}+(x+1) \sqrt [3]{x^3+2 x^2+2 x+1}+2 x+1\right )-\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x^3+2 x^2+2 x+1}}{\sqrt [3]{x^3+2 x^2+2 x+1}+2 x+2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 123, normalized size = 0.93 \begin {gather*} \sqrt {3} \arctan \left (-\frac {4 \, \sqrt {3} {\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} + \sqrt {3} {\left (x^{2} + x + 1\right )} - 2 \, \sqrt {3} {\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )}^{\frac {2}{3}}}{9 \, x^{2} + 17 \, x + 9}\right ) - \frac {1}{2} \, \log \left (-\frac {3 \, {\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} - x - 3 \, {\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )}^{\frac {2}{3}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)/x/(x^3+2*x^2+2*x+1)^(1/3),x, algorithm="fricas")

[Out]

sqrt(3)*arctan(-(4*sqrt(3)*(x^3 + 2*x^2 + 2*x + 1)^(1/3)*(x + 1) + sqrt(3)*(x^2 + x + 1) - 2*sqrt(3)*(x^3 + 2*
x^2 + 2*x + 1)^(2/3))/(9*x^2 + 17*x + 9)) - 1/2*log(-(3*(x^3 + 2*x^2 + 2*x + 1)^(1/3)*(x + 1) - x - 3*(x^3 + 2
*x^2 + 2*x + 1)^(2/3))/x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x - 1}{{\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)/x/(x^3+2*x^2+2*x+1)^(1/3),x, algorithm="giac")

[Out]

integrate((x - 1)/((x^3 + 2*x^2 + 2*x + 1)^(1/3)*x), x)

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maple [C]  time = 1.19, size = 455, normalized size = 3.45 \begin {gather*} \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+111 \left (x^{3}+2 x^{2}+2 x +1\right )^{\frac {2}{3}}+111 x \left (x^{3}+2 x^{2}+2 x +1\right )^{\frac {1}{3}}-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}-39 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +110 x^{2}+111 \left (x^{3}+2 x^{2}+2 x +1\right )^{\frac {1}{3}}+\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+187 x +110}{x}\right )-\ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+111 \left (x^{3}+2 x^{2}+2 x +1\right )^{\frac {2}{3}}+111 x \left (x^{3}+2 x^{2}+2 x +1\right )^{\frac {1}{3}}-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}+35 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +110 x^{2}+111 \left (x^{3}+2 x^{2}+2 x +1\right )^{\frac {1}{3}}+\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+150 x +110}{x}\right ) \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+\ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+111 \left (x^{3}+2 x^{2}+2 x +1\right )^{\frac {2}{3}}+111 x \left (x^{3}+2 x^{2}+2 x +1\right )^{\frac {1}{3}}-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2}+35 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +110 x^{2}+111 \left (x^{3}+2 x^{2}+2 x +1\right )^{\frac {1}{3}}+\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+150 x +110}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1+x)/x/(x^3+2*x^2+2*x+1)^(1/3),x)

[Out]

RootOf(_Z^2-_Z+1)*ln((-RootOf(_Z^2-_Z+1)^2*x^2+2*RootOf(_Z^2-_Z+1)^2*x+RootOf(_Z^2-_Z+1)*x^2+111*(x^3+2*x^2+2*
x+1)^(2/3)+111*x*(x^3+2*x^2+2*x+1)^(1/3)-RootOf(_Z^2-_Z+1)^2-39*RootOf(_Z^2-_Z+1)*x+110*x^2+111*(x^3+2*x^2+2*x
+1)^(1/3)+RootOf(_Z^2-_Z+1)+187*x+110)/x)-ln((-RootOf(_Z^2-_Z+1)^2*x^2+2*RootOf(_Z^2-_Z+1)^2*x+RootOf(_Z^2-_Z+
1)*x^2+111*(x^3+2*x^2+2*x+1)^(2/3)+111*x*(x^3+2*x^2+2*x+1)^(1/3)-RootOf(_Z^2-_Z+1)^2+35*RootOf(_Z^2-_Z+1)*x+11
0*x^2+111*(x^3+2*x^2+2*x+1)^(1/3)+RootOf(_Z^2-_Z+1)+150*x+110)/x)*RootOf(_Z^2-_Z+1)+ln((-RootOf(_Z^2-_Z+1)^2*x
^2+2*RootOf(_Z^2-_Z+1)^2*x+RootOf(_Z^2-_Z+1)*x^2+111*(x^3+2*x^2+2*x+1)^(2/3)+111*x*(x^3+2*x^2+2*x+1)^(1/3)-Roo
tOf(_Z^2-_Z+1)^2+35*RootOf(_Z^2-_Z+1)*x+110*x^2+111*(x^3+2*x^2+2*x+1)^(1/3)+RootOf(_Z^2-_Z+1)+150*x+110)/x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x - 1}{{\left (x^{3} + 2 \, x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)/x/(x^3+2*x^2+2*x+1)^(1/3),x, algorithm="maxima")

[Out]

integrate((x - 1)/((x^3 + 2*x^2 + 2*x + 1)^(1/3)*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x-1}{x\,{\left (x^3+2\,x^2+2\,x+1\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 1)/(x*(2*x + 2*x^2 + x^3 + 1)^(1/3)),x)

[Out]

int((x - 1)/(x*(2*x + 2*x^2 + x^3 + 1)^(1/3)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+x)/x/(x**3+2*x**2+2*x+1)**(1/3),x)

[Out]

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

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