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

Optimal. Leaf size=125 \[ -\frac {\log \left (2^{2/3} \sqrt [3]{x^2-2 x-1}+2\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-\sqrt [3]{2} \left (x^2-2 x-1\right )^{2/3}+2^{2/3} \sqrt [3]{x^2-2 x-1}-2\right )}{4 \sqrt [3]{2}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \sqrt [3]{x^2-2 x-1}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 84, normalized size of antiderivative = 0.67, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {694, 266, 56, 617, 204, 31} \begin {gather*} -\frac {3 \log \left (\sqrt [3]{(x-1)^2-2}+\sqrt [3]{2}\right )}{4 \sqrt [3]{2}}+\frac {\log (1-x)}{2 \sqrt [3]{2}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {1-2^{2/3} \sqrt [3]{(x-1)^2-2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[3]*ArcTan[(1 - 2^(2/3)*(-2 + (-1 + x)^2)^(1/3))/Sqrt[3]])/2^(1/3) - (3*Log[2^(1/3) + (-2 + (-1 + x)
^2)^(1/3)])/(4*2^(1/3)) + Log[1 - x]/(2*2^(1/3))

Rule 31

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

Rule 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[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] && Ne
gQ[(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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 694

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[x^m*(
a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& EqQ[2*c*d - b*e, 0]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 36, normalized size = 0.29 \begin {gather*} \frac {3}{8} \left ((x-1)^2-2\right )^{2/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {1}{2} \left (2-(x-1)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-2 + (-1 + x)^2)^(2/3)*Hypergeometric2F1[2/3, 1, 5/3, (2 - (-1 + x)^2)/2])/8

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IntegrateAlgebraic [A]  time = 0.21, size = 125, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2^{2/3} \sqrt [3]{-1-2 x+x^2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}-\frac {\log \left (2+2^{2/3} \sqrt [3]{-1-2 x+x^2}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-2+2^{2/3} \sqrt [3]{-1-2 x+x^2}-\sqrt [3]{2} \left (-1-2 x+x^2\right )^{2/3}\right )}{4 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[3]*ArcTan[1/Sqrt[3] - (2^(2/3)*(-1 - 2*x + x^2)^(1/3))/Sqrt[3]])/2^(1/3) - Log[2 + 2^(2/3)*(-1 - 2*
x + x^2)^(1/3)]/(2*2^(1/3)) + Log[-2 + 2^(2/3)*(-1 - 2*x + x^2)^(1/3) - 2^(1/3)*(-1 - 2*x + x^2)^(2/3)]/(4*2^(
1/3))

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fricas [A]  time = 0.45, size = 116, normalized size = 0.93 \begin {gather*} \frac {1}{4} \, \sqrt {3} 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{6}} {\left (2 \, \sqrt {2} \left (-1\right )^{\frac {1}{3}} {\left (x^{2} - 2 \, x - 1\right )}^{\frac {1}{3}} + 2^{\frac {5}{6}}\right )}\right ) - \frac {1}{8} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{2} - 2 \, x - 1\right )}^{\frac {1}{3}} - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} + {\left (x^{2} - 2 \, x - 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{4} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} + {\left (x^{2} - 2 \, x - 1\right )}^{\frac {1}{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 8.37, size = 802, normalized size = 6.42

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{3}+4\right ) \ln \left (-\frac {-2 \RootOf \left (\textit {\_Z}^{3}+4\right )^{3} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) x^{2}+\RootOf \left (\textit {\_Z}^{3}+4\right )^{2} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right )^{2} x^{2}+12 \left (x^{2}-2 x -1\right )^{\frac {2}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+4 \RootOf \left (\textit {\_Z}^{3}+4\right )^{3} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) x -2 \RootOf \left (\textit {\_Z}^{3}+4\right )^{2} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right )^{2} x -12 \left (x^{2}-2 x -1\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3}+4\right )^{2}-15 \left (x^{2}-2 x -1\right )^{\frac {1}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+4\right )-2 \RootOf \left (\textit {\_Z}^{3}+4\right ) x^{2}+\RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) x^{2}-9 \left (x^{2}-2 x -1\right )^{\frac {2}{3}}+4 \RootOf \left (\textit {\_Z}^{3}+4\right ) x -2 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) x +14 \RootOf \left (\textit {\_Z}^{3}+4\right )-7 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right )}{\left (-1+x \right )^{2}}\right )}{4}+\frac {\RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{3}+4\right )^{3} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) x^{2}-8 \RootOf \left (\textit {\_Z}^{3}+4\right )^{2} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right )^{2} x^{2}+24 \left (x^{2}-2 x -1\right )^{\frac {2}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+4\right )^{2}-2 \RootOf \left (\textit {\_Z}^{3}+4\right )^{3} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) x +16 \RootOf \left (\textit {\_Z}^{3}+4\right )^{2} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right )^{2} x -24 \left (x^{2}-2 x -1\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3}+4\right )^{2}-18 \left (x^{2}-2 x -1\right )^{\frac {1}{3}} \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}+4\right )-3 \RootOf \left (\textit {\_Z}^{3}+4\right ) x^{2}+24 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) x^{2}-30 \left (x^{2}-2 x -1\right )^{\frac {2}{3}}+6 \RootOf \left (\textit {\_Z}^{3}+4\right ) x -48 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right ) x +7 \RootOf \left (\textit {\_Z}^{3}+4\right )-56 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}+4\right )^{2}+2 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}+4\right )+4 \textit {\_Z}^{2}\right )}{\left (-1+x \right )^{2}}\right )}{2}\)	\(802\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*RootOf(_Z^3+4)*ln(-(-2*RootOf(_Z^3+4)^3*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^2+RootOf(_Z^
3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*x^2+12*(x^2-2*x-1)^(2/3)*RootOf(RootOf(_Z^3+4)^2+
2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^2+4*RootOf(_Z^3+4)^3*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*
_Z^2)*x-2*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*x-12*(x^2-2*x-1)^(1/3)*RootOf
(_Z^3+4)^2-15*(x^2-2*x-1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)-2*RootOf(_Z
^3+4)*x^2+RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^2-9*(x^2-2*x-1)^(2/3)+4*RootOf(_Z^3+4)*x-2*Roo
tOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x+14*RootOf(_Z^3+4)-7*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^
3+4)+4*_Z^2))/(-1+x)^2)+1/2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*ln((RootOf(_Z^3+4)^3*RootOf(Ro
otOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^2-8*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+
4*_Z^2)^2*x^2+24*(x^2-2*x-1)^(2/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^2-2*Root
Of(_Z^3+4)^3*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x+16*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2
+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*x-24*(x^2-2*x-1)^(1/3)*RootOf(_Z^3+4)^2-18*(x^2-2*x-1)^(1/3)*RootOf(RootOf(_Z^3
+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)-3*RootOf(_Z^3+4)*x^2+24*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_
Z^3+4)+4*_Z^2)*x^2-30*(x^2-2*x-1)^(2/3)+6*RootOf(_Z^3+4)*x-48*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z
^2)*x+7*RootOf(_Z^3+4)-56*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2))/(-1+x)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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