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

Optimal. Leaf size=86 \[ \frac {1}{2} \log \left (\sqrt [3]{x^2-2 x+2}-1\right )-\frac {1}{4} \log \left (\left (x^2-2 x+2\right )^{2/3}+\sqrt [3]{x^2-2 x+2}+1\right )+\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x^2-2 x+2}}{\sqrt {3}}+\frac {1}{\sqrt {3}}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 62, normalized size of antiderivative = 0.72, 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, 55, 618, 204, 31} \begin {gather*} \frac {3}{4} \log \left (1-\sqrt [3]{(x-1)^2+1}\right )-\frac {1}{2} \log (1-x)+\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{(x-1)^2+1}+1}{\sqrt {3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 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 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]{2-2 x+x^2}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{1+x^2}} \, dx,x,-1+x\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{1+x}} \, dx,x,(-1+x)^2\right )\\ &=-\frac {1}{2} \log (1-x)-\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1+(-1+x)^2}\right )+\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1+(-1+x)^2}\right )\\ &=\frac {3}{4} \log \left (1-\sqrt [3]{1+(-1+x)^2}\right )-\frac {1}{2} \log (1-x)-\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1+(-1+x)^2}\right )\\ &=\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {1+2 \sqrt [3]{1+(-1+x)^2}}{\sqrt {3}}\right )+\frac {3}{4} \log \left (1-\sqrt [3]{1+(-1+x)^2}\right )-\frac {1}{2} \log (1-x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 61, normalized size = 0.71 \begin {gather*} \frac {1}{2} \left (\frac {3}{2} \log \left (1-\sqrt [3]{(x-1)^2+1}\right )-\log (1-x)+\sqrt {3} \tan ^{-1}\left (\frac {2 \sqrt [3]{(x-1)^2+1}+1}{\sqrt {3}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.09, size = 86, normalized size = 1.00 \begin {gather*} \frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{2-2 x+x^2}}{\sqrt {3}}\right )+\frac {1}{2} \log \left (-1+\sqrt [3]{2-2 x+x^2}\right )-\frac {1}{4} \log \left (1+\sqrt [3]{2-2 x+x^2}+\left (2-2 x+x^2\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 68, normalized size = 0.79 \begin {gather*} \frac {1}{2} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - \frac {1}{4} \, \log \left ({\left (x^{2} - 2 \, x + 2\right )}^{\frac {2}{3}} + {\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{2} \, \log \left ({\left (x^{2} - 2 \, x + 2\right )}^{\frac {1}{3}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 + 2\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+2)^(1/3),x, algorithm="giac")

[Out]

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

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maple [C]  time = 1.68, size = 432, normalized size = 5.02

method	result	size
trager	\(\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^{2}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-2 x +2\right )^{\frac {2}{3}}+8 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x +9 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}-9 \left (x^{2}-2 x +2\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-2 x +2\right )^{\frac {1}{3}}-18 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x -2 x^{2}-9 \left (x^{2}-2 x +2\right )^{\frac {1}{3}}+28 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )+4 x -7}{\left (-1+x \right )^{2}}\right )}{2}-\frac {\ln \left (-\frac {4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-2 x +2\right )^{\frac {2}{3}}-8 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x +17 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+24 \left (x^{2}-2 x +2\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-2 x +2\right )^{\frac {1}{3}}-34 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x +15 x^{2}+24 \left (x^{2}-2 x +2\right )^{\frac {1}{3}}+28 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-30 x +35}{\left (-1+x \right )^{2}}\right ) \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )}{2}-\frac {\ln \left (-\frac {4 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x^{2}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-2 x +2\right )^{\frac {2}{3}}-8 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )^{2} x +17 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x^{2}+24 \left (x^{2}-2 x +2\right )^{\frac {2}{3}}+15 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) \left (x^{2}-2 x +2\right )^{\frac {1}{3}}-34 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right ) x +15 x^{2}+24 \left (x^{2}-2 x +2\right )^{\frac {1}{3}}+28 \RootOf \left (\textit {\_Z}^{2}+\textit {\_Z} +1\right )-30 x +35}{\left (-1+x \right )^{2}}\right )}{2}\)	\(432\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*RootOf(_Z^2+_Z+1)*ln((-4*RootOf(_Z^2+_Z+1)^2*x^2+15*RootOf(_Z^2+_Z+1)*(x^2-2*x+2)^(2/3)+8*RootOf(_Z^2+_Z+1
)^2*x+9*RootOf(_Z^2+_Z+1)*x^2-9*(x^2-2*x+2)^(2/3)+15*RootOf(_Z^2+_Z+1)*(x^2-2*x+2)^(1/3)-18*RootOf(_Z^2+_Z+1)*
x-2*x^2-9*(x^2-2*x+2)^(1/3)+28*RootOf(_Z^2+_Z+1)+4*x-7)/(-1+x)^2)-1/2*ln(-(4*RootOf(_Z^2+_Z+1)^2*x^2+15*RootOf
(_Z^2+_Z+1)*(x^2-2*x+2)^(2/3)-8*RootOf(_Z^2+_Z+1)^2*x+17*RootOf(_Z^2+_Z+1)*x^2+24*(x^2-2*x+2)^(2/3)+15*RootOf(
_Z^2+_Z+1)*(x^2-2*x+2)^(1/3)-34*RootOf(_Z^2+_Z+1)*x+15*x^2+24*(x^2-2*x+2)^(1/3)+28*RootOf(_Z^2+_Z+1)-30*x+35)/
(-1+x)^2)*RootOf(_Z^2+_Z+1)-1/2*ln(-(4*RootOf(_Z^2+_Z+1)^2*x^2+15*RootOf(_Z^2+_Z+1)*(x^2-2*x+2)^(2/3)-8*RootOf
(_Z^2+_Z+1)^2*x+17*RootOf(_Z^2+_Z+1)*x^2+24*(x^2-2*x+2)^(2/3)+15*RootOf(_Z^2+_Z+1)*(x^2-2*x+2)^(1/3)-34*RootOf
(_Z^2+_Z+1)*x+15*x^2+24*(x^2-2*x+2)^(1/3)+28*RootOf(_Z^2+_Z+1)-30*x+35)/(-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 + 2\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+2)^(1/3),x, algorithm="maxima")

[Out]

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

[Out]

int(1/((x - 1)*(x^2 - 2*x + 2)^(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 + 2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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