3.12.99 \(\int \frac {1}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx\)

Optimal. Leaf size=88 \[ -\frac {1}{4} \log \left (\sqrt [3]{x^2-4 x-4}+2\right )+\frac {1}{8} \log \left (\left (x^2-4 x-4\right )^{2/3}-2 \sqrt [3]{x^2-4 x-4}+4\right )-\frac {1}{4} \sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {\sqrt [3]{x^2-4 x-4}}{\sqrt {3}}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 60, normalized size of antiderivative = 0.68, 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, 618, 204, 31} \begin {gather*} -\frac {3}{8} \log \left (\sqrt [3]{(x-2)^2-8}+2\right )+\frac {1}{4} \log (2-x)-\frac {1}{4} \sqrt {3} \tan ^{-1}\left (\frac {1-\sqrt [3]{(x-2)^2-8}}{\sqrt {3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((-2 + x)*(-4 - 4*x + x^2)^(1/3)),x]

[Out]

-1/4*(Sqrt[3]*ArcTan[(1 - (-8 + (-2 + x)^2)^(1/3))/Sqrt[3]]) - (3*Log[2 + (-8 + (-2 + x)^2)^(1/3)])/8 + Log[2
- x]/4

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 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}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{x \sqrt [3]{-8+x^2}} \, dx,x,-2+x\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{-8+x} x} \, dx,x,(-2+x)^2\right )\\ &=\frac {1}{4} \log (2-x)-\frac {3}{8} \operatorname {Subst}\left (\int \frac {1}{2+x} \, dx,x,\sqrt [3]{-8+(-2+x)^2}\right )+\frac {3}{4} \operatorname {Subst}\left (\int \frac {1}{4-2 x+x^2} \, dx,x,\sqrt [3]{-8+(-2+x)^2}\right )\\ &=-\frac {3}{8} \log \left (2+\sqrt [3]{-8+(-2+x)^2}\right )+\frac {1}{4} \log (2-x)-\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{-12-x^2} \, dx,x,-2+2 \sqrt [3]{-8+(-2+x)^2}\right )\\ &=-\frac {1}{4} \sqrt {3} \tan ^{-1}\left (\frac {1-\sqrt [3]{-8+(-2+x)^2}}{\sqrt {3}}\right )-\frac {3}{8} \log \left (2+\sqrt [3]{-8+(-2+x)^2}\right )+\frac {1}{4} \log (2-x)\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 36, normalized size = 0.41 \begin {gather*} \frac {3}{32} \left ((x-2)^2-8\right )^{2/3} \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {1}{8} \left (8-(x-2)^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((-2 + x)*(-4 - 4*x + x^2)^(1/3)),x]

[Out]

(3*(-8 + (-2 + x)^2)^(2/3)*Hypergeometric2F1[2/3, 1, 5/3, (8 - (-2 + x)^2)/8])/32

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IntegrateAlgebraic [A]  time = 0.10, size = 88, normalized size = 1.00 \begin {gather*} -\frac {1}{4} \sqrt {3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {\sqrt [3]{-4-4 x+x^2}}{\sqrt {3}}\right )-\frac {1}{4} \log \left (2+\sqrt [3]{-4-4 x+x^2}\right )+\frac {1}{8} \log \left (4-2 \sqrt [3]{-4-4 x+x^2}+\left (-4-4 x+x^2\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((-2 + x)*(-4 - 4*x + x^2)^(1/3)),x]

[Out]

-1/4*(Sqrt[3]*ArcTan[1/Sqrt[3] - (-4 - 4*x + x^2)^(1/3)/Sqrt[3]]) - Log[2 + (-4 - 4*x + x^2)^(1/3)]/4 + Log[4
- 2*(-4 - 4*x + x^2)^(1/3) + (-4 - 4*x + x^2)^(2/3)]/8

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fricas [A]  time = 0.45, size = 70, normalized size = 0.80 \begin {gather*} \frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x^{2} - 4 \, x - 4\right )}^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + \frac {1}{8} \, \log \left ({\left (x^{2} - 4 \, x - 4\right )}^{\frac {2}{3}} - 2 \, {\left (x^{2} - 4 \, x - 4\right )}^{\frac {1}{3}} + 4\right ) - \frac {1}{4} \, \log \left ({\left (x^{2} - 4 \, x - 4\right )}^{\frac {1}{3}} + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2+x)/(x^2-4*x-4)^(1/3),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (x^{2} - 4 \, x - 4\right )}^{\frac {1}{3}} {\left (x - 2\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2+x)/(x^2-4*x-4)^(1/3),x, algorithm="giac")

[Out]

integrate(1/((x^2 - 4*x - 4)^(1/3)*(x - 2)), x)

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maple [C]  time = 1.96, size = 346, normalized size = 3.93

method	result	size
trager	\(-\frac {\ln \left (-\frac {4 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x^{2}+48 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {2}{3}}-16 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x +9 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x^{2}-9 \left (x^{2}-4 x -4\right )^{\frac {2}{3}}+60 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {1}{3}}-36 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x +2 x^{2}-48 \left (x^{2}-4 x -4\right )^{\frac {1}{3}}-28 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )-8 x -56}{\left (-2+x \right )^{2}}\right )}{4}+\frac {\RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \ln \left (\frac {-32 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x^{2}+96 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {2}{3}}+128 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x +20 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x^{2}-30 \left (x^{2}-4 x -4\right )^{\frac {2}{3}}+72 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {1}{3}}-80 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x +3 x^{2}-96 \left (x^{2}-4 x -4\right )^{\frac {1}{3}}-224 \RootOf \left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )-12 x -28}{\left (-2+x \right )^{2}}\right )}{2}\)	\(346\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-2+x)/(x^2-4*x-4)^(1/3),x,method=_RETURNVERBOSE)

[Out]

-1/4*ln(-(4*RootOf(4*_Z^2-2*_Z+1)^2*x^2+48*RootOf(4*_Z^2-2*_Z+1)*(x^2-4*x-4)^(2/3)-16*RootOf(4*_Z^2-2*_Z+1)^2*
x+9*RootOf(4*_Z^2-2*_Z+1)*x^2-9*(x^2-4*x-4)^(2/3)+60*RootOf(4*_Z^2-2*_Z+1)*(x^2-4*x-4)^(1/3)-36*RootOf(4*_Z^2-
2*_Z+1)*x+2*x^2-48*(x^2-4*x-4)^(1/3)-28*RootOf(4*_Z^2-2*_Z+1)-8*x-56)/(-2+x)^2)+1/2*RootOf(4*_Z^2-2*_Z+1)*ln((
-32*RootOf(4*_Z^2-2*_Z+1)^2*x^2+96*RootOf(4*_Z^2-2*_Z+1)*(x^2-4*x-4)^(2/3)+128*RootOf(4*_Z^2-2*_Z+1)^2*x+20*Ro
otOf(4*_Z^2-2*_Z+1)*x^2-30*(x^2-4*x-4)^(2/3)+72*RootOf(4*_Z^2-2*_Z+1)*(x^2-4*x-4)^(1/3)-80*RootOf(4*_Z^2-2*_Z+
1)*x+3*x^2-96*(x^2-4*x-4)^(1/3)-224*RootOf(4*_Z^2-2*_Z+1)-12*x-28)/(-2+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} - 4 \, x - 4\right )}^{\frac {1}{3}} {\left (x - 2\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2+x)/(x^2-4*x-4)^(1/3),x, algorithm="maxima")

[Out]

integrate(1/((x^2 - 4*x - 4)^(1/3)*(x - 2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\left (x-2\right )\,{\left (x^2-4\,x-4\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x - 2)*(x^2 - 4*x - 4)^(1/3)),x)

[Out]

int(1/((x - 2)*(x^2 - 4*x - 4)^(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 - 2\right ) \sqrt [3]{x^{2} - 4 x - 4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-2+x)/(x**2-4*x-4)**(1/3),x)

[Out]

Integral(1/((x - 2)*(x**2 - 4*x - 4)**(1/3)), x)

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