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

Optimal. Leaf size=67 \[ -\frac {1}{2} \log (x+1)-\frac {3}{2} \log \left (1-\frac {x-1}{\sqrt [3]{(x-1)^2 (x+1)}}\right )+\sqrt {3} \tan ^{-1}\left (\frac {\frac {2 (x-1)}{\sqrt [3]{(x-1)^2 (x+1)}}+1}{\sqrt {3}}\right ) \]

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Rubi [B]  time = 0.12, antiderivative size = 188, normalized size of antiderivative = 2.81, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2067, 2064, 60} \[ -\frac {(3-3 x)^{2/3} \sqrt [3]{x+1} \log \left (-\frac {8}{3} (x-1)\right )}{2\ 3^{2/3} \sqrt [3]{x^3-x^2-x+1}}-\frac {\sqrt [3]{3} (3-3 x)^{2/3} \sqrt [3]{x+1} \log \left (\frac {\sqrt [3]{3} \sqrt [3]{x+1}}{\sqrt [3]{3-3 x}}+1\right )}{2 \sqrt [3]{x^3-x^2-x+1}}-\frac {(3-3 x)^{2/3} \sqrt [3]{x+1} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{x+1}}{\sqrt [6]{3} \sqrt [3]{3-3 x}}\right )}{\sqrt [6]{3} \sqrt [3]{x^3-x^2-x+1}} \]

Antiderivative was successfully verified.

[In]

Int[((-1 + x)^2*(1 + x))^(-1/3),x]

[Out]

-(((3 - 3*x)^(2/3)*(1 + x)^(1/3)*ArcTan[1/Sqrt[3] - (2*(1 + x)^(1/3))/(3^(1/6)*(3 - 3*x)^(1/3))])/(3^(1/6)*(1
- x - x^2 + x^3)^(1/3))) - ((3 - 3*x)^(2/3)*(1 + x)^(1/3)*Log[(-8*(-1 + x))/3])/(2*3^(2/3)*(1 - x - x^2 + x^3)
^(1/3)) - (3^(1/3)*(3 - 3*x)^(2/3)*(1 + x)^(1/3)*Log[1 + (3^(1/3)*(1 + x)^(1/3))/(3 - 3*x)^(1/3)])/(2*(1 - x -
x^2 + x^3)^(1/3))

Rule 60

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-(d/b), 3]}, Simp[(Sq
rt[3]*q*ArcTan[1/Sqrt[3] - (2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/3))])/d, x] + (Simp[(3*q*Log[(q*(a + b*
x)^(1/3))/(c + d*x)^(1/3) + 1])/(2*d), x] + Simp[(q*Log[c + d*x])/(2*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ
[b*c - a*d, 0] && NegQ[d/b]

Rule 2064

Int[((a_.) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> Dist[(a + b*x + d*x^3)^p/((3*a - b*x)^p*(3*a + 2*b*
x)^(2*p)), Int[(3*a - b*x)^p*(3*a + 2*b*x)^(2*p), x], x] /; FreeQ[{a, b, d, p}, x] && EqQ[4*b^3 + 27*a^2*d, 0]
&&  !IntegerQ[p]

Rule 2067

Int[(P3_)^(p_), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = Coeff[P3, x, 2], d = Coeff[P3
, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c*d + 27*a*d^2)/(27*d^2) - ((c^2 - 3*b*d)*x)/(3*d) + d*x^3, x]^p, x], x,
x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[p, x] && PolyQ[P3, x, 3]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{\frac {16}{27}-\frac {4 x}{3}+x^3}} \, dx,x,-\frac {1}{3}+x\right )\\ &=\frac {\left (4\ 2^{2/3} (1-x)^{2/3} \sqrt [3]{1+x}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\frac {16}{9}-\frac {8 x}{3}\right )^{2/3} \sqrt [3]{\frac {16}{9}+\frac {4 x}{3}}} \, dx,x,-\frac {1}{3}+x\right )}{3 \sqrt [3]{1-x-x^2+x^3}}\\ &=-\frac {\sqrt {3} (1-x)^{2/3} \sqrt [3]{1+x} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1+x}}{\sqrt {3} \sqrt [3]{1-x}}\right )}{\sqrt [3]{1-x-x^2+x^3}}-\frac {(1-x)^{2/3} \sqrt [3]{1+x} \log (1-x)}{2 \sqrt [3]{1-x-x^2+x^3}}-\frac {3 (1-x)^{2/3} \sqrt [3]{1+x} \log \left (\frac {3 \left (\sqrt [3]{1-x}+\sqrt [3]{1+x}\right )}{\sqrt [3]{1-x}}\right )}{2 \sqrt [3]{1-x-x^2+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 49, normalized size = 0.73 \[ \frac {3 (x-1) \sqrt [3]{x+1} \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};\frac {1-x}{2}\right )}{\sqrt [3]{2} \sqrt [3]{(x-1)^2 (x+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-1 + x)*(1 + x)^(1/3)*Hypergeometric2F1[1/3, 1/3, 4/3, (1 - x)/2])/(2^(1/3)*((-1 + x)^2*(1 + x))^(1/3))

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IntegrateAlgebraic [A]  time = 0.26, size = 132, normalized size = 1.97 \[ -\log \left (\sqrt [3]{x^3-x^2-x+1}-x+1\right )+\frac {1}{2} \log \left (x^2+\left (x^3-x^2-x+1\right )^{2/3}+(x-1) \sqrt [3]{x^3-x^2-x+1}-2 x+1\right )-\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x^3-x^2-x+1}}{\sqrt [3]{x^3-x^2-x+1}+2 x-2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.90, size = 128, normalized size = 1.91 \[ -\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x - 1\right )} + 2 \, \sqrt {3} {\left (x^{3} - x^{2} - x + 1\right )}^{\frac {1}{3}}}{3 \, {\left (x - 1\right )}}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} - x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} - 2 \, x + {\left (x^{3} - x^{2} - x + 1\right )}^{\frac {2}{3}} + 1}{x^{2} - 2 \, x + 1}\right ) - \log \left (-\frac {x - {\left (x^{3} - x^{2} - x + 1\right )}^{\frac {1}{3}} - 1}{x - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.44, size = 370, normalized size = 5.52

method	result	size
trager	\(-\ln \left (\frac {4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}-4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}}-3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x -4 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}}+2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x +x^{2}+2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}}-1}{-1+x}\right )+\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (-\frac {2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}-2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x +3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}}-5 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+6 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -3 \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}}+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x +2 x^{2}-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}}-4 x +2}{-1+x}\right )\)	\(370\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln((4*RootOf(_Z^2-_Z+1)^2*x^2-4*RootOf(_Z^2-_Z+1)^2*x+3*RootOf(_Z^2-_Z+1)*(x^3-x^2-x+1)^(2/3)-3*RootOf(_Z^2-_
Z+1)*(x^3-x^2-x+1)^(1/3)*x-4*RootOf(_Z^2-_Z+1)*x^2+3*RootOf(_Z^2-_Z+1)*(x^3-x^2-x+1)^(1/3)+2*RootOf(_Z^2-_Z+1)
*x+3*(x^3-x^2-x+1)^(1/3)*x+x^2+2*RootOf(_Z^2-_Z+1)-3*(x^3-x^2-x+1)^(1/3)-1)/(-1+x))+RootOf(_Z^2-_Z+1)*ln(-(2*R
ootOf(_Z^2-_Z+1)^2*x^2-2*RootOf(_Z^2-_Z+1)^2*x+3*RootOf(_Z^2-_Z+1)*(x^3-x^2-x+1)^(2/3)-5*RootOf(_Z^2-_Z+1)*x^2
+6*RootOf(_Z^2-_Z+1)*x-3*(x^3-x^2-x+1)^(2/3)+3*(x^3-x^2-x+1)^(1/3)*x+2*x^2-RootOf(_Z^2-_Z+1)-3*(x^3-x^2-x+1)^(
1/3)-4*x+2)/(-1+x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x - 1)^2*(x + 1))^(1/3),x)

[Out]

int(1/((x - 1)^2*(x + 1))^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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