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

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

[Out]

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

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Rubi [B]  time = 0.120864, 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 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]

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 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]

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*}

Mathematica [C]  time = 0.0111695, 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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Maple [F]  time = 0.008, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt [3]{ \left ( -1+x \right ) ^{2} \left ( 1+x \right ) }}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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Fricas [B]  time = 1.71082, size = 331, normalized size = 4.94 \begin{align*} -\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 ) \end{align*}

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

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

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)