3.105 \(\int \frac {(1-x)^2}{(1-x^3)^{4/3}} \, dx\)

Optimal. Leaf size=39 \[ x^2 \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};x^3\right )+\frac {(1-2 x) x+1}{\sqrt [3]{1-x^3}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1854, 12, 364} \[ x^2 \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};x^3\right )+\frac {(1-2 x) x+1}{\sqrt [3]{1-x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - x)^2/(1 - x^3)^(4/3),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1854

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], i}, Simp[((a*Coeff[Pq, x, q] -
b*x*ExpandToSum[Pq - Coeff[Pq, x, q]*x^q, x])*(a + b*x^n)^(p + 1))/(a*b*n*(p + 1)), x] + Dist[1/(a*n*(p + 1))
, Int[Sum[(n*(p + 1) + i + 1)*Coeff[Pq, x, i]*x^i, {i, 0, q - 1}]*(a + b*x^n)^(p + 1), x], x] /; q == n - 1] /
; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {(1-x)^2}{\left (1-x^3\right )^{4/3}} \, dx &=\frac {1+(1-2 x) x}{\sqrt [3]{1-x^3}}-\int -\frac {2 x}{\sqrt [3]{1-x^3}} \, dx\\ &=\frac {1+(1-2 x) x}{\sqrt [3]{1-x^3}}+2 \int \frac {x}{\sqrt [3]{1-x^3}} \, dx\\ &=\frac {1+(1-2 x) x}{\sqrt [3]{1-x^3}}+x^2 \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};x^3\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 43, normalized size = 1.10 \[ x^2 \left (-\, _2F_1\left (\frac {2}{3},\frac {4}{3};\frac {5}{3};x^3\right )\right )+\frac {x}{\sqrt [3]{1-x^3}}+\frac {1}{\sqrt [3]{1-x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - x^3)^(-1/3) + x/(1 - x^3)^(1/3) - x^2*Hypergeometric2F1[2/3, 4/3, 5/3, x^3]

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{4} + 2 \, x^{3} + 3 \, x^{2} + 2 \, x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 34, normalized size = 0.87 \[ x^{2} \hypergeom \left (\left [\frac {1}{3}, \frac {2}{3}\right ], \left [\frac {5}{3}\right ], x^{3}\right )-\frac {\left (x -1\right ) \left (2 x +1\right )}{\left (-x^{3}+1\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {{\left (x-1\right )}^2}{{\left (1-x^3\right )}^{4/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 1)^2/(1 - x^3)^(4/3),x)

[Out]

int((x - 1)^2/(1 - x^3)^(4/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (x - 1\right )^{2}}{\left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {4}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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