3.2.9 \(\int \frac {(1-x^3)^{2/3}}{(1-x+x^2)^2} \, dx\) [109]

Optimal. Leaf size=234 \[ -\frac {\left (1-x^3\right )^{2/3}}{3 \left (1+x^3\right )}+\frac {x \left (1-x^3\right )^{2/3}}{3 \left (1+x^3\right )}+\frac {2 x^2 \left (1-x^3\right )^{2/3}}{3 \left (1+x^3\right )}-\frac {2^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {2^{2/3} \tan ^{-1}\left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{3} x^2 \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};x^3\right )-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{3 \sqrt [3]{2}}+\frac {\log \left (-\sqrt [3]{2} x-\sqrt [3]{1-x^3}\right )}{3 \sqrt [3]{2}} \]

[Out]

-1/3*(-x^3+1)^(2/3)/(x^3+1)+1/3*x*(-x^3+1)^(2/3)/(x^3+1)+2/3*x^2*(-x^3+1)^(2/3)/(x^3+1)+1/3*x^2*hypergeom([1/3
, 2/3],[5/3],x^3)-1/6*ln(2^(1/3)-(-x^3+1)^(1/3))*2^(2/3)+1/6*ln(-2^(1/3)*x-(-x^3+1)^(1/3))*2^(2/3)-1/9*arctan(
1/3*(1-2*2^(1/3)*x/(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)-1/9*arctan(1/3*(1+2^(2/3)*(-x^3+1)^(1/3))*3^(1/2))
*2^(2/3)*3^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {2183, 386, 384, 480, 21, 371, 455, 43, 57, 631, 210, 31} \begin {gather*} \frac {1}{3} x^2 \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};x^3\right )+\frac {\left (1-x^3\right )^{2/3} x}{3 \left (x^3+1\right )}-\frac {\left (1-x^3\right )^{2/3}}{3 \left (x^3+1\right )}-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{3 \sqrt [3]{2}}+\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{3 \sqrt [3]{2}}-\frac {2^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {2^{2/3} \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {2 \left (1-x^3\right )^{2/3} x^2}{3 \left (x^3+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 57

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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 386

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Simp[(-x)*(a + b*x^n)^(p + 1)*(
(c + d*x^n)^q/(a*n*(p + 1))), x] - Dist[c*(q/(a*(p + 1))), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 480

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-(e*x)^
(m + 1))*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*e*n*(p + 1))), x] + Dist[1/(a*n*(p + 1)), Int[(e*x)^m*(a + b*x^
n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m + n*(p + 1) + 1) + d*(m + n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ
[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[0, q, 1] && IntBinomialQ[a, b,
c, d, e, m, n, p, q, x]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2183

Int[(Px_.)*((c_) + (d_.)*(x_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^3)^(p_.), x_Symbol] :> Dist[1/c^q, Int[E
xpandIntegrand[(c^3 - d^3*x^3)^q*(a + b*x^3)^p, Px/(c - d*x)^q, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] &&
PolyQ[Px, x] && EqQ[d^2 - c*e, 0] && ILtQ[q, 0] && RationalQ[p] && EqQ[Denominator[p], 3]

Rubi steps

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

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Mathematica [F]
time = 20.25, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1-x^3\right )^{2/3}}{\left (1-x+x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Mathics [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 1.37, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^3+1)^(2/3)/(x^2-x+1)^2,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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {2}{3}}}{\left (x^{2} - x + 1\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F] N/A
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Could not integrate

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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