∫ 1−x2 ______________________________(1−x+x2 )(1−x3)2∕3 dx [1017]

3.11.17 \(\int \frac {1-x^2}{(1-x+x^2) (1-x^3)^{2/3}} \, dx\) [1017]

Optimal. Leaf size=103 \[ \frac {\sqrt {3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2^{2/3}}-\frac {\log \left (1+2 (1-x)^3-x^3\right )}{2\ 2^{2/3}}+\frac {3 \log \left (\sqrt [3]{2} (1-x)+\sqrt [3]{1-x^3}\right )}{2\ 2^{2/3}} \]

[Out]

-1/4*ln(1+2*(1-x)^3-x^3)*2^(1/3)+3/4*ln(2^(1/3)*(1-x)+(-x^3+1)^(1/3))*2^(1/3)+1/2*arctan(1/3*(1-2*2^(1/3)*(1-x

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

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(425\) vs. \(2(103)=206\).
time = 0.40, antiderivative size = 425, normalized size of antiderivative = 4.13, number of steps used = 42, number of rules used = 15, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.517, Rules used = {2183, 421, 251, 420, 493, 298, 31, 648, 631, 210, 642, 503, 455, 59, 494} \begin {gather*} \frac {\sqrt [3]{2} \text {ArcTan}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\text {ArcTan}\left (\frac {\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}-\frac {\text {ArcTan}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {\text {ArcTan}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {\log \left (x^3+1\right )}{3\ 2^{2/3}}+\frac {\log \left (2^{2/3}-\frac {1-x}{\sqrt [3]{1-x^3}}\right )}{3\ 2^{2/3}}-\frac {\log \left (\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{3\ 2^{2/3}}+\frac {1}{3} \sqrt [3]{2} \log \left (\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )-\frac {\log \left (\frac {(1-x)^2}{\left (1-x^3\right )^{2/3}}+\frac {2^{2/3} (1-x)}{\sqrt [3]{1-x^3}}+2 \sqrt [3]{2}\right )}{6\ 2^{2/3}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2\ 2^{2/3}}-\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2\ 2^{2/3}} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

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

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

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

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

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

Rule 31

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

Rule 59

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L

og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x

)^(1/3)], x] - Dist[3/(2*b*q^2), 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 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],

x] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p, 0] && !IntegerQ[1/n] && !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p

] || GtQ[a, 0])

Rule 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),

x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3

]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 420

Int[((a_) + (b_.)*(x_)^3)^(1/3)/((c_) + (d_.)*(x_)^3), x_Symbol] :> With[{q = Rt[b/a, 3]}, Dist[9*(a/(c*q)), S

ubst[Int[x/((4 - a*x^3)*(1 + 2*a*x^3)), x], x, (1 + q*x)/(a + b*x^3)^(1/3)], x]] /; FreeQ[{a, b, c, d}, x] &&

NeQ[b*c - a*d, 0] && EqQ[b*c + a*d, 0]

Rule 421

Int[1/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x^3)^

(2/3), x], x] - Dist[d/(b*c - a*d), Int[(a + b*x^3)^(1/3)/(c + d*x^3), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ

[b*c - a*d, 0] && EqQ[b*c + a*d, 0]

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 493

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[b/(b*c - a*d), I

nt[(e*x)^m/(a + b*x^n), x], x] - Dist[d/(b*c - a*d), Int[(e*x)^m/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,

m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0]

Rule 494

Int[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_)^(n_))^(q_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Dist[e^n/b, Int[

(e*x)^(m - n)*(c + d*x^n)^q, x], x] - Dist[a*(e^n/b), Int[(e*x)^(m - n)*((c + d*x^n)^q/(a + b*x^n)), x], x] /;

FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1] && IntBinomialQ[a, b

, c, d, e, m, n, -1, q, x]

Rule 503

Int[(x_)/(((a_) + (b_.)*(x_)^3)^(2/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Si

mp[-ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q^2), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/

(2*c*q^2), x] + Simp[Log[c + d*x^3]/(6*c*q^2), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +

c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +

b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &

& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && !NiceSqrtQ[b^2 - 4*a*c]

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

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Mathematica [A]
time = 1.15, size = 181, normalized size = 1.76 \begin {gather*} -\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \left (1-x^3\right )^{2/3}}{2^{2/3}+2^{2/3} x+2^{2/3} x^2-\left (1-x^3\right )^{2/3}}\right )-2 \log \left (2^{2/3}+2^{2/3} x+2^{2/3} x^2+2 \left (1-x^3\right )^{2/3}\right )+\log \left (-\left (\left (1+x+x^2\right ) \left (\sqrt [3]{2}+\sqrt [3]{2} x^2-\left (2-2 x^3\right )^{2/3}+2 \sqrt [3]{1-x^3}+x \left (\sqrt [3]{2}-2 \sqrt [3]{1-x^3}\right )\right )\right )\right )}{2\ 2^{2/3}} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 5.43, size = 690, normalized size = 6.70


method	result	size

trager	\(\text {Expression too large to display}\)	\(690\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*ln((2*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^4*x-2*(-x^3+1)^(2/3)*RootOf(Root

Of(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2+RootOf(_Z^3-2)^2*x^2-RootOf(_Z^3-2)^2*x-4*(-x^3+1)^(

1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x+RootOf(_Z^3-2)^2+4*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^

3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2))/(x^2-x+1))*RootOf(_Z^3-2)-ln((2*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2

)+4*_Z^2)*RootOf(_Z^3-2)^4*x-2*(-x^3+1)^(2/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-

2)^2+RootOf(_Z^3-2)^2*x^2-RootOf(_Z^3-2)^2*x-4*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z

^2)*x+RootOf(_Z^3-2)^2+4*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2))/(x^2-x+1))*RootOf

(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)+RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*ln(-(2*RootO

f(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^4*x-2*(-x^3+1)^(2/3)*RootOf(RootOf(_Z^3-2)^2+2*_

Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2-RootOf(_Z^3-2)^2*x^2+3*RootOf(_Z^3-2)^2*x-2*(-x^3+1)^(1/3)*RootOf(_Z

^3-2)*x-4*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x-RootOf(_Z^3-2)^2+2*(-x^3+1)^(1/

3)*RootOf(_Z^3-2)+4*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)-2*(-x^3+1)^(2/3))/(x^2-

x+1))

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (78) = 156\).
time = 3.24, size = 289, normalized size = 2.81 \begin {gather*} -\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {4^{\frac {1}{6}} \sqrt {3} {\left (2 \cdot 4^{\frac {2}{3}} {\left (x^{5} - x^{4} - 3 \, x^{3} + 3 \, x^{2} + x - 1\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 4 \, {\left (x^{4} - 4 \, x^{3} + 5 \, x^{2} - 4 \, x + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (x^{6} - 7 \, x^{5} + 10 \, x^{4} - 7 \, x^{3} + 10 \, x^{2} - 7 \, x + 1\right )}\right )}}{6 \, {\left (3 \, x^{6} - 9 \, x^{5} + 6 \, x^{4} - x^{3} + 6 \, x^{2} - 9 \, x + 3\right )}}\right ) - \frac {1}{24} \cdot 4^{\frac {2}{3}} \log \left (\frac {2 \cdot 4^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} {\left (x^{2} - 3 \, x + 1\right )} - 4^{\frac {2}{3}} {\left (x^{4} - 3 \, x^{2} + 1\right )} - 8 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x^{2} - x\right )}}{x^{4} - 2 \, x^{3} + 3 \, x^{2} - 2 \, x + 1}\right ) + \frac {1}{12} \cdot 4^{\frac {2}{3}} \log \left (-\frac {4^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} - 4^{\frac {1}{3}} {\left (x^{2} - x + 1\right )} - 2 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2} - x + 1}\right ) \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*4^(1/6)*sqrt(3)*arctan(1/6*4^(1/6)*sqrt(3)*(2*4^(2/3)*(x^5 - x^4 - 3*x^3 + 3*x^2 + x - 1)*(-x^3 + 1)^(1/3

) + 4*(x^4 - 4*x^3 + 5*x^2 - 4*x + 1)*(-x^3 + 1)^(2/3) + 4^(1/3)*(x^6 - 7*x^5 + 10*x^4 - 7*x^3 + 10*x^2 - 7*x

+ 1))/(3*x^6 - 9*x^5 + 6*x^4 - x^3 + 6*x^2 - 9*x + 3)) - 1/24*4^(2/3)*log((2*4^(1/3)*(-x^3 + 1)^(2/3)*(x^2 - 3

*x + 1) - 4^(2/3)*(x^4 - 3*x^2 + 1) - 8*(-x^3 + 1)^(1/3)*(x^2 - x))/(x^4 - 2*x^3 + 3*x^2 - 2*x + 1)) + 1/12*4^

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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Giac [F]
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^2+1)/(x^2-x+1)/(-x^3+1)^(2/3),x, algorithm="giac")

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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