3.1.100 \(\int \frac {1+x}{(1-x+x^2) \sqrt [3]{1-x^3}} \, dx\) [100]

Optimal. Leaf size=135 \[ \frac {\sqrt {3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2}}+\frac {\log \left (1+\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{2 \sqrt [3]{2}}-\frac {\log \left (1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{\sqrt [3]{2}} \]

[Out]

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

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Rubi [B] Leaf count is larger than twice the leaf count of optimal. \(383\) vs. \(2(135)=270\).
time = 0.17, antiderivative size = 383, normalized size of antiderivative = 2.84, number of steps used = 16, number of rules used = 12, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {2183, 384, 502, 2174, 206, 31, 648, 631, 210, 642, 455, 57} \begin {gather*} -\frac {\log \left (x^3+1\right )}{3 \sqrt [3]{2}}+\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 \sqrt [3]{2}}-\frac {1}{3} 2^{2/3} \log \left (\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2 \sqrt [3]{2}}-\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3}+x-1\right )}{2 \sqrt [3]{2}}+\frac {2^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\tan ^{-1}\left (\frac {\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\log \left ((1-x) (x+1)^2\right )}{6 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

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 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 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 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 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 502

Int[(x_)/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[b/a, 3]}, Dist[-q^2/(3
*d), Int[1/((1 - q*x)*(a + b*x^3)^(1/3)), x], x] + Dist[q/d, Subst[Int[1/(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 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 2174

Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Simp[Sqrt[3]*(ArcTan[(1 - 2^(1/3)*Rt[b,
3]*((c - d*x)/(d*(a + b*x^3)^(1/3))))/Sqrt[3]]/(2^(4/3)*Rt[b, 3]*c)), x] + (Simp[Log[(c + d*x)^2*(c - d*x)]/(2
^(7/3)*Rt[b, 3]*c), x] - Simp[(3*Log[Rt[b, 3]*(c - d*x) + 2^(2/3)*d*(a + b*x^3)^(1/3)])/(2^(7/3)*Rt[b, 3]*c),
x]) /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 + a*d^3, 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 {1+x}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}} \, dx &=\int \left (\frac {1-i \sqrt {3}}{\left (-1-i \sqrt {3}+2 x\right ) \sqrt [3]{1-x^3}}+\frac {1+i \sqrt {3}}{\left (-1+i \sqrt {3}+2 x\right ) \sqrt [3]{1-x^3}}\right ) \, dx\\ &=\left (1-i \sqrt {3}\right ) \int \frac {1}{\left (-1-i \sqrt {3}+2 x\right ) \sqrt [3]{1-x^3}} \, dx+\left (1+i \sqrt {3}\right ) \int \frac {1}{\left (-1+i \sqrt {3}+2 x\right ) \sqrt [3]{1-x^3}} \, dx\\ &=-\frac {\left (3-i \sqrt {3}\right ) \tan ^{-1}\left (\frac {2-\frac {\sqrt [3]{2} \left (1-i \sqrt {3}+2 x\right )}{\sqrt [3]{1-x^3}}}{2 \sqrt {3}}\right )}{2 \sqrt [3]{2} \left (i+\sqrt {3}\right )}+\frac {\left (3+i \sqrt {3}\right ) \tan ^{-1}\left (\frac {2-\frac {\sqrt [3]{2} \left (1+i \sqrt {3}+2 x\right )}{\sqrt [3]{1-x^3}}}{2 \sqrt {3}}\right )}{2 \sqrt [3]{2} \left (i-\sqrt {3}\right )}+\frac {\left (i-\sqrt {3}\right ) \log \left (-\left (1-i \sqrt {3}-2 x\right )^2 \left (1-i \sqrt {3}+2 x\right )\right )}{4 \sqrt [3]{2} \left (i+\sqrt {3}\right )}+\frac {\left (i+\sqrt {3}\right ) \log \left (-\left (1+i \sqrt {3}-2 x\right )^2 \left (1+i \sqrt {3}+2 x\right )\right )}{4 \sqrt [3]{2} \left (i-\sqrt {3}\right )}-\frac {3 \left (i-\sqrt {3}\right ) \log \left (1-i \sqrt {3}+2 x+2\ 2^{2/3} \sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2} \left (i+\sqrt {3}\right )}-\frac {3 \left (i+\sqrt {3}\right ) \log \left (1+i \sqrt {3}+2 x+2\ 2^{2/3} \sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2} \left (i-\sqrt {3}\right )}\\ \end {align*}

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Mathematica [A]
time = 0.91, size = 145, normalized size = 1.07 \begin {gather*} \frac {-2 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{1-x^3}}{-2 \sqrt [3]{2}+2 \sqrt [3]{2} x+\sqrt [3]{1-x^3}}\right )-2 \log \left (-\sqrt [3]{2}+\sqrt [3]{2} x-\sqrt [3]{1-x^3}\right )+\log \left (2^{2/3}-2\ 2^{2/3} x+2^{2/3} x^2+(-1+x) \sqrt [3]{2-2 x^3}+\left (1-x^3\right )^{2/3}\right )}{2 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded in comparison} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

cought exception: maximum recursion depth exceeded in comparison

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 5.21, size = 720, normalized size = 5.33

method	result	size
trager	\(\text {Expression too large to display}\)	\(720\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*ln((RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_
Z^3+4)+4*_Z^2)*x+RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)*(-x^3+1)^(2/3)-2*(-x^3+1)^
(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x+2*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*Root
Of(_Z^3+4)+4*_Z^2)-x^2+x-1)/(x^2-x+1))-1/2*ln(-(RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^
3+4)^3*x+RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^2*(-x^3+1)^(2/3)-2*(-x^3+1)^(1/3)*
RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)*x-(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2*x+2*(-x^3
+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)+(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2+x
^2*RootOf(_Z^3+4)-3*x*RootOf(_Z^3+4)-2*(-x^3+1)^(2/3)+RootOf(_Z^3+4))/(x^2-x+1))*RootOf(_Z^3+4)-ln(-(RootOf(Ro
otOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^3*x+RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z
^2)*RootOf(_Z^3+4)^2*(-x^3+1)^(2/3)-2*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootO
f(_Z^3+4)*x-(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2*x+2*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_
Z^2)*RootOf(_Z^3+4)+(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2+x^2*RootOf(_Z^3+4)-3*x*RootOf(_Z^3+4)-2*(-x^3+1)^(2/3)+Roo
tOf(_Z^3+4))/(x^2-x+1))*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (101) = 202\).
time = 5.36, size = 318, normalized size = 2.36 \begin {gather*} \frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {1}{6}} {\left (4 \cdot 2^{\frac {1}{6}} \left (-1\right )^{\frac {2}{3}} {\left (x^{4} - 4 \, x^{3} + 5 \, x^{2} - 4 \, x + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - 4 \, \sqrt {2} \left (-1\right )^{\frac {1}{3}} {\left (x^{5} - x^{4} - 3 \, x^{3} + 3 \, x^{2} + x - 1\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 2^{\frac {5}{6}} {\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}{12} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-\frac {2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} {\left (x^{2} - 3 \, x + 1\right )} + 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{4} - 3 \, x^{2} + 1\right )} + 4 \, {\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}{6} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-\frac {2 \cdot 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} + 2^{\frac {2}{3}} \left (-1\right )^{\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((1+x)/(x^2-x+1)/(-x^3+1)^(1/3),x, algorithm="fricas")

[Out]

1/6*sqrt(3)*2^(2/3)*(-1)^(1/3)*arctan(1/6*sqrt(3)*2^(1/6)*(4*2^(1/6)*(-1)^(2/3)*(x^4 - 4*x^3 + 5*x^2 - 4*x + 1
)*(-x^3 + 1)^(2/3) - 4*sqrt(2)*(-1)^(1/3)*(x^5 - x^4 - 3*x^3 + 3*x^2 + x - 1)*(-x^3 + 1)^(1/3) + 2^(5/6)*(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)) - 1/12*2^(2/3)*
(-1)^(1/3)*log(-(2^(2/3)*(-1)^(1/3)*(-x^3 + 1)^(2/3)*(x^2 - 3*x + 1) + 2^(1/3)*(-1)^(2/3)*(x^4 - 3*x^2 + 1) +
4*(-x^3 + 1)^(1/3)*(x^2 - x))/(x^4 - 2*x^3 + 3*x^2 - 2*x + 1)) + 1/6*2^(2/3)*(-1)^(1/3)*log(-(2*2^(1/3)*(-1)^(
2/3)*(-x^3 + 1)^(1/3)*(x - 1) + 2^(2/3)*(-1)^(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 + 1}{\sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x^{2} - x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Could not integrate

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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