3.7.20 \(\int \frac {x}{\sqrt [3]{1-x^3} (1+x^3)} \, dx\) [620]

Optimal. Leaf size=233 \[ \frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}+\frac {\log \left ((1-x) (1+x)^2\right )}{12 \sqrt [3]{2}}+\frac {\log \left (1+\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{6 \sqrt [3]{2}}-\frac {\log \left (1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{3 \sqrt [3]{2}}-\frac {\log \left (-1+x+2^{2/3} \sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}} \]

[Out]

1/24*ln((1-x)*(1+x)^2)*2^(2/3)+1/12*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)-
1/6*ln(1+2^(1/3)*(1-x)/(-x^3+1)^(1/3))*2^(2/3)-1/8*ln(-1+x+2^(2/3)*(-x^3+1)^(1/3))*2^(2/3)+1/6*arctan(1/3*(1-2
*2^(1/3)*(1-x)/(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)+1/12*arctan(1/3*(1+2^(1/3)*(1-x)/(-x^3+1)^(1/3))*3^(1/
2))*2^(2/3)*3^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {502, 2174, 206, 31, 648, 631, 210, 642} \begin {gather*} \frac {\text {ArcTan}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\text {ArcTan}\left (\frac {\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {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 )}{6 \sqrt [3]{2}}-\frac {\log \left (\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{3 \sqrt [3]{2}}-\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3}+x-1\right )}{4 \sqrt [3]{2}}+\frac {\log \left ((1-x) (x+1)^2\right )}{12 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/((1 - x^3)^(1/3)*(1 + x^3)),x]

[Out]

ArcTan[(1 - (2*2^(1/3)*(1 - x))/(1 - x^3)^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3]) + ArcTan[(1 + (2^(1/3)*(1 - x))/(1
- x^3)^(1/3))/Sqrt[3]]/(2*2^(1/3)*Sqrt[3]) + Log[(1 - x)*(1 + x)^2]/(12*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)]/(6*2^(1/3)) - Log[1 + (2^(1/3)*(1 - x))/(1 - x^3)^(1/3)
]/(3*2^(1/3)) - Log[-1 + x + 2^(2/3)*(1 - x^3)^(1/3)]/(4*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 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 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]

Rubi steps

\begin {align*} \int \frac {x}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx &=\frac {1}{2} x^2 F_1\left (\frac {2}{3};\frac {1}{3},1;\frac {5}{3};x^3,-x^3\right )\\ \end {align*}

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Mathematica [A]
time = 0.75, size = 283, normalized size = 1.21 \begin {gather*} \frac {-2 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{1-x^3}}{\sqrt [3]{2}-\sqrt [3]{2} x+\sqrt [3]{1-x^3}}\right )-4 \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 )-4 \log \left (-\sqrt [3]{2}+\sqrt [3]{2} x-\sqrt [3]{1-x^3}\right )-2 \log \left (-\sqrt [3]{2}+\sqrt [3]{2} x+2 \sqrt [3]{1-x^3}\right )+2 \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 )+\log \left (2^{2/3}-2\ 2^{2/3} x+2^{2/3} x^2-2 (-1+x) \sqrt [3]{2-2 x^3}+4 \left (1-x^3\right )^{2/3}\right )}{12 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[3]*ArcTan[(Sqrt[3]*(1 - x^3)^(1/3))/(2^(1/3) - 2^(1/3)*x + (1 - x^3)^(1/3))] - 4*Sqrt[3]*ArcTan[(Sqrt
[3]*(1 - x^3)^(1/3))/(-2*2^(1/3) + 2*2^(1/3)*x + (1 - x^3)^(1/3))] - 4*Log[-2^(1/3) + 2^(1/3)*x - (1 - x^3)^(1
/3)] - 2*Log[-2^(1/3) + 2^(1/3)*x + 2*(1 - x^3)^(1/3)] + 2*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)] + Log[2^(2/3) - 2*2^(2/3)*x + 2^(2/3)*x^2 - 2*(-1 + x)*(2 - 2*x^3)^(1/3)
+ 4*(1 - x^3)^(2/3)])/(12*2^(1/3))

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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x}{\left (-x^{3}+1\right )^{\frac {1}{3}} \left (x^{3}+1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (171) = 342\).
time = 7.98, size = 373, normalized size = 1.60 \begin {gather*} -\frac {1}{36} \, \sqrt {6} 2^{\frac {1}{6}} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {2^{\frac {1}{6}} {\left (24 \, \sqrt {6} 2^{\frac {2}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{14} - 2 \, x^{11} - 6 \, x^{8} - 2 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 12 \, \sqrt {6} \left (-1\right )^{\frac {1}{3}} {\left (x^{16} - 33 \, x^{13} + 110 \, x^{10} - 110 \, x^{7} + 33 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + \sqrt {6} 2^{\frac {1}{3}} {\left (x^{18} + 42 \, x^{15} - 417 \, x^{12} + 812 \, x^{9} - 417 \, x^{6} + 42 \, x^{3} + 1\right )}\right )}}{6 \, {\left (x^{18} - 102 \, x^{15} + 447 \, x^{12} - 628 \, x^{9} + 447 \, x^{6} - 102 \, x^{3} + 1\right )}}\right ) - \frac {1}{72} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-\frac {12 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{8} - 4 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{12} - 32 \, x^{9} + 78 \, x^{6} - 32 \, x^{3} + 1\right )} - 6 \, {\left (x^{10} - 11 \, x^{7} + 11 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x^{12} + 4 \, x^{9} + 6 \, x^{6} + 4 \, x^{3} + 1}\right ) + \frac {1}{36} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-\frac {12 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x^{2} - 6 \cdot 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{6} + 2 \, x^{3} + 1\right )}}{x^{6} + 2 \, x^{3} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*sqrt(6)*2^(1/6)*(-1)^(1/3)*arctan(1/6*2^(1/6)*(24*sqrt(6)*2^(2/3)*(-1)^(2/3)*(x^14 - 2*x^11 - 6*x^8 - 2*
x^5 + x^2)*(-x^3 + 1)^(2/3) + 12*sqrt(6)*(-1)^(1/3)*(x^16 - 33*x^13 + 110*x^10 - 110*x^7 + 33*x^4 - x)*(-x^3 +
1)^(1/3) + sqrt(6)*2^(1/3)*(x^18 + 42*x^15 - 417*x^12 + 812*x^9 - 417*x^6 + 42*x^3 + 1))/(x^18 - 102*x^15 + 4
47*x^12 - 628*x^9 + 447*x^6 - 102*x^3 + 1)) - 1/72*2^(2/3)*(-1)^(1/3)*log(-(12*2^(2/3)*(-1)^(1/3)*(x^8 - 4*x^5
+ x^2)*(-x^3 + 1)^(2/3) - 2^(1/3)*(-1)^(2/3)*(x^12 - 32*x^9 + 78*x^6 - 32*x^3 + 1) - 6*(x^10 - 11*x^7 + 11*x^
4 - x)*(-x^3 + 1)^(1/3))/(x^12 + 4*x^9 + 6*x^6 + 4*x^3 + 1)) + 1/36*2^(2/3)*(-1)^(1/3)*log(-(12*(-x^3 + 1)^(2/
3)*x^2 - 6*2^(1/3)*(-1)^(2/3)*(x^4 - x)*(-x^3 + 1)^(1/3) - 2^(2/3)*(-1)^(1/3)*(x^6 + 2*x^3 + 1))/(x^6 + 2*x^3
+ 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/((-(x - 1)*(x**2 + x + 1))**(1/3)*(x + 1)*(x**2 - x + 1)), 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/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((1 - x^3)^(1/3)*(x^3 + 1)),x)

[Out]

int(x/((1 - x^3)^(1/3)*(x^3 + 1)), x)

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