3.7.31 \(\int \frac {x}{(1-x^3)^{2/3} (1+x^3)} \, dx\) [631]

Optimal. Leaf size=88 \[ -\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {\log \left (1+x^3\right )}{6\ 2^{2/3}}-\frac {\log \left (-\sqrt [3]{2} x-\sqrt [3]{1-x^3}\right )}{2\ 2^{2/3}} \]

[Out]

1/12*ln(x^3+1)*2^(1/3)-1/4*ln(-2^(1/3)*x-(-x^3+1)^(1/3))*2^(1/3)-1/6*arctan(1/3*(1-2*2^(1/3)*x/(-x^3+1)^(1/3))
*3^(1/2))*2^(1/3)*3^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {503} \begin {gather*} -\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 {\log \left (x^3+1\right )}{6\ 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[x/((1 - x^3)^(2/3)*(1 + x^3)),x]

[Out]

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

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]

Rubi steps

\begin {align*} \int \frac {x}{\left (1-x^3\right )^{2/3} \left (1+x^3\right )} \, dx &=\text {Subst}\left (\int \frac {x}{1+2 x^3} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\\ &=-\frac {\text {Subst}\left (\int \frac {1}{1+\sqrt [3]{2} x} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )}{3 \sqrt [3]{2}}+\frac {\text {Subst}\left (\int \frac {1+\sqrt [3]{2} x}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )}{3 \sqrt [3]{2}}\\ &=-\frac {\log \left (1+\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}\right )}{3\ 2^{2/3}}+\frac {\text {Subst}\left (\int \frac {-\sqrt [3]{2}+2\ 2^{2/3} x}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )}{6\ 2^{2/3}}+\frac {\text {Subst}\left (\int \frac {1}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )}{2 \sqrt [3]{2}}\\ &=\frac {\log \left (1+\frac {2^{2/3} x^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}\right )}{6\ 2^{2/3}}-\frac {\log \left (1+\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}\right )}{3\ 2^{2/3}}+\frac {\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}\right )}{2^{2/3}}\\ &=-\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {\log \left (1+\frac {2^{2/3} x^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}\right )}{6\ 2^{2/3}}-\frac {\log \left (1+\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}\right )}{3\ 2^{2/3}}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 114, normalized size = 1.30 \begin {gather*} \frac {-2 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x-2^{2/3} \sqrt [3]{1-x^3}}\right )-2 \log \left (2 x+2^{2/3} \sqrt [3]{1-x^3}\right )+\log \left (-2 x^2+2^{2/3} x \sqrt [3]{1-x^3}-\sqrt [3]{2} \left (1-x^3\right )^{2/3}\right )}{6\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 2.50, size = 954, normalized size = 10.84

method	result	size
trager	\(\text {Expression too large to display}\)	\(954\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*ln((6*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2
)*RootOf(_Z^3+2)^4*x^3-18*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)^2*RootOf(_Z^3+2)^3*x^3+12*(-x^3
+1)^(2/3)*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)^2*x+RootOf(_Z^3+2)^2*x^3-3*RootO
f(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)*x^3+(-x^3+1)^(1/3)*RootOf(_Z^3+2)*x^2-24*(-x^3+
1)^(1/3)*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*x^2+x*(-x^3+1)^(2/3)-RootOf(_Z^3+2)^2+3*RootOf(R
ootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2))/(x+1)/(x^2-x+1))-1/6*ln(-(18*RootOf(RootOf(_Z^3+2
)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)^4*x^3+36*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)^
2*RootOf(_Z^3+2)^3*x^3+24*(-x^3+1)^(2/3)*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)^2
*x-9*RootOf(_Z^3+2)^2*x^3-18*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)*x^3-10*(-x^3+
1)^(1/3)*RootOf(_Z^3+2)*x^2-48*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*x^2-10*x*(-
x^3+1)^(2/3)+3*RootOf(_Z^3+2)^2+6*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2))/(x+1)/(
x^2-x+1))*RootOf(_Z^3+2)-ln(-(18*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)^4*x^3+36*
RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)^2*RootOf(_Z^3+2)^3*x^3+24*(-x^3+1)^(2/3)*RootOf(RootOf(_Z
^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)^2*x-9*RootOf(_Z^3+2)^2*x^3-18*RootOf(RootOf(_Z^3+2)^2+6*_Z
*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)*x^3-10*(-x^3+1)^(1/3)*RootOf(_Z^3+2)*x^2-48*(-x^3+1)^(1/3)*RootOf(Root
Of(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*x^2-10*x*(-x^3+1)^(2/3)+3*RootOf(_Z^3+2)^2+6*RootOf(RootOf(_Z^3+2)^2
+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2))/(x+1)/(x^2-x+1))*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*
_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(x/(-x^3+1)^(2/3)/(x^3+1),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (67) = 134\).
time = 5.66, size = 283, normalized size = 3.22 \begin {gather*} -\frac {1}{18} \cdot 4^{\frac {1}{6}} \sqrt {3} \left (-1\right )^{\frac {1}{3}} \arctan \left (-\frac {4^{\frac {1}{6}} {\left (6 \cdot 4^{\frac {2}{3}} \sqrt {3} \left (-1\right )^{\frac {2}{3}} {\left (19 \, x^{8} - 16 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 12 \, \sqrt {3} \left (-1\right )^{\frac {1}{3}} {\left (5 \, x^{7} + 4 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - 4^{\frac {1}{3}} \sqrt {3} {\left (71 \, x^{9} - 111 \, x^{6} + 33 \, x^{3} - 1\right )}\right )}}{6 \, {\left (109 \, x^{9} - 105 \, x^{6} + 3 \, x^{3} + 1\right )}}\right ) + \frac {1}{36} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-\frac {3 \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{3} + 1\right )} - 6 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x}{x^{3} + 1}\right ) - \frac {1}{72} \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (\frac {6 \cdot 4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (5 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (19 \, x^{6} - 16 \, x^{3} + 1\right )} - 24 \, {\left (2 \, x^{5} - x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{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)^(2/3)/(x^3+1),x, algorithm="fricas")

[Out]

-1/18*4^(1/6)*sqrt(3)*(-1)^(1/3)*arctan(-1/6*4^(1/6)*(6*4^(2/3)*sqrt(3)*(-1)^(2/3)*(19*x^8 - 16*x^5 + x^2)*(-x
^3 + 1)^(1/3) - 12*sqrt(3)*(-1)^(1/3)*(5*x^7 + 4*x^4 - x)*(-x^3 + 1)^(2/3) - 4^(1/3)*sqrt(3)*(71*x^9 - 111*x^6
+ 33*x^3 - 1))/(109*x^9 - 105*x^6 + 3*x^3 + 1)) + 1/36*4^(2/3)*(-1)^(1/3)*log(-(3*4^(2/3)*(-1)^(1/3)*(-x^3 +
1)^(1/3)*x^2 - 4^(1/3)*(-1)^(2/3)*(x^3 + 1) - 6*(-x^3 + 1)^(2/3)*x)/(x^3 + 1)) - 1/72*4^(2/3)*(-1)^(1/3)*log((
6*4^(1/3)*(-1)^(2/3)*(5*x^4 - x)*(-x^3 + 1)^(2/3) - 4^(2/3)*(-1)^(1/3)*(19*x^6 - 16*x^3 + 1) - 24*(2*x^5 - x^2
)*(-x^3 + 1)^(1/3))/(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}{\left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {2}{3}} \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)**(2/3)/(x**3+1),x)

[Out]

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

[Out]

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

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

[Out]

int(x/((1 - x^3)^(2/3)*(x^3 + 1)), x)

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