3.4.95 \(\int \frac {x}{(1-x^3)^{2/3} (1+x^3)} \, dx\)

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

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Rubi [A]  time = 0.06, antiderivative size = 122, normalized size of antiderivative = 1.39, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {494, 292, 31, 634, 617, 204, 628} \begin {gather*} \frac {\log \left (\frac {2^{2/3} x^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}+1\right )}{6\ 2^{2/3}}-\frac {\log \left (\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}+1\right )}{3\ 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}} \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 + (2^(2/3)*x^2)/(1 - x^3)^(2/
3) - (2^(1/3)*x)/(1 - x^3)^(1/3)]/(6*2^(2/3)) - Log[1 + (2^(1/3)*x)/(1 - x^3)^(1/3)]/(3*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 204

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

Rule 292

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 494

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[(k*a^(p + (m + 1)/n))/n, Subst[Int[(x^((k*(m + 1))/n - 1)*(c - (b*c - a*d)*x^k)^q)/(1 - b*x^k)^(p
+ q + (m + 1)/n + 1), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

Rule 617

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 628

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 634

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]

Rubi steps

\begin {align*} \int \frac {x}{\left (1-x^3\right )^{2/3} \left (1+x^3\right )} \, dx &=\operatorname {Subst}\left (\int \frac {x}{1+2 x^3} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\\ &=-\frac {\operatorname {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 {\operatorname {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 {\operatorname {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 {\operatorname {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 {\operatorname {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 [C]  time = 0.01, size = 38, normalized size = 0.43 \begin {gather*} \frac {x^2 \, _2F_1\left (\frac {2}{3},\frac {2}{3};\frac {5}{3};\frac {2 x^3}{x^3+1}\right )}{2 \left (x^3+1\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^2*Hypergeometric2F1[2/3, 2/3, 5/3, (2*x^3)/(1 + x^3)])/(2*(1 + x^3)^(2/3))

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IntegrateAlgebraic [A]  time = 0.29, size = 126, normalized size = 1.43 \begin {gather*} -\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3}+2 x\right )}{3\ 2^{2/3}}+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2^{2/3} \sqrt [3]{1-x^3}-x}\right )}{2^{2/3} \sqrt {3}}+\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3} x-\sqrt [3]{2} \left (1-x^3\right )^{2/3}-2 x^2\right )}{6\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[3]*x)/(-x + 2^(2/3)*(1 - x^3)^(1/3))]/(2^(2/3)*Sqrt[3]) - Log[2*x + 2^(2/3)*(1 - x^3)^(1/3)]/(3*2
^(2/3)) + 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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fricas [B]  time = 1.92, 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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}\,{d x} \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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maple [C]  time = 4.24, size = 938, normalized size = 10.66

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*RootOf(_Z^3+2)*ln((54*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)^2*RootOf(_Z^3+2)^3*x^3+6*RootOf
(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)^4*x^3-12*(-x^3+1)^(2/3)*RootOf(_Z^3+2)^2*RootOf(
RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*x-27*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf
(_Z^3+2)*x^3-3*x^3*RootOf(_Z^3+2)^2+6*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*(-x^3+1)^(1/3)*x^2-
4*RootOf(_Z^3+2)*(-x^3+1)^(1/3)*x^2+5*(-x^3+1)^(2/3)*x+9*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*
RootOf(_Z^3+2)+RootOf(_Z^3+2)^2)/(x+1)/(x^2-x+1))-1/6*ln(-(72*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_
Z^2)^2*RootOf(_Z^3+2)^3*x^3+18*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)^4*x^3-24*(-
x^3+1)^(2/3)*RootOf(_Z^3+2)^2*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*x+12*RootOf(RootOf(_Z^3+2)^
2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)*x^3+3*x^3*RootOf(_Z^3+2)^2-60*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootO
f(_Z^3+2)+36*_Z^2)*(-x^3+1)^(1/3)*x^2-8*RootOf(_Z^3+2)*(-x^3+1)^(1/3)*x^2-2*(-x^3+1)^(2/3)*x-12*RootOf(RootOf(
_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)-3*RootOf(_Z^3+2)^2)/(x+1)/(x^2-x+1))*RootOf(_Z^3+2)-ln(-
(72*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)^2*RootOf(_Z^3+2)^3*x^3+18*RootOf(RootOf(_Z^3+2)^2+6*_
Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)^4*x^3-24*(-x^3+1)^(2/3)*RootOf(_Z^3+2)^2*RootOf(RootOf(_Z^3+2)^2+6*_Z
*RootOf(_Z^3+2)+36*_Z^2)*x+12*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)*x^3+3*x^3*Ro
otOf(_Z^3+2)^2-60*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*(-x^3+1)^(1/3)*x^2-8*RootOf(_Z^3+2)*(-x
^3+1)^(1/3)*x^2-2*(-x^3+1)^(2/3)*x-12*RootOf(RootOf(_Z^3+2)^2+6*_Z*RootOf(_Z^3+2)+36*_Z^2)*RootOf(_Z^3+2)-3*Ro
otOf(_Z^3+2)^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*} \int \frac {x}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}\,{d x} \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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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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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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