3.4.94 \(\int \frac {x^4}{(1-x^3)^{2/3} (1+x^3)} \, dx\)

Optimal. Leaf size=139 \[ -\frac {\log \left (x^3+1\right )}{6\ 2^{2/3}}-\frac {1}{2} \log \left (-\sqrt [3]{1-x^3}-x\right )+\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2\ 2^{2/3}}-\frac {\tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}} \]

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Rubi [A]  time = 0.10, antiderivative size = 207, normalized size of antiderivative = 1.49, number of steps used = 14, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {494, 481, 292, 31, 634, 618, 204, 628, 617} \begin {gather*} \frac {1}{6} \log \left (\frac {x^2}{\left (1-x^3\right )^{2/3}}-\frac {x}{\sqrt [3]{1-x^3}}+1\right )-\frac {1}{3} \log \left (\frac {x}{\sqrt [3]{1-x^3}}+1\right )-\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 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {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^4/((1 - x^3)^(2/3)*(1 + x^3)),x]

[Out]

-(ArcTan[(1 - (2*x)/(1 - x^3)^(1/3))/Sqrt[3]]/Sqrt[3]) + ArcTan[(1 - (2*2^(1/3)*x)/(1 - x^3)^(1/3))/Sqrt[3]]/(
2^(2/3)*Sqrt[3]) + Log[1 + x^2/(1 - x^3)^(2/3) - x/(1 - x^3)^(1/3)]/6 - Log[1 + x/(1 - x^3)^(1/3)]/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 481

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> -Dist[(a*e^n)/(b*c -
a*d), Int[(e*x)^(m - n)/(a + b*x^n), x], x] + Dist[(c*e^n)/(b*c - a*d), Int[(e*x)^(m - n)/(c + d*x^n), x], x]
/; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1]

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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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^4}{\left (1-x^3\right )^{2/3} \left (1+x^3\right )} \, dx &=\operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^3\right ) \left (1+2 x^3\right )} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\\ &=\operatorname {Subst}\left (\int \frac {x}{1+x^3} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )-\operatorname {Subst}\left (\int \frac {x}{1+2 x^3} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\\ &=-\left (\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1+x}{1-x+x^2} \, 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 {1}{3} \log \left (1+\frac {x}{\sqrt [3]{1-x^3}}\right )+\frac {\log \left (1+\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}\right )}{3\ 2^{2/3}}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )-\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 {1}{6} \log \left (1+\frac {x^2}{\left (1-x^3\right )^{2/3}}-\frac {x}{\sqrt [3]{1-x^3}}\right )-\frac {1}{3} \log \left (1+\frac {x}{\sqrt [3]{1-x^3}}\right )-\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}}-\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+\frac {2 x}{\sqrt [3]{1-x^3}}\right )\\ &=\frac {\tan ^{-1}\left (\frac {-1+\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {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 {1}{6} \log \left (1+\frac {x^2}{\left (1-x^3\right )^{2/3}}-\frac {x}{\sqrt [3]{1-x^3}}\right )-\frac {1}{3} \log \left (1+\frac {x}{\sqrt [3]{1-x^3}}\right )-\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 = 0.05, size = 26, normalized size = 0.19 \begin {gather*} \frac {1}{5} x^5 F_1\left (\frac {5}{3};\frac {2}{3},1;\frac {8}{3};x^3,-x^3\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4/((1 - x^3)^(2/3)*(1 + x^3)),x]

[Out]

(x^5*AppellF1[5/3, 2/3, 1, 8/3, x^3, -x^3])/5

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IntegrateAlgebraic [A]  time = 0.38, size = 212, normalized size = 1.53 \begin {gather*} -\frac {1}{3} \log \left (\sqrt [3]{1-x^3}+x\right )+\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 \sqrt [3]{1-x^3}-x}\right )}{\sqrt {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 {1}{6} \log \left (-\sqrt [3]{1-x^3} x+\left (1-x^3\right )^{2/3}+x^2\right )-\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^4/((1 - x^3)^(2/3)*(1 + x^3)),x]

[Out]

ArcTan[(Sqrt[3]*x)/(-x + 2*(1 - x^3)^(1/3))]/Sqrt[3] - ArcTan[(Sqrt[3]*x)/(-x + 2^(2/3)*(1 - x^3)^(1/3))]/(2^(
2/3)*Sqrt[3]) - Log[x + (1 - x^3)^(1/3)]/3 + Log[2*x + 2^(2/3)*(1 - x^3)^(1/3)]/(3*2^(2/3)) + Log[x^2 - x*(1 -
x^3)^(1/3) + (1 - x^3)^(2/3)]/6 - 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 [A]  time = 0.46, size = 197, normalized size = 1.42 \begin {gather*} \frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (-\frac {4^{\frac {1}{6}} {\left (4^{\frac {1}{3}} \sqrt {3} x - 4^{\frac {2}{3}} \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{6 \, x}\right ) + \frac {1}{12} \cdot 4^{\frac {2}{3}} \log \left (\frac {4^{\frac {2}{3}} x + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{24} \cdot 4^{\frac {2}{3}} \log \left (\frac {2 \cdot 4^{\frac {1}{3}} x^{2} - 4^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x + 2 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) - \frac {1}{3} \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} x - 2 \, \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) - \frac {1}{3} \, \log \left (\frac {x + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x}\right ) + \frac {1}{6} \, \log \left (\frac {x^{2} - {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(-x^3+1)^(2/3)/(x^3+1),x, algorithm="fricas")

[Out]

1/6*4^(1/6)*sqrt(3)*arctan(-1/6*4^(1/6)*(4^(1/3)*sqrt(3)*x - 4^(2/3)*sqrt(3)*(-x^3 + 1)^(1/3))/x) + 1/12*4^(2/
3)*log((4^(2/3)*x + 2*(-x^3 + 1)^(1/3))/x) - 1/24*4^(2/3)*log((2*4^(1/3)*x^2 - 4^(2/3)*(-x^3 + 1)^(1/3)*x + 2*
(-x^3 + 1)^(2/3))/x^2) - 1/3*sqrt(3)*arctan(-1/3*(sqrt(3)*x - 2*sqrt(3)*(-x^3 + 1)^(1/3))/x) - 1/3*log((x + (-
x^3 + 1)^(1/3))/x) + 1/6*log((x^2 - (-x^3 + 1)^(1/3)*x + (-x^3 + 1)^(2/3))/x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{{\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^4/(-x^3+1)^(2/3)/(x^3+1),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.76, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\left (-x^{3}+1\right )^{\frac {2}{3}} \left (x^{3}+1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(-x^3+1)^(2/3)/(x^3+1),x)

[Out]

int(x^4/(-x^3+1)^(2/3)/(x^3+1),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{{\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^4/(-x^3+1)^(2/3)/(x^3+1),x, algorithm="maxima")

[Out]

integrate(x^4/((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^4}{{\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^4/((1 - x^3)^(2/3)*(x^3 + 1)),x)

[Out]

int(x^4/((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^{4}}{\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**4/(-x**3+1)**(2/3)/(x**3+1),x)

[Out]

Integral(x**4/((-(x - 1)*(x**2 + x + 1))**(2/3)*(x + 1)*(x**2 - x + 1)), x)

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