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

Optimal. Leaf size=135 \[ \frac{\log \left (x^3+1\right )}{6 \sqrt [3]{2}}-\frac{\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2 \sqrt [3]{2}}+\frac{1}{2} \log \left (\sqrt [3]{1-x^3}+x\right )-\frac{\tan ^{-1}\left (\frac{1-\frac{2 x}{\sqrt [3]{1-x^3}}}{\sqrt{3}}\right )}{\sqrt{3}}+\frac{\tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt{3}}\right )}{\sqrt [3]{2} \sqrt{3}} \]

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

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Rubi [A]  time = 0.104934, antiderivative size = 207, normalized size of antiderivative = 1.53, 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, 200, 31, 634, 618, 204, 628, 617} \[ -\frac{1}{6} \log \left (\frac{x^2}{\left (1-x^3\right )^{2/3}}-\frac{x}{\sqrt [3]{1-x^3}}+1\right )+\frac{1}{3} \log \left (\frac{x}{\sqrt [3]{1-x^3}}+1\right )+\frac{\log \left (\frac{2^{2/3} x^2}{\left (1-x^3\right )^{2/3}}-\frac{\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}+1\right )}{6 \sqrt [3]{2}}-\frac{\log \left (\frac{\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}+1\right )}{3 \sqrt [3]{2}}-\frac{\tan ^{-1}\left (\frac{1-\frac{2 x}{\sqrt [3]{1-x^3}}}{\sqrt{3}}\right )}{\sqrt{3}}+\frac{\tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt{3}}\right )}{\sqrt [3]{2} \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

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 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 200

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 31

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

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]

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 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 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 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]

Rubi steps

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

Mathematica [C]  time = 0.0180336, size = 26, normalized size = 0.19 \[ \frac{1}{4} x^4 F_1\left (\frac{4}{3};\frac{1}{3},1;\frac{7}{3};x^3,-x^3\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^4*AppellF1[4/3, 1/3, 1, 7/3, x^3, -x^3])/4

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Maple [F]  time = 0.037, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3}}{{x}^{3}+1}{\frac{1}{\sqrt [3]{-{x}^{3}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (x^{3} + 1\right )}{\left (-x^{3} + 1\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 11.7116, size = 1392, normalized size = 10.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*2^(2/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))*log(-1/8*(x*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^3 - 6*2^(1/3)
*x*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2 + 8*x - 24*(-x^3 + 1)^(1/3))/x) - 1/24*(2^(2/3)*(I*sqrt(3)*(-1)^(1/3)
- (-1)^(1/3)) - 2*sqrt(3/2)*sqrt(-2^(1/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2))*log(-3/8*(2^(2/3)*sqrt(3/2)
*sqrt(-2^(1/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2)*x*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3)) + 2^(1/3)*x*(I*sqr
t(3)*(-1)^(1/3) - (-1)^(1/3))^2 - 8*(-x^3 + 1)^(1/3))/x) - 1/24*(2^(2/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3)) +
2*sqrt(3/2)*sqrt(-2^(1/3)*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2))*log(3/8*(2^(2/3)*sqrt(3/2)*sqrt(-2^(1/3)*(I
*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^2)*x*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3)) - 2^(1/3)*x*(I*sqrt(3)*(-1)^(1/3) -
(-1)^(1/3))^2 + 8*(-x^3 + 1)^(1/3))/x) - 1/3*sqrt(3)*arctan(-1/3*(sqrt(3)*x - 2*sqrt(3)*(-x^3 + 1)^(1/3))/x)
+ 1/3*log(1/24*(x*(I*sqrt(3)*(-1)^(1/3) - (-1)^(1/3))^3 + 32*x + 24*(-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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (x^{3} + 1\right )}{\left (-x^{3} + 1\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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