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

Optimal. Leaf size=154 \[ \frac{1}{2 \sqrt [3]{1-x^3}}+\frac{\log \left (x^3+1\right )}{12 \sqrt [3]{2}}+\frac{1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )-\frac{\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}+\frac{\tan ^{-1}\left (\frac{2 \sqrt [3]{1-x^3}+1}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{\tan ^{-1}\left (\frac{2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt{3}}\right )}{2 \sqrt [3]{2} \sqrt{3}}-\frac{\log (x)}{2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.110702, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {446, 85, 156, 55, 618, 204, 31, 617} \[ \frac{1}{2 \sqrt [3]{1-x^3}}+\frac{\log \left (x^3+1\right )}{12 \sqrt [3]{2}}+\frac{1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )-\frac{\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}+\frac{\tan ^{-1}\left (\frac{2 \sqrt [3]{1-x^3}+1}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{\tan ^{-1}\left (\frac{2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt{3}}\right )}{2 \sqrt [3]{2} \sqrt{3}}-\frac{\log (x)}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 85

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[(f*(e + f*x)^(p +
1))/((p + 1)*(b*e - a*f)*(d*e - c*f)), x] + Dist[1/((b*e - a*f)*(d*e - c*f)), Int[((b*d*e - b*c*f - a*d*f - b
*d*f*x)*(e + f*x)^(p + 1))/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[p, -1]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

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 31

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

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{1}{x \left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{(1-x)^{4/3} x (1+x)} \, dx,x,x^3\right )\\ &=\frac{1}{2 \sqrt [3]{1-x^3}}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{2+x}{\sqrt [3]{1-x} x (1+x)} \, dx,x,x^3\right )\\ &=\frac{1}{2 \sqrt [3]{1-x^3}}-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} (1+x)} \, dx,x,x^3\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} x} \, dx,x,x^3\right )\\ &=\frac{1}{2 \sqrt [3]{1-x^3}}-\frac{\log (x)}{2}+\frac{\log \left (1+x^3\right )}{12 \sqrt [3]{2}}-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx,x,\sqrt [3]{1-x^3}\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,\sqrt [3]{1-x^3}\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,\sqrt [3]{1-x^3}\right )+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{2}-x} \, dx,x,\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}\\ &=\frac{1}{2 \sqrt [3]{1-x^3}}-\frac{\log (x)}{2}+\frac{\log \left (1+x^3\right )}{12 \sqrt [3]{2}}+\frac{1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )-\frac{\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2^{2/3} \sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}-\operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1-x^3}\right )\\ &=\frac{1}{2 \sqrt [3]{1-x^3}}+\frac{\tan ^{-1}\left (\frac{1+2 \sqrt [3]{1-x^3}}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{\tan ^{-1}\left (\frac{1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt{3}}\right )}{2 \sqrt [3]{2} \sqrt{3}}-\frac{\log (x)}{2}+\frac{\log \left (1+x^3\right )}{12 \sqrt [3]{2}}+\frac{1}{2} \log \left (1-\sqrt [3]{1-x^3}\right )-\frac{\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}\\ \end{align*}

Mathematica [C]  time = 0.0150435, size = 54, normalized size = 0.35 \[ \frac{2 \, _2F_1\left (-\frac{1}{3},1;\frac{2}{3};1-x^3\right )-\, _2F_1\left (-\frac{1}{3},1;\frac{2}{3};\frac{1}{2} \left (1-x^3\right )\right )}{2 \sqrt [3]{1-x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ({x}^{3}+1 \right ) x} \left ( -{x}^{3}+1 \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{3} + 1\right )}{\left (-x^{3} + 1\right )}^{\frac{4}{3}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.80998, size = 701, normalized size = 4.55 \begin{align*} -\frac{2 \, \sqrt{6} 2^{\frac{1}{6}} \left (-1\right )^{\frac{1}{3}}{\left (x^{3} - 1\right )} \arctan \left (\frac{1}{6} \cdot 2^{\frac{1}{6}}{\left (2 \, \sqrt{6} \left (-1\right )^{\frac{1}{3}}{\left (-x^{3} + 1\right )}^{\frac{1}{3}} - \sqrt{6} 2^{\frac{1}{3}}\right )}\right ) + 2^{\frac{2}{3}} \left (-1\right )^{\frac{1}{3}}{\left (x^{3} - 1\right )} \log \left (2^{\frac{1}{3}} \left (-1\right )^{\frac{2}{3}}{\left (-x^{3} + 1\right )}^{\frac{1}{3}} - 2^{\frac{2}{3}} \left (-1\right )^{\frac{1}{3}} +{\left (-x^{3} + 1\right )}^{\frac{2}{3}}\right ) - 2 \cdot 2^{\frac{2}{3}} \left (-1\right )^{\frac{1}{3}}{\left (x^{3} - 1\right )} \log \left (-2^{\frac{1}{3}} \left (-1\right )^{\frac{2}{3}} +{\left (-x^{3} + 1\right )}^{\frac{1}{3}}\right ) - 8 \, \sqrt{3}{\left (x^{3} - 1\right )} \arctan \left (\frac{2}{3} \, \sqrt{3}{\left (-x^{3} + 1\right )}^{\frac{1}{3}} + \frac{1}{3} \, \sqrt{3}\right ) + 4 \,{\left (x^{3} - 1\right )} \log \left ({\left (-x^{3} + 1\right )}^{\frac{2}{3}} +{\left (-x^{3} + 1\right )}^{\frac{1}{3}} + 1\right ) - 8 \,{\left (x^{3} - 1\right )} \log \left ({\left (-x^{3} + 1\right )}^{\frac{1}{3}} - 1\right ) + 12 \,{\left (-x^{3} + 1\right )}^{\frac{2}{3}}}{24 \,{\left (x^{3} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(2*sqrt(6)*2^(1/6)*(-1)^(1/3)*(x^3 - 1)*arctan(1/6*2^(1/6)*(2*sqrt(6)*(-1)^(1/3)*(-x^3 + 1)^(1/3) - sqrt
(6)*2^(1/3))) + 2^(2/3)*(-1)^(1/3)*(x^3 - 1)*log(2^(1/3)*(-1)^(2/3)*(-x^3 + 1)^(1/3) - 2^(2/3)*(-1)^(1/3) + (-
x^3 + 1)^(2/3)) - 2*2^(2/3)*(-1)^(1/3)*(x^3 - 1)*log(-2^(1/3)*(-1)^(2/3) + (-x^3 + 1)^(1/3)) - 8*sqrt(3)*(x^3
- 1)*arctan(2/3*sqrt(3)*(-x^3 + 1)^(1/3) + 1/3*sqrt(3)) + 4*(x^3 - 1)*log((-x^3 + 1)^(2/3) + (-x^3 + 1)^(1/3)
+ 1) - 8*(x^3 - 1)*log((-x^3 + 1)^(1/3) - 1) + 12*(-x^3 + 1)^(2/3))/(x^3 - 1)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac{4}{3}} \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(1/x/(-x**3+1)**(4/3)/(x**3+1),x)

[Out]

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

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError