3.633 \(\int \frac{1}{x^5 (1-x^3)^{2/3} (1+x^3)} \, dx\)

Optimal. Leaf size=124 \[ \frac{\sqrt [3]{1-x^3}}{4 x}-\frac{\sqrt [3]{1-x^3}}{4 x^4}+\frac{\log \left (x^3+1\right )}{6\ 2^{2/3}}-\frac{\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2\ 2^{2/3}}-\frac{\tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt{3}}\right )}{2^{2/3} \sqrt{3}} \]

[Out]

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

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Rubi [A]  time = 0.0844884, antiderivative size = 140, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {494, 461, 292, 31, 634, 617, 204, 628} \[ -\frac{\left (1-x^3\right )^{4/3}}{4 x^4}+\frac{\log \left (\frac{2^{2/3} x^2}{\left (1-x^3\right )^{2/3}}-\frac{\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}+1\right )}{6\ 2^{2/3}}-\frac{\log \left (\frac{\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}+1\right )}{3\ 2^{2/3}}-\frac{\tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt{3}}\right )}{2^{2/3} \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 - x^3)^(4/3)/(4*x^4) - 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 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 461

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

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{x^5 \left (1-x^3\right )^{2/3} \left (1+x^3\right )} \, dx &=\operatorname{Subst}\left (\int \frac{\left (1+x^3\right )^2}{x^5 \left (1+2 x^3\right )} \, dx,x,\frac{x}{\sqrt [3]{1-x^3}}\right )\\ &=\operatorname{Subst}\left (\int \left (\frac{1}{x^5}+\frac{x}{1+2 x^3}\right ) \, dx,x,\frac{x}{\sqrt [3]{1-x^3}}\right )\\ &=-\frac{\left (1-x^3\right )^{4/3}}{4 x^4}+\operatorname{Subst}\left (\int \frac{x}{1+2 x^3} \, dx,x,\frac{x}{\sqrt [3]{1-x^3}}\right )\\ &=-\frac{\left (1-x^3\right )^{4/3}}{4 x^4}-\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{\left (1-x^3\right )^{4/3}}{4 x^4}-\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{\left (1-x^3\right )^{4/3}}{4 x^4}+\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{\left (1-x^3\right )^{4/3}}{4 x^4}-\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*}

Mathematica [C]  time = 8.51815, size = 145, normalized size = 1.17 \[ -\frac{81 \left (x^3+1\right )^2 x^3 \text{HypergeometricPFQ}\left (\left \{\frac{2}{3},2,2,2\right \},\left \{1,1,\frac{8}{3}\right \},\frac{2 x^3}{x^3-1}\right )+216 \left (x^9+x^6\right ) \text{HypergeometricPFQ}\left (\left \{\frac{2}{3},2,2\right \},\left \{1,\frac{8}{3}\right \},\frac{2 x^3}{x^3-1}\right )+5 \left (\left (9 x^9-20 x^6-13 x^3+4\right ) \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{2 x^3}{x^3-1}\right )+9 x^9+x^6-9 x^3-1\right )}{60 x^4 \left (1-x^3\right )^{5/3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(5*(-1 - 9*x^3 + x^6 + 9*x^9 + (4 - 13*x^3 - 20*x^6 + 9*x^9)*Hypergeometric2F1[2/3, 1, 5/3, (2*x^3)/(-1 + x^3
)]) + 216*(x^6 + x^9)*HypergeometricPFQ[{2/3, 2, 2}, {1, 8/3}, (2*x^3)/(-1 + x^3)] + 81*x^3*(1 + x^3)^2*Hyperg
eometricPFQ[{2/3, 2, 2, 2}, {1, 1, 8/3}, (2*x^3)/(-1 + x^3)])/(60*x^4*(1 - x^3)^(5/3))

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Maple [F]  time = 0.053, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{5} \left ({x}^{3}+1 \right ) } \left ( -{x}^{3}+1 \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(-x^3+1)^(2/3)/(x^3+1),x)

[Out]

int(1/x^5/(-x^3+1)^(2/3)/(x^3+1),x)

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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{2}{3}} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-x^3+1)^(2/3)/(x^3+1),x, algorithm="maxima")

[Out]

integrate(1/((x^3 + 1)*(-x^3 + 1)^(2/3)*x^5), x)

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Fricas [B]  time = 13.0231, size = 856, normalized size = 6.9 \begin{align*} -\frac{4 \cdot 4^{\frac{1}{6}} \sqrt{3} \left (-1\right )^{\frac{1}{3}} x^{4} \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 ) - 2 \cdot 4^{\frac{2}{3}} \left (-1\right )^{\frac{1}{3}} x^{4} \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 ) + 4^{\frac{2}{3}} \left (-1\right )^{\frac{1}{3}} x^{4} \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 ) - 18 \,{\left (x^{3} - 1\right )}{\left (-x^{3} + 1\right )}^{\frac{1}{3}}}{72 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-x^3+1)^(2/3)/(x^3+1),x, algorithm="fricas")

[Out]

-1/72*(4*4^(1/6)*sqrt(3)*(-1)^(1/3)*x^4*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)) - 2*4^(2/3)*(-1)^(1/3)*x^4*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)) + 4^(2/3)*(-1)^(1/3)*x^
4*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)) - 18*(x^3 - 1)*(-x^3 + 1)^(1/3))/x^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{5} \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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(-x**3+1)**(2/3)/(x**3+1),x)

[Out]

Integral(1/(x**5*(-(x - 1)*(x**2 + x + 1))**(2/3)*(x + 1)*(x**2 - x + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{3} + 1\right )}{\left (-x^{3} + 1\right )}^{\frac{2}{3}} x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(-x^3+1)^(2/3)/(x^3+1),x, algorithm="giac")

[Out]

integrate(1/((x^3 + 1)*(-x^3 + 1)^(2/3)*x^5), x)