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

Optimal. Leaf size=103 \[ -\frac {\sqrt [3]{1-x^3}}{x}-\frac {\log \left (x^3+1\right )}{6\ 2^{2/3}}+\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2\ 2^{2/3}}+\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 137, normalized size of antiderivative = 1.33, number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {494, 453, 292, 31, 634, 617, 204, 628} \begin {gather*} -\frac {\sqrt [3]{1-x^3}}{x}-\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[1/(x^2*(1 - x^3)^(2/3)*(1 + x^3)),x]

[Out]

-((1 - x^3)^(1/3)/x) + 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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -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 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 {1}{x^2 \left (1-x^3\right )^{2/3} \left (1+x^3\right )} \, dx &=\operatorname {Subst}\left (\int \frac {1+x^3}{x^2 \left (1+2 x^3\right )} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\\ &=-\frac {\sqrt [3]{1-x^3}}{x}-\operatorname {Subst}\left (\int \frac {x}{1+2 x^3} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\\ &=-\frac {\sqrt [3]{1-x^3}}{x}+\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 {\sqrt [3]{1-x^3}}{x}+\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 {\sqrt [3]{1-x^3}}{x}-\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 {\sqrt [3]{1-x^3}}{x}+\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 = 0.55, size = 81, normalized size = 0.79 \begin {gather*} -\frac {5 \left (-3 x^6+x^3+2\right ) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {2 x^3}{x^3-1}\right )-12 x^3 \left (x^3+1\right ) \, _2F_1\left (\frac {5}{3},2;\frac {8}{3};\frac {2 x^3}{x^3-1}\right )}{10 x \left (1-x^3\right )^{5/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/10*(5*(2 + x^3 - 3*x^6)*Hypergeometric2F1[2/3, 1, 5/3, (2*x^3)/(-1 + x^3)] - 12*x^3*(1 + x^3)*Hypergeometri
c2F1[5/3, 2, 8/3, (2*x^3)/(-1 + x^3)])/(x*(1 - x^3)^(5/3))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.30, size = 143, normalized size = 1.39 \begin {gather*} -\frac {\sqrt [3]{1-x^3}}{x}+\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[1/(x^2*(1 - x^3)^(2/3)*(1 + x^3)),x]

[Out]

-((1 - x^3)^(1/3)/x) - 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))

________________________________________________________________________________________

fricas [B]  time = 1.94, size = 272, normalized size = 2.64 \begin {gather*} \frac {4 \cdot 4^{\frac {1}{6}} \sqrt {3} x \arctan \left (\frac {4^{\frac {1}{6}} {\left (6 \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (19 \, x^{8} - 16 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 12 \, \sqrt {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}} x \log \left (\frac {3 \cdot 4^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x^{2} + 6 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x + 4^{\frac {1}{3}} {\left (x^{3} + 1\right )}}{x^{3} + 1}\right ) - 4^{\frac {2}{3}} x \log \left (\frac {6 \cdot 4^{\frac {1}{3}} {\left (5 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 4^{\frac {2}{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 ) - 72 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{72 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(4*4^(1/6)*sqrt(3)*x*arctan(1/6*4^(1/6)*(6*4^(2/3)*sqrt(3)*(19*x^8 - 16*x^5 + x^2)*(-x^3 + 1)^(1/3) + 12*
sqrt(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)*x*log((3*4^(2/3)*(-x^3 + 1)^(1/3)*x^2 + 6*(-x^3 + 1)^(2/3)*x + 4^(1/3)*(x^3 + 1
))/(x^3 + 1)) - 4^(2/3)*x*log((6*4^(1/3)*(5*x^4 - x)*(-x^3 + 1)^(2/3) + 4^(2/3)*(19*x^6 - 16*x^3 + 1) - 24*(2*
x^5 - x^2)*(-x^3 + 1)^(1/3))/(x^6 + 2*x^3 + 1)) - 72*(-x^3 + 1)^(1/3))/x

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 4.06, size = 1387, normalized size = 13.47 \begin {gather*} \text {Expression too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^3-1)/x/(-x^3+1)^(2/3)+(-1/6*ln((18*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^2*
x^6+6*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^3*x^6+9*(x^6-2*x^3+1)^(2/3)*RootOf(R
ootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^2*x^2-9*(x^6-2*x^3+1)^(1/3)*RootOf(_Z^3-2)*RootOf(
RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^4-18*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*Ro
otOf(_Z^3-2)^2*x^3-6*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^3*x^3-3*RootOf(RootOf
(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^6-RootOf(_Z^3-2)*x^6+3*(x^6-2*x^3+1)^(2/3)*x^2+9*(x^6-2*x^3+1)^(1/3)
*RootOf(_Z^3-2)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x+6*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_
Z^3-2)+36*_Z^2)*x^3+2*RootOf(_Z^3-2)*x^3-3*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)-RootOf(_Z^3-2)
)/(x-1)/(x^2+x+1)/(x+1)/(x^2-x+1))*RootOf(_Z^3-2)-ln((18*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^
2*RootOf(_Z^3-2)^2*x^6+6*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^3*x^6+9*(x^6-2*x^
3+1)^(2/3)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^2*x^2-9*(x^6-2*x^3+1)^(1/3)*Roo
tOf(_Z^3-2)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^4-18*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z
^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^2*x^3-6*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^3*
x^3-3*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^6-RootOf(_Z^3-2)*x^6+3*(x^6-2*x^3+1)^(2/3)*x^2+9*
(x^6-2*x^3+1)^(1/3)*RootOf(_Z^3-2)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x+6*RootOf(RootOf(_Z^3
-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^3+2*RootOf(_Z^3-2)*x^3-3*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_
Z^2)-RootOf(_Z^3-2))/(x-1)/(x^2+x+1)/(x+1)/(x^2-x+1))*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)+1/6
*RootOf(_Z^3-2)*ln(-(-36*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^2*x^6-12*RootOf
(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^3*x^6+18*(x^6-2*x^3+1)^(2/3)*RootOf(RootOf(_Z^3-
2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^2*x^2+3*(x^6-2*x^3+1)^(1/3)*RootOf(_Z^3-2)^2*x^4+36*RootOf(Ro
otOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^2*x^3+12*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3
-2)+36*_Z^2)*RootOf(_Z^3-2)^3*x^3-18*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^6-6*RootOf(_Z^3-2)
*x^6-3*(x^6-2*x^3+1)^(1/3)*RootOf(_Z^3-2)^2*x+24*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*x^3+8*Ro
otOf(_Z^3-2)*x^3-6*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)-2*RootOf(_Z^3-2))/(x-1)/(x^2+x+1)/(x+1
)/(x^2-x+1)))/(-x^3+1)^(2/3)*((x^3-1)^2)^(1/3)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^2\,{\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(1/(x^2*(1 - x^3)^(2/3)*(x^3 + 1)),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________