3.617 \(\int \frac {1}{x^9 \sqrt [3]{1-x^3} (1+x^3)} \, dx\)

Optimal. Leaf size=141 \[ \frac {\log \left (x^3+1\right )}{6 \sqrt [3]{2}}-\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\left (1-x^3\right )^{2/3}}{8 x^8}+\frac {\left (1-x^3\right )^{2/3}}{20 x^5}-\frac {17 \left (1-x^3\right )^{2/3}}{40 x^2} \]

[Out]

-1/8*(-x^3+1)^(2/3)/x^8+1/20*(-x^3+1)^(2/3)/x^5-17/40*(-x^3+1)^(2/3)/x^2+1/12*ln(x^3+1)*2^(2/3)-1/4*ln(-2^(1/3
)*x-(-x^3+1)^(1/3))*2^(2/3)+1/6*arctan(1/3*(1-2*2^(1/3)*x/(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 175, normalized size of antiderivative = 1.24, 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, 200, 31, 634, 617, 204, 628} \[ -\frac {\left (1-x^3\right )^{8/3}}{8 x^8}-\frac {\left (1-x^3\right )^{5/3}}{5 x^5}-\frac {\left (1-x^3\right )^{2/3}}{2 x^2}+\frac {\log \left (\frac {2^{2/3} x^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}+1\right )}{6 \sqrt [3]{2}}-\frac {\log \left (\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}+1\right )}{3 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^9*(1 - x^3)^(1/3)*(1 + x^3)),x]

[Out]

-(1 - x^3)^(2/3)/(2*x^2) - (1 - x^3)^(5/3)/(5*x^5) - (1 - x^3)^(8/3)/(8*x^8) + ArcTan[(1 - (2*2^(1/3)*x)/(1 -
x^3)^(1/3))/Sqrt[3]]/(2^(1/3)*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^(1/3)) - Log[1 + (2^(1/3)*x)/(1 - x^3)^(1/3)]/(3*2^(1/3))

Rule 31

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

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 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 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 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^9 \sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx &=\operatorname {Subst}\left (\int \frac {\left (1+x^3\right )^3}{x^9 \left (1+2 x^3\right )} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\\ &=\operatorname {Subst}\left (\int \left (\frac {1}{x^9}+\frac {1}{x^6}+\frac {1}{x^3}+\frac {1}{-1-2 x^3}\right ) \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\\ &=-\frac {\left (1-x^3\right )^{2/3}}{2 x^2}-\frac {\left (1-x^3\right )^{5/3}}{5 x^5}-\frac {\left (1-x^3\right )^{8/3}}{8 x^8}+\operatorname {Subst}\left (\int \frac {1}{-1-2 x^3} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\\ &=-\frac {\left (1-x^3\right )^{2/3}}{2 x^2}-\frac {\left (1-x^3\right )^{5/3}}{5 x^5}-\frac {\left (1-x^3\right )^{8/3}}{8 x^8}+\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+\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 {\left (1-x^3\right )^{2/3}}{2 x^2}-\frac {\left (1-x^3\right )^{5/3}}{5 x^5}-\frac {\left (1-x^3\right )^{8/3}}{8 x^8}-\frac {\log \left (1+\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}\right )}{3 \sqrt [3]{2}}-\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 {\left (1-x^3\right )^{2/3}}{2 x^2}-\frac {\left (1-x^3\right )^{5/3}}{5 x^5}-\frac {\left (1-x^3\right )^{8/3}}{8 x^8}+\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}}\\ &=-\frac {\left (1-x^3\right )^{2/3}}{2 x^2}-\frac {\left (1-x^3\right )^{5/3}}{5 x^5}-\frac {\left (1-x^3\right )^{8/3}}{8 x^8}+\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 {\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*}

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Mathematica [A]  time = 5.11, size = 133, normalized size = 0.94 \[ \frac {-2 \log \left (\frac {\sqrt [3]{2} x}{\sqrt [3]{x^3-1}}+1\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{2} x}{\sqrt [3]{x^3-1}}-1}{\sqrt {3}}\right )+\log \left (-\frac {\sqrt [3]{2} x}{\sqrt [3]{x^3-1}}+\frac {2^{2/3} x^2}{\left (x^3-1\right )^{2/3}}+1\right )}{6 \sqrt [3]{2}}-\frac {\left (1-x^3\right )^{2/3} \left (17 x^6-2 x^3+5\right )}{40 x^8} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^9*(1 - x^3)^(1/3)*(1 + x^3)),x]

[Out]

-1/40*((1 - x^3)^(2/3)*(5 - 2*x^3 + 17*x^6))/x^8 + (-2*Sqrt[3]*ArcTan[(-1 + (2*2^(1/3)*x)/(-1 + x^3)^(1/3))/Sq
rt[3]] + Log[1 + (2^(2/3)*x^2)/(-1 + x^3)^(2/3) - (2^(1/3)*x)/(-1 + x^3)^(1/3)] - 2*Log[1 + (2^(1/3)*x)/(-1 +
x^3)^(1/3)])/(6*2^(1/3))

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fricas [B]  time = 4.95, size = 320, normalized size = 2.27 \[ -\frac {20 \, \sqrt {6} 2^{\frac {1}{6}} \left (-1\right )^{\frac {1}{3}} x^{8} \arctan \left (\frac {2^{\frac {1}{6}} {\left (6 \, \sqrt {6} 2^{\frac {2}{3}} \left (-1\right )^{\frac {2}{3}} {\left (5 \, x^{7} + 4 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - 12 \, \sqrt {6} \left (-1\right )^{\frac {1}{3}} {\left (19 \, x^{8} - 16 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - \sqrt {6} 2^{\frac {1}{3}} {\left (71 \, x^{9} - 111 \, x^{6} + 33 \, x^{3} - 1\right )}\right )}}{6 \, {\left (109 \, x^{9} - 105 \, x^{6} + 3 \, x^{3} + 1\right )}}\right ) - 20 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} x^{8} \log \left (\frac {6 \cdot 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{3} + 1\right )} + 6 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x}{x^{3} + 1}\right ) + 10 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} x^{8} \log \left (-\frac {3 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (5 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (19 \, x^{6} - 16 \, x^{3} + 1\right )} + 12 \, {\left (2 \, x^{5} - x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x^{6} + 2 \, x^{3} + 1}\right ) + 9 \, {\left (17 \, x^{6} - 2 \, x^{3} + 5\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{360 \, x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="fricas")

[Out]

-1/360*(20*sqrt(6)*2^(1/6)*(-1)^(1/3)*x^8*arctan(1/6*2^(1/6)*(6*sqrt(6)*2^(2/3)*(-1)^(2/3)*(5*x^7 + 4*x^4 - x)
*(-x^3 + 1)^(2/3) - 12*sqrt(6)*(-1)^(1/3)*(19*x^8 - 16*x^5 + x^2)*(-x^3 + 1)^(1/3) - sqrt(6)*2^(1/3)*(71*x^9 -
111*x^6 + 33*x^3 - 1))/(109*x^9 - 105*x^6 + 3*x^3 + 1)) - 20*2^(2/3)*(-1)^(1/3)*x^8*log((6*2^(1/3)*(-1)^(2/3)
*(-x^3 + 1)^(1/3)*x^2 - 2^(2/3)*(-1)^(1/3)*(x^3 + 1) + 6*(-x^3 + 1)^(2/3)*x)/(x^3 + 1)) + 10*2^(2/3)*(-1)^(1/3
)*x^8*log(-(3*2^(2/3)*(-1)^(1/3)*(5*x^4 - x)*(-x^3 + 1)^(2/3) - 2^(1/3)*(-1)^(2/3)*(19*x^6 - 16*x^3 + 1) + 12*
(2*x^5 - x^2)*(-x^3 + 1)^(1/3))/(x^6 + 2*x^3 + 1)) + 9*(17*x^6 - 2*x^3 + 5)*(-x^3 + 1)^(2/3))/x^8

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x^{9}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="giac")

[Out]

integrate(1/((x^3 + 1)*(-x^3 + 1)^(1/3)*x^9), x)

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maple [C]  time = 4.07, size = 963, normalized size = 6.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^9/(-x^3+1)^(1/3)/(x^3+1),x)

[Out]

1/40*(17*x^9-19*x^6+7*x^3-5)/x^8/(-x^3+1)^(1/3)+RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*ln(-(-18*
RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)^2*RootOf(_Z^3+4)^2*x^3-9*RootOf(RootOf(_Z^3+4)^2+6*_Z*Roo
tOf(_Z^3+4)+36*_Z^2)*RootOf(_Z^3+4)^3*x^3+12*(-x^3+1)^(2/3)*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^
2)*RootOf(_Z^3+4)^2*x+24*(-x^3+1)^(1/3)*RootOf(_Z^3+4)*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*x^
2+5*(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2*x^2+18*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*x^3+9*RootOf(_
Z^3+4)*x^3-10*(-x^3+1)^(2/3)*x-6*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)-3*RootOf(_Z^3+4))/(x+1)/
(x^2-x+1))-1/6*ln((18*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)^2*RootOf(_Z^3+4)^2*x^3-6*RootOf(Roo
tOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*RootOf(_Z^3+4)^3*x^3+12*(-x^3+1)^(2/3)*RootOf(RootOf(_Z^3+4)^2+6*_Z
*RootOf(_Z^3+4)+36*_Z^2)*RootOf(_Z^3+4)^2*x+24*(-x^3+1)^(1/3)*RootOf(_Z^3+4)*RootOf(RootOf(_Z^3+4)^2+6*_Z*Root
Of(_Z^3+4)+36*_Z^2)*x^2-(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2*x^2+6*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_
Z^2)*x^3-2*RootOf(_Z^3+4)*x^3+2*(-x^3+1)^(2/3)*x-6*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)+2*Root
Of(_Z^3+4))/(x+1)/(x^2-x+1))*RootOf(_Z^3+4)-ln((18*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)^2*Root
Of(_Z^3+4)^2*x^3-6*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*RootOf(_Z^3+4)^3*x^3+12*(-x^3+1)^(2/3)
*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*RootOf(_Z^3+4)^2*x+24*(-x^3+1)^(1/3)*RootOf(_Z^3+4)*Root
Of(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*x^2-(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2*x^2+6*RootOf(RootOf(_Z^3+
4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*x^3-2*RootOf(_Z^3+4)*x^3+2*(-x^3+1)^(2/3)*x-6*RootOf(RootOf(_Z^3+4)^2+6*_Z*R
ootOf(_Z^3+4)+36*_Z^2)+2*RootOf(_Z^3+4))/(x+1)/(x^2-x+1))*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x^{9}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="maxima")

[Out]

integrate(1/((x^3 + 1)*(-x^3 + 1)^(1/3)*x^9), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^9\,{\left (1-x^3\right )}^{1/3}\,\left (x^3+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^9*(1 - x^3)^(1/3)*(x^3 + 1)),x)

[Out]

int(1/(x^9*(1 - x^3)^(1/3)*(x^3 + 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{9} \sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**9/(-x**3+1)**(1/3)/(x**3+1),x)

[Out]

Integral(1/(x**9*(-(x - 1)*(x**2 + x + 1))**(1/3)*(x + 1)*(x**2 - x + 1)), x)

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