3.640 \(\int \frac {x^8}{(1-x^3)^{4/3} (1+x^3)} \, dx\)

Optimal. Leaf size=115 \[ \frac {1}{2} \left (1-x^3\right )^{2/3}+\frac {1}{2 \sqrt [3]{1-x^3}}-\frac {\log \left (x^3+1\right )}{12 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}} \]

[Out]

1/2/(-x^3+1)^(1/3)+1/2*(-x^3+1)^(2/3)-1/24*ln(x^3+1)*2^(2/3)+1/8*ln(2^(1/3)-(-x^3+1)^(1/3))*2^(2/3)+1/12*arcta
n(1/3*(1+2^(2/3)*(-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 = 115, normalized size of antiderivative = 1.00, 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 = {446, 87, 627, 51, 55, 617, 204, 31} \[ \frac {1}{2} \left (1-x^3\right )^{2/3}+\frac {1}{2 \sqrt [3]{1-x^3}}-\frac {\log \left (x^3+1\right )}{12 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[x^8/((1 - x^3)^(4/3)*(1 + x^3)),x]

[Out]

1/(2*(1 - x^3)^(1/3)) + (1 - x^3)^(2/3)/2 + ArcTan[(1 + 2^(2/3)*(1 - x^3)^(1/3))/Sqrt[3]]/(2*2^(1/3)*Sqrt[3])
- Log[1 + x^3]/(12*2^(1/3)) + Log[2^(1/3) - (1 - x^3)^(1/3)]/(4*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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, 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 87

Int[(((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegr
and[(e + f*x)^FractionalPart[p], ((c + d*x)^n*(e + f*x)^IntegerPart[p])/(a + b*x), x], x] /; FreeQ[{a, b, c, d
, e, f}, x] && IGtQ[n, 0] && LtQ[p, -1] && FractionQ[p]

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 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 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 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rubi steps

\begin {align*} \int \frac {x^8}{\left (1-x^3\right )^{4/3} \left (1+x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2}{(1-x)^{4/3} (1+x)} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (-\frac {1}{\sqrt [3]{1-x}}+\frac {1}{\sqrt [3]{1-x} \left (1-x^2\right )}\right ) \, dx,x,x^3\right )\\ &=\frac {1}{2} \left (1-x^3\right )^{2/3}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} \left (1-x^2\right )} \, dx,x,x^3\right )\\ &=\frac {1}{2} \left (1-x^3\right )^{2/3}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{(1-x)^{4/3} (1+x)} \, dx,x,x^3\right )\\ &=\frac {1}{2 \sqrt [3]{1-x^3}}+\frac {1}{2} \left (1-x^3\right )^{2/3}+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (1+x)} \, dx,x,x^3\right )\\ &=\frac {1}{2 \sqrt [3]{1-x^3}}+\frac {1}{2} \left (1-x^3\right )^{2/3}-\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 {\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 {1}{2} \left (1-x^3\right )^{2/3}-\frac {\log \left (1+x^3\right )}{12 \sqrt [3]{2}}+\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}}\\ &=\frac {1}{2 \sqrt [3]{1-x^3}}+\frac {1}{2} \left (1-x^3\right )^{2/3}+\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 \left (1+x^3\right )}{12 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 41, normalized size = 0.36 \[ \frac {\, _2F_1\left (-\frac {1}{3},1;\frac {2}{3};\frac {1}{2} \left (1-x^3\right )\right )-x^3+1}{2 \sqrt [3]{1-x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^8/((1 - x^3)^(4/3)*(1 + x^3)),x]

[Out]

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

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fricas [A]  time = 0.74, size = 130, normalized size = 1.13 \[ \frac {2 \, \sqrt {6} 2^{\frac {1}{6}} {\left (x^{3} - 1\right )} \arctan \left (\frac {1}{6} \cdot 2^{\frac {1}{6}} {\left (\sqrt {6} 2^{\frac {1}{3}} + 2 \, \sqrt {6} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - 2^{\frac {2}{3}} {\left (x^{3} - 1\right )} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} {\left (x^{3} - 1\right )} \log \left (-2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) + 12 \, {\left (x^{3} - 2\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{24 \, {\left (x^{3} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 109, normalized size = 0.95 \[ \frac {1}{12} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{24} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \right |}\right ) + \frac {1}{2} \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + \frac {1}{2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="giac")

[Out]

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

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maple [C]  time = 3.78, size = 672, normalized size = 5.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(-x^3+1)^(4/3)/(x^3+1),x)

[Out]

-1/2*(x^3-2)/(-x^3+1)^(1/3)+1/2*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+15*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z
^2)*RootOf(_Z^3-4)^3*x^3-6*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*x^3-5*RootOf(_Z^3-4)*x^3+21*Ro
otOf(_Z^3-4)^2*(-x^3+1)^(1/3)+42*(-x^3+1)^(2/3)+42*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)+35*Roo
tOf(_Z^3-4))/(x+1)/(x^2-x+1))-1/12*ln((18*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)^2*RootOf(_Z^3-4
)^2*x^3-12*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*RootOf(_Z^3-4)^3*x^3+18*RootOf(RootOf(_Z^3-4)^
2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*x^3-12*RootOf(_Z^3-4)*x^3+21*RootOf(_Z^3-4)^2*(-x^3+1)^(1/3)+42*(-x^3+1)^(2/3)-
42*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)+28*RootOf(_Z^3-4))/(x+1)/(x^2-x+1))*RootOf(_Z^3-4)-1/2
*ln((18*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)^2*RootOf(_Z^3-4)^2*x^3-12*RootOf(RootOf(_Z^3-4)^2
+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*RootOf(_Z^3-4)^3*x^3+18*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*x^3
-12*RootOf(_Z^3-4)*x^3+21*RootOf(_Z^3-4)^2*(-x^3+1)^(1/3)+42*(-x^3+1)^(2/3)-42*RootOf(RootOf(_Z^3-4)^2+6*_Z*Ro
otOf(_Z^3-4)+36*_Z^2)+28*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 [A]  time = 1.27, size = 108, normalized size = 0.94 \[ \frac {1}{12} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{24} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) + \frac {1}{2} \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + \frac {1}{2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(-x^3+1)^(4/3)/(x^3+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.89, size = 128, normalized size = 1.11 \[ \frac {2^{2/3}\,\ln \left (\frac {{\left (1-x^3\right )}^{1/3}}{4}-\frac {2^{1/3}}{4}\right )}{12}+\frac {1}{2\,{\left (1-x^3\right )}^{1/3}}+\frac {{\left (1-x^3\right )}^{2/3}}{2}+\frac {2^{2/3}\,\ln \left (\frac {{\left (1-x^3\right )}^{1/3}}{4}-\frac {2^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{24}-\frac {2^{2/3}\,\ln \left (\frac {{\left (1-x^3\right )}^{1/3}}{4}-\frac {2^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{24} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/((1 - x^3)^(4/3)*(x^3 + 1)),x)

[Out]

(2^(2/3)*log((1 - x^3)^(1/3)/4 - 2^(1/3)/4))/12 + 1/(2*(1 - x^3)^(1/3)) + (1 - x^3)^(2/3)/2 + (2^(2/3)*log((1
- x^3)^(1/3)/4 - (2^(1/3)*(3^(1/2)*1i - 1)^2)/16)*(3^(1/2)*1i - 1))/24 - (2^(2/3)*log((1 - x^3)^(1/3)/4 - (2^(
1/3)*(3^(1/2)*1i + 1)^2)/16)*(3^(1/2)*1i + 1))/24

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{8}}{\left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {4}{3}} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(-x**3+1)**(4/3)/(x**3+1),x)

[Out]

Integral(x**8/((-(x - 1)*(x**2 + x + 1))**(4/3)*(x + 1)*(x**2 - x + 1)), x)

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