3.4.80 \(\int \frac {x^6}{\sqrt [3]{1-x^3} (1+x^3)} \, dx\)

Optimal. Leaf size=154 \[ -\frac {1}{3} \left (1-x^3\right )^{2/3} x-\frac {\log \left (x^3+1\right )}{6 \sqrt [3]{2}}+\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2 \sqrt [3]{2}}-\frac {1}{3} \log \left (\sqrt [3]{1-x^3}+x\right )+\frac {2 \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}} \]

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Rubi [A]  time = 0.15, antiderivative size = 226, normalized size of antiderivative = 1.47, number of steps used = 15, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {494, 470, 522, 200, 31, 634, 618, 204, 628, 617} \begin {gather*} -\frac {1}{3} \left (1-x^3\right )^{2/3} x+\frac {1}{9} \log \left (\frac {x^2}{\left (1-x^3\right )^{2/3}}-\frac {x}{\sqrt [3]{1-x^3}}+1\right )-\frac {2}{9} \log \left (\frac {x}{\sqrt [3]{1-x^3}}+1\right )-\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 {2 \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3}}-\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6/((1 - x^3)^(1/3)*(1 + x^3)),x]

[Out]

-(x*(1 - x^3)^(2/3))/3 + (2*ArcTan[(1 - (2*x)/(1 - x^3)^(1/3))/Sqrt[3]])/(3*Sqrt[3]) - ArcTan[(1 - (2*2^(1/3)*
x)/(1 - x^3)^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3]) + Log[1 + x^2/(1 - x^3)^(2/3) - x/(1 - x^3)^(1/3)]/9 - (2*Log[1
+ x/(1 - x^3)^(1/3)])/9 - 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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

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 522

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

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 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 {x^6}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx &=\operatorname {Subst}\left (\int \frac {x^6}{\left (1+x^3\right )^2 \left (1+2 x^3\right )} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\\ &=-\frac {1}{3} x \left (1-x^3\right )^{2/3}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1-x^3}{\left (1+x^3\right ) \left (1+2 x^3\right )} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\\ &=-\frac {1}{3} x \left (1-x^3\right )^{2/3}-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{1+x^3} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )+\operatorname {Subst}\left (\int \frac {1}{1+2 x^3} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )\\ &=-\frac {1}{3} x \left (1-x^3\right )^{2/3}-\frac {2}{9} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )-\frac {2}{9} \operatorname {Subst}\left (\int \frac {2-x}{1-x+x^2} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )+\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 {1}{3} x \left (1-x^3\right )^{2/3}-\frac {2}{9} \log \left (1+\frac {x}{\sqrt [3]{1-x^3}}\right )+\frac {\log \left (1+\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}\right )}{3 \sqrt [3]{2}}+\frac {1}{9} \operatorname {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\frac {x}{\sqrt [3]{1-x^3}}\right )+\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 {1}{3} x \left (1-x^3\right )^{2/3}+\frac {1}{9} \log \left (1+\frac {x^2}{\left (1-x^3\right )^{2/3}}-\frac {x}{\sqrt [3]{1-x^3}}\right )-\frac {2}{9} \log \left (1+\frac {x}{\sqrt [3]{1-x^3}}\right )-\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 {2}{3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+\frac {2 x}{\sqrt [3]{1-x^3}}\right )+\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 {1}{3} x \left (1-x^3\right )^{2/3}+\frac {2 \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3}}-\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 {1}{9} \log \left (1+\frac {x^2}{\left (1-x^3\right )^{2/3}}-\frac {x}{\sqrt [3]{1-x^3}}\right )-\frac {2}{9} \log \left (1+\frac {x}{\sqrt [3]{1-x^3}}\right )-\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 [C]  time = 0.16, size = 144, normalized size = 0.94 \begin {gather*} \frac {1}{36} \left (-6 x^4 F_1\left (\frac {4}{3};\frac {1}{3},1;\frac {7}{3};x^3,-x^3\right )-12 \left (1-x^3\right )^{2/3} x+2^{2/3} \left (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 )\right )\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x^6/((1 - x^3)^(1/3)*(1 + x^3)),x]

[Out]

(-12*x*(1 - x^3)^(2/3) - 6*x^4*AppellF1[4/3, 1/3, 1, 7/3, x^3, -x^3] + 2^(2/3)*(2*Sqrt[3]*ArcTan[(-1 + (2*2^(1
/3)*x)/(-1 + x^3)^(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)] + 2
*Log[1 + (2^(1/3)*x)/(-1 + x^3)^(1/3)]))/36

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IntegrateAlgebraic [A]  time = 0.54, size = 230, normalized size = 1.49 \begin {gather*} -\frac {1}{3} \left (1-x^3\right )^{2/3} x-\frac {2}{9} \log \left (\sqrt [3]{1-x^3}+x\right )+\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3}+2 x\right )}{3 \sqrt [3]{2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{1-x^3}-x}\right )}{3 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2^{2/3} \sqrt [3]{1-x^3}-x}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {1}{9} \log \left (-\sqrt [3]{1-x^3} x+\left (1-x^3\right )^{2/3}+x^2\right )-\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 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^6/((1 - x^3)^(1/3)*(1 + x^3)),x]

[Out]

-1/3*(x*(1 - x^3)^(2/3)) - (2*ArcTan[(Sqrt[3]*x)/(-x + 2*(1 - x^3)^(1/3))])/(3*Sqrt[3]) + ArcTan[(Sqrt[3]*x)/(
-x + 2^(2/3)*(1 - x^3)^(1/3))]/(2^(1/3)*Sqrt[3]) - (2*Log[x + (1 - x^3)^(1/3)])/9 + Log[2*x + 2^(2/3)*(1 - x^3
)^(1/3)]/(3*2^(1/3)) + Log[x^2 - x*(1 - x^3)^(1/3) + (1 - x^3)^(2/3)]/9 - Log[-2*x^2 + 2^(2/3)*x*(1 - x^3)^(1/
3) - 2^(1/3)*(1 - x^3)^(2/3)]/(6*2^(1/3))

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fricas [A]  time = 0.47, size = 201, normalized size = 1.31 \begin {gather*} -\frac {1}{3} \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x - \frac {1}{6} \, \sqrt {6} 2^{\frac {1}{6}} \arctan \left (-\frac {2^{\frac {1}{6}} {\left (\sqrt {6} 2^{\frac {1}{3}} x - 2 \, \sqrt {6} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{6 \, x}\right ) + \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left (\frac {2^{\frac {1}{3}} x + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (\frac {2^{\frac {2}{3}} x^{2} - 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) + \frac {2}{9} \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} x - 2 \, \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{3 \, x}\right ) - \frac {2}{9} \, \log \left (\frac {x + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x}\right ) + \frac {1}{9} \, \log \left (\frac {x^{2} - {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="fricas")

[Out]

-1/3*(-x^3 + 1)^(2/3)*x - 1/6*sqrt(6)*2^(1/6)*arctan(-1/6*2^(1/6)*(sqrt(6)*2^(1/3)*x - 2*sqrt(6)*(-x^3 + 1)^(1
/3))/x) + 1/6*2^(2/3)*log((2^(1/3)*x + (-x^3 + 1)^(1/3))/x) - 1/12*2^(2/3)*log((2^(2/3)*x^2 - 2^(1/3)*(-x^3 +
1)^(1/3)*x + (-x^3 + 1)^(2/3))/x^2) + 2/9*sqrt(3)*arctan(-1/3*(sqrt(3)*x - 2*sqrt(3)*(-x^3 + 1)^(1/3))/x) - 2/
9*log((x + (-x^3 + 1)^(1/3))/x) + 1/9*log((x^2 - (-x^3 + 1)^(1/3)*x + (-x^3 + 1)^(2/3))/x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6}}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="giac")

[Out]

integrate(x^6/((x^3 + 1)*(-x^3 + 1)^(1/3)), x)

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maple [F]  time = 1.92, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6}}{\left (-x^{3}+1\right )^{\frac {1}{3}} \left (x^{3}+1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(-x^3+1)^(1/3)/(x^3+1),x)

[Out]

int(x^6/(-x^3+1)^(1/3)/(x^3+1),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6}}{{\left (x^{3} + 1\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(-x^3+1)^(1/3)/(x^3+1),x, algorithm="maxima")

[Out]

integrate(x^6/((x^3 + 1)*(-x^3 + 1)^(1/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^6}{{\left (1-x^3\right )}^{1/3}\,\left (x^3+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/((1 - x^3)^(1/3)*(x^3 + 1)),x)

[Out]

int(x^6/((1 - x^3)^(1/3)*(x^3 + 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{6}}{\sqrt [3]{- \left (x - 1\right ) \left (x^{2} + x + 1\right )} \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(x**6/(-x**3+1)**(1/3)/(x**3+1),x)

[Out]

Integral(x**6/((-(x - 1)*(x**2 + x + 1))**(1/3)*(x + 1)*(x**2 - x + 1)), x)

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