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

Optimal. Leaf size=85 \[ \frac {1}{3} \log \left (\sqrt [3]{2 x^3-1}-x\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{2 x^3-1}+x}\right )}{\sqrt {3}}-\frac {1}{6} \log \left (\sqrt [3]{2 x^3-1} x+\left (2 x^3-1\right )^{2/3}+x^2\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 86, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {494, 292, 31, 634, 618, 204, 628} \begin {gather*} \frac {1}{3} \log \left (1-\frac {x}{\sqrt [3]{2 x^3-1}}\right )+\frac {\tan ^{-1}\left (\frac {\frac {2 x}{\sqrt [3]{2 x^3-1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{6} \log \left (\frac {x}{\sqrt [3]{2 x^3-1}}+\frac {x^2}{\left (2 x^3-1\right )^{2/3}}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.01, size = 40, normalized size = 0.47 \begin {gather*} -\frac {x^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};-\frac {x^3}{1-2 x^3}\right )}{2 \left (2 x^3-1\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(x^2*Hypergeometric2F1[2/3, 1, 5/3, -(x^3/(1 - 2*x^3))])/(-1 + 2*x^3)^(2/3)

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IntegrateAlgebraic [A]  time = 0.19, size = 85, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+2 x^3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (-x+\sqrt [3]{-1+2 x^3}\right )-\frac {1}{6} \log \left (x^2+x \sqrt [3]{-1+2 x^3}+\left (-1+2 x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.90, size = 104, normalized size = 1.22 \begin {gather*} \frac {1}{3} \, \sqrt {3} \arctan \left (-\frac {4 \, \sqrt {3} {\left (2 \, x^{3} - 1\right )}^{\frac {1}{3}} x^{2} - 2 \, \sqrt {3} {\left (2 \, x^{3} - 1\right )}^{\frac {2}{3}} x + \sqrt {3} {\left (2 \, x^{3} - 1\right )}}{10 \, x^{3} - 1}\right ) + \frac {1}{6} \, \log \left (\frac {x^{3} + 3 \, {\left (2 \, x^{3} - 1\right )}^{\frac {1}{3}} x^{2} - 3 \, {\left (2 \, x^{3} - 1\right )}^{\frac {2}{3}} x - 1}{x^{3} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*arctan(-(4*sqrt(3)*(2*x^3 - 1)^(1/3)*x^2 - 2*sqrt(3)*(2*x^3 - 1)^(2/3)*x + sqrt(3)*(2*x^3 - 1))/(1
0*x^3 - 1)) + 1/6*log((x^3 + 3*(2*x^3 - 1)^(1/3)*x^2 - 3*(2*x^3 - 1)^(2/3)*x - 1)/(x^3 - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.88, size = 447, normalized size = 5.26

method	result	size
trager	\(\frac {\ln \left (-\frac {-27 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{3}+15 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (2 x^{3}-1\right )^{\frac {2}{3}} x -3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (2 x^{3}-1\right )^{\frac {1}{3}} x^{2}-30 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{3}+4 \left (2 x^{3}-1\right )^{\frac {2}{3}} x -5 \left (2 x^{3}-1\right )^{\frac {1}{3}} x^{2}-3 x^{3}+9 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )+1}{\left (-1+x \right ) \left (x^{2}+x +1\right )}\right )}{3}-\frac {\ln \left (\frac {-9 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{3}+15 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (2 x^{3}-1\right )^{\frac {2}{3}} x -12 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (2 x^{3}-1\right )^{\frac {1}{3}} x^{2}-3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{3}+\left (2 x^{3}-1\right )^{\frac {2}{3}} x -5 \left (2 x^{3}-1\right )^{\frac {1}{3}} x^{2}+6 x^{3}-3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-3}{\left (-1+x \right ) \left (x^{2}+x +1\right )}\right )}{3}-\ln \left (\frac {-9 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{3}+15 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (2 x^{3}-1\right )^{\frac {2}{3}} x -12 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (2 x^{3}-1\right )^{\frac {1}{3}} x^{2}-3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{3}+\left (2 x^{3}-1\right )^{\frac {2}{3}} x -5 \left (2 x^{3}-1\right )^{\frac {1}{3}} x^{2}+6 x^{3}-3 \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-3}{\left (-1+x \right ) \left (x^{2}+x +1\right )}\right ) \RootOf \left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )\)	\(447\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(x^3-1)/(2*x^3-1)^(2/3),x,method=_RETURNVERBOSE)

[Out]

1/3*ln(-(-27*RootOf(9*_Z^2+3*_Z+1)^2*x^3+15*RootOf(9*_Z^2+3*_Z+1)*(2*x^3-1)^(2/3)*x-3*RootOf(9*_Z^2+3*_Z+1)*(2
*x^3-1)^(1/3)*x^2-30*RootOf(9*_Z^2+3*_Z+1)*x^3+4*(2*x^3-1)^(2/3)*x-5*(2*x^3-1)^(1/3)*x^2-3*x^3+9*RootOf(9*_Z^2
+3*_Z+1)+1)/(-1+x)/(x^2+x+1))-1/3*ln((-9*RootOf(9*_Z^2+3*_Z+1)^2*x^3+15*RootOf(9*_Z^2+3*_Z+1)*(2*x^3-1)^(2/3)*
x-12*RootOf(9*_Z^2+3*_Z+1)*(2*x^3-1)^(1/3)*x^2-3*RootOf(9*_Z^2+3*_Z+1)*x^3+(2*x^3-1)^(2/3)*x-5*(2*x^3-1)^(1/3)
*x^2+6*x^3-3*RootOf(9*_Z^2+3*_Z+1)-3)/(-1+x)/(x^2+x+1))-ln((-9*RootOf(9*_Z^2+3*_Z+1)^2*x^3+15*RootOf(9*_Z^2+3*
_Z+1)*(2*x^3-1)^(2/3)*x-12*RootOf(9*_Z^2+3*_Z+1)*(2*x^3-1)^(1/3)*x^2-3*RootOf(9*_Z^2+3*_Z+1)*x^3+(2*x^3-1)^(2/
3)*x-5*(2*x^3-1)^(1/3)*x^2+6*x^3-3*RootOf(9*_Z^2+3*_Z+1)-3)/(-1+x)/(x^2+x+1))*RootOf(9*_Z^2+3*_Z+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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