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

Optimal. Leaf size=110 \[ \frac {1}{4} \log \left (2 \sqrt [3]{x^3-x}+x+1\right )+\frac {1}{4} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x^3-x}}{\sqrt [3]{x^3-x}-x-1}\right )-\frac {1}{8} \log \left (4 \left (x^3-x\right )^{2/3}+(-2 x-2) \sqrt [3]{x^3-x}+x^2+2 x+1\right ) \]

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Rubi [C]  time = 0.22, antiderivative size = 220, normalized size of antiderivative = 2.00, number of steps used = 14, number of rules used = 13, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {2056, 959, 466, 465, 377, 200, 31, 634, 618, 204, 628, 511, 510} \begin {gather*} \frac {9 \sqrt [3]{1-x^2} x^2 F_1\left (\frac {5}{6};1,\frac {1}{3};\frac {11}{6};9 x^2,x^2\right )}{5 \sqrt [3]{x^3-x}}-\frac {\sqrt [3]{x^2-1} \sqrt [3]{x} \log \left (\frac {4 x^{4/3}}{\left (x^2-1\right )^{2/3}}-\frac {2 x^{2/3}}{\sqrt [3]{x^2-1}}+1\right )}{8 \sqrt [3]{x^3-x}}+\frac {\sqrt [3]{x^2-1} \sqrt [3]{x} \log \left (\frac {2 x^{2/3}}{\sqrt [3]{x^2-1}}+1\right )}{4 \sqrt [3]{x^3-x}}-\frac {\sqrt {3} \sqrt [3]{x^2-1} \sqrt [3]{x} \tan ^{-1}\left (\frac {1-\frac {4 x^{2/3}}{\sqrt [3]{x^2-1}}}{\sqrt {3}}\right )}{4 \sqrt [3]{x^3-x}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[1/((1 - 3*x)*(-x + x^3)^(1/3)),x]

[Out]

(9*x^2*(1 - x^2)^(1/3)*AppellF1[5/6, 1, 1/3, 11/6, 9*x^2, x^2])/(5*(-x + x^3)^(1/3)) - (Sqrt[3]*x^(1/3)*(-1 +
x^2)^(1/3)*ArcTan[(1 - (4*x^(2/3))/(-1 + x^2)^(1/3))/Sqrt[3]])/(4*(-x + x^3)^(1/3)) - (x^(1/3)*(-1 + x^2)^(1/3
)*Log[1 + (4*x^(4/3))/(-1 + x^2)^(2/3) - (2*x^(2/3))/(-1 + x^2)^(1/3)])/(8*(-x + x^3)^(1/3)) + (x^(1/3)*(-1 +
x^2)^(1/3)*Log[1 + (2*x^(2/3))/(-1 + x^2)^(1/3)])/(4*(-x + x^3)^(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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 465

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 511

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

Rule 959

Int[(((g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[(d*(g*x)^n)/x^n, In
t[(x^n*(a + c*x^2)^p)/(d^2 - e^2*x^2), x], x] - Dist[(e*(g*x)^n)/x^n, Int[(x^(n + 1)*(a + c*x^2)^p)/(d^2 - e^2
*x^2), x], x] /; FreeQ[{a, c, d, e, g, n, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !IntegersQ[n, 2
*p]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rubi steps

\begin {align*} \int \frac {1}{(1-3 x) \sqrt [3]{-x+x^3}} \, dx &=\frac {\left (\sqrt [3]{x} \sqrt [3]{-1+x^2}\right ) \int \frac {1}{(1-3 x) \sqrt [3]{x} \sqrt [3]{-1+x^2}} \, dx}{\sqrt [3]{-x+x^3}}\\ &=\frac {\left (\sqrt [3]{x} \sqrt [3]{-1+x^2}\right ) \int \frac {1}{\sqrt [3]{x} \left (1-9 x^2\right ) \sqrt [3]{-1+x^2}} \, dx}{\sqrt [3]{-x+x^3}}+\frac {\left (3 \sqrt [3]{x} \sqrt [3]{-1+x^2}\right ) \int \frac {x^{2/3}}{\left (1-9 x^2\right ) \sqrt [3]{-1+x^2}} \, dx}{\sqrt [3]{-x+x^3}}\\ &=\frac {\left (3 \sqrt [3]{x} \sqrt [3]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1-9 x^6\right ) \sqrt [3]{-1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{-x+x^3}}+\frac {\left (9 \sqrt [3]{x} \sqrt [3]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (1-9 x^6\right ) \sqrt [3]{-1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{-x+x^3}}\\ &=\frac {\left (9 \sqrt [3]{x} \sqrt [3]{1-x^2}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (1-9 x^6\right ) \sqrt [3]{1-x^6}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{-x+x^3}}+\frac {\left (3 \sqrt [3]{x} \sqrt [3]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-9 x^3\right ) \sqrt [3]{-1+x^3}} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{-x+x^3}}\\ &=\frac {9 x^2 \sqrt [3]{1-x^2} F_1\left (\frac {5}{6};1,\frac {1}{3};\frac {11}{6};9 x^2,x^2\right )}{5 \sqrt [3]{-x+x^3}}+\frac {\left (3 \sqrt [3]{x} \sqrt [3]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+8 x^3} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{2 \sqrt [3]{-x+x^3}}\\ &=\frac {9 x^2 \sqrt [3]{1-x^2} F_1\left (\frac {5}{6};1,\frac {1}{3};\frac {11}{6};9 x^2,x^2\right )}{5 \sqrt [3]{-x+x^3}}+\frac {\left (\sqrt [3]{x} \sqrt [3]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+2 x} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{2 \sqrt [3]{-x+x^3}}+\frac {\left (\sqrt [3]{x} \sqrt [3]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {2-2 x}{1-2 x+4 x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{2 \sqrt [3]{-x+x^3}}\\ &=\frac {9 x^2 \sqrt [3]{1-x^2} F_1\left (\frac {5}{6};1,\frac {1}{3};\frac {11}{6};9 x^2,x^2\right )}{5 \sqrt [3]{-x+x^3}}+\frac {\sqrt [3]{x} \sqrt [3]{-1+x^2} \log \left (1+\frac {2 x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{4 \sqrt [3]{-x+x^3}}-\frac {\left (\sqrt [3]{x} \sqrt [3]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {-2+8 x}{1-2 x+4 x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{8 \sqrt [3]{-x+x^3}}+\frac {\left (3 \sqrt [3]{x} \sqrt [3]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-2 x+4 x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{4 \sqrt [3]{-x+x^3}}\\ &=\frac {9 x^2 \sqrt [3]{1-x^2} F_1\left (\frac {5}{6};1,\frac {1}{3};\frac {11}{6};9 x^2,x^2\right )}{5 \sqrt [3]{-x+x^3}}-\frac {\sqrt [3]{x} \sqrt [3]{-1+x^2} \log \left (1+\frac {4 x^{4/3}}{\left (-1+x^2\right )^{2/3}}-\frac {2 x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{8 \sqrt [3]{-x+x^3}}+\frac {\sqrt [3]{x} \sqrt [3]{-1+x^2} \log \left (1+\frac {2 x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{4 \sqrt [3]{-x+x^3}}-\frac {\left (3 \sqrt [3]{x} \sqrt [3]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-12-x^2} \, dx,x,-2+\frac {8 x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{2 \sqrt [3]{-x+x^3}}\\ &=\frac {9 x^2 \sqrt [3]{1-x^2} F_1\left (\frac {5}{6};1,\frac {1}{3};\frac {11}{6};9 x^2,x^2\right )}{5 \sqrt [3]{-x+x^3}}-\frac {\sqrt {3} \sqrt [3]{x} \sqrt [3]{-1+x^2} \tan ^{-1}\left (\frac {1-\frac {4 x^{2/3}}{\sqrt [3]{-1+x^2}}}{\sqrt {3}}\right )}{4 \sqrt [3]{-x+x^3}}-\frac {\sqrt [3]{x} \sqrt [3]{-1+x^2} \log \left (1+\frac {4 x^{4/3}}{\left (-1+x^2\right )^{2/3}}-\frac {2 x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{8 \sqrt [3]{-x+x^3}}+\frac {\sqrt [3]{x} \sqrt [3]{-1+x^2} \log \left (1+\frac {2 x^{2/3}}{\sqrt [3]{-1+x^2}}\right )}{4 \sqrt [3]{-x+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.30, size = 81, normalized size = 0.74 \begin {gather*} \frac {3 \sqrt [3]{\frac {\frac {1}{x}-1}{\frac {1}{x}-3}} \sqrt [3]{\frac {\frac {1}{x}+1}{\frac {1}{x}-3}} x F_1\left (\frac {2}{3};\frac {1}{3},\frac {1}{3};\frac {5}{3};-\frac {4}{\frac {1}{x}-3},-\frac {2}{\frac {1}{x}-3}\right )}{2 \sqrt [3]{x \left (x^2-1\right )}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*((-1 + x^(-1))/(-3 + x^(-1)))^(1/3)*((1 + x^(-1))/(-3 + x^(-1)))^(1/3)*x*AppellF1[2/3, 1/3, 1/3, 5/3, -4/(-
3 + x^(-1)), -2/(-3 + x^(-1))])/(2*(x*(-1 + x^2))^(1/3))

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IntegrateAlgebraic [A]  time = 0.29, size = 110, normalized size = 1.00 \begin {gather*} \frac {1}{4} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{-x+x^3}}{-1-x+\sqrt [3]{-x+x^3}}\right )+\frac {1}{4} \log \left (1+x+2 \sqrt [3]{-x+x^3}\right )-\frac {1}{8} \log \left (1+2 x+x^2+(-2-2 x) \sqrt [3]{-x+x^3}+4 \left (-x+x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*(-x + x^3)^(1/3))/(-1 - x + (-x + x^3)^(1/3))])/4 + Log[1 + x + 2*(-x + x^3)^(1/3)]/4
- Log[1 + 2*x + x^2 + (-2 - 2*x)*(-x + x^3)^(1/3) + 4*(-x + x^3)^(2/3)]/8

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fricas [A]  time = 0.87, size = 117, normalized size = 1.06 \begin {gather*} \frac {1}{4} \, \sqrt {3} \arctan \left (\frac {286273 \, \sqrt {3} {\left (x^{3} - x\right )}^{\frac {1}{3}} {\left (x + 1\right )} + \sqrt {3} {\left (635653 \, x^{2} - 434719 \, x + 66978\right )} + 539695 \, \sqrt {3} {\left (x^{3} - x\right )}^{\frac {2}{3}}}{1293894 \, x^{2} - 1974837 \, x - 226981}\right ) + \frac {1}{8} \, \log \left (\frac {9 \, x^{2} + 6 \, {\left (x^{3} - x\right )}^{\frac {1}{3}} {\left (x + 1\right )} - 6 \, x + 12 \, {\left (x^{3} - x\right )}^{\frac {2}{3}} + 1}{9 \, x^{2} - 6 \, x + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(3)*arctan((286273*sqrt(3)*(x^3 - x)^(1/3)*(x + 1) + sqrt(3)*(635653*x^2 - 434719*x + 66978) + 539695*
sqrt(3)*(x^3 - x)^(2/3))/(1293894*x^2 - 1974837*x - 226981)) + 1/8*log((9*x^2 + 6*(x^3 - x)^(1/3)*(x + 1) - 6*
x + 12*(x^3 - x)^(2/3) + 1)/(9*x^2 - 6*x + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 2.33, size = 442, normalized size = 4.02

method	result	size
trager	\(\frac {\ln \left (-\frac {35769664 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{2}-66429376 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x [phone] \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}-x \right )^{\frac {2}{3}}+976188 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}-x \right )^{\frac {1}{3}} x [phone] \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x^{2}-10219904 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2}+976188 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}-x \right )^{\frac {1}{3}}-79246712 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x [phone] \left (x^{3}-x \right )^{\frac {2}{3}}+7997400 x \left (x^{3}-x \right )^{\frac {1}{3}}[phone] x^{2}-7450356 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )+7997400 \left (x^{3}-x \right )^{\frac {1}{3}}-23140462 x -1140727}{\left (-1+3 x \right )^{2}}\right )}{4}+\frac {\RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \ln \left (\frac {18251632 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{2}-33895888 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x [phone] \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}-x \right )^{\frac {2}{3}}+7509306 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}-x \right )^{\frac {1}{3}} x -1650049 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x^{2}-5214752 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2}+7509306 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{3}-x \right )^{\frac {1}{3}}-10414826 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x +488094 \left (x^{3}-x \right )^{\frac {2}{3}}+3998700 x \left (x^{3}-x \right )^{\frac {1}{3}}-5189795 x^{2}-4021625 \RootOf \left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )+3998700 \left (x^{3}-x \right )^{\frac {1}{3}}+3034034 x -718587}{\left (-1+3 x \right )^{2}}\right )}{2}\)	\(442\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*ln(-(35769664*RootOf(4*_Z^2+2*_Z+1)^2*x^2-66429376*RootOf(4*_Z^2+2*_Z+1)^2*x+31989600*RootOf(4*_Z^2+2*_Z+1
)*(x^3-x)^(2/3)+976188*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(1/3)*x+56895644*RootOf(4*_Z^2+2*_Z+1)*x^2-10219904*RootO
f(4*_Z^2+2*_Z+1)^2+976188*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(1/3)-79246712*RootOf(4*_Z^2+2*_Z+1)*x+15018612*(x^3-x
)^(2/3)+7997400*x*(x^3-x)^(1/3)[phone]*x^2-7450356*RootOf(4*_Z^2+2*_Z+1)+7997400*(x^3-x)^(1/3)-23140462*x-11
40727)/(-1+3*x)^2)+1/2*RootOf(4*_Z^2+2*_Z+1)*ln((18251632*RootOf(4*_Z^2+2*_Z+1)^2*x^2-33895888*RootOf(4*_Z^2+2
*_Z+1)^2*x+15994800*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(2/3)+7509306*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(1/3)*x-1650049*
RootOf(4*_Z^2+2*_Z+1)*x^2-5214752*RootOf(4*_Z^2+2*_Z+1)^2+7509306*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(1/3)-10414826
*RootOf(4*_Z^2+2*_Z+1)*x+488094*(x^3-x)^(2/3)+3998700*x*(x^3-x)^(1/3)-5189795*x^2-4021625*RootOf(4*_Z^2+2*_Z+1
)+3998700*(x^3-x)^(1/3)+3034034*x-718587)/(-1+3*x)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/((x^3 - x)^(1/3)*(3*x - 1)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {1}{3 x \sqrt [3]{x^{3} - x} - \sqrt [3]{x^{3} - x}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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