3.17.6 \(\int \frac {1}{(1-3 x) \sqrt [3]{-x+x^3}} \, dx\) [1606]

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

[Out]

1/4*3^(1/2)*arctan(3^(1/2)*(x^3-x)^(1/3)/(-1-x+(x^3-x)^(1/3)))+1/4*ln(1+x+2*(x^3-x)^(1/3))-1/8*ln(1+2*x+x^2+(-
2-2*x)*(x^3-x)^(1/3)+4*(x^3-x)^(2/3))

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Rubi [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 0.11, antiderivative size = 192, normalized size of antiderivative = 1.75, number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {2081, 973, 477, 476, 384, 525, 524} \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} \sqrt [3]{x^2-1} \sqrt [3]{x} \text {ArcTan}\left (\frac {1-\frac {4 x^{2/3}}{\sqrt [3]{x^2-1}}}{\sqrt {3}}\right )}{4 \sqrt [3]{x^3-x}}-\frac {\sqrt [3]{x^2-1} \sqrt [3]{x} \log \left (1-9 x^2\right )}{8 \sqrt [3]{x^3-x}}+\frac {3 \sqrt [3]{x^2-1} \sqrt [3]{x} \log \left (2 x^{2/3}+\sqrt [3]{x^2-1}\right )}{8 \sqrt [3]{x^3-x}} \end {gather*}

Antiderivative was successfully verified.

[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 - 9*x^2])/(8*(-x + x^3)^(1/3)) + (3*x^(1/3)*(-1 + x^2)^(1/3)*Log[2*x^(2/3) + (-1 + x^2)^(1/3)])/(8*(-x
+ x^3)^(1/3))

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 476

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 477

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[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 973

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 2081

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 [A]
time = 20.35, size = 126, normalized size = 1.15 \begin {gather*} -\frac {\sqrt [3]{-1+\frac {1}{x^2}} x \left (2 \sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{-1+\frac {1}{x^2}}}{1+\sqrt [3]{-1+\frac {1}{x^2}}+\frac {1}{x}}\right )+2 \log \left (-1+2 \sqrt [3]{-1+\frac {1}{x^2}}-\frac {1}{x}\right )-\log \left (1+4 \left (-1+\frac {1}{x^2}\right )^{2/3}+2 \sqrt [3]{-1+\frac {1}{x^2}} \left (1+\frac {1}{x}\right )+\frac {1}{x^2}+\frac {2}{x}\right )\right )}{8 \sqrt [3]{x \left (-1+x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.93, size = 656, normalized size = 5.96

method	result	size
trager	\(\text {Expression too large to display}\)	\(656\)

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-21125980*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)[phone]*RootOf(4*_Z^2+2*_Z+1)*x-976188*(x^3-x)^
(2/3)+7509306*x*(x^3-x)^(1/3)+212875*x^2-2769548*RootOf(4*_Z^2+2*_Z+1)+7509306*(x^3-x)^(1/3)-124450*x+29475)/(
-1+3*x)^2)-1/4*ln((36503264*RootOf(4*_Z^2+2*_Z+1)^2*x^2-67791776*RootOf(4*_Z^2+2*_Z+1)^2*x-31989600*RootOf(4*_
Z^2+2*_Z+1)*(x^3-x)^(2/3)-15018612*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(1/3)*x+39803362*RootOf(4*_Z^2+2*_Z+1)*x^2-10
429504*RootOf(4*_Z^2+2*_Z+1)^2-15018612*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(1/3)-46962124*RootOf(4*_Z^2+2*_Z+1)*x-1
5018612*(x^3-x)^(2/3)+488094*x*(x^3-x)^(1/3)+396275*x^2-2386254*RootOf(4*_Z^2+2*_Z+1)+488094*(x^3-x)^(1/3)-465
050*x-22925)/(-1+3*x)^2)-1/2*ln((36503264*RootOf(4*_Z^2+2*_Z+1)^2*x^2-67791776*RootOf(4*_Z^2+2*_Z+1)^2*x-31989
600*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(2/3)-15018612*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(1/3)*x+39803362*RootOf(4*_Z^2+
2*_Z+1)*x^2-10429504*RootOf(4*_Z^2+2*_Z+1)^2-15018612*RootOf(4*_Z^2+2*_Z+1)*(x^3-x)^(1/3)-46962124*RootOf(4*_Z
^2+2*_Z+1)*x-15018612*(x^3-x)^(2/3)+488094*x*(x^3-x)^(1/3)+396275*x^2-2386254*RootOf(4*_Z^2+2*_Z+1)+488094*(x^
3-x)^(1/3)-465050*x-22925)/(-1+3*x)^2)*RootOf(4*_Z^2+2*_Z+1)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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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Fricas [A]
time = 0.64, 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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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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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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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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