3.22.65 \(\int \frac {1}{(-1+x^2) \sqrt [3]{-x^2+x^3}} \, dx\)

Optimal. Leaf size=159 \[ -\frac {3 \left (x^3-x^2\right )^{2/3}}{2 (x-1) x}+\frac {\log \left (2^{2/3} \sqrt [3]{x^3-x^2}-2 x\right )}{2 \sqrt [3]{2}}-\frac {\log \left (2 x^2+2^{2/3} \sqrt [3]{x^3-x^2} x+\sqrt [3]{2} \left (x^3-x^2\right )^{2/3}\right )}{4 \sqrt [3]{2}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2^{2/3} \sqrt [3]{x^3-x^2}+x}\right )}{2 \sqrt [3]{2}} \]

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Rubi [A]  time = 0.08, antiderivative size = 182, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2056, 848, 96, 91} \begin {gather*} -\frac {3 x}{2 \sqrt [3]{x^3-x^2}}+\frac {3 \sqrt [3]{x-1} x^{2/3} \log \left (\frac {\sqrt [3]{x-1}}{\sqrt [3]{2}}-\sqrt [3]{x}\right )}{4 \sqrt [3]{2} \sqrt [3]{x^3-x^2}}-\frac {\sqrt [3]{x-1} x^{2/3} \log (x+1)}{4 \sqrt [3]{2} \sqrt [3]{x^3-x^2}}+\frac {\sqrt {3} \sqrt [3]{x-1} x^{2/3} \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{x-1}}{\sqrt {3} \sqrt [3]{x}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt [3]{x^3-x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 91

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

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

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

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Mathematica [C]  time = 0.01, size = 41, normalized size = 0.26 \begin {gather*} -\frac {3 \left ((x-1) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {x-1}{2 x}\right )+4 x\right )}{8 \sqrt [3]{(x-1) x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(4*x + (-1 + x)*Hypergeometric2F1[2/3, 1, 5/3, (-1 + x)/(2*x)]))/(8*((-1 + x)*x^2)^(1/3))

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IntegrateAlgebraic [A]  time = 0.44, size = 159, normalized size = 1.00 \begin {gather*} -\frac {3 \left (-x^2+x^3\right )^{2/3}}{2 (-1+x) x}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{-x^2+x^3}}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-2 x+2^{2/3} \sqrt [3]{-x^2+x^3}\right )}{2 \sqrt [3]{2}}-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{-x^2+x^3}+\sqrt [3]{2} \left (-x^2+x^3\right )^{2/3}\right )}{4 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 163, normalized size = 1.03 \begin {gather*} \frac {2 \, \sqrt {3} 2^{\frac {2}{3}} {\left (x^{2} - x\right )} \arctan \left (\frac {\sqrt {3} 2^{\frac {1}{6}} {\left (2^{\frac {5}{6}} x + 2 \, \sqrt {2} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}\right )}}{6 \, x}\right ) + 2 \cdot 2^{\frac {2}{3}} {\left (x^{2} - x\right )} \log \left (-\frac {2^{\frac {1}{3}} x - {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - 2^{\frac {2}{3}} {\left (x^{2} - x\right )} \log \left (\frac {2^{\frac {2}{3}} x^{2} + 2^{\frac {1}{3}} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} x + {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 12 \, {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{8 \, {\left (x^{2} - x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*sqrt(3)*2^(2/3)*(x^2 - x)*arctan(1/6*sqrt(3)*2^(1/6)*(2^(5/6)*x + 2*sqrt(2)*(x^3 - x^2)^(1/3))/x) + 2*2
^(2/3)*(x^2 - x)*log(-(2^(1/3)*x - (x^3 - x^2)^(1/3))/x) - 2^(2/3)*(x^2 - x)*log((2^(2/3)*x^2 + 2^(1/3)*(x^3 -
x^2)^(1/3)*x + (x^3 - x^2)^(2/3))/x^2) - 12*(x^3 - x^2)^(2/3))/(x^2 - x)

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giac [A]  time = 0.20, size = 98, normalized size = 0.62 \begin {gather*} \frac {1}{4} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{8} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{4} \cdot 2^{\frac {2}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} \right |}\right ) - \frac {3}{2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(3)*2^(2/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3) + 2*(-1/x + 1)^(1/3))) - 1/8*2^(2/3)*log(2^(2/3) + 2^(
1/3)*(-1/x + 1)^(1/3) + (-1/x + 1)^(2/3)) + 1/4*2^(2/3)*log(abs(-2^(1/3) + (-1/x + 1)^(1/3))) - 3/2/(-1/x + 1)
^(1/3)

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maple [C]  time = 5.10, size = 1072, normalized size = 6.74

method	result	size
risch	\(\text {Expression too large to display}\)	\(1072\)
trager	\(\text {Expression too large to display}\)	\(1159\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*x/((-1+x)*x^2)^(1/3)-1/4*ln((3*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)^2*RootOf(_Z^3-4)^2*x^2+2*R
ootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)^3*x^2-6*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4
)+_Z^2)^2*RootOf(_Z^3-4)^2*x-4*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)^3*x+24*(x^3-x^2)
^(2/3)*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)^2+48*(x^3-x^2)^(1/3)*RootOf(RootOf(_Z^3-
4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)*x+18*(x^3-x^2)^(1/3)*RootOf(_Z^3-4)^2*x+78*RootOf(RootOf(_Z^3-4)^2
+_Z*RootOf(_Z^3-4)+_Z^2)*x^2+52*RootOf(_Z^3-4)*x^2-30*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*x-20*Roo
tOf(_Z^3-4)*x+36*(x^3-x^2)^(2/3))/x/(1+x))*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)-1/4*ln((3*RootOf(Ro
otOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)^2*RootOf(_Z^3-4)^2*x^2+2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2
)*RootOf(_Z^3-4)^3*x^2-6*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)^2*RootOf(_Z^3-4)^2*x-4*RootOf(RootOf(
_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)^3*x+24*(x^3-x^2)^(2/3)*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3
-4)+_Z^2)*RootOf(_Z^3-4)^2+48*(x^3-x^2)^(1/3)*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)*x
+18*(x^3-x^2)^(1/3)*RootOf(_Z^3-4)^2*x+78*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*x^2+52*RootOf(_Z^3-4
)*x^2-30*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*x-20*RootOf(_Z^3-4)*x+36*(x^3-x^2)^(2/3))/x/(1+x))*Ro
otOf(_Z^3-4)+1/4*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*ln((3*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-
4)+_Z^2)^2*RootOf(_Z^3-4)^2*x^2+RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)^3*x^2-6*RootOf(
RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)^2*RootOf(_Z^3-4)^2*x-2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2
)*RootOf(_Z^3-4)^3*x-24*(x^3-x^2)^(2/3)*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)^2-48*(x
^3-x^2)^(1/3)*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)*x-30*(x^3-x^2)^(1/3)*RootOf(_Z^3-
4)^2*x-66*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*x^2-22*RootOf(_Z^3-4)*x^2+6*RootOf(RootOf(_Z^3-4)^2+
_Z*RootOf(_Z^3-4)+_Z^2)*x+2*RootOf(_Z^3-4)*x-60*(x^3-x^2)^(2/3))/x/(1+x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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