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

Optimal. Leaf size=219 \[ -\log \left (\sqrt [3]{x^3-x^2}-x\right )+\frac {\log \left (2^{2/3} \sqrt [3]{x^3-x^2}-2 x\right )}{2^{2/3}}+\frac {1}{2} \log \left (x^2+\sqrt [3]{x^3-x^2} x+\left (x^3-x^2\right )^{2/3}\right )-\frac {\log \left (2 x^2+2^{2/3} \sqrt [3]{x^3-x^2} x+\sqrt [3]{2} \left (x^3-x^2\right )^{2/3}\right )}{2\ 2^{2/3}}-\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{x^3-x^2}+x}\right )+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2^{2/3} \sqrt [3]{x^3-x^2}+x}\right )}{2^{2/3}} \]

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Rubi [A]  time = 0.08, antiderivative size = 298, normalized size of antiderivative = 1.36, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2056, 848, 105, 59, 91} \begin {gather*} \frac {3 \sqrt [3]{x^3-x^2} \log \left (\sqrt [3]{2} \sqrt [3]{x}-\sqrt [3]{x-1}\right )}{2\ 2^{2/3} \sqrt [3]{x-1} x^{2/3}}-\frac {3 \sqrt [3]{x^3-x^2} \log \left (\frac {\sqrt [3]{x}}{\sqrt [3]{x-1}}-1\right )}{2 \sqrt [3]{x-1} x^{2/3}}-\frac {\sqrt [3]{x^3-x^2} \log (x-1)}{2 \sqrt [3]{x-1} x^{2/3}}-\frac {\sqrt [3]{x^3-x^2} \log (x+1)}{2\ 2^{2/3} \sqrt [3]{x-1} x^{2/3}}-\frac {\sqrt {3} \sqrt [3]{x^3-x^2} \tan ^{-1}\left (\frac {2 \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{x-1}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [3]{x-1} x^{2/3}}+\frac {\sqrt {3} \sqrt [3]{x^3-x^2} \tan ^{-1}\left (\frac {2 \sqrt [3]{2} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{x-1}}+\frac {1}{\sqrt {3}}\right )}{2^{2/3} \sqrt [3]{x-1} x^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x^2 + x^3)^(1/3)/(-1 + x^2),x]

[Out]

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

Rule 59

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

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 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -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 {\sqrt [3]{-x^2+x^3}}{-1+x^2} \, dx &=\frac {\sqrt [3]{-x^2+x^3} \int \frac {\sqrt [3]{-1+x} x^{2/3}}{-1+x^2} \, dx}{\sqrt [3]{-1+x} x^{2/3}}\\ &=\frac {\sqrt [3]{-x^2+x^3} \int \frac {x^{2/3}}{(-1+x)^{2/3} (1+x)} \, dx}{\sqrt [3]{-1+x} x^{2/3}}\\ &=\frac {\sqrt [3]{-x^2+x^3} \int \frac {1}{(-1+x)^{2/3} \sqrt [3]{x}} \, dx}{\sqrt [3]{-1+x} x^{2/3}}-\frac {\sqrt [3]{-x^2+x^3} \int \frac {1}{(-1+x)^{2/3} \sqrt [3]{x} (1+x)} \, dx}{\sqrt [3]{-1+x} x^{2/3}}\\ &=-\frac {\sqrt {3} \sqrt [3]{-x^2+x^3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{-1+x}}\right )}{\sqrt [3]{-1+x} x^{2/3}}+\frac {\sqrt {3} \sqrt [3]{-x^2+x^3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{2} \sqrt [3]{x}}{\sqrt {3} \sqrt [3]{-1+x}}\right )}{2^{2/3} \sqrt [3]{-1+x} x^{2/3}}+\frac {3 \sqrt [3]{-x^2+x^3} \log \left (-\sqrt [3]{-1+x}+\sqrt [3]{2} \sqrt [3]{x}\right )}{2\ 2^{2/3} \sqrt [3]{-1+x} x^{2/3}}-\frac {3 \sqrt [3]{-x^2+x^3} \log \left (-1+\frac {\sqrt [3]{x}}{\sqrt [3]{-1+x}}\right )}{2 \sqrt [3]{-1+x} x^{2/3}}-\frac {\sqrt [3]{-x^2+x^3} \log (-1+x)}{2 \sqrt [3]{-1+x} x^{2/3}}-\frac {\sqrt [3]{-x^2+x^3} \log (1+x)}{2\ 2^{2/3} \sqrt [3]{-1+x} x^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 61, normalized size = 0.28 \begin {gather*} \frac {3 \sqrt [3]{(x-1) x^2} \left (2 \sqrt [3]{x} \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};1-x\right )-\, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {x-1}{2 x}\right )\right )}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.45, size = 219, normalized size = 1.00 \begin {gather*} -\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-x^2+x^3}}\right )+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{-x^2+x^3}}\right )}{2^{2/3}}-\log \left (-x+\sqrt [3]{-x^2+x^3}\right )+\frac {\log \left (-2 x+2^{2/3} \sqrt [3]{-x^2+x^3}\right )}{2^{2/3}}+\frac {1}{2} \log \left (x^2+x \sqrt [3]{-x^2+x^3}+\left (-x^2+x^3\right )^{2/3}\right )-\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 )}{2\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 210, normalized size = 0.96 \begin {gather*} -\frac {1}{2} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {4^{\frac {1}{6}} \sqrt {3} {\left (4^{\frac {1}{3}} x + 4^{\frac {2}{3}} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}\right )}}{6 \, x}\right ) + \frac {1}{4} \cdot 4^{\frac {2}{3}} \log \left (-\frac {4^{\frac {2}{3}} x - 2 \, {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{8} \cdot 4^{\frac {2}{3}} \log \left (\frac {2 \cdot 4^{\frac {1}{3}} x^{2} + 4^{\frac {2}{3}} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} x + 2 \, {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) + \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) - \log \left (-\frac {x - {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} x + {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 148, normalized size = 0.68 \begin {gather*} -\frac {1}{2} \, \sqrt {3} 2^{\frac {1}{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 ) + \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{4} \cdot 2^{\frac {1}{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}{2} \cdot 2^{\frac {1}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} \right |}\right ) + \frac {1}{2} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) - \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 5.73, size = 1679, normalized size = 7.67

method	result	size
trager	\(\text {Expression too large to display}\)	\(1679\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*RootOf(_Z^3-2)*ln((-6*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^4*x^2-4*RootOf(Ro
otOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^3*x^2+12*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-
2)+4*_Z^2)*RootOf(_Z^3-2)^4*x+8*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^3*x+48*Ro
otOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2*(x^3-x^2)^(2/3)-39*RootOf(_Z^3-2)^2*x^2-26*
RootOf(_Z^3-2)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x^2+30*(x^3-x^2)^(1/3)*RootOf(_Z^3-2)*x-36*
(x^3-x^2)^(1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x+15*RootOf(_Z^3-2)^2*x+10*RootOf(RootOf(_
Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)*x+18*(x^3-x^2)^(2/3))/x/(1+x))+RootOf(RootOf(_Z^3-2)^2+2*_
Z*RootOf(_Z^3-2)+4*_Z^2)*ln(-(6*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^4*x^2+8*Roo
tOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^3*x^2-12*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf
(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^4*x-16*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^3*
x+48*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2*(x^3-x^2)^(2/3)-33*RootOf(_Z^3-2)^2*
x^2-44*RootOf(_Z^3-2)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x^2+18*(x^3-x^2)^(1/3)*RootOf(_Z^3-2
)*x-60*(x^3-x^2)^(1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x+3*RootOf(_Z^3-2)^2*x+4*RootOf(Roo
tOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)*x+30*(x^3-x^2)^(2/3))/x/(1+x))-RootOf(_Z^3-2)^2*RootO
f(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*ln((5*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*Roo
tOf(_Z^3-2)^4*x^2-10*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^4*x+24*RootOf(RootOf
(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2*(x^3-x^2)^(2/3)+24*(x^3-x^2)^(1/3)*RootOf(RootOf(_Z^3-
2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2*x+29*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*Roo
tOf(_Z^3-2)^2*x^2-23*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2*x+15*(x^3-x^2)^(2/3)
+15*x*(x^3-x^2)^(1/3)+20*x^2-12*x)/x)+RootOf(_Z^3-2)^2*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*ln(
-(-5*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^4*x^2+10*RootOf(RootOf(_Z^3-2)^2+2*_
Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^4*x+24*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z
^3-2)^2*(x^3-x^2)^(2/3)+24*(x^3-x^2)^(1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^
2*x+19*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2*x^2-3*RootOf(RootOf(_Z^3-2)^2+2*_Z
*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2*x+9*(x^3-x^2)^(2/3)+9*x*(x^3-x^2)^(1/3)+4*x^2-x)/x)+ln(-(-5*RootOf(Ro
otOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^4*x^2+10*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-
2)+4*_Z^2)^2*RootOf(_Z^3-2)^4*x+24*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2*(x^3-x
^2)^(2/3)+24*(x^3-x^2)^(1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2*x+19*RootOf(
RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2*x^2-3*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2
)+4*_Z^2)*RootOf(_Z^3-2)^2*x+9*(x^3-x^2)^(2/3)+9*x*(x^3-x^2)^(1/3)+4*x^2-x)/x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 - x^2)^(1/3)/(x^2 - 1),x)

[Out]

int((x^3 - x^2)^(1/3)/(x^2 - 1), x)

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

[Out]

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

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