\(\int \frac {\sqrt [3]{-x^2+x^3}}{-1+x^2} \, dx\) [2570]

Optimal result

Integrand size = 21, antiderivative size = 219 \[ \int \frac {\sqrt [3]{-x^2+x^3}}{-1+x^2} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-x^2+x^3}}\right )+\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{-x^2+x^3}}\right )}{2^{2/3}}-\log \left (-x+\sqrt [3]{-x^2+x^3}\right )+\frac {\log \left (-2 x+2^{2/3} \sqrt [3]{-x^2+x^3}\right )}{2^{2/3}}+\frac {1}{2} \log \left (x^2+x \sqrt [3]{-x^2+x^3}+\left (-x^2+x^3\right )^{2/3}\right )-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{-x^2+x^3}+\sqrt [3]{2} \left (-x^2+x^3\right )^{2/3}\right )}{2\ 2^{2/3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.18, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2081, 862, 129, 494, 337, 503} \[ \int \frac {\sqrt [3]{-x^2+x^3}}{-1+x^2} \, dx=-\frac {\sqrt {3} \sqrt [3]{x^3-x^2} \arctan \left (\frac {\frac {2 \sqrt [3]{x}}{\sqrt [3]{x-1}}+1}{\sqrt {3}}\right )}{\sqrt [3]{x-1} x^{2/3}}+\frac {\sqrt {3} \sqrt [3]{x^3-x^2} \arctan \left (\frac {\frac {2 \sqrt [3]{2} \sqrt [3]{x}}{\sqrt [3]{x-1}}+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt [3]{x-1} x^{2/3}}-\frac {3 \sqrt [3]{x^3-x^2} \log \left (\sqrt [3]{x-1}-\sqrt [3]{x}\right )}{2 \sqrt [3]{x-1} x^{2/3}}+\frac {3 \sqrt [3]{x^3-x^2} \log \left (\sqrt [3]{x-1}-\sqrt [3]{2} \sqrt [3]{x}\right )}{2\ 2^{2/3} \sqrt [3]{x-1} x^{2/3}}-\frac {\sqrt [3]{x^3-x^2} \log (x+1)}{2\ 2^{2/3} \sqrt [3]{x-1} x^{2/3}} \]

[In]

Int[(-x^2 + x^3)^(1/3)/(-1 + x^2),x]

[Out]

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

Rule 129

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 337

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

Rule 494

Int[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_)^(n_))^(q_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Dist[e^n/b, Int[
(e*x)^(m - n)*(c + d*x^n)^q, x], x] - Dist[a*(e^n/b), Int[(e*x)^(m - n)*((c + d*x^n)^q/(a + b*x^n)), x], x] /;
FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1] && IntBinomialQ[a, b
, c, d, e, m, n, -1, q, x]

Rule 503

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

Rule 862

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt [3]{-x^2+x^3}}{-1+x^2} \, dx=-\frac {(-1+x)^{2/3} x^{4/3} \left (4 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{2 \sqrt [3]{-1+x}+\sqrt [3]{x}}\right )-2 \sqrt [3]{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{2^{2/3} \sqrt [3]{-1+x}+\sqrt [3]{x}}\right )-2 \sqrt [3]{2} \log \left (2^{2/3} \sqrt [3]{-1+x}-2 \sqrt [3]{x}\right )+4 \log \left (\sqrt [3]{-1+x}-\sqrt [3]{x}\right )-2 \log \left ((-1+x)^{2/3}+\sqrt [3]{-1+x} \sqrt [3]{x}+x^{2/3}\right )+\sqrt [3]{2} \log \left (\sqrt [3]{2} (-1+x)^{2/3}+2^{2/3} \sqrt [3]{-1+x} \sqrt [3]{x}+2 x^{2/3}\right )\right )}{4 \left ((-1+x) x^2\right )^{2/3}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 3.30 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(\frac {2^{\frac {1}{3}} \ln \left (\frac {-2^{\frac {1}{3}} x +\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}}{x}\right )}{2}-\frac {2^{\frac {1}{3}} \ln \left (\frac {2^{\frac {2}{3}} x^{2}+2^{\frac {1}{3}} \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x^{2}\right )^{\frac {2}{3}}}{x^{2}}\right )}{4}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}+x \right )}{3 x}\right )}{2}-\ln \left (\frac {-x +\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}}{x}\right )+\frac {\ln \left (\frac {\left (\left (-1+x \right ) x^{2}\right )^{\frac {2}{3}}+x \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}+x^{2}}{x^{2}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\left (2 \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}+x \right ) \sqrt {3}}{3 x}\right )\)	\(178\)
trager	\(\text {Expression too large to display}\)	\(2020\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt [3]{-x^2+x^3}}{-1+x^2} \, dx=-\frac {1}{2} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {4^{\frac {1}{6}} \sqrt {3} {\left (4^{\frac {1}{3}} x + 4^{\frac {2}{3}} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}\right )}}{6 \, x}\right ) + \frac {1}{4} \cdot 4^{\frac {2}{3}} \log \left (-\frac {4^{\frac {2}{3}} x - 2 \, {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - \frac {1}{8} \cdot 4^{\frac {2}{3}} \log \left (\frac {2 \cdot 4^{\frac {1}{3}} x^{2} + 4^{\frac {2}{3}} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} x + 2 \, {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) + \sqrt {3} \arctan \left (\frac {\sqrt {3} x + 2 \, \sqrt {3} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{3 \, x}\right ) - \log \left (-\frac {x - {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} x + {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) \]

[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)

Sympy [F]

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

[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)

Maxima [F]

\[ \int \frac {\sqrt [3]{-x^2+x^3}}{-1+x^2} \, dx=\int { \frac {{\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x^{2} - 1} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt [3]{-x^2+x^3}}{-1+x^2} \, dx=-\frac {1}{2} \, \sqrt {3} 2^{\frac {1}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{4} \cdot 2^{\frac {1}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{2} \cdot 2^{\frac {1}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} \right |}\right ) + \frac {1}{2} \, \log \left ({\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + 1\right ) - \log \left ({\left | {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]

[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))

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [3]{-x^2+x^3}}{-1+x^2} \, dx=\int \frac {{\left (x^3-x^2\right )}^{1/3}}{x^2-1} \,d x \]

[In]

int((x^3 - x^2)^(1/3)/(x^2 - 1),x)

[Out]

int((x^3 - x^2)^(1/3)/(x^2 - 1), x)