\(\int \frac {(2+x^3) \sqrt [3]{-x+x^4}}{(-1+x^3)^2} \, dx\) [135]

Optimal result

Integrand size = 24, antiderivative size = 18 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=-\frac {3 x^2}{2 \left (-x+x^4\right )^{2/3}} \]

[Out]

-3/2*x^2/(x^4-x)^(2/3)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2081, 460} \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=\frac {3 x \sqrt [3]{x^4-x}}{2 \left (1-x^3\right )} \]

[In]

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

[Out]

(3*x*(-x + x^4)^(1/3))/(2*(1 - x^3))

Rule 460

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

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+x^4} \int \frac {\sqrt [3]{x} \left (2+x^3\right )}{\left (-1+x^3\right )^{5/3}} \, dx}{\sqrt [3]{x} \sqrt [3]{-1+x^3}} \\ & = \frac {3 x \sqrt [3]{-x+x^4}}{2 \left (1-x^3\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.35 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=-\frac {3 x^2}{2 \left (x \left (-1+x^3\right )\right )^{2/3}} \]

[In]

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

[Out]

(-3*x^2)/(2*(x*(-1 + x^3))^(2/3))

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(-\frac {3 x^{2}}{2 \left (x^{4}-x \right )^{\frac {2}{3}}}\)	\(15\)
gosper	\(-\frac {3 x \left (x^{4}-x \right )^{\frac {1}{3}}}{2 \left (x^{3}-1\right )}\)	\(20\)
trager	\(-\frac {3 x \left (x^{4}-x \right )^{\frac {1}{3}}}{2 \left (x^{3}-1\right )}\)	\(20\)
risch	\(-\frac {3 {\left (x \left (x^{3}-1\right )\right )}^{\frac {1}{3}} x}{2 \left (x^{3}-1\right )}\)	\(20\)
meijerg	\(\frac {3 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} x^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [\frac {4}{9}, \frac {5}{3}\right ], \left [\frac {13}{9}\right ], x^{3}\right )}{2 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}}}+\frac {3 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} x^{\frac {13}{3}} \operatorname {hypergeom}\left (\left [\frac {13}{9}, \frac {5}{3}\right ], \left [\frac {22}{9}\right ], x^{3}\right )}{13 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}}}\)	\(66\)

[In]

int((x^3+2)*(x^4-x)^(1/3)/(x^3-1)^2,x,method=_RETURNVERBOSE)

[Out]

-3/2*x^2/(x^4-x)^(2/3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=-\frac {3 \, {\left (x^{4} - x\right )}^{\frac {1}{3}} x}{2 \, {\left (x^{3} - 1\right )}} \]

[In]

integrate((x^3+2)*(x^4-x)^(1/3)/(x^3-1)^2,x, algorithm="fricas")

[Out]

-3/2*(x^4 - x)^(1/3)*x/(x^3 - 1)

Sympy [F]

\[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=\int \frac {\sqrt [3]{x \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x^{3} + 2\right )}{\left (x - 1\right )^{2} \left (x^{2} + x + 1\right )^{2}}\, dx \]

[In]

integrate((x**3+2)*(x**4-x)**(1/3)/(x**3-1)**2,x)

[Out]

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

Maxima [F]

\[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=\int { \frac {{\left (x^{4} - x\right )}^{\frac {1}{3}} {\left (x^{3} + 2\right )}}{{\left (x^{3} - 1\right )}^{2}} \,d x } \]

[In]

integrate((x^3+2)*(x^4-x)^(1/3)/(x^3-1)^2,x, algorithm="maxima")

[Out]

integrate((x^4 - x)^(1/3)*(x^3 + 2)/(x^3 - 1)^2, x)

Giac [F]

\[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=\int { \frac {{\left (x^{4} - x\right )}^{\frac {1}{3}} {\left (x^{3} + 2\right )}}{{\left (x^{3} - 1\right )}^{2}} \,d x } \]

[In]

integrate((x^3+2)*(x^4-x)^(1/3)/(x^3-1)^2,x, algorithm="giac")

[Out]

integrate((x^4 - x)^(1/3)*(x^3 + 2)/(x^3 - 1)^2, x)

Mupad [B] (verification not implemented)

Time = 5.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{-x+x^4}}{\left (-1+x^3\right )^2} \, dx=-\frac {3\,x\,{\left (x^4-x\right )}^{1/3}}{2\,\left (x^3-1\right )} \]

[In]

int(((x^4 - x)^(1/3)*(x^3 + 2))/(x^3 - 1)^2,x)

[Out]

-(3*x*(x^4 - x)^(1/3))/(2*(x^3 - 1))