\(\int \frac {(-2+x^3) \sqrt [3]{x+x^4}}{(1+x^3)^2} \, dx\) [91]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 16 \[ \int \frac {\left (-2+x^3\right ) \sqrt [3]{x+x^4}}{\left (1+x^3\right )^2} \, dx=-\frac {3 x^2}{2 \left (x+x^4\right )^{2/3}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2081, 460} \[ \int \frac {\left (-2+x^3\right ) \sqrt [3]{x+x^4}}{\left (1+x^3\right )^2} \, dx=-\frac {3 x \sqrt [3]{x^4+x}}{2 \left (x^3+1\right )} \]

[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 = 6.63 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.31 \[ \int \frac {\left (-2+x^3\right ) \sqrt [3]{x+x^4}}{\left (1+x^3\right )^2} \, dx=-\frac {3 x \sqrt [3]{x+x^4}}{2 \left (1+x^3\right )} \]

[In]

Integrate[((-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))

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method result size
pseudoelliptic \(-\frac {3 x^{2}}{2 \left (x^{4}+x \right )^{\frac {2}{3}}}\) \(13\)
gosper \(-\frac {3 x \left (x^{4}+x \right )^{\frac {1}{3}}}{2 \left (x^{3}+1\right )}\) \(18\)
trager \(-\frac {3 x \left (x^{4}+x \right )^{\frac {1}{3}}}{2 \left (x^{3}+1\right )}\) \(18\)
risch \(-\frac {3 {\left (x \left (x^{3}+1\right )\right )}^{\frac {1}{3}} x}{2 \left (x^{3}+1\right )}\) \(20\)
meijerg \(-\frac {3 x^{\frac {4}{3}} \operatorname {hypergeom}\left (\left [\frac {4}{9}, \frac {5}{3}\right ], \left [\frac {13}{9}\right ], -x^{3}\right )}{2}+\frac {3 x^{\frac {13}{3}} \operatorname {hypergeom}\left (\left [\frac {13}{9}, \frac {5}{3}\right ], \left [\frac {22}{9}\right ], -x^{3}\right )}{13}\) \(34\)

[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 = 0.25 (sec) , antiderivative size = 17, 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 = 4.97 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.19 \[ \int \frac {\left (-2+x^3\right ) \sqrt [3]{x+x^4}}{\left (1+x^3\right )^2} \, dx=-\frac {3\,x\,{\left (x^4+x\right )}^{1/3}}{2\,\left (x^3+1\right )} \]

[In]

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

[Out]

-(3*x*(x + x^4)^(1/3))/(2*(x^3 + 1))