\(\int \frac {(1+2 x^3) \sqrt [3]{-x^2+x^5}}{x^3} \, dx\) [180]

Optimal result

Integrand size = 24, antiderivative size = 20 \[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\frac {3 \left (-x^2+x^5\right )^{4/3}}{4 x^4} \]

[Out]

3/4*(x^5-x^2)^(4/3)/x^4

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {1604} \[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\frac {3 \left (x^5-x^2\right )^{4/3}}{4 x^4} \]

[In]

Int[((1 + 2*x^3)*(-x^2 + x^5)^(1/3))/x^3,x]

[Out]

(3*(-x^2 + x^5)^(4/3))/(4*x^4)

Rule 1604

Int[(Pp_)*(Qq_)^(m_.)*(Rr_)^(n_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x], r = Expon[Rr, x]}, S
imp[Coeff[Pp, x, p]*x^(p - q - r + 1)*Qq^(m + 1)*(Rr^(n + 1)/((p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x,
r])), x] /; NeQ[p + m*q + n*r + 1, 0] && EqQ[(p + m*q + n*r + 1)*Coeff[Qq, x, q]*Coeff[Rr, x, r]*Pp, Coeff[Pp
, x, p]*x^(p - q - r)*((p - q - r + 1)*Qq*Rr + (m + 1)*x*Rr*D[Qq, x] + (n + 1)*x*Qq*D[Rr, x])]] /; FreeQ[{m, n
}, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && PolyQ[Rr, x] && NeQ[m, -1] && NeQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {3 \left (-x^2+x^5\right )^{4/3}}{4 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\frac {3 \left (x^2 \left (-1+x^3\right )\right )^{4/3}}{4 x^4} \]

[In]

Integrate[((1 + 2*x^3)*(-x^2 + x^5)^(1/3))/x^3,x]

[Out]

(3*(x^2*(-1 + x^3))^(4/3))/(4*x^4)

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10

method	result	size
trager	\(\frac {3 \left (x^{3}-1\right ) \left (x^{5}-x^{2}\right )^{\frac {1}{3}}}{4 x^{2}}\)	\(22\)
pseudoelliptic	\(\frac {3 \left (x^{3}-1\right ) \left (x^{5}-x^{2}\right )^{\frac {1}{3}}}{4 x^{2}}\)	\(22\)
gosper	\(\frac {3 \left (x -1\right ) \left (x^{2}+x +1\right ) \left (x^{5}-x^{2}\right )^{\frac {1}{3}}}{4 x^{2}}\)	\(26\)
risch	\(\frac {3 \left (x^{2} \left (x^{3}-1\right )\right )^{\frac {1}{3}} \left (x^{6}-2 x^{3}+1\right )}{4 x^{2} \left (x^{3}-1\right )}\)	\(34\)
meijerg	\(-\frac {3 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} \operatorname {hypergeom}\left (\left [-\frac {4}{9}, -\frac {1}{3}\right ], \left [\frac {5}{9}\right ], x^{3}\right )}{4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}} x^{\frac {4}{3}}}+\frac {6 \operatorname {signum}\left (x^{3}-1\right )^{\frac {1}{3}} x^{\frac {5}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{3}, \frac {5}{9}\right ], \left [\frac {14}{9}\right ], x^{3}\right )}{5 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {1}{3}}}\)	\(66\)

[In]

int((2*x^3+1)*(x^5-x^2)^(1/3)/x^3,x,method=_RETURNVERBOSE)

[Out]

3/4*(x^3-1)/x^2*(x^5-x^2)^(1/3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\frac {3 \, {\left (x^{5} - x^{2}\right )}^{\frac {1}{3}} {\left (x^{3} - 1\right )}}{4 \, x^{2}} \]

[In]

integrate((2*x^3+1)*(x^5-x^2)^(1/3)/x^3,x, algorithm="fricas")

[Out]

3/4*(x^5 - x^2)^(1/3)*(x^3 - 1)/x^2

Sympy [F]

\[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\int \frac {\sqrt [3]{x^{2} \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (2 x^{3} + 1\right )}{x^{3}}\, dx \]

[In]

integrate((2*x**3+1)*(x**5-x**2)**(1/3)/x**3,x)

[Out]

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

Maxima [F]

\[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\int { \frac {{\left (x^{5} - x^{2}\right )}^{\frac {1}{3}} {\left (2 \, x^{3} + 1\right )}}{x^{3}} \,d x } \]

[In]

integrate((2*x^3+1)*(x^5-x^2)^(1/3)/x^3,x, algorithm="maxima")

[Out]

integrate((x^5 - x^2)^(1/3)*(2*x^3 + 1)/x^3, x)

Giac [F]

\[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\int { \frac {{\left (x^{5} - x^{2}\right )}^{\frac {1}{3}} {\left (2 \, x^{3} + 1\right )}}{x^{3}} \,d x } \]

[In]

integrate((2*x^3+1)*(x^5-x^2)^(1/3)/x^3,x, algorithm="giac")

[Out]

integrate((x^5 - x^2)^(1/3)*(2*x^3 + 1)/x^3, x)

Mupad [B] (verification not implemented)

Time = 5.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\left (1+2 x^3\right ) \sqrt [3]{-x^2+x^5}}{x^3} \, dx=\frac {3\,\left (x^3-1\right )\,{\left (x^5-x^2\right )}^{1/3}}{4\,x^2} \]

[In]

int(((x^5 - x^2)^(1/3)*(2*x^3 + 1))/x^3,x)

[Out]

(3*(x^3 - 1)*(x^5 - x^2)^(1/3))/(4*x^2)