\(\int \frac {\sqrt [3]{-x+x^3}}{x^4} \, dx\) [124]

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int \frac {\sqrt [3]{-x+x^3}}{x^4} \, dx=\frac {3 \left (-x+x^3\right )^{4/3}}{8 x^4} \]

[Out]

3/8*(x^3-x)^(4/3)/x^4

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2039} \[ \int \frac {\sqrt [3]{-x+x^3}}{x^4} \, dx=\frac {3 \left (x^3-x\right )^{4/3}}{8 x^4} \]

[In]

Int[(-x + x^3)^(1/3)/x^4,x]

[Out]

(3*(-x + x^3)^(4/3))/(8*x^4)

Rule 2039

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {3 \left (-x+x^3\right )^{4/3}}{8 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [3]{-x+x^3}}{x^4} \, dx=\frac {3 \left (x \left (-1+x^2\right )\right )^{4/3}}{8 x^4} \]

[In]

Integrate[(-x + x^3)^(1/3)/x^4,x]

[Out]

(3*(x*(-1 + x^2))^(4/3))/(8*x^4)

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11

method	result	size
trager	\(\frac {3 \left (x^{2}-1\right ) \left (x^{3}-x \right )^{\frac {1}{3}}}{8 x^{3}}\)	\(20\)
pseudoelliptic	\(\frac {3 \left (x^{2}-1\right ) \left (x^{3}-x \right )^{\frac {1}{3}}}{8 x^{3}}\)	\(20\)
gosper	\(\frac {3 \left (1+x \right ) \left (x -1\right ) \left (x^{3}-x \right )^{\frac {1}{3}}}{8 x^{3}}\)	\(21\)
risch	\(\frac {3 {\left (x \left (x^{2}-1\right )\right )}^{\frac {1}{3}} \left (x^{4}-2 x^{2}+1\right )}{8 x^{3} \left (x^{2}-1\right )}\)	\(32\)
meijerg	\(-\frac {3 \operatorname {signum}\left (x^{2}-1\right )^{\frac {1}{3}} \left (-x^{2}+1\right )^{\frac {4}{3}}}{8 {\left (-\operatorname {signum}\left (x^{2}-1\right )\right )}^{\frac {1}{3}} x^{\frac {8}{3}}}\)	\(33\)

[In]

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

[Out]

3/8*(x^2-1)/x^3*(x^3-x)^(1/3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt [3]{-x+x^3}}{x^4} \, dx=\frac {3 \, {\left (x^{3} - x\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )}}{8 \, x^{3}} \]

[In]

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

[Out]

3/8*(x^3 - x)^(1/3)*(x^2 - 1)/x^3

Sympy [F]

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

[In]

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

[Out]

Integral((x*(x - 1)*(x + 1))**(1/3)/x**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt [3]{-x+x^3}}{x^4} \, dx=\frac {3 \, {\left (x^{3} - x\right )} {\left (x + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )}^{\frac {1}{3}}}{8 \, x^{\frac {11}{3}}} \]

[In]

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

[Out]

3/8*(x^3 - x)*(x + 1)^(1/3)*(x - 1)^(1/3)/x^(11/3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt [3]{-x+x^3}}{x^4} \, dx=\frac {3}{8} \, {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {4}{3}} \]

[In]

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

[Out]

3/8*(-1/x^2 + 1)^(4/3)

Mupad [B] (verification not implemented)

Time = 4.97 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.72 \[ \int \frac {\sqrt [3]{-x+x^3}}{x^4} \, dx=-\frac {3\,{\left (x^3-x\right )}^{1/3}-3\,x^2\,{\left (x^3-x\right )}^{1/3}}{8\,x^3} \]

[In]

int((x^3 - x)^(1/3)/x^4,x)

[Out]

-(3*(x^3 - x)^(1/3) - 3*x^2*(x^3 - x)^(1/3))/(8*x^3)