\(\int \frac {1}{x^3 \sqrt [3]{-x^2+x^6}} \, dx\) [184]

Optimal result

Integrand size = 17, antiderivative size = 20 \[ \int \frac {1}{x^3 \sqrt [3]{-x^2+x^6}} \, dx=\frac {3 \left (-x^2+x^6\right )^{2/3}}{8 x^4} \]

[Out]

3/8*(x^6-x^2)^(2/3)/x^4

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2039} \[ \int \frac {1}{x^3 \sqrt [3]{-x^2+x^6}} \, dx=\frac {3 \left (x^6-x^2\right )^{2/3}}{8 x^4} \]

[In]

Int[1/(x^3*(-x^2 + x^6)^(1/3)),x]

[Out]

(3*(-x^2 + x^6)^(2/3))/(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^2+x^6\right )^{2/3}}{8 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 \sqrt [3]{-x^2+x^6}} \, dx=\frac {3 \left (x^2 \left (-1+x^4\right )\right )^{2/3}}{8 x^4} \]

[In]

Integrate[1/(x^3*(-x^2 + x^6)^(1/3)),x]

[Out]

(3*(x^2*(-1 + x^4))^(2/3))/(8*x^4)

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
trager	\(\frac {3 \left (x^{6}-x^{2}\right )^{\frac {2}{3}}}{8 x^{4}}\)	\(17\)
pseudoelliptic	\(\frac {3 \left (x^{6}-x^{2}\right )^{\frac {2}{3}}}{8 x^{4}}\)	\(17\)
risch	\(\frac {\frac {3 x^{4}}{8}-\frac {3}{8}}{x^{2} \left (x^{2} \left (x^{4}-1\right )\right )^{\frac {1}{3}}}\)	\(22\)
gosper	\(\frac {3 \left (x^{2}+1\right ) \left (x -1\right ) \left (1+x \right )}{8 x^{2} \left (x^{6}-x^{2}\right )^{\frac {1}{3}}}\)	\(28\)
meijerg	\(-\frac {3 {\left (-\operatorname {signum}\left (x^{4}-1\right )\right )}^{\frac {1}{3}} \left (-x^{4}+1\right )^{\frac {2}{3}}}{8 \operatorname {signum}\left (x^{4}-1\right )^{\frac {1}{3}} x^{\frac {8}{3}}}\)	\(33\)

[In]

int(1/x^3/(x^6-x^2)^(1/3),x,method=_RETURNVERBOSE)

[Out]

3/8*(x^6-x^2)^(2/3)/x^4

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^3 \sqrt [3]{-x^2+x^6}} \, dx=\frac {3 \, {\left (x^{6} - x^{2}\right )}^{\frac {2}{3}}}{8 \, x^{4}} \]

[In]

integrate(1/x^3/(x^6-x^2)^(1/3),x, algorithm="fricas")

[Out]

3/8*(x^6 - x^2)^(2/3)/x^4

Sympy [F]

\[ \int \frac {1}{x^3 \sqrt [3]{-x^2+x^6}} \, dx=\int \frac {1}{x^{3} \sqrt [3]{x^{2} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )}}\, dx \]

[In]

integrate(1/x**3/(x**6-x**2)**(1/3),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.50 \[ \int \frac {1}{x^3 \sqrt [3]{-x^2+x^6}} \, dx=\frac {3 \, {\left (x^{6} - x^{2}\right )}}{8 \, {\left (x^{2} + 1\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )}^{\frac {1}{3}} {\left (x^{2}\right )}^{\frac {7}{3}}} \]

[In]

integrate(1/x^3/(x^6-x^2)^(1/3),x, algorithm="maxima")

[Out]

3/8*(x^6 - x^2)/((x^2 + 1)^(1/3)*(x^2 - 1)^(1/3)*(x^2)^(7/3))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int \frac {1}{x^3 \sqrt [3]{-x^2+x^6}} \, dx=\frac {3}{8} \, {\left (-\frac {1}{x^{4}} + 1\right )}^{\frac {2}{3}} \]

[In]

integrate(1/x^3/(x^6-x^2)^(1/3),x, algorithm="giac")

[Out]

3/8*(-1/x^4 + 1)^(2/3)

Mupad [B] (verification not implemented)

Time = 5.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x^3 \sqrt [3]{-x^2+x^6}} \, dx=\frac {3\,{\left (x^6-x^2\right )}^{2/3}}{8\,x^4} \]

[In]

int(1/(x^3*(x^6 - x^2)^(1/3)),x)

[Out]

(3*(x^6 - x^2)^(2/3))/(8*x^4)