\(\int \frac {(1+x^3)^{2/3}}{x^6} \, dx\) [73]

Optimal result

Integrand size = 13, antiderivative size = 16 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx=-\frac {\left (1+x^3\right )^{5/3}}{5 x^5} \]

[Out]

-1/5*(x^3+1)^(5/3)/x^5

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {270} \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx=-\frac {\left (x^3+1\right )^{5/3}}{5 x^5} \]

[In]

Int[(1 + x^3)^(2/3)/x^6,x]

[Out]

-1/5*(1 + x^3)^(5/3)/x^5

Rule 270

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\left (1+x^3\right )^{5/3}}{5 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx=-\frac {\left (1+x^3\right )^{5/3}}{5 x^5} \]

[In]

Integrate[(1 + x^3)^(2/3)/x^6,x]

[Out]

-1/5*(1 + x^3)^(5/3)/x^5

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
trager	\(-\frac {\left (x^{3}+1\right )^{\frac {5}{3}}}{5 x^{5}}\)	\(13\)
meijerg	\(-\frac {\left (x^{3}+1\right )^{\frac {5}{3}}}{5 x^{5}}\)	\(13\)
pseudoelliptic	\(-\frac {\left (x^{3}+1\right )^{\frac {5}{3}}}{5 x^{5}}\)	\(13\)
risch	\(-\frac {x^{6}+2 x^{3}+1}{5 x^{5} \left (x^{3}+1\right )^{\frac {1}{3}}}\)	\(23\)
gosper	\(-\frac {\left (1+x \right ) \left (x^{2}-x +1\right ) \left (x^{3}+1\right )^{\frac {2}{3}}}{5 x^{5}}\)	\(24\)

[In]

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

[Out]

-1/5*(x^3+1)^(5/3)/x^5

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx=-\frac {{\left (x^{3} + 1\right )}^{\frac {5}{3}}}{5 \, x^{5}} \]

[In]

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

[Out]

-1/5*(x^3 + 1)^(5/3)/x^5

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (14) = 28\).

Time = 0.42 (sec) , antiderivative size = 53, normalized size of antiderivative = 3.31 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx=\frac {\left (1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 \Gamma \left (- \frac {2}{3}\right )} + \frac {\left (1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 x^{3} \Gamma \left (- \frac {2}{3}\right )} \]

[In]

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

[Out]

(1 + x**(-3))**(2/3)*gamma(-5/3)/(3*gamma(-2/3)) + (1 + x**(-3))**(2/3)*gamma(-5/3)/(3*x**3*gamma(-2/3))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx=-\frac {{\left (x^{3} + 1\right )}^{\frac {5}{3}}}{5 \, x^{5}} \]

[In]

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

[Out]

-1/5*(x^3 + 1)^(5/3)/x^5

Giac [F]

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

[In]

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

[Out]

integrate((x^3 + 1)^(2/3)/x^6, x)

Mupad [B] (verification not implemented)

Time = 5.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx=-\frac {{\left (x^3+1\right )}^{2/3}+x^3\,{\left (x^3+1\right )}^{2/3}}{5\,x^5} \]

[In]

int((x^3 + 1)^(2/3)/x^6,x)

[Out]

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