Integrand size = 18, antiderivative size = 28 \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^9} \, dx=\frac {\left (-1+x^3\right )^{2/3} \left (-5-2 x^3+7 x^6\right )}{20 x^8} \] Output:
1/20*(x^3-1)^(2/3)*(7*x^6-2*x^3-5)/x^8
Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^9} \, dx=\frac {\left (-1+x^3\right )^{2/3} \left (-5-2 x^3+7 x^6\right )}{20 x^8} \] Input:
Integrate[((-1 + x^3)^(2/3)*(2 + x^3))/x^9,x]
Output:
((-1 + x^3)^(2/3)*(-5 - 2*x^3 + 7*x^6))/(20*x^8)
Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {955, 796}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (x^3-1\right )^{2/3} \left (x^3+2\right )}{x^9} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle \frac {7}{4} \int \frac {\left (x^3-1\right )^{2/3}}{x^6}dx+\frac {\left (x^3-1\right )^{5/3}}{4 x^8}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {\left (x^3-1\right )^{5/3}}{4 x^8}+\frac {7 \left (x^3-1\right )^{5/3}}{20 x^5}\) |
Input:
Int[((-1 + x^3)^(2/3)*(2 + x^3))/x^9,x]
Output:
(-1 + x^3)^(5/3)/(4*x^8) + (7*(-1 + x^3)^(5/3))/(20*x^5)
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]
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] + Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71
method | result | size |
pseudoelliptic | \(\frac {\left (7 x^{3}+5\right ) \left (x^{3}-1\right )^{\frac {5}{3}}}{20 x^{8}}\) | \(20\) |
trager | \(\frac {\left (x^{3}-1\right )^{\frac {2}{3}} \left (7 x^{6}-2 x^{3}-5\right )}{20 x^{8}}\) | \(25\) |
gosper | \(\frac {\left (-1+x \right ) \left (x^{2}+x +1\right ) \left (7 x^{3}+5\right ) \left (x^{3}-1\right )^{\frac {2}{3}}}{20 x^{8}}\) | \(29\) |
orering | \(\frac {\left (-1+x \right ) \left (x^{2}+x +1\right ) \left (7 x^{3}+5\right ) \left (x^{3}-1\right )^{\frac {2}{3}}}{20 x^{8}}\) | \(29\) |
risch | \(\frac {7 x^{9}-9 x^{6}-3 x^{3}+5}{20 x^{8} \left (x^{3}-1\right )^{\frac {1}{3}}}\) | \(30\) |
meijerg | \(-\frac {\operatorname {signum}\left (x^{3}-1\right )^{\frac {2}{3}} \left (-x^{3}+1\right )^{\frac {5}{3}}}{5 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {2}{3}} x^{5}}-\frac {\operatorname {signum}\left (x^{3}-1\right )^{\frac {2}{3}} \left (-\frac {3}{5} x^{6}-\frac {2}{5} x^{3}+1\right ) \left (-x^{3}+1\right )^{\frac {2}{3}}}{4 {\left (-\operatorname {signum}\left (x^{3}-1\right )\right )}^{\frac {2}{3}} x^{8}}\) | \(78\) |
Input:
int((x^3-1)^(2/3)*(x^3+2)/x^9,x,method=_RETURNVERBOSE)
Output:
1/20*(7*x^3+5)/x^8*(x^3-1)^(5/3)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^9} \, dx=\frac {{\left (7 \, x^{6} - 2 \, x^{3} - 5\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{20 \, x^{8}} \] Input:
integrate((x^3-1)^(2/3)*(x^3+2)/x^9,x, algorithm="fricas")
Output:
1/20*(7*x^6 - 2*x^3 - 5)*(x^3 - 1)^(2/3)/x^8
Result contains complex when optimal does not.
Time = 1.53 (sec) , antiderivative size = 428, normalized size of antiderivative = 15.29 \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^9} \, dx=\begin {cases} \frac {\left (-1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} e^{- \frac {i \pi }{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}} e^{- \frac {i \pi }{3}} \Gamma \left (- \frac {5}{3}\right )}{3 x^{3} \Gamma \left (- \frac {2}{3}\right )} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\- \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 )} & \text {otherwise} \end {cases} + 2 \left (\begin {cases} \frac {3 x^{6} \left (-1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} e^{\frac {2 i \pi }{3}} \Gamma \left (- \frac {8}{3}\right )}{9 x^{6} \Gamma \left (- \frac {2}{3}\right ) - 9 x^{3} \Gamma \left (- \frac {2}{3}\right )} - \frac {x^{3} \left (-1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} e^{\frac {2 i \pi }{3}} \Gamma \left (- \frac {8}{3}\right )}{9 x^{6} \Gamma \left (- \frac {2}{3}\right ) - 9 x^{3} \Gamma \left (- \frac {2}{3}\right )} + \frac {5 \left (-1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} e^{\frac {2 i \pi }{3}} \Gamma \left (- \frac {8}{3}\right )}{9 x^{9} \Gamma \left (- \frac {2}{3}\right ) - 9 x^{6} \Gamma \left (- \frac {2}{3}\right )} - \frac {7 \left (-1 + \frac {1}{x^{3}}\right )^{\frac {2}{3}} e^{\frac {2 i \pi }{3}} \Gamma \left (- \frac {8}{3}\right )}{9 x^{6} \Gamma \left (- \frac {2}{3}\right ) - 9 x^{3} \Gamma \left (- \frac {2}{3}\right )} & \text {for}\: \frac {1}{\left |{x^{3}}\right |} > 1 \\\frac {\left (1 - \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{3 \Gamma \left (- \frac {2}{3}\right )} + \frac {2 \left (1 - \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{9 x^{3} \Gamma \left (- \frac {2}{3}\right )} - \frac {5 \left (1 - \frac {1}{x^{3}}\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{9 x^{6} \Gamma \left (- \frac {2}{3}\right )} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((x**3-1)**(2/3)*(x**3+2)/x**9,x)
Output:
Piecewise(((-1 + x**(-3))**(2/3)*exp(-I*pi/3)*gamma(-5/3)/(3*gamma(-2/3)) - (-1 + x**(-3))**(2/3)*exp(-I*pi/3)*gamma(-5/3)/(3*x**3*gamma(-2/3)), 1/A bs(x**3) > 1), (-(1 - 1/x**3)**(2/3)*gamma(-5/3)/(3*gamma(-2/3)) + (1 - 1/ x**3)**(2/3)*gamma(-5/3)/(3*x**3*gamma(-2/3)), True)) + 2*Piecewise((3*x** 6*(-1 + x**(-3))**(2/3)*exp(2*I*pi/3)*gamma(-8/3)/(9*x**6*gamma(-2/3) - 9* x**3*gamma(-2/3)) - x**3*(-1 + x**(-3))**(2/3)*exp(2*I*pi/3)*gamma(-8/3)/( 9*x**6*gamma(-2/3) - 9*x**3*gamma(-2/3)) + 5*(-1 + x**(-3))**(2/3)*exp(2*I *pi/3)*gamma(-8/3)/(9*x**9*gamma(-2/3) - 9*x**6*gamma(-2/3)) - 7*(-1 + x** (-3))**(2/3)*exp(2*I*pi/3)*gamma(-8/3)/(9*x**6*gamma(-2/3) - 9*x**3*gamma( -2/3)), 1/Abs(x**3) > 1), ((1 - 1/x**3)**(2/3)*gamma(-8/3)/(3*gamma(-2/3)) + 2*(1 - 1/x**3)**(2/3)*gamma(-8/3)/(9*x**3*gamma(-2/3)) - 5*(1 - 1/x**3) **(2/3)*gamma(-8/3)/(9*x**6*gamma(-2/3)), True))
Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^9} \, dx=\frac {3 \, {\left (x^{3} - 1\right )}^{\frac {5}{3}}}{5 \, x^{5}} - \frac {{\left (x^{3} - 1\right )}^{\frac {8}{3}}}{4 \, x^{8}} \] Input:
integrate((x^3-1)^(2/3)*(x^3+2)/x^9,x, algorithm="maxima")
Output:
3/5*(x^3 - 1)^(5/3)/x^5 - 1/4*(x^3 - 1)^(8/3)/x^8
\[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^9} \, dx=\int { \frac {{\left (x^{3} + 2\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{9}} \,d x } \] Input:
integrate((x^3-1)^(2/3)*(x^3+2)/x^9,x, algorithm="giac")
Output:
integrate((x^3 + 2)*(x^3 - 1)^(2/3)/x^9, x)
Time = 7.47 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^9} \, dx=-\frac {{\left (x^3-1\right )}^{2/3}\,\left (-7\,x^6+2\,x^3+5\right )}{20\,x^8} \] Input:
int(((x^3 - 1)^(2/3)*(x^3 + 2))/x^9,x)
Output:
-((x^3 - 1)^(2/3)*(2*x^3 - 7*x^6 + 5))/(20*x^8)
Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^9} \, dx=\frac {\left (x^{3}-1\right )^{\frac {2}{3}} \left (7 x^{6}-2 x^{3}-5\right )}{20 x^{8}} \] Input:
int((x^3-1)^(2/3)*(x^3+2)/x^9,x)
Output:
((x**3 - 1)**(2/3)*(7*x**6 - 2*x**3 - 5))/(20*x**8)