Integrand size = 15, antiderivative size = 68 \[ \int \frac {1}{x^9 \sqrt [3]{a+b x^3}} \, dx=-\frac {\left (a+b x^3\right )^{2/3}}{8 a x^8}+\frac {3 b \left (a+b x^3\right )^{2/3}}{20 a^2 x^5}-\frac {9 b^2 \left (a+b x^3\right )^{2/3}}{40 a^3 x^2} \] Output:
-1/8*(b*x^3+a)^(2/3)/a/x^8+3/20*b*(b*x^3+a)^(2/3)/a^2/x^5-9/40*b^2*(b*x^3+ a)^(2/3)/a^3/x^2
Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^9 \sqrt [3]{a+b x^3}} \, dx=\frac {\left (a+b x^3\right )^{2/3} \left (-5 a^2+6 a b x^3-9 b^2 x^6\right )}{40 a^3 x^8} \] Input:
Integrate[1/(x^9*(a + b*x^3)^(1/3)),x]
Output:
((a + b*x^3)^(2/3)*(-5*a^2 + 6*a*b*x^3 - 9*b^2*x^6))/(40*a^3*x^8)
Time = 0.31 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {803, 803, 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.
\(\displaystyle \int \frac {1}{x^9 \sqrt [3]{a+b x^3}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {3 b \int \frac {1}{x^6 \sqrt [3]{b x^3+a}}dx}{4 a}-\frac {\left (a+b x^3\right )^{2/3}}{8 a x^8}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {3 b \left (-\frac {3 b \int \frac {1}{x^3 \sqrt [3]{b x^3+a}}dx}{5 a}-\frac {\left (a+b x^3\right )^{2/3}}{5 a x^5}\right )}{4 a}-\frac {\left (a+b x^3\right )^{2/3}}{8 a x^8}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {3 b \left (\frac {3 b \left (a+b x^3\right )^{2/3}}{10 a^2 x^2}-\frac {\left (a+b x^3\right )^{2/3}}{5 a x^5}\right )}{4 a}-\frac {\left (a+b x^3\right )^{2/3}}{8 a x^8}\) |
Input:
Int[1/(x^9*(a + b*x^3)^(1/3)),x]
Output:
-1/8*(a + b*x^3)^(2/3)/(a*x^8) - (3*b*(-1/5*(a + b*x^3)^(2/3)/(a*x^5) + (3 *b*(a + b*x^3)^(2/3))/(10*a^2*x^2)))/(4*a)
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[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*(m + 1 ))) Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x] && I LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
Time = 0.50 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.57
method | result | size |
gosper | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (9 b^{2} x^{6}-6 a b \,x^{3}+5 a^{2}\right )}{40 a^{3} x^{8}}\) | \(39\) |
trager | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (9 b^{2} x^{6}-6 a b \,x^{3}+5 a^{2}\right )}{40 a^{3} x^{8}}\) | \(39\) |
risch | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (9 b^{2} x^{6}-6 a b \,x^{3}+5 a^{2}\right )}{40 a^{3} x^{8}}\) | \(39\) |
pseudoelliptic | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (9 b^{2} x^{6}-6 a b \,x^{3}+5 a^{2}\right )}{40 a^{3} x^{8}}\) | \(39\) |
orering | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (9 b^{2} x^{6}-6 a b \,x^{3}+5 a^{2}\right )}{40 a^{3} x^{8}}\) | \(39\) |
Input:
int(1/x^9/(b*x^3+a)^(1/3),x,method=_RETURNVERBOSE)
Output:
-1/40*(b*x^3+a)^(2/3)*(9*b^2*x^6-6*a*b*x^3+5*a^2)/a^3/x^8
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x^9 \sqrt [3]{a+b x^3}} \, dx=-\frac {{\left (9 \, b^{2} x^{6} - 6 \, a b x^{3} + 5 \, a^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{40 \, a^{3} x^{8}} \] Input:
integrate(1/x^9/(b*x^3+a)^(1/3),x, algorithm="fricas")
Output:
-1/40*(9*b^2*x^6 - 6*a*b*x^3 + 5*a^2)*(b*x^3 + a)^(2/3)/(a^3*x^8)
Leaf count of result is larger than twice the leaf count of optimal. 406 vs. \(2 (61) = 122\).
Time = 0.77 (sec) , antiderivative size = 406, normalized size of antiderivative = 5.97 \[ \int \frac {1}{x^9 \sqrt [3]{a+b x^3}} \, dx=\frac {10 a^{4} b^{\frac {14}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {1}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {1}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {1}{3}\right )} + \frac {8 a^{3} b^{\frac {17}{3}} x^{3} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {1}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {1}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {1}{3}\right )} + \frac {4 a^{2} b^{\frac {20}{3}} x^{6} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {1}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {1}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {1}{3}\right )} + \frac {24 a b^{\frac {23}{3}} x^{9} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {1}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {1}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {1}{3}\right )} + \frac {18 b^{\frac {26}{3}} x^{12} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {1}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {1}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {1}{3}\right )} \] Input:
integrate(1/x**9/(b*x**3+a)**(1/3),x)
Output:
10*a**4*b**(14/3)*(a/(b*x**3) + 1)**(2/3)*gamma(-8/3)/(27*a**5*b**4*x**6*g amma(1/3) + 54*a**4*b**5*x**9*gamma(1/3) + 27*a**3*b**6*x**12*gamma(1/3)) + 8*a**3*b**(17/3)*x**3*(a/(b*x**3) + 1)**(2/3)*gamma(-8/3)/(27*a**5*b**4* x**6*gamma(1/3) + 54*a**4*b**5*x**9*gamma(1/3) + 27*a**3*b**6*x**12*gamma( 1/3)) + 4*a**2*b**(20/3)*x**6*(a/(b*x**3) + 1)**(2/3)*gamma(-8/3)/(27*a**5 *b**4*x**6*gamma(1/3) + 54*a**4*b**5*x**9*gamma(1/3) + 27*a**3*b**6*x**12* gamma(1/3)) + 24*a*b**(23/3)*x**9*(a/(b*x**3) + 1)**(2/3)*gamma(-8/3)/(27* a**5*b**4*x**6*gamma(1/3) + 54*a**4*b**5*x**9*gamma(1/3) + 27*a**3*b**6*x* *12*gamma(1/3)) + 18*b**(26/3)*x**12*(a/(b*x**3) + 1)**(2/3)*gamma(-8/3)/( 27*a**5*b**4*x**6*gamma(1/3) + 54*a**4*b**5*x**9*gamma(1/3) + 27*a**3*b**6 *x**12*gamma(1/3))
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^9 \sqrt [3]{a+b x^3}} \, dx=-\frac {\frac {20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{2}}{x^{2}} - \frac {16 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b}{x^{5}} + \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}}}{x^{8}}}{40 \, a^{3}} \] Input:
integrate(1/x^9/(b*x^3+a)^(1/3),x, algorithm="maxima")
Output:
-1/40*(20*(b*x^3 + a)^(2/3)*b^2/x^2 - 16*(b*x^3 + a)^(5/3)*b/x^5 + 5*(b*x^ 3 + a)^(8/3)/x^8)/a^3
\[ \int \frac {1}{x^9 \sqrt [3]{a+b x^3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{9}} \,d x } \] Input:
integrate(1/x^9/(b*x^3+a)^(1/3),x, algorithm="giac")
Output:
integrate(1/((b*x^3 + a)^(1/3)*x^9), x)
Time = 0.33 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x^9 \sqrt [3]{a+b x^3}} \, dx=-\frac {{\left (b\,x^3+a\right )}^{2/3}\,\left (5\,a^2-6\,a\,b\,x^3+9\,b^2\,x^6\right )}{40\,a^3\,x^8} \] Input:
int(1/(x^9*(a + b*x^3)^(1/3)),x)
Output:
-((a + b*x^3)^(2/3)*(5*a^2 + 9*b^2*x^6 - 6*a*b*x^3))/(40*a^3*x^8)
\[ \int \frac {1}{x^9 \sqrt [3]{a+b x^3}} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{9}}d x \] Input:
int(1/x^9/(b*x^3+a)^(1/3),x)
Output:
int(1/((a + b*x**3)**(1/3)*x**9),x)