Integrand size = 15, antiderivative size = 68 \[ \int \frac {1}{x^8 \left (a+b x^3\right )^{2/3}} \, dx=-\frac {\sqrt [3]{a+b x^3}}{7 a x^7}+\frac {3 b \sqrt [3]{a+b x^3}}{14 a^2 x^4}-\frac {9 b^2 \sqrt [3]{a+b x^3}}{14 a^3 x} \] Output:
-1/7*(b*x^3+a)^(1/3)/a/x^7+3/14*b*(b*x^3+a)^(1/3)/a^2/x^4-9/14*b^2*(b*x^3+ a)^(1/3)/a^3/x
Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^8 \left (a+b x^3\right )^{2/3}} \, dx=\frac {\sqrt [3]{a+b x^3} \left (-2 a^2+3 a b x^3-9 b^2 x^6\right )}{14 a^3 x^7} \] Input:
Integrate[1/(x^8*(a + b*x^3)^(2/3)),x]
Output:
((a + b*x^3)^(1/3)*(-2*a^2 + 3*a*b*x^3 - 9*b^2*x^6))/(14*a^3*x^7)
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.
\(\displaystyle \int \frac {1}{x^8 \left (a+b x^3\right )^{2/3}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {6 b \int \frac {1}{x^5 \left (b x^3+a\right )^{2/3}}dx}{7 a}-\frac {\sqrt [3]{a+b x^3}}{7 a x^7}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {6 b \left (-\frac {3 b \int \frac {1}{x^2 \left (b x^3+a\right )^{2/3}}dx}{4 a}-\frac {\sqrt [3]{a+b x^3}}{4 a x^4}\right )}{7 a}-\frac {\sqrt [3]{a+b x^3}}{7 a x^7}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {6 b \left (\frac {3 b \sqrt [3]{a+b x^3}}{4 a^2 x}-\frac {\sqrt [3]{a+b x^3}}{4 a x^4}\right )}{7 a}-\frac {\sqrt [3]{a+b x^3}}{7 a x^7}\) |
Input:
Int[1/(x^8*(a + b*x^3)^(2/3)),x]
Output:
-1/7*(a + b*x^3)^(1/3)/(a*x^7) - (6*b*(-1/4*(a + b*x^3)^(1/3)/(a*x^4) + (3 *b*(a + b*x^3)^(1/3))/(4*a^2*x)))/(7*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 = 0.80 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.57
method | result | size |
gosper | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (9 b^{2} x^{6}-3 a b \,x^{3}+2 a^{2}\right )}{14 a^{3} x^{7}}\) | \(39\) |
trager | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (9 b^{2} x^{6}-3 a b \,x^{3}+2 a^{2}\right )}{14 a^{3} x^{7}}\) | \(39\) |
risch | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (9 b^{2} x^{6}-3 a b \,x^{3}+2 a^{2}\right )}{14 a^{3} x^{7}}\) | \(39\) |
pseudoelliptic | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (9 b^{2} x^{6}-3 a b \,x^{3}+2 a^{2}\right )}{14 a^{3} x^{7}}\) | \(39\) |
orering | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {1}{3}} \left (9 b^{2} x^{6}-3 a b \,x^{3}+2 a^{2}\right )}{14 a^{3} x^{7}}\) | \(39\) |
Input:
int(1/x^8/(b*x^3+a)^(2/3),x,method=_RETURNVERBOSE)
Output:
-1/14*(b*x^3+a)^(1/3)*(9*b^2*x^6-3*a*b*x^3+2*a^2)/a^3/x^7
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x^8 \left (a+b x^3\right )^{2/3}} \, dx=-\frac {{\left (9 \, b^{2} x^{6} - 3 \, a b x^{3} + 2 \, a^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{14 \, a^{3} x^{7}} \] Input:
integrate(1/x^8/(b*x^3+a)^(2/3),x, algorithm="fricas")
Output:
-1/14*(9*b^2*x^6 - 3*a*b*x^3 + 2*a^2)*(b*x^3 + a)^(1/3)/(a^3*x^7)
Leaf count of result is larger than twice the leaf count of optimal. 406 vs. \(2 (60) = 120\).
Time = 0.79 (sec) , antiderivative size = 406, normalized size of antiderivative = 5.97 \[ \int \frac {1}{x^8 \left (a+b x^3\right )^{2/3}} \, dx=\frac {4 a^{4} b^{\frac {13}{3}} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {7}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {2}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {2}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {2}{3}\right )} + \frac {2 a^{3} b^{\frac {16}{3}} x^{3} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {7}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {2}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {2}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {2}{3}\right )} + \frac {10 a^{2} b^{\frac {19}{3}} x^{6} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {7}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {2}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {2}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {2}{3}\right )} + \frac {30 a b^{\frac {22}{3}} x^{9} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {7}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {2}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {2}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {2}{3}\right )} + \frac {18 b^{\frac {25}{3}} x^{12} \sqrt [3]{\frac {a}{b x^{3}} + 1} \Gamma \left (- \frac {7}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {2}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {2}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {2}{3}\right )} \] Input:
integrate(1/x**8/(b*x**3+a)**(2/3),x)
Output:
4*a**4*b**(13/3)*(a/(b*x**3) + 1)**(1/3)*gamma(-7/3)/(27*a**5*b**4*x**6*ga mma(2/3) + 54*a**4*b**5*x**9*gamma(2/3) + 27*a**3*b**6*x**12*gamma(2/3)) + 2*a**3*b**(16/3)*x**3*(a/(b*x**3) + 1)**(1/3)*gamma(-7/3)/(27*a**5*b**4*x **6*gamma(2/3) + 54*a**4*b**5*x**9*gamma(2/3) + 27*a**3*b**6*x**12*gamma(2 /3)) + 10*a**2*b**(19/3)*x**6*(a/(b*x**3) + 1)**(1/3)*gamma(-7/3)/(27*a**5 *b**4*x**6*gamma(2/3) + 54*a**4*b**5*x**9*gamma(2/3) + 27*a**3*b**6*x**12* gamma(2/3)) + 30*a*b**(22/3)*x**9*(a/(b*x**3) + 1)**(1/3)*gamma(-7/3)/(27* a**5*b**4*x**6*gamma(2/3) + 54*a**4*b**5*x**9*gamma(2/3) + 27*a**3*b**6*x* *12*gamma(2/3)) + 18*b**(25/3)*x**12*(a/(b*x**3) + 1)**(1/3)*gamma(-7/3)/( 27*a**5*b**4*x**6*gamma(2/3) + 54*a**4*b**5*x**9*gamma(2/3) + 27*a**3*b**6 *x**12*gamma(2/3))
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {1}{x^8 \left (a+b x^3\right )^{2/3}} \, dx=-\frac {\frac {14 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{2}}{x} - \frac {7 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} b}{x^{4}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {7}{3}}}{x^{7}}}{14 \, a^{3}} \] Input:
integrate(1/x^8/(b*x^3+a)^(2/3),x, algorithm="maxima")
Output:
-1/14*(14*(b*x^3 + a)^(1/3)*b^2/x - 7*(b*x^3 + a)^(4/3)*b/x^4 + 2*(b*x^3 + a)^(7/3)/x^7)/a^3
\[ \int \frac {1}{x^8 \left (a+b x^3\right )^{2/3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )}^{\frac {2}{3}} x^{8}} \,d x } \] Input:
integrate(1/x^8/(b*x^3+a)^(2/3),x, algorithm="giac")
Output:
integrate(1/((b*x^3 + a)^(2/3)*x^8), x)
Time = 0.33 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x^8 \left (a+b x^3\right )^{2/3}} \, dx=-\frac {{\left (b\,x^3+a\right )}^{1/3}\,\left (2\,a^2-3\,a\,b\,x^3+9\,b^2\,x^6\right )}{14\,a^3\,x^7} \] Input:
int(1/(x^8*(a + b*x^3)^(2/3)),x)
Output:
-((a + b*x^3)^(1/3)*(2*a^2 + 9*b^2*x^6 - 3*a*b*x^3))/(14*a^3*x^7)
\[ \int \frac {1}{x^8 \left (a+b x^3\right )^{2/3}} \, dx=\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} x^{8}}d x \] Input:
int(1/x^8/(b*x^3+a)^(2/3),x)
Output:
int(1/((a + b*x**3)**(2/3)*x**8),x)