Integrand size = 22, antiderivative size = 84 \[ \int \frac {A+B x^3}{x^9 \sqrt [3]{a+b x^3}} \, dx=-\frac {A \left (a+b x^3\right )^{2/3}}{8 a x^8}+\frac {(3 A b-4 a B) \left (a+b x^3\right )^{2/3}}{20 a^2 x^5}-\frac {3 b (3 A b-4 a B) \left (a+b x^3\right )^{2/3}}{40 a^3 x^2} \] Output:
-1/8*A*(b*x^3+a)^(2/3)/a/x^8+1/20*(3*A*b-4*B*a)*(b*x^3+a)^(2/3)/a^2/x^5-3/ 40*b*(3*A*b-4*B*a)*(b*x^3+a)^(2/3)/a^3/x^2
Time = 0.32 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.74 \[ \int \frac {A+B x^3}{x^9 \sqrt [3]{a+b x^3}} \, dx=\frac {\left (a+b x^3\right )^{2/3} \left (-5 a^2 A+6 a A b x^3-8 a^2 B x^3-9 A b^2 x^6+12 a b B x^6\right )}{40 a^3 x^8} \] Input:
Integrate[(A + B*x^3)/(x^9*(a + b*x^3)^(1/3)),x]
Output:
((a + b*x^3)^(2/3)*(-5*a^2*A + 6*a*A*b*x^3 - 8*a^2*B*x^3 - 9*A*b^2*x^6 + 1 2*a*b*B*x^6))/(40*a^3*x^8)
Time = 0.35 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {955, 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.
\(\displaystyle \int \frac {A+B x^3}{x^9 \sqrt [3]{a+b x^3}} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle -\frac {(3 A b-4 a B) \int \frac {1}{x^6 \sqrt [3]{b x^3+a}}dx}{4 a}-\frac {A \left (a+b x^3\right )^{2/3}}{8 a x^8}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle -\frac {(3 A b-4 a B) \left (-\frac {3 b \int \frac {1}{x^3 \sqrt [3]{b x^3+a}}dx}{5 a}-\frac {\left (a+b x^3\right )^{2/3}}{5 a x^5}\right )}{4 a}-\frac {A \left (a+b x^3\right )^{2/3}}{8 a x^8}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle -\frac {\left (\frac {3 b \left (a+b x^3\right )^{2/3}}{10 a^2 x^2}-\frac {\left (a+b x^3\right )^{2/3}}{5 a x^5}\right ) (3 A b-4 a B)}{4 a}-\frac {A \left (a+b x^3\right )^{2/3}}{8 a x^8}\) |
Input:
Int[(A + B*x^3)/(x^9*(a + b*x^3)^(1/3)),x]
Output:
-1/8*(A*(a + b*x^3)^(2/3))/(a*x^8) - ((3*A*b - 4*a*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]
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 = 0.86 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.65
method | result | size |
pseudoelliptic | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (\left (\frac {8 B \,x^{3}}{5}+A \right ) a^{2}-\frac {6 b \,x^{3} \left (2 B \,x^{3}+A \right ) a}{5}+\frac {9 A \,b^{2} x^{6}}{5}\right )}{8 a^{3} x^{8}}\) | \(55\) |
gosper | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (9 A \,b^{2} x^{6}-12 B a b \,x^{6}-6 a A b \,x^{3}+8 B \,a^{2} x^{3}+5 a^{2} A \right )}{40 a^{3} x^{8}}\) | \(59\) |
trager | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (9 A \,b^{2} x^{6}-12 B a b \,x^{6}-6 a A b \,x^{3}+8 B \,a^{2} x^{3}+5 a^{2} A \right )}{40 a^{3} x^{8}}\) | \(59\) |
risch | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (9 A \,b^{2} x^{6}-12 B a b \,x^{6}-6 a A b \,x^{3}+8 B \,a^{2} x^{3}+5 a^{2} A \right )}{40 a^{3} x^{8}}\) | \(59\) |
orering | \(-\frac {\left (b \,x^{3}+a \right )^{\frac {2}{3}} \left (9 A \,b^{2} x^{6}-12 B a b \,x^{6}-6 a A b \,x^{3}+8 B \,a^{2} x^{3}+5 a^{2} A \right )}{40 a^{3} x^{8}}\) | \(59\) |
Input:
int((B*x^3+A)/x^9/(b*x^3+a)^(1/3),x,method=_RETURNVERBOSE)
Output:
-1/8*(b*x^3+a)^(2/3)*((8/5*B*x^3+A)*a^2-6/5*b*x^3*(2*B*x^3+A)*a+9/5*A*b^2* x^6)/a^3/x^8
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \frac {A+B x^3}{x^9 \sqrt [3]{a+b x^3}} \, dx=\frac {{\left (3 \, {\left (4 \, B a b - 3 \, A b^{2}\right )} x^{6} - 2 \, {\left (4 \, B a^{2} - 3 \, A a b\right )} x^{3} - 5 \, A a^{2}\right )} {\left (b x^{3} + a\right )}^{\frac {2}{3}}}{40 \, a^{3} x^{8}} \] Input:
integrate((B*x^3+A)/x^9/(b*x^3+a)^(1/3),x, algorithm="fricas")
Output:
1/40*(3*(4*B*a*b - 3*A*b^2)*x^6 - 2*(4*B*a^2 - 3*A*a*b)*x^3 - 5*A*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. 490 vs. \(2 (78) = 156\).
Time = 1.49 (sec) , antiderivative size = 490, normalized size of antiderivative = 5.83 \[ \int \frac {A+B x^3}{x^9 \sqrt [3]{a+b x^3}} \, dx=\frac {10 A a^{4} b^{\frac {14}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {1}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {1}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {1}{3}\right )} + \frac {8 A a^{3} b^{\frac {17}{3}} x^{3} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {1}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {1}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {1}{3}\right )} + \frac {4 A a^{2} b^{\frac {20}{3}} x^{6} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {1}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {1}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {1}{3}\right )} + \frac {24 A a b^{\frac {23}{3}} x^{9} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {1}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {1}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {1}{3}\right )} + \frac {18 A b^{\frac {26}{3}} x^{12} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {8}{3}\right )}{27 a^{5} b^{4} x^{6} \Gamma \left (\frac {1}{3}\right ) + 54 a^{4} b^{5} x^{9} \Gamma \left (\frac {1}{3}\right ) + 27 a^{3} b^{6} x^{12} \Gamma \left (\frac {1}{3}\right )} - \frac {2 B b^{\frac {2}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{9 a x^{3} \Gamma \left (\frac {1}{3}\right )} + \frac {B b^{\frac {5}{3}} \left (\frac {a}{b x^{3}} + 1\right )^{\frac {2}{3}} \Gamma \left (- \frac {5}{3}\right )}{3 a^{2} \Gamma \left (\frac {1}{3}\right )} \] Input:
integrate((B*x**3+A)/x**9/(b*x**3+a)**(1/3),x)
Output:
10*A*a**4*b**(14/3)*(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) ) + 8*A*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*ga mma(1/3)) + 4*A*a**2*b**(20/3)*x**6*(a/(b*x**3) + 1)**(2/3)*gamma(-8/3)/(2 7*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*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*A*b**(26/3)*x**12*(a/(b*x**3) + 1)**(2/3)*gam ma(-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)) - 2*B*b**(2/3)*(a/(b*x**3) + 1)**(2/3)*gamma( -5/3)/(9*a*x**3*gamma(1/3)) + B*b**(5/3)*(a/(b*x**3) + 1)**(2/3)*gamma(-5/ 3)/(3*a**2*gamma(1/3))
Time = 0.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x^3}{x^9 \sqrt [3]{a+b x^3}} \, dx=\frac {B {\left (\frac {5 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b}{x^{2}} - \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}}}{x^{5}}\right )}}{10 \, a^{2}} - \frac {A {\left (\frac {20 \, {\left (b x^{3} + a\right )}^{\frac {2}{3}} b^{2}}{x^{2}} - \frac {16 \, {\left (b x^{3} + a\right )}^{\frac {5}{3}} b}{x^{5}} + \frac {5 \, {\left (b x^{3} + a\right )}^{\frac {8}{3}}}{x^{8}}\right )}}{40 \, a^{3}} \] Input:
integrate((B*x^3+A)/x^9/(b*x^3+a)^(1/3),x, algorithm="maxima")
Output:
1/10*B*(5*(b*x^3 + a)^(2/3)*b/x^2 - 2*(b*x^3 + a)^(5/3)/x^5)/a^2 - 1/40*A* (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
\[ \int \frac {A+B x^3}{x^9 \sqrt [3]{a+b x^3}} \, dx=\int { \frac {B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} x^{9}} \,d x } \] Input:
integrate((B*x^3+A)/x^9/(b*x^3+a)^(1/3),x, algorithm="giac")
Output:
integrate((B*x^3 + A)/((b*x^3 + a)^(1/3)*x^9), x)
Time = 0.98 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.69 \[ \int \frac {A+B x^3}{x^9 \sqrt [3]{a+b x^3}} \, dx=-\frac {{\left (b\,x^3+a\right )}^{2/3}\,\left (8\,B\,a^2\,x^3+5\,A\,a^2-12\,B\,a\,b\,x^6-6\,A\,a\,b\,x^3+9\,A\,b^2\,x^6\right )}{40\,a^3\,x^8} \] Input:
int((A + B*x^3)/(x^9*(a + b*x^3)^(1/3)),x)
Output:
-((a + b*x^3)^(2/3)*(5*A*a^2 + 8*B*a^2*x^3 + 9*A*b^2*x^6 - 6*A*a*b*x^3 - 1 2*B*a*b*x^6))/(40*a^3*x^8)
\[ \int \frac {A+B x^3}{x^9 \sqrt [3]{a+b x^3}} \, dx=\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{9}}d x \right ) a +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} x^{6}}d x \right ) b \] Input:
int((B*x^3+A)/x^9/(b*x^3+a)^(1/3),x)
Output:
int(1/((a + b*x**3)**(1/3)*x**9),x)*a + int(1/((a + b*x**3)**(1/3)*x**6),x )*b