Integrand size = 20, antiderivative size = 112 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^3} \, dx=-\frac {a^5 A}{2 x^2}+a^4 (5 A b+a B) x+\frac {5}{4} a^3 b (2 A b+a B) x^4+\frac {10}{7} a^2 b^2 (A b+a B) x^7+\frac {1}{2} a b^3 (A b+2 a B) x^{10}+\frac {1}{13} b^4 (A b+5 a B) x^{13}+\frac {1}{16} b^5 B x^{16} \]
-1/2*a^5*A/x^2+a^4*(5*A*b+B*a)*x+5/4*a^3*b*(2*A*b+B*a)*x^4+10/7*a^2*b^2*(A *b+B*a)*x^7+1/2*a*b^3*(A*b+2*B*a)*x^10+1/13*b^4*(A*b+5*B*a)*x^13+1/16*b^5* B*x^16
Time = 0.03 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^3} \, dx=-\frac {a^5 A}{2 x^2}+a^4 (5 A b+a B) x+\frac {5}{4} a^3 b (2 A b+a B) x^4+\frac {10}{7} a^2 b^2 (A b+a B) x^7+\frac {1}{2} a b^3 (A b+2 a B) x^{10}+\frac {1}{13} b^4 (A b+5 a B) x^{13}+\frac {1}{16} b^5 B x^{16} \]
-1/2*(a^5*A)/x^2 + a^4*(5*A*b + a*B)*x + (5*a^3*b*(2*A*b + a*B)*x^4)/4 + ( 10*a^2*b^2*(A*b + a*B)*x^7)/7 + (a*b^3*(A*b + 2*a*B)*x^10)/2 + (b^4*(A*b + 5*a*B)*x^13)/13 + (b^5*B*x^16)/16
Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {950, 2009}
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.
\(\displaystyle \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 950 |
\(\displaystyle \int \left (\frac {a^5 A}{x^3}+a^4 (a B+5 A b)+5 a^3 b x^3 (a B+2 A b)+10 a^2 b^2 x^6 (a B+A b)+b^4 x^{12} (5 a B+A b)+5 a b^3 x^9 (2 a B+A b)+b^5 B x^{15}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^5 A}{2 x^2}+a^4 x (a B+5 A b)+\frac {5}{4} a^3 b x^4 (a B+2 A b)+\frac {10}{7} a^2 b^2 x^7 (a B+A b)+\frac {1}{13} b^4 x^{13} (5 a B+A b)+\frac {1}{2} a b^3 x^{10} (2 a B+A b)+\frac {1}{16} b^5 B x^{16}\) |
-1/2*(a^5*A)/x^2 + a^4*(5*A*b + a*B)*x + (5*a^3*b*(2*A*b + a*B)*x^4)/4 + ( 10*a^2*b^2*(A*b + a*B)*x^7)/7 + (a*b^3*(A*b + 2*a*B)*x^10)/2 + (b^4*(A*b + 5*a*B)*x^13)/13 + (b^5*B*x^16)/16
3.1.35.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^ n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && IGt Q[p, 0] && IGtQ[q, 0]
Time = 4.19 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.07
method | result | size |
default | \(\frac {b^{5} B \,x^{16}}{16}+\frac {A \,b^{5} x^{13}}{13}+\frac {5 B a \,b^{4} x^{13}}{13}+\frac {A a \,b^{4} x^{10}}{2}+B \,a^{2} b^{3} x^{10}+\frac {10 a^{2} A \,b^{3} x^{7}}{7}+\frac {10 B \,a^{3} b^{2} x^{7}}{7}+\frac {5 a^{3} A \,b^{2} x^{4}}{2}+\frac {5 B \,a^{4} b \,x^{4}}{4}+5 a^{4} A b x +a^{5} B x -\frac {a^{5} A}{2 x^{2}}\) | \(120\) |
norman | \(\frac {-\frac {a^{5} A}{2}+\left (5 a^{4} b A +a^{5} B \right ) x^{3}+\left (\frac {5}{2} a^{3} b^{2} A +\frac {5}{4} a^{4} b B \right ) x^{6}+\left (\frac {10}{7} a^{2} b^{3} A +\frac {10}{7} a^{3} b^{2} B \right ) x^{9}+\left (\frac {1}{2} a \,b^{4} A +a^{2} b^{3} B \right ) x^{12}+\left (\frac {1}{13} b^{5} A +\frac {5}{13} a \,b^{4} B \right ) x^{15}+\frac {b^{5} B \,x^{18}}{16}}{x^{2}}\) | \(120\) |
risch | \(\frac {b^{5} B \,x^{16}}{16}+\frac {A \,b^{5} x^{13}}{13}+\frac {5 B a \,b^{4} x^{13}}{13}+\frac {A a \,b^{4} x^{10}}{2}+B \,a^{2} b^{3} x^{10}+\frac {10 a^{2} A \,b^{3} x^{7}}{7}+\frac {10 B \,a^{3} b^{2} x^{7}}{7}+\frac {5 a^{3} A \,b^{2} x^{4}}{2}+\frac {5 B \,a^{4} b \,x^{4}}{4}+5 a^{4} A b x +a^{5} B x -\frac {a^{5} A}{2 x^{2}}\) | \(120\) |
gosper | \(-\frac {-91 b^{5} B \,x^{18}-112 A \,b^{5} x^{15}-560 B a \,b^{4} x^{15}-728 a A \,b^{4} x^{12}-1456 B \,a^{2} b^{3} x^{12}-2080 a^{2} A \,b^{3} x^{9}-2080 B \,a^{3} b^{2} x^{9}-3640 a^{3} A \,b^{2} x^{6}-1820 B \,a^{4} b \,x^{6}-7280 a^{4} A b \,x^{3}-1456 B \,a^{5} x^{3}+728 a^{5} A}{1456 x^{2}}\) | \(128\) |
parallelrisch | \(\frac {91 b^{5} B \,x^{18}+112 A \,b^{5} x^{15}+560 B a \,b^{4} x^{15}+728 a A \,b^{4} x^{12}+1456 B \,a^{2} b^{3} x^{12}+2080 a^{2} A \,b^{3} x^{9}+2080 B \,a^{3} b^{2} x^{9}+3640 a^{3} A \,b^{2} x^{6}+1820 B \,a^{4} b \,x^{6}+7280 a^{4} A b \,x^{3}+1456 B \,a^{5} x^{3}-728 a^{5} A}{1456 x^{2}}\) | \(128\) |
1/16*b^5*B*x^16+1/13*A*b^5*x^13+5/13*B*a*b^4*x^13+1/2*A*a*b^4*x^10+B*a^2*b ^3*x^10+10/7*a^2*A*b^3*x^7+10/7*B*a^3*b^2*x^7+5/2*a^3*A*b^2*x^4+5/4*B*a^4* b*x^4+5*a^4*A*b*x+a^5*B*x-1/2*a^5*A/x^2
Time = 0.24 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.08 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^3} \, dx=\frac {91 \, B b^{5} x^{18} + 112 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{15} + 728 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{12} + 2080 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{9} + 1820 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{6} - 728 \, A a^{5} + 1456 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{3}}{1456 \, x^{2}} \]
1/1456*(91*B*b^5*x^18 + 112*(5*B*a*b^4 + A*b^5)*x^15 + 728*(2*B*a^2*b^3 + A*a*b^4)*x^12 + 2080*(B*a^3*b^2 + A*a^2*b^3)*x^9 + 1820*(B*a^4*b + 2*A*a^3 *b^2)*x^6 - 728*A*a^5 + 1456*(B*a^5 + 5*A*a^4*b)*x^3)/x^2
Time = 0.11 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^3} \, dx=- \frac {A a^{5}}{2 x^{2}} + \frac {B b^{5} x^{16}}{16} + x^{13} \left (\frac {A b^{5}}{13} + \frac {5 B a b^{4}}{13}\right ) + x^{10} \left (\frac {A a b^{4}}{2} + B a^{2} b^{3}\right ) + x^{7} \cdot \left (\frac {10 A a^{2} b^{3}}{7} + \frac {10 B a^{3} b^{2}}{7}\right ) + x^{4} \cdot \left (\frac {5 A a^{3} b^{2}}{2} + \frac {5 B a^{4} b}{4}\right ) + x \left (5 A a^{4} b + B a^{5}\right ) \]
-A*a**5/(2*x**2) + B*b**5*x**16/16 + x**13*(A*b**5/13 + 5*B*a*b**4/13) + x **10*(A*a*b**4/2 + B*a**2*b**3) + x**7*(10*A*a**2*b**3/7 + 10*B*a**3*b**2/ 7) + x**4*(5*A*a**3*b**2/2 + 5*B*a**4*b/4) + x*(5*A*a**4*b + B*a**5)
Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^3} \, dx=\frac {1}{16} \, B b^{5} x^{16} + \frac {1}{13} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{13} + \frac {1}{2} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{10} + \frac {10}{7} \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{7} + \frac {5}{4} \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} - \frac {A a^{5}}{2 \, x^{2}} + {\left (B a^{5} + 5 \, A a^{4} b\right )} x \]
1/16*B*b^5*x^16 + 1/13*(5*B*a*b^4 + A*b^5)*x^13 + 1/2*(2*B*a^2*b^3 + A*a*b ^4)*x^10 + 10/7*(B*a^3*b^2 + A*a^2*b^3)*x^7 + 5/4*(B*a^4*b + 2*A*a^3*b^2)* x^4 - 1/2*A*a^5/x^2 + (B*a^5 + 5*A*a^4*b)*x
Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^3} \, dx=\frac {1}{16} \, B b^{5} x^{16} + \frac {5}{13} \, B a b^{4} x^{13} + \frac {1}{13} \, A b^{5} x^{13} + B a^{2} b^{3} x^{10} + \frac {1}{2} \, A a b^{4} x^{10} + \frac {10}{7} \, B a^{3} b^{2} x^{7} + \frac {10}{7} \, A a^{2} b^{3} x^{7} + \frac {5}{4} \, B a^{4} b x^{4} + \frac {5}{2} \, A a^{3} b^{2} x^{4} + B a^{5} x + 5 \, A a^{4} b x - \frac {A a^{5}}{2 \, x^{2}} \]
1/16*B*b^5*x^16 + 5/13*B*a*b^4*x^13 + 1/13*A*b^5*x^13 + B*a^2*b^3*x^10 + 1 /2*A*a*b^4*x^10 + 10/7*B*a^3*b^2*x^7 + 10/7*A*a^2*b^3*x^7 + 5/4*B*a^4*b*x^ 4 + 5/2*A*a^3*b^2*x^4 + B*a^5*x + 5*A*a^4*b*x - 1/2*A*a^5/x^2
Time = 0.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^3} \, dx=x\,\left (B\,a^5+5\,A\,b\,a^4\right )+x^{13}\,\left (\frac {A\,b^5}{13}+\frac {5\,B\,a\,b^4}{13}\right )-\frac {A\,a^5}{2\,x^2}+\frac {B\,b^5\,x^{16}}{16}+\frac {10\,a^2\,b^2\,x^7\,\left (A\,b+B\,a\right )}{7}+\frac {5\,a^3\,b\,x^4\,\left (2\,A\,b+B\,a\right )}{4}+\frac {a\,b^3\,x^{10}\,\left (A\,b+2\,B\,a\right )}{2} \]