Integrand size = 20, antiderivative size = 117 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{21}} \, dx=-\frac {a^5 A}{20 x^{20}}-\frac {a^4 (5 A b+a B)}{17 x^{17}}-\frac {5 a^3 b (2 A b+a B)}{14 x^{14}}-\frac {10 a^2 b^2 (A b+a B)}{11 x^{11}}-\frac {5 a b^3 (A b+2 a B)}{8 x^8}-\frac {b^4 (A b+5 a B)}{5 x^5}-\frac {b^5 B}{2 x^2} \] Output:
-1/20*a^5*A/x^20-1/17*a^4*(5*A*b+B*a)/x^17-5/14*a^3*b*(2*A*b+B*a)/x^14-10/ 11*a^2*b^2*(A*b+B*a)/x^11-5/8*a*b^3*(A*b+2*B*a)/x^8-1/5*b^4*(A*b+5*B*a)/x^ 5-1/2*b^5*B/x^2
Time = 0.04 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{21}} \, dx=-\frac {5236 b^5 x^{15} \left (2 A+5 B x^3\right )+6545 a b^4 x^{12} \left (5 A+8 B x^3\right )+5950 a^2 b^3 x^9 \left (8 A+11 B x^3\right )+3400 a^3 b^2 x^6 \left (11 A+14 B x^3\right )+1100 a^4 b x^3 \left (14 A+17 B x^3\right )+154 a^5 \left (17 A+20 B x^3\right )}{52360 x^{20}} \] Input:
Integrate[((a + b*x^3)^5*(A + B*x^3))/x^21,x]
Output:
-1/52360*(5236*b^5*x^15*(2*A + 5*B*x^3) + 6545*a*b^4*x^12*(5*A + 8*B*x^3) + 5950*a^2*b^3*x^9*(8*A + 11*B*x^3) + 3400*a^3*b^2*x^6*(11*A + 14*B*x^3) + 1100*a^4*b*x^3*(14*A + 17*B*x^3) + 154*a^5*(17*A + 20*B*x^3))/x^20
Time = 0.43 (sec) , antiderivative size = 117, 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.
\(\displaystyle \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{21}} \, dx\) |
\(\Big \downarrow \) 950 |
\(\displaystyle \int \left (\frac {a^5 A}{x^{21}}+\frac {a^4 (a B+5 A b)}{x^{18}}+\frac {5 a^3 b (a B+2 A b)}{x^{15}}+\frac {10 a^2 b^2 (a B+A b)}{x^{12}}+\frac {b^4 (5 a B+A b)}{x^6}+\frac {5 a b^3 (2 a B+A b)}{x^9}+\frac {b^5 B}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^5 A}{20 x^{20}}-\frac {a^4 (a B+5 A b)}{17 x^{17}}-\frac {5 a^3 b (a B+2 A b)}{14 x^{14}}-\frac {10 a^2 b^2 (a B+A b)}{11 x^{11}}-\frac {b^4 (5 a B+A b)}{5 x^5}-\frac {5 a b^3 (2 a B+A b)}{8 x^8}-\frac {b^5 B}{2 x^2}\) |
Input:
Int[((a + b*x^3)^5*(A + B*x^3))/x^21,x]
Output:
-1/20*(a^5*A)/x^20 - (a^4*(5*A*b + a*B))/(17*x^17) - (5*a^3*b*(2*A*b + a*B ))/(14*x^14) - (10*a^2*b^2*(A*b + a*B))/(11*x^11) - (5*a*b^3*(A*b + 2*a*B) )/(8*x^8) - (b^4*(A*b + 5*a*B))/(5*x^5) - (b^5*B)/(2*x^2)
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 = 0.55 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.89
method | result | size |
default | \(-\frac {a^{5} A}{20 x^{20}}-\frac {a^{4} \left (5 A b +B a \right )}{17 x^{17}}-\frac {5 a^{3} b \left (2 A b +B a \right )}{14 x^{14}}-\frac {10 a^{2} b^{2} \left (A b +B a \right )}{11 x^{11}}-\frac {5 a \,b^{3} \left (A b +2 B a \right )}{8 x^{8}}-\frac {b^{4} \left (A b +5 B a \right )}{5 x^{5}}-\frac {b^{5} B}{2 x^{2}}\) | \(104\) |
norman | \(\frac {-\frac {a^{5} A}{20}+\left (-\frac {5}{17} a^{4} b A -\frac {1}{17} a^{5} B \right ) x^{3}+\left (-\frac {5}{7} a^{3} b^{2} A -\frac {5}{14} a^{4} b B \right ) x^{6}+\left (-\frac {10}{11} a^{2} b^{3} A -\frac {10}{11} a^{3} b^{2} B \right ) x^{9}+\left (-\frac {5}{8} a \,b^{4} A -\frac {5}{4} a^{2} b^{3} B \right ) x^{12}+\left (-\frac {1}{5} b^{5} A -a \,b^{4} B \right ) x^{15}-\frac {b^{5} B \,x^{18}}{2}}{x^{20}}\) | \(122\) |
risch | \(\frac {-\frac {a^{5} A}{20}+\left (-\frac {5}{17} a^{4} b A -\frac {1}{17} a^{5} B \right ) x^{3}+\left (-\frac {5}{7} a^{3} b^{2} A -\frac {5}{14} a^{4} b B \right ) x^{6}+\left (-\frac {10}{11} a^{2} b^{3} A -\frac {10}{11} a^{3} b^{2} B \right ) x^{9}+\left (-\frac {5}{8} a \,b^{4} A -\frac {5}{4} a^{2} b^{3} B \right ) x^{12}+\left (-\frac {1}{5} b^{5} A -a \,b^{4} B \right ) x^{15}-\frac {b^{5} B \,x^{18}}{2}}{x^{20}}\) | \(122\) |
gosper | \(-\frac {26180 b^{5} B \,x^{18}+10472 A \,b^{5} x^{15}+52360 B a \,b^{4} x^{15}+32725 a A \,b^{4} x^{12}+65450 B \,a^{2} b^{3} x^{12}+47600 a^{2} A \,b^{3} x^{9}+47600 B \,a^{3} b^{2} x^{9}+37400 a^{3} A \,b^{2} x^{6}+18700 B \,a^{4} b \,x^{6}+15400 a^{4} A b \,x^{3}+3080 B \,a^{5} x^{3}+2618 a^{5} A}{52360 x^{20}}\) | \(128\) |
parallelrisch | \(-\frac {26180 b^{5} B \,x^{18}+10472 A \,b^{5} x^{15}+52360 B a \,b^{4} x^{15}+32725 a A \,b^{4} x^{12}+65450 B \,a^{2} b^{3} x^{12}+47600 a^{2} A \,b^{3} x^{9}+47600 B \,a^{3} b^{2} x^{9}+37400 a^{3} A \,b^{2} x^{6}+18700 B \,a^{4} b \,x^{6}+15400 a^{4} A b \,x^{3}+3080 B \,a^{5} x^{3}+2618 a^{5} A}{52360 x^{20}}\) | \(128\) |
orering | \(-\frac {26180 b^{5} B \,x^{18}+10472 A \,b^{5} x^{15}+52360 B a \,b^{4} x^{15}+32725 a A \,b^{4} x^{12}+65450 B \,a^{2} b^{3} x^{12}+47600 a^{2} A \,b^{3} x^{9}+47600 B \,a^{3} b^{2} x^{9}+37400 a^{3} A \,b^{2} x^{6}+18700 B \,a^{4} b \,x^{6}+15400 a^{4} A b \,x^{3}+3080 B \,a^{5} x^{3}+2618 a^{5} A}{52360 x^{20}}\) | \(128\) |
Input:
int((b*x^3+a)^5*(B*x^3+A)/x^21,x,method=_RETURNVERBOSE)
Output:
-1/20*a^5*A/x^20-1/17*a^4*(5*A*b+B*a)/x^17-5/14*a^3*b*(2*A*b+B*a)/x^14-10/ 11*a^2*b^2*(A*b+B*a)/x^11-5/8*a*b^3*(A*b+2*B*a)/x^8-1/5*b^4*(A*b+5*B*a)/x^ 5-1/2*b^5*B/x^2
Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{21}} \, dx=-\frac {26180 \, B b^{5} x^{18} + 10472 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{15} + 32725 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{12} + 47600 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{9} + 18700 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{6} + 2618 \, A a^{5} + 3080 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{3}}{52360 \, x^{20}} \] Input:
integrate((b*x^3+a)^5*(B*x^3+A)/x^21,x, algorithm="fricas")
Output:
-1/52360*(26180*B*b^5*x^18 + 10472*(5*B*a*b^4 + A*b^5)*x^15 + 32725*(2*B*a ^2*b^3 + A*a*b^4)*x^12 + 47600*(B*a^3*b^2 + A*a^2*b^3)*x^9 + 18700*(B*a^4* b + 2*A*a^3*b^2)*x^6 + 2618*A*a^5 + 3080*(B*a^5 + 5*A*a^4*b)*x^3)/x^20
Timed out. \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{21}} \, dx=\text {Timed out} \] Input:
integrate((b*x**3+a)**5*(B*x**3+A)/x**21,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{21}} \, dx=-\frac {26180 \, B b^{5} x^{18} + 10472 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{15} + 32725 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{12} + 47600 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{9} + 18700 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{6} + 2618 \, A a^{5} + 3080 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{3}}{52360 \, x^{20}} \] Input:
integrate((b*x^3+a)^5*(B*x^3+A)/x^21,x, algorithm="maxima")
Output:
-1/52360*(26180*B*b^5*x^18 + 10472*(5*B*a*b^4 + A*b^5)*x^15 + 32725*(2*B*a ^2*b^3 + A*a*b^4)*x^12 + 47600*(B*a^3*b^2 + A*a^2*b^3)*x^9 + 18700*(B*a^4* b + 2*A*a^3*b^2)*x^6 + 2618*A*a^5 + 3080*(B*a^5 + 5*A*a^4*b)*x^3)/x^20
Time = 0.12 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{21}} \, dx=-\frac {26180 \, B b^{5} x^{18} + 52360 \, B a b^{4} x^{15} + 10472 \, A b^{5} x^{15} + 65450 \, B a^{2} b^{3} x^{12} + 32725 \, A a b^{4} x^{12} + 47600 \, B a^{3} b^{2} x^{9} + 47600 \, A a^{2} b^{3} x^{9} + 18700 \, B a^{4} b x^{6} + 37400 \, A a^{3} b^{2} x^{6} + 3080 \, B a^{5} x^{3} + 15400 \, A a^{4} b x^{3} + 2618 \, A a^{5}}{52360 \, x^{20}} \] Input:
integrate((b*x^3+a)^5*(B*x^3+A)/x^21,x, algorithm="giac")
Output:
-1/52360*(26180*B*b^5*x^18 + 52360*B*a*b^4*x^15 + 10472*A*b^5*x^15 + 65450 *B*a^2*b^3*x^12 + 32725*A*a*b^4*x^12 + 47600*B*a^3*b^2*x^9 + 47600*A*a^2*b ^3*x^9 + 18700*B*a^4*b*x^6 + 37400*A*a^3*b^2*x^6 + 3080*B*a^5*x^3 + 15400* A*a^4*b*x^3 + 2618*A*a^5)/x^20
Time = 0.73 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{21}} \, dx=-\frac {\frac {A\,a^5}{20}+x^{12}\,\left (\frac {5\,B\,a^2\,b^3}{4}+\frac {5\,A\,a\,b^4}{8}\right )+x^6\,\left (\frac {5\,B\,a^4\,b}{14}+\frac {5\,A\,a^3\,b^2}{7}\right )+x^3\,\left (\frac {B\,a^5}{17}+\frac {5\,A\,b\,a^4}{17}\right )+x^{15}\,\left (\frac {A\,b^5}{5}+B\,a\,b^4\right )+x^9\,\left (\frac {10\,B\,a^3\,b^2}{11}+\frac {10\,A\,a^2\,b^3}{11}\right )+\frac {B\,b^5\,x^{18}}{2}}{x^{20}} \] Input:
int(((A + B*x^3)*(a + b*x^3)^5)/x^21,x)
Output:
-((A*a^5)/20 + x^12*((5*B*a^2*b^3)/4 + (5*A*a*b^4)/8) + x^6*((5*A*a^3*b^2) /7 + (5*B*a^4*b)/14) + x^3*((B*a^5)/17 + (5*A*a^4*b)/17) + x^15*((A*b^5)/5 + B*a*b^4) + x^9*((10*A*a^2*b^3)/11 + (10*B*a^3*b^2)/11) + (B*b^5*x^18)/2 )/x^20
Time = 0.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.60 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{21}} \, dx=\frac {-26180 b^{6} x^{18}-62832 a \,b^{5} x^{15}-98175 a^{2} b^{4} x^{12}-95200 a^{3} b^{3} x^{9}-56100 a^{4} b^{2} x^{6}-18480 a^{5} b \,x^{3}-2618 a^{6}}{52360 x^{20}} \] Input:
int((b*x^3+a)^5*(B*x^3+A)/x^21,x)
Output:
( - 2618*a**6 - 18480*a**5*b*x**3 - 56100*a**4*b**2*x**6 - 95200*a**3*b**3 *x**9 - 98175*a**2*b**4*x**12 - 62832*a*b**5*x**15 - 26180*b**6*x**18)/(52 360*x**20)