Integrand size = 20, antiderivative size = 113 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{16}} \, dx=-\frac {a^5 A}{15 x^{15}}-\frac {a^4 (5 A b+a B)}{12 x^{12}}-\frac {5 a^3 b (2 A b+a B)}{9 x^9}-\frac {5 a^2 b^2 (A b+a B)}{3 x^6}-\frac {5 a b^3 (A b+2 a B)}{3 x^3}+\frac {1}{3} b^5 B x^3+b^4 (A b+5 a B) \log (x) \]
-1/15*a^5*A/x^15-1/12*a^4*(5*A*b+B*a)/x^12-5/9*a^3*b*(2*A*b+B*a)/x^9-5/3*a ^2*b^2*(A*b+B*a)/x^6-5/3*a*b^3*(A*b+2*B*a)/x^3+1/3*b^5*B*x^3+b^4*(A*b+5*B* a)*ln(x)
Time = 0.06 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{16}} \, dx=-\frac {300 a A b^4 x^{12}-60 b^5 B x^{18}+300 a^2 b^3 x^9 \left (A+2 B x^3\right )+100 a^3 b^2 x^6 \left (2 A+3 B x^3\right )+25 a^4 b x^3 \left (3 A+4 B x^3\right )+3 a^5 \left (4 A+5 B x^3\right )}{180 x^{15}}+b^4 (A b+5 a B) \log (x) \]
-1/180*(300*a*A*b^4*x^12 - 60*b^5*B*x^18 + 300*a^2*b^3*x^9*(A + 2*B*x^3) + 100*a^3*b^2*x^6*(2*A + 3*B*x^3) + 25*a^4*b*x^3*(3*A + 4*B*x^3) + 3*a^5*(4 *A + 5*B*x^3))/x^15 + b^4*(A*b + 5*a*B)*Log[x]
Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {948, 85, 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.
\(\displaystyle \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{16}} \, dx\) |
\(\Big \downarrow \) 948 |
\(\displaystyle \frac {1}{3} \int \frac {\left (b x^3+a\right )^5 \left (B x^3+A\right )}{x^{18}}dx^3\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {1}{3} \int \left (\frac {A a^5}{x^{18}}+\frac {(5 A b+a B) a^4}{x^{15}}+\frac {5 b (2 A b+a B) a^3}{x^{12}}+\frac {10 b^2 (A b+a B) a^2}{x^9}+\frac {5 b^3 (A b+2 a B) a}{x^6}+b^5 B+\frac {b^4 (A b+5 a B)}{x^3}\right )dx^3\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (-\frac {a^5 A}{5 x^{15}}-\frac {a^4 (a B+5 A b)}{4 x^{12}}-\frac {5 a^3 b (a B+2 A b)}{3 x^9}-\frac {5 a^2 b^2 (a B+A b)}{x^6}+b^4 \log \left (x^3\right ) (5 a B+A b)-\frac {5 a b^3 (2 a B+A b)}{x^3}+b^5 B x^3\right )\) |
(-1/5*(a^5*A)/x^15 - (a^4*(5*A*b + a*B))/(4*x^12) - (5*a^3*b*(2*A*b + a*B) )/(3*x^9) - (5*a^2*b^2*(A*b + a*B))/x^6 - (5*a*b^3*(A*b + 2*a*B))/x^3 + b^ 5*B*x^3 + b^4*(A*b + 5*a*B)*Log[x^3])/3
3.1.48.3.1 Defintions of rubi rules used
Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 4.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {a^{5} A}{15 x^{15}}-\frac {a^{4} \left (5 A b +B a \right )}{12 x^{12}}-\frac {5 a^{3} b \left (2 A b +B a \right )}{9 x^{9}}-\frac {5 a^{2} b^{2} \left (A b +B a \right )}{3 x^{6}}-\frac {5 a \,b^{3} \left (A b +2 B a \right )}{3 x^{3}}+\frac {b^{5} B \,x^{3}}{3}+b^{4} \left (A b +5 B a \right ) \ln \left (x \right )\) | \(102\) |
norman | \(\frac {\left (-\frac {5}{3} a \,b^{4} A -\frac {10}{3} a^{2} b^{3} B \right ) x^{12}+\left (-\frac {5}{3} a^{2} b^{3} A -\frac {5}{3} a^{3} b^{2} B \right ) x^{9}+\left (-\frac {10}{9} a^{3} b^{2} A -\frac {5}{9} a^{4} b B \right ) x^{6}+\left (-\frac {5}{12} a^{4} b A -\frac {1}{12} a^{5} B \right ) x^{3}-\frac {a^{5} A}{15}+\frac {b^{5} B \,x^{18}}{3}}{x^{15}}+\left (b^{5} A +5 a \,b^{4} B \right ) \ln \left (x \right )\) | \(121\) |
risch | \(\frac {b^{5} B \,x^{3}}{3}+\frac {-\frac {a^{5} A}{15}+\left (-\frac {5}{12} a^{4} b A -\frac {1}{12} a^{5} B \right ) x^{3}+\left (-\frac {10}{9} a^{3} b^{2} A -\frac {5}{9} a^{4} b B \right ) x^{6}+\left (-\frac {5}{3} a^{2} b^{3} A -\frac {5}{3} a^{3} b^{2} B \right ) x^{9}+\left (-\frac {5}{3} a \,b^{4} A -\frac {10}{3} a^{2} b^{3} B \right ) x^{12}}{x^{15}}+A \ln \left (x \right ) b^{5}+5 B \ln \left (x \right ) a \,b^{4}\) | \(121\) |
parallelrisch | \(\frac {60 b^{5} B \,x^{18}+180 A \ln \left (x \right ) x^{15} b^{5}+900 B \ln \left (x \right ) x^{15} a \,b^{4}-300 a A \,b^{4} x^{12}-600 B \,a^{2} b^{3} x^{12}-300 a^{2} A \,b^{3} x^{9}-300 B \,a^{3} b^{2} x^{9}-200 a^{3} A \,b^{2} x^{6}-100 B \,a^{4} b \,x^{6}-75 a^{4} A b \,x^{3}-15 B \,a^{5} x^{3}-12 a^{5} A}{180 x^{15}}\) | \(132\) |
-1/15*a^5*A/x^15-1/12*a^4*(5*A*b+B*a)/x^12-5/9*a^3*b*(2*A*b+B*a)/x^9-5/3*a ^2*b^2*(A*b+B*a)/x^6-5/3*a*b^3*(A*b+2*B*a)/x^3+1/3*b^5*B*x^3+b^4*(A*b+5*B* a)*ln(x)
Time = 0.23 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{16}} \, dx=\frac {60 \, B b^{5} x^{18} + 180 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{15} \log \left (x\right ) - 300 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{12} - 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{9} - 100 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{6} - 12 \, A a^{5} - 15 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{3}}{180 \, x^{15}} \]
1/180*(60*B*b^5*x^18 + 180*(5*B*a*b^4 + A*b^5)*x^15*log(x) - 300*(2*B*a^2* b^3 + A*a*b^4)*x^12 - 300*(B*a^3*b^2 + A*a^2*b^3)*x^9 - 100*(B*a^4*b + 2*A *a^3*b^2)*x^6 - 12*A*a^5 - 15*(B*a^5 + 5*A*a^4*b)*x^3)/x^15
Timed out. \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{16}} \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{16}} \, dx=\frac {1}{3} \, B b^{5} x^{3} + \frac {1}{3} \, {\left (5 \, B a b^{4} + A b^{5}\right )} \log \left (x^{3}\right ) - \frac {300 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{12} + 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{9} + 100 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{6} + 12 \, A a^{5} + 15 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{3}}{180 \, x^{15}} \]
1/3*B*b^5*x^3 + 1/3*(5*B*a*b^4 + A*b^5)*log(x^3) - 1/180*(300*(2*B*a^2*b^3 + A*a*b^4)*x^12 + 300*(B*a^3*b^2 + A*a^2*b^3)*x^9 + 100*(B*a^4*b + 2*A*a^ 3*b^2)*x^6 + 12*A*a^5 + 15*(B*a^5 + 5*A*a^4*b)*x^3)/x^15
Time = 0.29 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.28 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{16}} \, dx=\frac {1}{3} \, B b^{5} x^{3} + {\left (5 \, B a b^{4} + A b^{5}\right )} \log \left ({\left | x \right |}\right ) - \frac {685 \, B a b^{4} x^{15} + 137 \, A b^{5} x^{15} + 600 \, B a^{2} b^{3} x^{12} + 300 \, A a b^{4} x^{12} + 300 \, B a^{3} b^{2} x^{9} + 300 \, A a^{2} b^{3} x^{9} + 100 \, B a^{4} b x^{6} + 200 \, A a^{3} b^{2} x^{6} + 15 \, B a^{5} x^{3} + 75 \, A a^{4} b x^{3} + 12 \, A a^{5}}{180 \, x^{15}} \]
1/3*B*b^5*x^3 + (5*B*a*b^4 + A*b^5)*log(abs(x)) - 1/180*(685*B*a*b^4*x^15 + 137*A*b^5*x^15 + 600*B*a^2*b^3*x^12 + 300*A*a*b^4*x^12 + 300*B*a^3*b^2*x ^9 + 300*A*a^2*b^3*x^9 + 100*B*a^4*b*x^6 + 200*A*a^3*b^2*x^6 + 15*B*a^5*x^ 3 + 75*A*a^4*b*x^3 + 12*A*a^5)/x^15
Time = 6.51 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^3\right )^5 \left (A+B x^3\right )}{x^{16}} \, dx=\ln \left (x\right )\,\left (A\,b^5+5\,B\,a\,b^4\right )-\frac {\frac {A\,a^5}{15}+x^{12}\,\left (\frac {10\,B\,a^2\,b^3}{3}+\frac {5\,A\,a\,b^4}{3}\right )+x^6\,\left (\frac {5\,B\,a^4\,b}{9}+\frac {10\,A\,a^3\,b^2}{9}\right )+x^3\,\left (\frac {B\,a^5}{12}+\frac {5\,A\,b\,a^4}{12}\right )+x^9\,\left (\frac {5\,B\,a^3\,b^2}{3}+\frac {5\,A\,a^2\,b^3}{3}\right )}{x^{15}}+\frac {B\,b^5\,x^3}{3} \]