Integrand size = 20, antiderivative size = 113 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{11}} \, dx=-\frac {a^5 A}{10 x^{10}}-\frac {a^4 (5 A b+a B)}{8 x^8}-\frac {5 a^3 b (2 A b+a B)}{6 x^6}-\frac {5 a^2 b^2 (A b+a B)}{2 x^4}-\frac {5 a b^3 (A b+2 a B)}{2 x^2}+\frac {1}{2} b^5 B x^2+b^4 (A b+5 a B) \log (x) \]
-1/10*a^5*A/x^10-1/8*a^4*(5*A*b+B*a)/x^8-5/6*a^3*b*(2*A*b+B*a)/x^6-5/2*a^2 *b^2*(A*b+B*a)/x^4-5/2*a*b^3*(A*b+2*B*a)/x^2+1/2*b^5*B*x^2+b^4*(A*b+5*B*a) *ln(x)
Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{11}} \, dx=-\frac {300 a A b^4 x^8-60 b^5 B x^{12}+300 a^2 b^3 x^6 \left (A+2 B x^2\right )+100 a^3 b^2 x^4 \left (2 A+3 B x^2\right )+25 a^4 b x^2 \left (3 A+4 B x^2\right )+3 a^5 \left (4 A+5 B x^2\right )}{120 x^{10}}+b^4 (A b+5 a B) \log (x) \]
-1/120*(300*a*A*b^4*x^8 - 60*b^5*B*x^12 + 300*a^2*b^3*x^6*(A + 2*B*x^2) + 100*a^3*b^2*x^4*(2*A + 3*B*x^2) + 25*a^4*b*x^2*(3*A + 4*B*x^2) + 3*a^5*(4* A + 5*B*x^2))/x^10 + b^4*(A*b + 5*a*B)*Log[x]
Time = 0.26 (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 = {354, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{11}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^5 \left (B x^2+A\right )}{x^{12}}dx^2\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {1}{2} \int \left (\frac {A a^5}{x^{12}}+\frac {(5 A b+a B) a^4}{x^{10}}+\frac {5 b (2 A b+a B) a^3}{x^8}+\frac {10 b^2 (A b+a B) a^2}{x^6}+\frac {5 b^3 (A b+2 a B) a}{x^4}+b^5 B+\frac {b^4 (A b+5 a B)}{x^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {a^5 A}{5 x^{10}}-\frac {a^4 (a B+5 A b)}{4 x^8}-\frac {5 a^3 b (a B+2 A b)}{3 x^6}-\frac {5 a^2 b^2 (a B+A b)}{x^4}+b^4 \log \left (x^2\right ) (5 a B+A b)-\frac {5 a b^3 (2 a B+A b)}{x^2}+b^5 B x^2\right )\) |
(-1/5*(a^5*A)/x^10 - (a^4*(5*A*b + a*B))/(4*x^8) - (5*a^3*b*(2*A*b + a*B)) /(3*x^6) - (5*a^2*b^2*(A*b + a*B))/x^4 - (5*a*b^3*(A*b + 2*a*B))/x^2 + b^5 *B*x^2 + b^4*(A*b + 5*a*B)*Log[x^2])/2
3.1.43.3.1 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_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 2.58 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {a^{5} A}{10 x^{10}}-\frac {a^{4} \left (5 A b +B a \right )}{8 x^{8}}-\frac {5 a^{3} b \left (2 A b +B a \right )}{6 x^{6}}-\frac {5 a^{2} b^{2} \left (A b +B a \right )}{2 x^{4}}-\frac {5 a \,b^{3} \left (A b +2 B a \right )}{2 x^{2}}+\frac {b^{5} B \,x^{2}}{2}+b^{4} \left (A b +5 B a \right ) \ln \left (x \right )\) | \(102\) |
norman | \(\frac {\left (-\frac {5}{2} a \,b^{4} A -5 a^{2} b^{3} B \right ) x^{8}+\left (-\frac {5}{2} a^{2} b^{3} A -\frac {5}{2} a^{3} b^{2} B \right ) x^{6}+\left (-\frac {5}{3} a^{3} b^{2} A -\frac {5}{6} a^{4} b B \right ) x^{4}+\left (-\frac {5}{8} a^{4} b A -\frac {1}{8} a^{5} B \right ) x^{2}-\frac {a^{5} A}{10}+\frac {b^{5} B \,x^{12}}{2}}{x^{10}}+\left (b^{5} A +5 a \,b^{4} B \right ) \ln \left (x \right )\) | \(121\) |
risch | \(\frac {b^{5} B \,x^{2}}{2}+\frac {\left (-\frac {5}{2} a \,b^{4} A -5 a^{2} b^{3} B \right ) x^{8}+\left (-\frac {5}{2} a^{2} b^{3} A -\frac {5}{2} a^{3} b^{2} B \right ) x^{6}+\left (-\frac {5}{3} a^{3} b^{2} A -\frac {5}{6} a^{4} b B \right ) x^{4}+\left (-\frac {5}{8} a^{4} b A -\frac {1}{8} a^{5} B \right ) x^{2}-\frac {a^{5} A}{10}}{x^{10}}+A \ln \left (x \right ) b^{5}+5 B \ln \left (x \right ) a \,b^{4}\) | \(121\) |
parallelrisch | \(\frac {60 b^{5} B \,x^{12}+120 A \ln \left (x \right ) x^{10} b^{5}+600 B \ln \left (x \right ) x^{10} a \,b^{4}-300 a A \,b^{4} x^{8}-600 B \,a^{2} b^{3} x^{8}-300 a^{2} A \,b^{3} x^{6}-300 B \,a^{3} b^{2} x^{6}-200 a^{3} A \,b^{2} x^{4}-100 B \,a^{4} b \,x^{4}-75 a^{4} A b \,x^{2}-15 a^{5} B \,x^{2}-12 a^{5} A}{120 x^{10}}\) | \(132\) |
-1/10*a^5*A/x^10-1/8*a^4*(5*A*b+B*a)/x^8-5/6*a^3*b*(2*A*b+B*a)/x^6-5/2*a^2 *b^2*(A*b+B*a)/x^4-5/2*a*b^3*(A*b+2*B*a)/x^2+1/2*b^5*B*x^2+b^4*(A*b+5*B*a) *ln(x)
Time = 0.30 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{11}} \, dx=\frac {60 \, B b^{5} x^{12} + 120 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{10} \log \left (x\right ) - 300 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{8} - 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{6} - 12 \, A a^{5} - 100 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} - 15 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2}}{120 \, x^{10}} \]
1/120*(60*B*b^5*x^12 + 120*(5*B*a*b^4 + A*b^5)*x^10*log(x) - 300*(2*B*a^2* b^3 + A*a*b^4)*x^8 - 300*(B*a^3*b^2 + A*a^2*b^3)*x^6 - 12*A*a^5 - 100*(B*a ^4*b + 2*A*a^3*b^2)*x^4 - 15*(B*a^5 + 5*A*a^4*b)*x^2)/x^10
Time = 5.02 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{11}} \, dx=\frac {B b^{5} x^{2}}{2} + b^{4} \left (A b + 5 B a\right ) \log {\left (x \right )} + \frac {- 12 A a^{5} + x^{8} \left (- 300 A a b^{4} - 600 B a^{2} b^{3}\right ) + x^{6} \left (- 300 A a^{2} b^{3} - 300 B a^{3} b^{2}\right ) + x^{4} \left (- 200 A a^{3} b^{2} - 100 B a^{4} b\right ) + x^{2} \left (- 75 A a^{4} b - 15 B a^{5}\right )}{120 x^{10}} \]
B*b**5*x**2/2 + b**4*(A*b + 5*B*a)*log(x) + (-12*A*a**5 + x**8*(-300*A*a*b **4 - 600*B*a**2*b**3) + x**6*(-300*A*a**2*b**3 - 300*B*a**3*b**2) + x**4* (-200*A*a**3*b**2 - 100*B*a**4*b) + x**2*(-75*A*a**4*b - 15*B*a**5))/(120* x**10)
Time = 0.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{11}} \, dx=\frac {1}{2} \, B b^{5} x^{2} + \frac {1}{2} \, {\left (5 \, B a b^{4} + A b^{5}\right )} \log \left (x^{2}\right ) - \frac {300 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{8} + 300 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{6} + 12 \, A a^{5} + 100 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} + 15 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2}}{120 \, x^{10}} \]
1/2*B*b^5*x^2 + 1/2*(5*B*a*b^4 + A*b^5)*log(x^2) - 1/120*(300*(2*B*a^2*b^3 + A*a*b^4)*x^8 + 300*(B*a^3*b^2 + A*a^2*b^3)*x^6 + 12*A*a^5 + 100*(B*a^4* b + 2*A*a^3*b^2)*x^4 + 15*(B*a^5 + 5*A*a^4*b)*x^2)/x^10
Time = 0.30 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{11}} \, dx=\frac {1}{2} \, B b^{5} x^{2} + \frac {1}{2} \, {\left (5 \, B a b^{4} + A b^{5}\right )} \log \left (x^{2}\right ) - \frac {685 \, B a b^{4} x^{10} + 137 \, A b^{5} x^{10} + 600 \, B a^{2} b^{3} x^{8} + 300 \, A a b^{4} x^{8} + 300 \, B a^{3} b^{2} x^{6} + 300 \, A a^{2} b^{3} x^{6} + 100 \, B a^{4} b x^{4} + 200 \, A a^{3} b^{2} x^{4} + 15 \, B a^{5} x^{2} + 75 \, A a^{4} b x^{2} + 12 \, A a^{5}}{120 \, x^{10}} \]
1/2*B*b^5*x^2 + 1/2*(5*B*a*b^4 + A*b^5)*log(x^2) - 1/120*(685*B*a*b^4*x^10 + 137*A*b^5*x^10 + 600*B*a^2*b^3*x^8 + 300*A*a*b^4*x^8 + 300*B*a^3*b^2*x^ 6 + 300*A*a^2*b^3*x^6 + 100*B*a^4*b*x^4 + 200*A*a^3*b^2*x^4 + 15*B*a^5*x^2 + 75*A*a^4*b*x^2 + 12*A*a^5)/x^10
Time = 4.95 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.07 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{11}} \, dx=\ln \left (x\right )\,\left (A\,b^5+5\,B\,a\,b^4\right )-\frac {\frac {A\,a^5}{10}+x^8\,\left (5\,B\,a^2\,b^3+\frac {5\,A\,a\,b^4}{2}\right )+x^4\,\left (\frac {5\,B\,a^4\,b}{6}+\frac {5\,A\,a^3\,b^2}{3}\right )+x^2\,\left (\frac {B\,a^5}{8}+\frac {5\,A\,b\,a^4}{8}\right )+x^6\,\left (\frac {5\,B\,a^3\,b^2}{2}+\frac {5\,A\,a^2\,b^3}{2}\right )}{x^{10}}+\frac {B\,b^5\,x^2}{2} \]