Integrand size = 20, antiderivative size = 112 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^9} \, dx=-\frac {a^5 A}{8 x^8}-\frac {a^4 (5 A b+a B)}{6 x^6}-\frac {5 a^3 b (2 A b+a B)}{4 x^4}-\frac {5 a^2 b^2 (A b+a B)}{x^2}+\frac {1}{2} b^4 (A b+5 a B) x^2+\frac {1}{4} b^5 B x^4+5 a b^3 (A b+2 a B) \log (x) \]
-1/8*a^5*A/x^8-1/6*a^4*(5*A*b+B*a)/x^6-5/4*a^3*b*(2*A*b+B*a)/x^4-5*a^2*b^2 *(A*b+B*a)/x^2+1/2*b^4*(A*b+5*B*a)*x^2+1/4*b^5*B*x^4+5*a*b^3*(A*b+2*B*a)*l n(x)
Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^9} \, dx=-\frac {120 a^2 A b^3 x^6-60 a b^4 B x^{10}-6 b^5 x^{10} \left (2 A+B x^2\right )+60 a^3 b^2 x^4 \left (A+2 B x^2\right )+10 a^4 b x^2 \left (2 A+3 B x^2\right )+a^5 \left (3 A+4 B x^2\right )}{24 x^8}+5 a b^3 (A b+2 a B) \log (x) \]
-1/24*(120*a^2*A*b^3*x^6 - 60*a*b^4*B*x^10 - 6*b^5*x^10*(2*A + B*x^2) + 60 *a^3*b^2*x^4*(A + 2*B*x^2) + 10*a^4*b*x^2*(2*A + 3*B*x^2) + a^5*(3*A + 4*B *x^2))/x^8 + 5*a*b^3*(A*b + 2*a*B)*Log[x]
Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^9} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^5 \left (B x^2+A\right )}{x^{10}}dx^2\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {1}{2} \int \left (\frac {A a^5}{x^{10}}+\frac {(5 A b+a B) a^4}{x^8}+\frac {5 b (2 A b+a B) a^3}{x^6}+\frac {10 b^2 (A b+a B) a^2}{x^4}+\frac {5 b^3 (A b+2 a B) a}{x^2}+b^5 B x^2+b^4 (A b+5 a B)\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {a^5 A}{4 x^8}-\frac {a^4 (a B+5 A b)}{3 x^6}-\frac {5 a^3 b (a B+2 A b)}{2 x^4}-\frac {10 a^2 b^2 (a B+A b)}{x^2}+b^4 x^2 (5 a B+A b)+5 a b^3 \log \left (x^2\right ) (2 a B+A b)+\frac {1}{2} b^5 B x^4\right )\) |
(-1/4*(a^5*A)/x^8 - (a^4*(5*A*b + a*B))/(3*x^6) - (5*a^3*b*(2*A*b + a*B))/ (2*x^4) - (10*a^2*b^2*(A*b + a*B))/x^2 + b^4*(A*b + 5*a*B)*x^2 + (b^5*B*x^ 4)/2 + 5*a*b^3*(A*b + 2*a*B)*Log[x^2])/2
3.1.41.3.1 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 = 2.60 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.95
method | result | size |
default | \(\frac {b^{5} B \,x^{4}}{4}+\frac {A \,b^{5} x^{2}}{2}+\frac {5 B a \,b^{4} x^{2}}{2}+5 a \,b^{3} \left (A b +2 B a \right ) \ln \left (x \right )-\frac {a^{4} \left (5 A b +B a \right )}{6 x^{6}}-\frac {a^{5} A}{8 x^{8}}-\frac {5 a^{2} b^{2} \left (A b +B a \right )}{x^{2}}-\frac {5 a^{3} b \left (2 A b +B a \right )}{4 x^{4}}\) | \(106\) |
norman | \(\frac {\left (\frac {1}{2} b^{5} A +\frac {5}{2} a \,b^{4} B \right ) x^{10}+\left (-\frac {5}{2} a^{3} b^{2} A -\frac {5}{4} a^{4} b B \right ) x^{4}+\left (-\frac {5}{6} a^{4} b A -\frac {1}{6} a^{5} B \right ) x^{2}+\left (-5 a^{2} b^{3} A -5 a^{3} b^{2} B \right ) x^{6}-\frac {a^{5} A}{8}+\frac {b^{5} B \,x^{12}}{4}}{x^{8}}+\left (5 a \,b^{4} A +10 a^{2} b^{3} B \right ) \ln \left (x \right )\) | \(122\) |
parallelrisch | \(\frac {6 b^{5} B \,x^{12}+12 A \,b^{5} x^{10}+60 B a \,b^{4} x^{10}+120 A \ln \left (x \right ) x^{8} a \,b^{4}+240 B \ln \left (x \right ) x^{8} a^{2} b^{3}-120 a^{2} A \,b^{3} x^{6}-120 B \,a^{3} b^{2} x^{6}-60 a^{3} A \,b^{2} x^{4}-30 B \,a^{4} b \,x^{4}-20 a^{4} A b \,x^{2}-4 a^{5} B \,x^{2}-3 a^{5} A}{24 x^{8}}\) | \(132\) |
risch | \(\frac {b^{5} B \,x^{4}}{4}+\frac {A \,b^{5} x^{2}}{2}+\frac {5 B a \,b^{4} x^{2}}{2}+\frac {b^{5} A^{2}}{4 B}+\frac {5 a \,b^{4} A}{2}+\frac {25 a^{2} b^{3} B}{4}+\frac {\left (-5 a^{2} b^{3} A -5 a^{3} b^{2} B \right ) x^{6}+\left (-\frac {5}{2} a^{3} b^{2} A -\frac {5}{4} a^{4} b B \right ) x^{4}+\left (-\frac {5}{6} a^{4} b A -\frac {1}{6} a^{5} B \right ) x^{2}-\frac {a^{5} A}{8}}{x^{8}}+5 A \ln \left (x \right ) a \,b^{4}+10 B \ln \left (x \right ) a^{2} b^{3}\) | \(150\) |
1/4*b^5*B*x^4+1/2*A*b^5*x^2+5/2*B*a*b^4*x^2+5*a*b^3*(A*b+2*B*a)*ln(x)-1/6* a^4*(5*A*b+B*a)/x^6-1/8*a^5*A/x^8-5*a^2*b^2*(A*b+B*a)/x^2-5/4*a^3*b*(2*A*b +B*a)/x^4
Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^9} \, dx=\frac {6 \, B b^{5} x^{12} + 12 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{10} + 120 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{8} \log \left (x\right ) - 120 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{6} - 3 \, A a^{5} - 30 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} - 4 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2}}{24 \, x^{8}} \]
1/24*(6*B*b^5*x^12 + 12*(5*B*a*b^4 + A*b^5)*x^10 + 120*(2*B*a^2*b^3 + A*a* b^4)*x^8*log(x) - 120*(B*a^3*b^2 + A*a^2*b^3)*x^6 - 3*A*a^5 - 30*(B*a^4*b + 2*A*a^3*b^2)*x^4 - 4*(B*a^5 + 5*A*a^4*b)*x^2)/x^8
Time = 1.52 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^9} \, dx=\frac {B b^{5} x^{4}}{4} + 5 a b^{3} \left (A b + 2 B a\right ) \log {\left (x \right )} + x^{2} \left (\frac {A b^{5}}{2} + \frac {5 B a b^{4}}{2}\right ) + \frac {- 3 A a^{5} + x^{6} \left (- 120 A a^{2} b^{3} - 120 B a^{3} b^{2}\right ) + x^{4} \left (- 60 A a^{3} b^{2} - 30 B a^{4} b\right ) + x^{2} \left (- 20 A a^{4} b - 4 B a^{5}\right )}{24 x^{8}} \]
B*b**5*x**4/4 + 5*a*b**3*(A*b + 2*B*a)*log(x) + x**2*(A*b**5/2 + 5*B*a*b** 4/2) + (-3*A*a**5 + x**6*(-120*A*a**2*b**3 - 120*B*a**3*b**2) + x**4*(-60* A*a**3*b**2 - 30*B*a**4*b) + x**2*(-20*A*a**4*b - 4*B*a**5))/(24*x**8)
Time = 0.19 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.10 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^9} \, dx=\frac {1}{4} \, B b^{5} x^{4} + \frac {1}{2} \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{2} + \frac {5}{2} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} \log \left (x^{2}\right ) - \frac {120 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{6} + 3 \, A a^{5} + 30 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} + 4 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2}}{24 \, x^{8}} \]
1/4*B*b^5*x^4 + 1/2*(5*B*a*b^4 + A*b^5)*x^2 + 5/2*(2*B*a^2*b^3 + A*a*b^4)* log(x^2) - 1/24*(120*(B*a^3*b^2 + A*a^2*b^3)*x^6 + 3*A*a^5 + 30*(B*a^4*b + 2*A*a^3*b^2)*x^4 + 4*(B*a^5 + 5*A*a^4*b)*x^2)/x^8
Time = 0.29 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.34 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^9} \, dx=\frac {1}{4} \, B b^{5} x^{4} + \frac {5}{2} \, B a b^{4} x^{2} + \frac {1}{2} \, A b^{5} x^{2} + \frac {5}{2} \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} \log \left (x^{2}\right ) - \frac {250 \, B a^{2} b^{3} x^{8} + 125 \, A a b^{4} x^{8} + 120 \, B a^{3} b^{2} x^{6} + 120 \, A a^{2} b^{3} x^{6} + 30 \, B a^{4} b x^{4} + 60 \, A a^{3} b^{2} x^{4} + 4 \, B a^{5} x^{2} + 20 \, A a^{4} b x^{2} + 3 \, A a^{5}}{24 \, x^{8}} \]
1/4*B*b^5*x^4 + 5/2*B*a*b^4*x^2 + 1/2*A*b^5*x^2 + 5/2*(2*B*a^2*b^3 + A*a*b ^4)*log(x^2) - 1/24*(250*B*a^2*b^3*x^8 + 125*A*a*b^4*x^8 + 120*B*a^3*b^2*x ^6 + 120*A*a^2*b^3*x^6 + 30*B*a^4*b*x^4 + 60*A*a^3*b^2*x^4 + 4*B*a^5*x^2 + 20*A*a^4*b*x^2 + 3*A*a^5)/x^8
Time = 4.93 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.09 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^9} \, dx=\ln \left (x\right )\,\left (10\,B\,a^2\,b^3+5\,A\,a\,b^4\right )-\frac {\frac {A\,a^5}{8}+x^4\,\left (\frac {5\,B\,a^4\,b}{4}+\frac {5\,A\,a^3\,b^2}{2}\right )+x^2\,\left (\frac {B\,a^5}{6}+\frac {5\,A\,b\,a^4}{6}\right )+x^6\,\left (5\,B\,a^3\,b^2+5\,A\,a^2\,b^3\right )}{x^8}+x^2\,\left (\frac {A\,b^5}{2}+\frac {5\,B\,a\,b^4}{2}\right )+\frac {B\,b^5\,x^4}{4} \]