Integrand size = 20, antiderivative size = 91 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{13}} \, dx=-\frac {a^5 B}{10 x^{10}}-\frac {5 a^4 b B}{8 x^8}-\frac {5 a^3 b^2 B}{3 x^6}-\frac {5 a^2 b^3 B}{2 x^4}-\frac {5 a b^4 B}{2 x^2}-\frac {A \left (a+b x^2\right )^6}{12 a x^{12}}+b^5 B \log (x) \]
-1/10*a^5*B/x^10-5/8*a^4*b*B/x^8-5/3*a^3*b^2*B/x^6-5/2*a^2*b^3*B/x^4-5/2*a *b^4*B/x^2-1/12*A*(b*x^2+a)^6/a/x^12+b^5*B*ln(x)
Time = 0.04 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.30 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{13}} \, dx=-\frac {60 A b^5 x^{10}+150 a b^4 x^8 \left (A+2 B x^2\right )+100 a^2 b^3 x^6 \left (2 A+3 B x^2\right )+50 a^3 b^2 x^4 \left (3 A+4 B x^2\right )+15 a^4 b x^2 \left (4 A+5 B x^2\right )+2 a^5 \left (5 A+6 B x^2\right )}{120 x^{12}}+b^5 B \log (x) \]
-1/120*(60*A*b^5*x^10 + 150*a*b^4*x^8*(A + 2*B*x^2) + 100*a^2*b^3*x^6*(2*A + 3*B*x^2) + 50*a^3*b^2*x^4*(3*A + 4*B*x^2) + 15*a^4*b*x^2*(4*A + 5*B*x^2 ) + 2*a^5*(5*A + 6*B*x^2))/x^12 + b^5*B*Log[x]
Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {354, 87, 49, 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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{13}} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^2+a\right )^5 \left (B x^2+A\right )}{x^{14}}dx^2\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (B \int \frac {\left (b x^2+a\right )^5}{x^{12}}dx^2-\frac {A \left (a+b x^2\right )^6}{6 a x^{12}}\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} \left (B \int \left (\frac {a^5}{x^{12}}+\frac {5 b a^4}{x^{10}}+\frac {10 b^2 a^3}{x^8}+\frac {10 b^3 a^2}{x^6}+\frac {5 b^4 a}{x^4}+\frac {b^5}{x^2}\right )dx^2-\frac {A \left (a+b x^2\right )^6}{6 a x^{12}}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (B \left (-\frac {a^5}{5 x^{10}}-\frac {5 a^4 b}{4 x^8}-\frac {10 a^3 b^2}{3 x^6}-\frac {5 a^2 b^3}{x^4}-\frac {5 a b^4}{x^2}+b^5 \log \left (x^2\right )\right )-\frac {A \left (a+b x^2\right )^6}{6 a x^{12}}\right )\) |
(-1/6*(A*(a + b*x^2)^6)/(a*x^12) + B*(-1/5*a^5/x^10 - (5*a^4*b)/(4*x^8) - (10*a^3*b^2)/(3*x^6) - (5*a^2*b^3)/x^4 - (5*a*b^4)/x^2 + b^5*Log[x^2]))/2
3.1.45.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 2.57 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.12
method | result | size |
default | \(b^{5} B \ln \left (x \right )-\frac {5 a^{2} b^{2} \left (A b +B a \right )}{3 x^{6}}-\frac {a^{5} A}{12 x^{12}}-\frac {5 a^{3} b \left (2 A b +B a \right )}{8 x^{8}}-\frac {a^{4} \left (5 A b +B a \right )}{10 x^{10}}-\frac {b^{4} \left (A b +5 B a \right )}{2 x^{2}}-\frac {5 a \,b^{3} \left (A b +2 B a \right )}{4 x^{4}}\) | \(102\) |
norman | \(\frac {\left (-\frac {1}{2} b^{5} A -\frac {5}{2} a \,b^{4} B \right ) x^{10}+\left (-\frac {5}{4} a \,b^{4} A -\frac {5}{2} a^{2} b^{3} B \right ) x^{8}+\left (-\frac {5}{3} a^{2} b^{3} A -\frac {5}{3} a^{3} b^{2} B \right ) x^{6}+\left (-\frac {5}{4} a^{3} b^{2} A -\frac {5}{8} a^{4} b B \right ) x^{4}+\left (-\frac {1}{2} a^{4} b A -\frac {1}{10} a^{5} B \right ) x^{2}-\frac {a^{5} A}{12}}{x^{12}}+b^{5} B \ln \left (x \right )\) | \(121\) |
risch | \(\frac {\left (-\frac {1}{2} b^{5} A -\frac {5}{2} a \,b^{4} B \right ) x^{10}+\left (-\frac {5}{4} a \,b^{4} A -\frac {5}{2} a^{2} b^{3} B \right ) x^{8}+\left (-\frac {5}{3} a^{2} b^{3} A -\frac {5}{3} a^{3} b^{2} B \right ) x^{6}+\left (-\frac {5}{4} a^{3} b^{2} A -\frac {5}{8} a^{4} b B \right ) x^{4}+\left (-\frac {1}{2} a^{4} b A -\frac {1}{10} a^{5} B \right ) x^{2}-\frac {a^{5} A}{12}}{x^{12}}+b^{5} B \ln \left (x \right )\) | \(121\) |
parallelrisch | \(-\frac {-120 b^{5} B \ln \left (x \right ) x^{12}+60 A \,b^{5} x^{10}+300 B a \,b^{4} x^{10}+150 a A \,b^{4} x^{8}+300 B \,a^{2} b^{3} x^{8}+200 a^{2} A \,b^{3} x^{6}+200 B \,a^{3} b^{2} x^{6}+150 a^{3} A \,b^{2} x^{4}+75 B \,a^{4} b \,x^{4}+60 a^{4} A b \,x^{2}+12 a^{5} B \,x^{2}+10 a^{5} A}{120 x^{12}}\) | \(130\) |
b^5*B*ln(x)-5/3*a^2*b^2*(A*b+B*a)/x^6-1/12*a^5*A/x^12-5/8*a^3*b*(2*A*b+B*a )/x^8-1/10*a^4*(5*A*b+B*a)/x^10-1/2*b^4*(A*b+5*B*a)/x^2-5/4*a*b^3*(A*b+2*B *a)/x^4
Time = 0.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{13}} \, dx=\frac {120 \, B b^{5} x^{12} \log \left (x\right ) - 60 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{10} - 150 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{8} - 200 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{6} - 10 \, A a^{5} - 75 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} - 12 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2}}{120 \, x^{12}} \]
1/120*(120*B*b^5*x^12*log(x) - 60*(5*B*a*b^4 + A*b^5)*x^10 - 150*(2*B*a^2* b^3 + A*a*b^4)*x^8 - 200*(B*a^3*b^2 + A*a^2*b^3)*x^6 - 10*A*a^5 - 75*(B*a^ 4*b + 2*A*a^3*b^2)*x^4 - 12*(B*a^5 + 5*A*a^4*b)*x^2)/x^12
Time = 27.10 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.46 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{13}} \, dx=B b^{5} \log {\left (x \right )} + \frac {- 10 A a^{5} + x^{10} \left (- 60 A b^{5} - 300 B a b^{4}\right ) + x^{8} \left (- 150 A a b^{4} - 300 B a^{2} b^{3}\right ) + x^{6} \left (- 200 A a^{2} b^{3} - 200 B a^{3} b^{2}\right ) + x^{4} \left (- 150 A a^{3} b^{2} - 75 B a^{4} b\right ) + x^{2} \left (- 60 A a^{4} b - 12 B a^{5}\right )}{120 x^{12}} \]
B*b**5*log(x) + (-10*A*a**5 + x**10*(-60*A*b**5 - 300*B*a*b**4) + x**8*(-1 50*A*a*b**4 - 300*B*a**2*b**3) + x**6*(-200*A*a**2*b**3 - 200*B*a**3*b**2) + x**4*(-150*A*a**3*b**2 - 75*B*a**4*b) + x**2*(-60*A*a**4*b - 12*B*a**5) )/(120*x**12)
Time = 0.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.35 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{13}} \, dx=\frac {1}{2} \, B b^{5} \log \left (x^{2}\right ) - \frac {60 \, {\left (5 \, B a b^{4} + A b^{5}\right )} x^{10} + 150 \, {\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} x^{8} + 200 \, {\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} x^{6} + 10 \, A a^{5} + 75 \, {\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} x^{4} + 12 \, {\left (B a^{5} + 5 \, A a^{4} b\right )} x^{2}}{120 \, x^{12}} \]
1/2*B*b^5*log(x^2) - 1/120*(60*(5*B*a*b^4 + A*b^5)*x^10 + 150*(2*B*a^2*b^3 + A*a*b^4)*x^8 + 200*(B*a^3*b^2 + A*a^2*b^3)*x^6 + 10*A*a^5 + 75*(B*a^4*b + 2*A*a^3*b^2)*x^4 + 12*(B*a^5 + 5*A*a^4*b)*x^2)/x^12
Time = 0.30 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{13}} \, dx=\frac {1}{2} \, B b^{5} \log \left (x^{2}\right ) - \frac {147 \, B b^{5} x^{12} + 300 \, B a b^{4} x^{10} + 60 \, A b^{5} x^{10} + 300 \, B a^{2} b^{3} x^{8} + 150 \, A a b^{4} x^{8} + 200 \, B a^{3} b^{2} x^{6} + 200 \, A a^{2} b^{3} x^{6} + 75 \, B a^{4} b x^{4} + 150 \, A a^{3} b^{2} x^{4} + 12 \, B a^{5} x^{2} + 60 \, A a^{4} b x^{2} + 10 \, A a^{5}}{120 \, x^{12}} \]
1/2*B*b^5*log(x^2) - 1/120*(147*B*b^5*x^12 + 300*B*a*b^4*x^10 + 60*A*b^5*x ^10 + 300*B*a^2*b^3*x^8 + 150*A*a*b^4*x^8 + 200*B*a^3*b^2*x^6 + 200*A*a^2* b^3*x^6 + 75*B*a^4*b*x^4 + 150*A*a^3*b^2*x^4 + 12*B*a^5*x^2 + 60*A*a^4*b*x ^2 + 10*A*a^5)/x^12
Time = 0.10 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.33 \[ \int \frac {\left (a+b x^2\right )^5 \left (A+B x^2\right )}{x^{13}} \, dx=B\,b^5\,\ln \left (x\right )-\frac {\frac {A\,a^5}{12}+x^8\,\left (\frac {5\,B\,a^2\,b^3}{2}+\frac {5\,A\,a\,b^4}{4}\right )+x^4\,\left (\frac {5\,B\,a^4\,b}{8}+\frac {5\,A\,a^3\,b^2}{4}\right )+x^2\,\left (\frac {B\,a^5}{10}+\frac {A\,b\,a^4}{2}\right )+x^{10}\,\left (\frac {A\,b^5}{2}+\frac {5\,B\,a\,b^4}{2}\right )+x^6\,\left (\frac {5\,B\,a^3\,b^2}{3}+\frac {5\,A\,a^2\,b^3}{3}\right )}{x^{12}} \]