Integrand size = 20, antiderivative size = 126 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {a^2 (3 A b-4 a B) x^2}{2 b^5}-\frac {a (2 A b-3 a B) x^4}{4 b^4}+\frac {(A b-2 a B) x^6}{6 b^3}+\frac {B x^8}{8 b^2}-\frac {a^4 (A b-a B)}{2 b^6 \left (a+b x^2\right )}-\frac {a^3 (4 A b-5 a B) \log \left (a+b x^2\right )}{2 b^6} \]
1/2*a^2*(3*A*b-4*B*a)*x^2/b^5-1/4*a*(2*A*b-3*B*a)*x^4/b^4+1/6*(A*b-2*B*a)* x^6/b^3+1/8*B*x^8/b^2-1/2*a^4*(A*b-B*a)/b^6/(b*x^2+a)-1/2*a^3*(4*A*b-5*B*a )*ln(b*x^2+a)/b^6
Time = 0.05 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {-12 a^2 b (-3 A b+4 a B) x^2+6 a b^2 (-2 A b+3 a B) x^4+4 b^3 (A b-2 a B) x^6+3 b^4 B x^8+\frac {12 a^4 (-A b+a B)}{a+b x^2}+12 a^3 (-4 A b+5 a B) \log \left (a+b x^2\right )}{24 b^6} \]
(-12*a^2*b*(-3*A*b + 4*a*B)*x^2 + 6*a*b^2*(-2*A*b + 3*a*B)*x^4 + 4*b^3*(A* b - 2*a*B)*x^6 + 3*b^4*B*x^8 + (12*a^4*(-(A*b) + a*B))/(a + b*x^2) + 12*a^ 3*(-4*A*b + 5*a*B)*Log[a + b*x^2])/(24*b^6)
Time = 0.31 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {354, 86, 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.
\(\displaystyle \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^8 \left (B x^2+A\right )}{\left (b x^2+a\right )^2}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {B x^6}{b^2}+\frac {(A b-2 a B) x^4}{b^3}+\frac {a (3 a B-2 A b) x^2}{b^4}-\frac {a^2 (4 a B-3 A b)}{b^5}+\frac {a^3 (5 a B-4 A b)}{b^5 \left (b x^2+a\right )}-\frac {a^4 (a B-A b)}{b^5 \left (b x^2+a\right )^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {a^4 (A b-a B)}{b^6 \left (a+b x^2\right )}-\frac {a^3 (4 A b-5 a B) \log \left (a+b x^2\right )}{b^6}+\frac {a^2 x^2 (3 A b-4 a B)}{b^5}-\frac {a x^4 (2 A b-3 a B)}{2 b^4}+\frac {x^6 (A b-2 a B)}{3 b^3}+\frac {B x^8}{4 b^2}\right )\) |
((a^2*(3*A*b - 4*a*B)*x^2)/b^5 - (a*(2*A*b - 3*a*B)*x^4)/(2*b^4) + ((A*b - 2*a*B)*x^6)/(3*b^3) + (B*x^8)/(4*b^2) - (a^4*(A*b - a*B))/(b^6*(a + b*x^2 )) - (a^3*(4*A*b - 5*a*B)*Log[a + b*x^2])/b^6)/2
3.1.71.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 2.49 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.97
method | result | size |
norman | \(\frac {\frac {B \,x^{10}}{8 b}-\frac {a \left (4 A \,a^{3} b -5 B \,a^{4}\right )}{2 b^{6}}+\frac {\left (4 A b -5 B a \right ) x^{8}}{24 b^{2}}-\frac {a \left (4 A b -5 B a \right ) x^{6}}{12 b^{3}}+\frac {a^{2} \left (4 A b -5 B a \right ) x^{4}}{4 b^{4}}}{b \,x^{2}+a}-\frac {a^{3} \left (4 A b -5 B a \right ) \ln \left (b \,x^{2}+a \right )}{2 b^{6}}\) | \(122\) |
default | \(\frac {\frac {b^{3} B \,x^{8}}{8}+\frac {\left (b^{3} A -2 a \,b^{2} B \right ) x^{6}}{6}+\frac {\left (-2 a \,b^{2} A +3 a^{2} b B \right ) x^{4}}{4}+\frac {\left (3 a^{2} b A -4 a^{3} B \right ) x^{2}}{2}}{b^{5}}-\frac {a^{3} \left (\frac {\left (4 A b -5 B a \right ) \ln \left (b \,x^{2}+a \right )}{b}+\frac {a \left (A b -B a \right )}{b \left (b \,x^{2}+a \right )}\right )}{2 b^{5}}\) | \(125\) |
risch | \(\frac {B \,x^{8}}{8 b^{2}}+\frac {x^{6} A}{6 b^{2}}-\frac {x^{6} a B}{3 b^{3}}-\frac {A a \,x^{4}}{2 b^{3}}+\frac {3 B \,a^{2} x^{4}}{4 b^{4}}+\frac {3 A \,a^{2} x^{2}}{2 b^{4}}-\frac {2 B \,a^{3} x^{2}}{b^{5}}-\frac {a^{4} A}{2 b^{5} \left (b \,x^{2}+a \right )}+\frac {a^{5} B}{2 b^{6} \left (b \,x^{2}+a \right )}-\frac {2 a^{3} \ln \left (b \,x^{2}+a \right ) A}{b^{5}}+\frac {5 a^{4} \ln \left (b \,x^{2}+a \right ) B}{2 b^{6}}\) | \(146\) |
parallelrisch | \(-\frac {-3 b^{5} B \,x^{10}-4 A \,b^{5} x^{8}+5 B a \,b^{4} x^{8}+8 A a \,b^{4} x^{6}-10 B \,a^{2} b^{3} x^{6}-24 A \,a^{2} b^{3} x^{4}+30 B \,a^{3} b^{2} x^{4}+48 A \ln \left (b \,x^{2}+a \right ) x^{2} a^{3} b^{2}-60 B \ln \left (b \,x^{2}+a \right ) x^{2} a^{4} b +48 A \ln \left (b \,x^{2}+a \right ) a^{4} b -60 B \ln \left (b \,x^{2}+a \right ) a^{5}+48 a^{4} b A -60 a^{5} B}{24 b^{6} \left (b \,x^{2}+a \right )}\) | \(170\) |
(1/8*B/b*x^10-1/2*a*(4*A*a^3*b-5*B*a^4)/b^6+1/24*(4*A*b-5*B*a)/b^2*x^8-1/1 2*a*(4*A*b-5*B*a)/b^3*x^6+1/4*a^2*(4*A*b-5*B*a)/b^4*x^4)/(b*x^2+a)-1/2*a^3 *(4*A*b-5*B*a)*ln(b*x^2+a)/b^6
Time = 0.25 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.37 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {3 \, B b^{5} x^{10} - {\left (5 \, B a b^{4} - 4 \, A b^{5}\right )} x^{8} + 2 \, {\left (5 \, B a^{2} b^{3} - 4 \, A a b^{4}\right )} x^{6} + 12 \, B a^{5} - 12 \, A a^{4} b - 6 \, {\left (5 \, B a^{3} b^{2} - 4 \, A a^{2} b^{3}\right )} x^{4} - 12 \, {\left (4 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x^{2} + 12 \, {\left (5 \, B a^{5} - 4 \, A a^{4} b + {\left (5 \, B a^{4} b - 4 \, A a^{3} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{24 \, {\left (b^{7} x^{2} + a b^{6}\right )}} \]
1/24*(3*B*b^5*x^10 - (5*B*a*b^4 - 4*A*b^5)*x^8 + 2*(5*B*a^2*b^3 - 4*A*a*b^ 4)*x^6 + 12*B*a^5 - 12*A*a^4*b - 6*(5*B*a^3*b^2 - 4*A*a^2*b^3)*x^4 - 12*(4 *B*a^4*b - 3*A*a^3*b^2)*x^2 + 12*(5*B*a^5 - 4*A*a^4*b + (5*B*a^4*b - 4*A*a ^3*b^2)*x^2)*log(b*x^2 + a))/(b^7*x^2 + a*b^6)
Time = 0.44 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.04 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B x^{8}}{8 b^{2}} + \frac {a^{3} \left (- 4 A b + 5 B a\right ) \log {\left (a + b x^{2} \right )}}{2 b^{6}} + x^{6} \left (\frac {A}{6 b^{2}} - \frac {B a}{3 b^{3}}\right ) + x^{4} \left (- \frac {A a}{2 b^{3}} + \frac {3 B a^{2}}{4 b^{4}}\right ) + x^{2} \cdot \left (\frac {3 A a^{2}}{2 b^{4}} - \frac {2 B a^{3}}{b^{5}}\right ) + \frac {- A a^{4} b + B a^{5}}{2 a b^{6} + 2 b^{7} x^{2}} \]
B*x**8/(8*b**2) + a**3*(-4*A*b + 5*B*a)*log(a + b*x**2)/(2*b**6) + x**6*(A /(6*b**2) - B*a/(3*b**3)) + x**4*(-A*a/(2*b**3) + 3*B*a**2/(4*b**4)) + x** 2*(3*A*a**2/(2*b**4) - 2*B*a**3/b**5) + (-A*a**4*b + B*a**5)/(2*a*b**6 + 2 *b**7*x**2)
Time = 0.19 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.04 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B a^{5} - A a^{4} b}{2 \, {\left (b^{7} x^{2} + a b^{6}\right )}} + \frac {3 \, B b^{3} x^{8} - 4 \, {\left (2 \, B a b^{2} - A b^{3}\right )} x^{6} + 6 \, {\left (3 \, B a^{2} b - 2 \, A a b^{2}\right )} x^{4} - 12 \, {\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} x^{2}}{24 \, b^{5}} + \frac {{\left (5 \, B a^{4} - 4 \, A a^{3} b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{6}} \]
1/2*(B*a^5 - A*a^4*b)/(b^7*x^2 + a*b^6) + 1/24*(3*B*b^3*x^8 - 4*(2*B*a*b^2 - A*b^3)*x^6 + 6*(3*B*a^2*b - 2*A*a*b^2)*x^4 - 12*(4*B*a^3 - 3*A*a^2*b)*x ^2)/b^5 + 1/2*(5*B*a^4 - 4*A*a^3*b)*log(b*x^2 + a)/b^6
Time = 0.29 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.26 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {{\left (5 \, B a^{4} - 4 \, A a^{3} b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{6}} - \frac {5 \, B a^{4} b x^{2} - 4 \, A a^{3} b^{2} x^{2} + 4 \, B a^{5} - 3 \, A a^{4} b}{2 \, {\left (b x^{2} + a\right )} b^{6}} + \frac {3 \, B b^{6} x^{8} - 8 \, B a b^{5} x^{6} + 4 \, A b^{6} x^{6} + 18 \, B a^{2} b^{4} x^{4} - 12 \, A a b^{5} x^{4} - 48 \, B a^{3} b^{3} x^{2} + 36 \, A a^{2} b^{4} x^{2}}{24 \, b^{8}} \]
1/2*(5*B*a^4 - 4*A*a^3*b)*log(abs(b*x^2 + a))/b^6 - 1/2*(5*B*a^4*b*x^2 - 4 *A*a^3*b^2*x^2 + 4*B*a^5 - 3*A*a^4*b)/((b*x^2 + a)*b^6) + 1/24*(3*B*b^6*x^ 8 - 8*B*a*b^5*x^6 + 4*A*b^6*x^6 + 18*B*a^2*b^4*x^4 - 12*A*a*b^5*x^4 - 48*B *a^3*b^3*x^2 + 36*A*a^2*b^4*x^2)/b^8
Time = 5.02 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.44 \[ \int \frac {x^9 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=x^2\,\left (\frac {a\,\left (\frac {2\,a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{b^4}\right )}{b}-\frac {a^2\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{2\,b^2}\right )+x^6\,\left (\frac {A}{6\,b^2}-\frac {B\,a}{3\,b^3}\right )-x^4\,\left (\frac {a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{2\,b}+\frac {B\,a^2}{4\,b^4}\right )+\frac {B\,x^8}{8\,b^2}+\frac {\ln \left (b\,x^2+a\right )\,\left (5\,B\,a^4-4\,A\,a^3\,b\right )}{2\,b^6}+\frac {B\,a^5-A\,a^4\,b}{2\,b\,\left (b^6\,x^2+a\,b^5\right )} \]
x^2*((a*((2*a*(A/b^2 - (2*B*a)/b^3))/b + (B*a^2)/b^4))/b - (a^2*(A/b^2 - ( 2*B*a)/b^3))/(2*b^2)) + x^6*(A/(6*b^2) - (B*a)/(3*b^3)) - x^4*((a*(A/b^2 - (2*B*a)/b^3))/(2*b) + (B*a^2)/(4*b^4)) + (B*x^8)/(8*b^2) + (log(a + b*x^2 )*(5*B*a^4 - 4*A*a^3*b))/(2*b^6) + (B*a^5 - A*a^4*b)/(2*b*(a*b^5 + b^6*x^2 ))