Integrand size = 20, antiderivative size = 104 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=-\frac {a (2 A b-3 a B) x^2}{2 b^4}+\frac {(A b-2 a B) x^4}{4 b^3}+\frac {B x^6}{6 b^2}+\frac {a^3 (A b-a B)}{2 b^5 \left (a+b x^2\right )}+\frac {a^2 (3 A b-4 a B) \log \left (a+b x^2\right )}{2 b^5} \]
-1/2*a*(2*A*b-3*B*a)*x^2/b^4+1/4*(A*b-2*B*a)*x^4/b^3+1/6*B*x^6/b^2+1/2*a^3 *(A*b-B*a)/b^5/(b*x^2+a)+1/2*a^2*(3*A*b-4*B*a)*ln(b*x^2+a)/b^5
Time = 0.04 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {6 a b (-2 A b+3 a B) x^2+3 b^2 (A b-2 a B) x^4+2 b^3 B x^6+\frac {6 a^3 (A b-a B)}{a+b x^2}+6 a^2 (3 A b-4 a B) \log \left (a+b x^2\right )}{12 b^5} \]
(6*a*b*(-2*A*b + 3*a*B)*x^2 + 3*b^2*(A*b - 2*a*B)*x^4 + 2*b^3*B*x^6 + (6*a ^3*(A*b - a*B))/(a + b*x^2) + 6*a^2*(3*A*b - 4*a*B)*Log[a + b*x^2])/(12*b^ 5)
Time = 0.27 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.96, 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.
\(\displaystyle \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {x^6 \left (B x^2+A\right )}{\left (b x^2+a\right )^2}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {B x^4}{b^2}+\frac {(A b-2 a B) x^2}{b^3}+\frac {a (3 a B-2 A b)}{b^4}-\frac {a^2 (4 a B-3 A b)}{b^4 \left (b x^2+a\right )}+\frac {a^3 (a B-A b)}{b^4 \left (b x^2+a\right )^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {a^3 (A b-a B)}{b^5 \left (a+b x^2\right )}+\frac {a^2 (3 A b-4 a B) \log \left (a+b x^2\right )}{b^5}-\frac {a x^2 (2 A b-3 a B)}{b^4}+\frac {x^4 (A b-2 a B)}{2 b^3}+\frac {B x^6}{3 b^2}\right )\) |
(-((a*(2*A*b - 3*a*B)*x^2)/b^4) + ((A*b - 2*a*B)*x^4)/(2*b^3) + (B*x^6)/(3 *b^2) + (a^3*(A*b - a*B))/(b^5*(a + b*x^2)) + (a^2*(3*A*b - 4*a*B)*Log[a + b*x^2])/b^5)/2
3.1.73.3.1 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 = 2.60 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.98
method | result | size |
norman | \(\frac {\frac {B \,x^{8}}{6 b}+\frac {a \left (3 a^{2} b A -4 a^{3} B \right )}{2 b^{5}}+\frac {\left (3 A b -4 B a \right ) x^{6}}{12 b^{2}}-\frac {a \left (3 A b -4 B a \right ) x^{4}}{4 b^{3}}}{b \,x^{2}+a}+\frac {a^{2} \left (3 A b -4 B a \right ) \ln \left (b \,x^{2}+a \right )}{2 b^{5}}\) | \(102\) |
default | \(-\frac {-\frac {b^{2} B \,x^{6}}{6}+\frac {\left (-b^{2} A +2 a b B \right ) x^{4}}{4}+\frac {\left (2 a b A -3 a^{2} B \right ) x^{2}}{2}}{b^{4}}+\frac {a^{2} \left (\frac {\left (3 A b -4 B a \right ) \ln \left (b \,x^{2}+a \right )}{b}+\frac {a \left (A b -B a \right )}{b \left (b \,x^{2}+a \right )}\right )}{2 b^{4}}\) | \(103\) |
risch | \(\frac {B \,x^{6}}{6 b^{2}}+\frac {A \,x^{4}}{4 b^{2}}-\frac {B a \,x^{4}}{2 b^{3}}-\frac {a A \,x^{2}}{b^{3}}+\frac {3 a^{2} B \,x^{2}}{2 b^{4}}+\frac {a^{3} A}{2 b^{4} \left (b \,x^{2}+a \right )}-\frac {a^{4} B}{2 b^{5} \left (b \,x^{2}+a \right )}+\frac {3 a^{2} \ln \left (b \,x^{2}+a \right ) A}{2 b^{4}}-\frac {2 a^{3} \ln \left (b \,x^{2}+a \right ) B}{b^{5}}\) | \(122\) |
parallelrisch | \(\frac {2 B \,x^{8} b^{4}+3 A \,x^{6} b^{4}-4 B \,x^{6} a \,b^{3}-9 A \,x^{4} a \,b^{3}+12 B \,x^{4} a^{2} b^{2}+18 A \ln \left (b \,x^{2}+a \right ) x^{2} a^{2} b^{2}-24 B \ln \left (b \,x^{2}+a \right ) x^{2} a^{3} b +18 A \ln \left (b \,x^{2}+a \right ) a^{3} b -24 B \ln \left (b \,x^{2}+a \right ) a^{4}+18 A \,a^{3} b -24 B \,a^{4}}{12 b^{5} \left (b \,x^{2}+a \right )}\) | \(146\) |
(1/6*B/b*x^8+1/2*a*(3*A*a^2*b-4*B*a^3)/b^5+1/12*(3*A*b-4*B*a)/b^2*x^6-1/4* a*(3*A*b-4*B*a)/b^3*x^4)/(b*x^2+a)+1/2*a^2*(3*A*b-4*B*a)*ln(b*x^2+a)/b^5
Time = 0.28 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.42 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {2 \, B b^{4} x^{8} - {\left (4 \, B a b^{3} - 3 \, A b^{4}\right )} x^{6} - 6 \, B a^{4} + 6 \, A a^{3} b + 3 \, {\left (4 \, B a^{2} b^{2} - 3 \, A a b^{3}\right )} x^{4} + 6 \, {\left (3 \, B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{2} - 6 \, {\left (4 \, B a^{4} - 3 \, A a^{3} b + {\left (4 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2}\right )} \log \left (b x^{2} + a\right )}{12 \, {\left (b^{6} x^{2} + a b^{5}\right )}} \]
1/12*(2*B*b^4*x^8 - (4*B*a*b^3 - 3*A*b^4)*x^6 - 6*B*a^4 + 6*A*a^3*b + 3*(4 *B*a^2*b^2 - 3*A*a*b^3)*x^4 + 6*(3*B*a^3*b - 2*A*a^2*b^2)*x^2 - 6*(4*B*a^4 - 3*A*a^3*b + (4*B*a^3*b - 3*A*a^2*b^2)*x^2)*log(b*x^2 + a))/(b^6*x^2 + a *b^5)
Time = 0.42 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=\frac {B x^{6}}{6 b^{2}} - \frac {a^{2} \left (- 3 A b + 4 B a\right ) \log {\left (a + b x^{2} \right )}}{2 b^{5}} + x^{4} \left (\frac {A}{4 b^{2}} - \frac {B a}{2 b^{3}}\right ) + x^{2} \left (- \frac {A a}{b^{3}} + \frac {3 B a^{2}}{2 b^{4}}\right ) + \frac {A a^{3} b - B a^{4}}{2 a b^{5} + 2 b^{6} x^{2}} \]
B*x**6/(6*b**2) - a**2*(-3*A*b + 4*B*a)*log(a + b*x**2)/(2*b**5) + x**4*(A /(4*b**2) - B*a/(2*b**3)) + x**2*(-A*a/b**3 + 3*B*a**2/(2*b**4)) + (A*a**3 *b - B*a**4)/(2*a*b**5 + 2*b**6*x**2)
Time = 0.20 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=-\frac {B a^{4} - A a^{3} b}{2 \, {\left (b^{6} x^{2} + a b^{5}\right )}} + \frac {2 \, B b^{2} x^{6} - 3 \, {\left (2 \, B a b - A b^{2}\right )} x^{4} + 6 \, {\left (3 \, B a^{2} - 2 \, A a b\right )} x^{2}}{12 \, b^{4}} - \frac {{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left (b x^{2} + a\right )}{2 \, b^{5}} \]
-1/2*(B*a^4 - A*a^3*b)/(b^6*x^2 + a*b^5) + 1/12*(2*B*b^2*x^6 - 3*(2*B*a*b - A*b^2)*x^4 + 6*(3*B*a^2 - 2*A*a*b)*x^2)/b^4 - 1/2*(4*B*a^3 - 3*A*a^2*b)* log(b*x^2 + a)/b^5
Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.30 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=-\frac {{\left (4 \, B a^{3} - 3 \, A a^{2} b\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{5}} + \frac {2 \, B b^{4} x^{6} - 6 \, B a b^{3} x^{4} + 3 \, A b^{4} x^{4} + 18 \, B a^{2} b^{2} x^{2} - 12 \, A a b^{3} x^{2}}{12 \, b^{6}} + \frac {4 \, B a^{3} b x^{2} - 3 \, A a^{2} b^{2} x^{2} + 3 \, B a^{4} - 2 \, A a^{3} b}{2 \, {\left (b x^{2} + a\right )} b^{5}} \]
-1/2*(4*B*a^3 - 3*A*a^2*b)*log(abs(b*x^2 + a))/b^5 + 1/12*(2*B*b^4*x^6 - 6 *B*a*b^3*x^4 + 3*A*b^4*x^4 + 18*B*a^2*b^2*x^2 - 12*A*a*b^3*x^2)/b^6 + 1/2* (4*B*a^3*b*x^2 - 3*A*a^2*b^2*x^2 + 3*B*a^4 - 2*A*a^3*b)/((b*x^2 + a)*b^5)
Time = 4.97 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.16 \[ \int \frac {x^7 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx=x^4\,\left (\frac {A}{4\,b^2}-\frac {B\,a}{2\,b^3}\right )-x^2\,\left (\frac {a\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{b}+\frac {B\,a^2}{2\,b^4}\right )+\frac {B\,x^6}{6\,b^2}-\frac {\ln \left (b\,x^2+a\right )\,\left (4\,B\,a^3-3\,A\,a^2\,b\right )}{2\,b^5}-\frac {B\,a^4-A\,a^3\,b}{2\,b\,\left (b^5\,x^2+a\,b^4\right )} \]