Integrand size = 20, antiderivative size = 97 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^2} \, dx=-\frac {A}{4 a^2 x^4}+\frac {2 A b-a B}{2 a^3 x^2}+\frac {b (A b-a B)}{2 a^3 \left (a+b x^2\right )}+\frac {b (3 A b-2 a B) \log (x)}{a^4}-\frac {b (3 A b-2 a B) \log \left (a+b x^2\right )}{2 a^4} \]
-1/4*A/a^2/x^4+1/2*(2*A*b-B*a)/a^3/x^2+1/2*b*(A*b-B*a)/a^3/(b*x^2+a)+b*(3* A*b-2*B*a)*ln(x)/a^4-1/2*b*(3*A*b-2*B*a)*ln(b*x^2+a)/a^4
Time = 0.06 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.88 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^2} \, dx=-\frac {\frac {a^2 A}{x^4}+\frac {2 a (-2 A b+a B)}{x^2}+\frac {2 a b (-A b+a B)}{a+b x^2}-4 b (3 A b-2 a B) \log (x)+2 b (3 A b-2 a B) \log \left (a+b x^2\right )}{4 a^4} \]
-1/4*((a^2*A)/x^4 + (2*a*(-2*A*b + a*B))/x^2 + (2*a*b*(-(A*b) + a*B))/(a + b*x^2) - 4*b*(3*A*b - 2*a*B)*Log[x] + 2*b*(3*A*b - 2*a*B)*Log[a + b*x^2]) /a^4
Time = 0.26 (sec) , antiderivative size = 95, 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.
\(\displaystyle \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {B x^2+A}{x^6 \left (b x^2+a\right )^2}dx^2\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {1}{2} \int \left (\frac {(2 a B-3 A b) b^2}{a^4 \left (b x^2+a\right )}+\frac {(a B-A b) b^2}{a^3 \left (b x^2+a\right )^2}-\frac {(2 a B-3 A b) b}{a^4 x^2}+\frac {a B-2 A b}{a^3 x^4}+\frac {A}{a^2 x^6}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {b \log \left (x^2\right ) (3 A b-2 a B)}{a^4}-\frac {b (3 A b-2 a B) \log \left (a+b x^2\right )}{a^4}+\frac {b (A b-a B)}{a^3 \left (a+b x^2\right )}+\frac {2 A b-a B}{a^3 x^2}-\frac {A}{2 a^2 x^4}\right )\) |
(-1/2*A/(a^2*x^4) + (2*A*b - a*B)/(a^3*x^2) + (b*(A*b - a*B))/(a^3*(a + b* x^2)) + (b*(3*A*b - 2*a*B)*Log[x^2])/a^4 - (b*(3*A*b - 2*a*B)*Log[a + b*x^ 2])/a^4)/2
3.1.85.3.1 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 = 2.51 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99
method | result | size |
default | \(-\frac {A}{4 a^{2} x^{4}}-\frac {-2 A b +B a}{2 x^{2} a^{3}}+\frac {b \left (3 A b -2 B a \right ) \ln \left (x \right )}{a^{4}}-\frac {b^{2} \left (\frac {\left (3 A b -2 B a \right ) \ln \left (b \,x^{2}+a \right )}{b}-\frac {a \left (A b -B a \right )}{b \left (b \,x^{2}+a \right )}\right )}{2 a^{4}}\) | \(96\) |
norman | \(\frac {-\frac {A}{4 a}+\frac {\left (3 A b -2 B a \right ) x^{2}}{4 a^{2}}-\frac {b \left (3 b^{2} A -2 a b B \right ) x^{6}}{2 a^{4}}}{x^{4} \left (b \,x^{2}+a \right )}+\frac {b \left (3 A b -2 B a \right ) \ln \left (x \right )}{a^{4}}-\frac {b \left (3 A b -2 B a \right ) \ln \left (b \,x^{2}+a \right )}{2 a^{4}}\) | \(99\) |
risch | \(\frac {\frac {b \left (3 A b -2 B a \right ) x^{4}}{2 a^{3}}+\frac {\left (3 A b -2 B a \right ) x^{2}}{4 a^{2}}-\frac {A}{4 a}}{x^{4} \left (b \,x^{2}+a \right )}+\frac {3 b^{2} \ln \left (x \right ) A}{a^{4}}-\frac {2 b \ln \left (x \right ) B}{a^{3}}-\frac {3 b^{2} \ln \left (b \,x^{2}+a \right ) A}{2 a^{4}}+\frac {b \ln \left (b \,x^{2}+a \right ) B}{a^{3}}\) | \(108\) |
parallelrisch | \(\frac {12 A \ln \left (x \right ) x^{6} b^{3}-6 A \ln \left (b \,x^{2}+a \right ) x^{6} b^{3}-8 B \ln \left (x \right ) x^{6} a \,b^{2}+4 B \ln \left (b \,x^{2}+a \right ) x^{6} a \,b^{2}-6 A \,x^{6} b^{3}+4 B \,x^{6} a \,b^{2}+12 A \ln \left (x \right ) x^{4} a \,b^{2}-6 A \ln \left (b \,x^{2}+a \right ) x^{4} a \,b^{2}-8 B \ln \left (x \right ) x^{4} a^{2} b +4 B \ln \left (b \,x^{2}+a \right ) x^{4} a^{2} b +3 A \,a^{2} b \,x^{2}-2 B \,a^{3} x^{2}-a^{3} A}{4 a^{4} x^{4} \left (b \,x^{2}+a \right )}\) | \(181\) |
-1/4*A/a^2/x^4-1/2*(-2*A*b+B*a)/x^2/a^3+b*(3*A*b-2*B*a)*ln(x)/a^4-1/2/a^4* b^2*((3*A*b-2*B*a)/b*ln(b*x^2+a)-a*(A*b-B*a)/b/(b*x^2+a))
Time = 0.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.59 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^2} \, dx=-\frac {2 \, {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{4} + A a^{3} + {\left (2 \, B a^{3} - 3 \, A a^{2} b\right )} x^{2} - 2 \, {\left ({\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} x^{6} + {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{4}\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left ({\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} x^{6} + {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} x^{4}\right )} \log \left (x\right )}{4 \, {\left (a^{4} b x^{6} + a^{5} x^{4}\right )}} \]
-1/4*(2*(2*B*a^2*b - 3*A*a*b^2)*x^4 + A*a^3 + (2*B*a^3 - 3*A*a^2*b)*x^2 - 2*((2*B*a*b^2 - 3*A*b^3)*x^6 + (2*B*a^2*b - 3*A*a*b^2)*x^4)*log(b*x^2 + a) + 4*((2*B*a*b^2 - 3*A*b^3)*x^6 + (2*B*a^2*b - 3*A*a*b^2)*x^4)*log(x))/(a^ 4*b*x^6 + a^5*x^4)
Time = 0.58 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^2} \, dx=\frac {- A a^{2} + x^{4} \cdot \left (6 A b^{2} - 4 B a b\right ) + x^{2} \cdot \left (3 A a b - 2 B a^{2}\right )}{4 a^{4} x^{4} + 4 a^{3} b x^{6}} - \frac {b \left (- 3 A b + 2 B a\right ) \log {\left (x \right )}}{a^{4}} + \frac {b \left (- 3 A b + 2 B a\right ) \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{4}} \]
(-A*a**2 + x**4*(6*A*b**2 - 4*B*a*b) + x**2*(3*A*a*b - 2*B*a**2))/(4*a**4* x**4 + 4*a**3*b*x**6) - b*(-3*A*b + 2*B*a)*log(x)/a**4 + b*(-3*A*b + 2*B*a )*log(a/b + x**2)/(2*a**4)
Time = 0.22 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^2} \, dx=-\frac {2 \, {\left (2 \, B a b - 3 \, A b^{2}\right )} x^{4} + A a^{2} + {\left (2 \, B a^{2} - 3 \, A a b\right )} x^{2}}{4 \, {\left (a^{3} b x^{6} + a^{4} x^{4}\right )}} + \frac {{\left (2 \, B a b - 3 \, A b^{2}\right )} \log \left (b x^{2} + a\right )}{2 \, a^{4}} - \frac {{\left (2 \, B a b - 3 \, A b^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{4}} \]
-1/4*(2*(2*B*a*b - 3*A*b^2)*x^4 + A*a^2 + (2*B*a^2 - 3*A*a*b)*x^2)/(a^3*b* x^6 + a^4*x^4) + 1/2*(2*B*a*b - 3*A*b^2)*log(b*x^2 + a)/a^4 - 1/2*(2*B*a*b - 3*A*b^2)*log(x^2)/a^4
Time = 0.28 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.55 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^2} \, dx=-\frac {{\left (2 \, B a b - 3 \, A b^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{4}} + \frac {{\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{4} b} - \frac {2 \, B a b^{2} x^{2} - 3 \, A b^{3} x^{2} + 3 \, B a^{2} b - 4 \, A a b^{2}}{2 \, {\left (b x^{2} + a\right )} a^{4}} + \frac {6 \, B a b x^{4} - 9 \, A b^{2} x^{4} - 2 \, B a^{2} x^{2} + 4 \, A a b x^{2} - A a^{2}}{4 \, a^{4} x^{4}} \]
-1/2*(2*B*a*b - 3*A*b^2)*log(x^2)/a^4 + 1/2*(2*B*a*b^2 - 3*A*b^3)*log(abs( b*x^2 + a))/(a^4*b) - 1/2*(2*B*a*b^2*x^2 - 3*A*b^3*x^2 + 3*B*a^2*b - 4*A*a *b^2)/((b*x^2 + a)*a^4) + 1/4*(6*B*a*b*x^4 - 9*A*b^2*x^4 - 2*B*a^2*x^2 + 4 *A*a*b*x^2 - A*a^2)/(a^4*x^4)
Time = 4.84 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x^2}{x^5 \left (a+b x^2\right )^2} \, dx=\frac {\frac {x^2\,\left (3\,A\,b-2\,B\,a\right )}{4\,a^2}-\frac {A}{4\,a}+\frac {b\,x^4\,\left (3\,A\,b-2\,B\,a\right )}{2\,a^3}}{b\,x^6+a\,x^4}-\frac {\ln \left (b\,x^2+a\right )\,\left (3\,A\,b^2-2\,B\,a\,b\right )}{2\,a^4}+\frac {\ln \left (x\right )\,\left (3\,A\,b^2-2\,B\,a\,b\right )}{a^4} \]