Integrand size = 24, antiderivative size = 101 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {1}{4 a^4 x^4}+\frac {2 b}{a^5 x^2}+\frac {b^2}{6 a^3 \left (a+b x^2\right )^3}+\frac {3 b^2}{4 a^4 \left (a+b x^2\right )^2}+\frac {3 b^2}{a^5 \left (a+b x^2\right )}+\frac {10 b^2 \log (x)}{a^6}-\frac {5 b^2 \log \left (a+b x^2\right )}{a^6} \]
-1/4/a^4/x^4+2*b/a^5/x^2+1/6*b^2/a^3/(b*x^2+a)^3+3/4*b^2/a^4/(b*x^2+a)^2+3 *b^2/a^5/(b*x^2+a)+10*b^2*ln(x)/a^6-5*b^2*ln(b*x^2+a)/a^6
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {a \left (-3 a^4+15 a^3 b x^2+110 a^2 b^2 x^4+150 a b^3 x^6+60 b^4 x^8\right )}{x^4 \left (a+b x^2\right )^3}+120 b^2 \log (x)-60 b^2 \log \left (a+b x^2\right )}{12 a^6} \]
((a*(-3*a^4 + 15*a^3*b*x^2 + 110*a^2*b^2*x^4 + 150*a*b^3*x^6 + 60*b^4*x^8) )/(x^4*(a + b*x^2)^3) + 120*b^2*Log[x] - 60*b^2*Log[a + b*x^2])/(12*a^6)
Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1380, 27, 243, 54, 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.
\(\displaystyle \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle b^4 \int \frac {1}{b^4 x^5 \left (b x^2+a\right )^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1}{x^5 \left (a+b x^2\right )^4}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^6 \left (b x^2+a\right )^4}dx^2\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {10 b^3}{a^6 \left (b x^2+a\right )}-\frac {6 b^3}{a^5 \left (b x^2+a\right )^2}-\frac {3 b^3}{a^4 \left (b x^2+a\right )^3}-\frac {b^3}{a^3 \left (b x^2+a\right )^4}+\frac {10 b^2}{a^6 x^2}-\frac {4 b}{a^5 x^4}+\frac {1}{a^4 x^6}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {10 b^2 \log \left (x^2\right )}{a^6}-\frac {10 b^2 \log \left (a+b x^2\right )}{a^6}+\frac {6 b^2}{a^5 \left (a+b x^2\right )}+\frac {4 b}{a^5 x^2}+\frac {3 b^2}{2 a^4 \left (a+b x^2\right )^2}-\frac {1}{2 a^4 x^4}+\frac {b^2}{3 a^3 \left (a+b x^2\right )^3}\right )\) |
(-1/2*1/(a^4*x^4) + (4*b)/(a^5*x^2) + b^2/(3*a^3*(a + b*x^2)^3) + (3*b^2)/ (2*a^4*(a + b*x^2)^2) + (6*b^2)/(a^5*(a + b*x^2)) + (10*b^2*Log[x^2])/a^6 - (10*b^2*Log[a + b*x^2])/a^6)/2
3.6.1.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {-\frac {1}{4 a}+\frac {5 b \,x^{2}}{4 a^{2}}-\frac {15 b^{3} x^{6}}{a^{4}}-\frac {45 b^{4} x^{8}}{2 a^{5}}-\frac {55 b^{5} x^{10}}{6 a^{6}}}{x^{4} \left (b \,x^{2}+a \right )^{3}}+\frac {10 b^{2} \ln \left (x \right )}{a^{6}}-\frac {5 b^{2} \ln \left (b \,x^{2}+a \right )}{a^{6}}\) | \(89\) |
default | \(-\frac {1}{4 a^{4} x^{4}}+\frac {2 b}{a^{5} x^{2}}+\frac {10 b^{2} \ln \left (x \right )}{a^{6}}-\frac {b^{3} \left (\frac {10 \ln \left (b \,x^{2}+a \right )}{b}-\frac {a^{3}}{3 b \left (b \,x^{2}+a \right )^{3}}-\frac {3 a^{2}}{2 b \left (b \,x^{2}+a \right )^{2}}-\frac {6 a}{b \left (b \,x^{2}+a \right )}\right )}{2 a^{6}}\) | \(100\) |
risch | \(\frac {\frac {5 b^{4} x^{8}}{a^{5}}+\frac {25 b^{3} x^{6}}{2 a^{4}}+\frac {55 b^{2} x^{4}}{6 a^{3}}+\frac {5 b \,x^{2}}{4 a^{2}}-\frac {1}{4 a}}{x^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \left (b \,x^{2}+a \right )}+\frac {10 b^{2} \ln \left (x \right )}{a^{6}}-\frac {5 b^{2} \ln \left (b \,x^{2}+a \right )}{a^{6}}\) | \(109\) |
parallelrisch | \(\frac {120 \ln \left (x \right ) x^{10} b^{5}-60 \ln \left (b \,x^{2}+a \right ) x^{10} b^{5}-110 x^{10} b^{5}+360 \ln \left (x \right ) x^{8} a \,b^{4}-180 \ln \left (b \,x^{2}+a \right ) x^{8} a \,b^{4}-270 a \,x^{8} b^{4}+360 \ln \left (x \right ) x^{6} a^{2} b^{3}-180 \ln \left (b \,x^{2}+a \right ) x^{6} a^{2} b^{3}-180 a^{2} x^{6} b^{3}+120 \ln \left (x \right ) x^{4} a^{3} b^{2}-60 \ln \left (b \,x^{2}+a \right ) x^{4} a^{3} b^{2}+15 x^{2} a^{4} b -3 a^{5}}{12 a^{6} x^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \left (b \,x^{2}+a \right )}\) | \(199\) |
(-1/4/a+5/4*b/a^2*x^2-15*b^3/a^4*x^6-45/2*b^4/a^5*x^8-55/6*b^5/a^6*x^10)/x ^4/(b*x^2+a)^3+10*b^2*ln(x)/a^6-5*b^2*ln(b*x^2+a)/a^6
Time = 0.25 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.76 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {60 \, a b^{4} x^{8} + 150 \, a^{2} b^{3} x^{6} + 110 \, a^{3} b^{2} x^{4} + 15 \, a^{4} b x^{2} - 3 \, a^{5} - 60 \, {\left (b^{5} x^{10} + 3 \, a b^{4} x^{8} + 3 \, a^{2} b^{3} x^{6} + a^{3} b^{2} x^{4}\right )} \log \left (b x^{2} + a\right ) + 120 \, {\left (b^{5} x^{10} + 3 \, a b^{4} x^{8} + 3 \, a^{2} b^{3} x^{6} + a^{3} b^{2} x^{4}\right )} \log \left (x\right )}{12 \, {\left (a^{6} b^{3} x^{10} + 3 \, a^{7} b^{2} x^{8} + 3 \, a^{8} b x^{6} + a^{9} x^{4}\right )}} \]
1/12*(60*a*b^4*x^8 + 150*a^2*b^3*x^6 + 110*a^3*b^2*x^4 + 15*a^4*b*x^2 - 3* a^5 - 60*(b^5*x^10 + 3*a*b^4*x^8 + 3*a^2*b^3*x^6 + a^3*b^2*x^4)*log(b*x^2 + a) + 120*(b^5*x^10 + 3*a*b^4*x^8 + 3*a^2*b^3*x^6 + a^3*b^2*x^4)*log(x))/ (a^6*b^3*x^10 + 3*a^7*b^2*x^8 + 3*a^8*b*x^6 + a^9*x^4)
Time = 0.31 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {- 3 a^{4} + 15 a^{3} b x^{2} + 110 a^{2} b^{2} x^{4} + 150 a b^{3} x^{6} + 60 b^{4} x^{8}}{12 a^{8} x^{4} + 36 a^{7} b x^{6} + 36 a^{6} b^{2} x^{8} + 12 a^{5} b^{3} x^{10}} + \frac {10 b^{2} \log {\left (x \right )}}{a^{6}} - \frac {5 b^{2} \log {\left (\frac {a}{b} + x^{2} \right )}}{a^{6}} \]
(-3*a**4 + 15*a**3*b*x**2 + 110*a**2*b**2*x**4 + 150*a*b**3*x**6 + 60*b**4 *x**8)/(12*a**8*x**4 + 36*a**7*b*x**6 + 36*a**6*b**2*x**8 + 12*a**5*b**3*x **10) + 10*b**2*log(x)/a**6 - 5*b**2*log(a/b + x**2)/a**6
Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.13 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {60 \, b^{4} x^{8} + 150 \, a b^{3} x^{6} + 110 \, a^{2} b^{2} x^{4} + 15 \, a^{3} b x^{2} - 3 \, a^{4}}{12 \, {\left (a^{5} b^{3} x^{10} + 3 \, a^{6} b^{2} x^{8} + 3 \, a^{7} b x^{6} + a^{8} x^{4}\right )}} - \frac {5 \, b^{2} \log \left (b x^{2} + a\right )}{a^{6}} + \frac {5 \, b^{2} \log \left (x^{2}\right )}{a^{6}} \]
1/12*(60*b^4*x^8 + 150*a*b^3*x^6 + 110*a^2*b^2*x^4 + 15*a^3*b*x^2 - 3*a^4) /(a^5*b^3*x^10 + 3*a^6*b^2*x^8 + 3*a^7*b*x^6 + a^8*x^4) - 5*b^2*log(b*x^2 + a)/a^6 + 5*b^2*log(x^2)/a^6
Time = 0.26 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {5 \, b^{2} \log \left (x^{2}\right )}{a^{6}} - \frac {5 \, b^{2} \log \left ({\left | b x^{2} + a \right |}\right )}{a^{6}} + \frac {110 \, b^{5} x^{6} + 366 \, a b^{4} x^{4} + 411 \, a^{2} b^{3} x^{2} + 157 \, a^{3} b^{2}}{12 \, {\left (b x^{2} + a\right )}^{3} a^{6}} - \frac {30 \, b^{2} x^{4} - 8 \, a b x^{2} + a^{2}}{4 \, a^{6} x^{4}} \]
5*b^2*log(x^2)/a^6 - 5*b^2*log(abs(b*x^2 + a))/a^6 + 1/12*(110*b^5*x^6 + 3 66*a*b^4*x^4 + 411*a^2*b^3*x^2 + 157*a^3*b^2)/((b*x^2 + a)^3*a^6) - 1/4*(3 0*b^2*x^4 - 8*a*b*x^2 + a^2)/(a^6*x^4)
Time = 13.35 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {5\,b\,x^2}{4\,a^2}-\frac {1}{4\,a}+\frac {55\,b^2\,x^4}{6\,a^3}+\frac {25\,b^3\,x^6}{2\,a^4}+\frac {5\,b^4\,x^8}{a^5}}{a^3\,x^4+3\,a^2\,b\,x^6+3\,a\,b^2\,x^8+b^3\,x^{10}}-\frac {5\,b^2\,\ln \left (b\,x^2+a\right )}{a^6}+\frac {10\,b^2\,\ln \left (x\right )}{a^6} \]