Integrand size = 24, antiderivative size = 66 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=-\frac {1}{4 a^2 x^4}+\frac {b}{a^3 x^2}+\frac {b^2}{2 a^3 \left (a+b x^2\right )}+\frac {3 b^2 \log (x)}{a^4}-\frac {3 b^2 \log \left (a+b x^2\right )}{2 a^4} \] Output:
-1/4/a^2/x^4+b/a^3/x^2+1/2*b^2/a^3/(b*x^2+a)+3*b^2*ln(x)/a^4-3/2*b^2*ln(b* x^2+a)/a^4
Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.86 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {a \left (-\frac {a}{x^4}+\frac {4 b}{x^2}+\frac {2 b^2}{a+b x^2}\right )+12 b^2 \log (x)-6 b^2 \log \left (a+b x^2\right )}{4 a^4} \] Input:
Integrate[1/(x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]
Output:
(a*(-(a/x^4) + (4*b)/x^2 + (2*b^2)/(a + b*x^2)) + 12*b^2*Log[x] - 6*b^2*Lo g[a + b*x^2])/(4*a^4)
Time = 0.38 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.03, 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.
\(\displaystyle \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle b^2 \int \frac {1}{b^2 x^5 \left (b x^2+a\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1}{x^5 \left (a+b x^2\right )^2}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^6 \left (b x^2+a\right )^2}dx^2\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {3 b^3}{a^4 \left (b x^2+a\right )}-\frac {b^3}{a^3 \left (b x^2+a\right )^2}+\frac {3 b^2}{a^4 x^2}-\frac {2 b}{a^3 x^4}+\frac {1}{a^2 x^6}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {3 b^2 \log \left (x^2\right )}{a^4}-\frac {3 b^2 \log \left (a+b x^2\right )}{a^4}+\frac {b^2}{a^3 \left (a+b x^2\right )}+\frac {2 b}{a^3 x^2}-\frac {1}{2 a^2 x^4}\right )\) |
Input:
Int[1/(x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)),x]
Output:
(-1/2*1/(a^2*x^4) + (2*b)/(a^3*x^2) + b^2/(a^3*(a + b*x^2)) + (3*b^2*Log[x ^2])/a^4 - (3*b^2*Log[a + b*x^2])/a^4)/2
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 = 0.10 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.98
method | result | size |
default | \(-\frac {b^{3} \left (\frac {3 \ln \left (b \,x^{2}+a \right )}{b}-\frac {a}{b \left (b \,x^{2}+a \right )}\right )}{2 a^{4}}-\frac {1}{4 a^{2} x^{4}}+\frac {b}{a^{3} x^{2}}+\frac {3 b^{2} \ln \left (x \right )}{a^{4}}\) | \(65\) |
norman | \(\frac {-\frac {1}{4 a}+\frac {3 b \,x^{2}}{4 a^{2}}-\frac {3 b^{3} x^{6}}{2 a^{4}}}{x^{4} \left (b \,x^{2}+a \right )}+\frac {3 b^{2} \ln \left (x \right )}{a^{4}}-\frac {3 b^{2} \ln \left (b \,x^{2}+a \right )}{2 a^{4}}\) | \(67\) |
risch | \(\frac {\frac {3 b^{2} x^{4}}{2 a^{3}}+\frac {3 b \,x^{2}}{4 a^{2}}-\frac {1}{4 a}}{x^{4} \left (b \,x^{2}+a \right )}+\frac {3 b^{2} \ln \left (x \right )}{a^{4}}-\frac {3 b^{2} \ln \left (b \,x^{2}+a \right )}{2 a^{4}}\) | \(67\) |
parallelrisch | \(\frac {12 \ln \left (x \right ) x^{6} b^{3}-6 \ln \left (b \,x^{2}+a \right ) x^{6} b^{3}-6 b^{3} x^{6}+12 \ln \left (x \right ) x^{4} a \,b^{2}-6 \ln \left (b \,x^{2}+a \right ) x^{4} a \,b^{2}+3 a^{2} b \,x^{2}-a^{3}}{4 a^{4} x^{4} \left (b \,x^{2}+a \right )}\) | \(95\) |
Input:
int(1/x^5/(b^2*x^4+2*a*b*x^2+a^2),x,method=_RETURNVERBOSE)
Output:
-1/2*b^3/a^4*(3/b*ln(b*x^2+a)-a/b/(b*x^2+a))-1/4/a^2/x^4+b/a^3/x^2+3*b^2*l n(x)/a^4
Time = 0.12 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.36 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {6 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} - a^{3} - 6 \, {\left (b^{3} x^{6} + a b^{2} x^{4}\right )} \log \left (b x^{2} + a\right ) + 12 \, {\left (b^{3} x^{6} + a b^{2} x^{4}\right )} \log \left (x\right )}{4 \, {\left (a^{4} b x^{6} + a^{5} x^{4}\right )}} \] Input:
integrate(1/x^5/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="fricas")
Output:
1/4*(6*a*b^2*x^4 + 3*a^2*b*x^2 - a^3 - 6*(b^3*x^6 + a*b^2*x^4)*log(b*x^2 + a) + 12*(b^3*x^6 + a*b^2*x^4)*log(x))/(a^4*b*x^6 + a^5*x^4)
Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {- a^{2} + 3 a b x^{2} + 6 b^{2} x^{4}}{4 a^{4} x^{4} + 4 a^{3} b x^{6}} + \frac {3 b^{2} \log {\left (x \right )}}{a^{4}} - \frac {3 b^{2} \log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{4}} \] Input:
integrate(1/x**5/(b**2*x**4+2*a*b*x**2+a**2),x)
Output:
(-a**2 + 3*a*b*x**2 + 6*b**2*x**4)/(4*a**4*x**4 + 4*a**3*b*x**6) + 3*b**2* log(x)/a**4 - 3*b**2*log(a/b + x**2)/(2*a**4)
Time = 0.03 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {6 \, b^{2} x^{4} + 3 \, a b x^{2} - a^{2}}{4 \, {\left (a^{3} b x^{6} + a^{4} x^{4}\right )}} - \frac {3 \, b^{2} \log \left (b x^{2} + a\right )}{2 \, a^{4}} + \frac {3 \, b^{2} \log \left (x^{2}\right )}{2 \, a^{4}} \] Input:
integrate(1/x^5/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="maxima")
Output:
1/4*(6*b^2*x^4 + 3*a*b*x^2 - a^2)/(a^3*b*x^6 + a^4*x^4) - 3/2*b^2*log(b*x^ 2 + a)/a^4 + 3/2*b^2*log(x^2)/a^4
Time = 0.13 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.30 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {3 \, b^{2} \log \left (x^{2}\right )}{2 \, a^{4}} - \frac {3 \, b^{2} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{4}} + \frac {3 \, b^{3} x^{2} + 4 \, a b^{2}}{2 \, {\left (b x^{2} + a\right )} a^{4}} - \frac {9 \, b^{2} x^{4} - 4 \, a b x^{2} + a^{2}}{4 \, a^{4} x^{4}} \] Input:
integrate(1/x^5/(b^2*x^4+2*a*b*x^2+a^2),x, algorithm="giac")
Output:
3/2*b^2*log(x^2)/a^4 - 3/2*b^2*log(abs(b*x^2 + a))/a^4 + 1/2*(3*b^3*x^2 + 4*a*b^2)/((b*x^2 + a)*a^4) - 1/4*(9*b^2*x^4 - 4*a*b*x^2 + a^2)/(a^4*x^4)
Time = 0.07 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {\frac {3\,b\,x^2}{4\,a^2}-\frac {1}{4\,a}+\frac {3\,b^2\,x^4}{2\,a^3}}{b\,x^6+a\,x^4}-\frac {3\,b^2\,\ln \left (b\,x^2+a\right )}{2\,a^4}+\frac {3\,b^2\,\ln \left (x\right )}{a^4} \] Input:
int(1/(x^5*(a^2 + b^2*x^4 + 2*a*b*x^2)),x)
Output:
((3*b*x^2)/(4*a^2) - 1/(4*a) + (3*b^2*x^4)/(2*a^3))/(a*x^4 + b*x^6) - (3*b ^2*log(a + b*x^2))/(2*a^4) + (3*b^2*log(x))/a^4
Time = 0.16 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.42 \[ \int \frac {1}{x^5 \left (a^2+2 a b x^2+b^2 x^4\right )} \, dx=\frac {-6 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} x^{4}-6 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{3} x^{6}+12 \,\mathrm {log}\left (x \right ) a \,b^{2} x^{4}+12 \,\mathrm {log}\left (x \right ) b^{3} x^{6}-a^{3}+3 a^{2} b \,x^{2}-6 b^{3} x^{6}}{4 a^{4} x^{4} \left (b \,x^{2}+a \right )} \] Input:
int(1/x^5/(b^2*x^4+2*a*b*x^2+a^2),x)
Output:
( - 6*log(a + b*x**2)*a*b**2*x**4 - 6*log(a + b*x**2)*b**3*x**6 + 12*log(x )*a*b**2*x**4 + 12*log(x)*b**3*x**6 - a**3 + 3*a**2*b*x**2 - 6*b**3*x**6)/ (4*a**4*x**4*(a + b*x**2))