Integrand size = 24, antiderivative size = 70 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {1}{6 a \left (a+b x^2\right )^3}+\frac {1}{4 a^2 \left (a+b x^2\right )^2}+\frac {1}{2 a^3 \left (a+b x^2\right )}+\frac {\log (x)}{a^4}-\frac {\log \left (a+b x^2\right )}{2 a^4} \] Output:
1/6/a/(b*x^2+a)^3+1/4/a^2/(b*x^2+a)^2+1/2/a^3/(b*x^2+a)+ln(x)/a^4-1/2*ln(b *x^2+a)/a^4
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {a \left (11 a^2+15 a b x^2+6 b^2 x^4\right )}{\left (a+b x^2\right )^3}+12 \log (x)-6 \log \left (a+b x^2\right )}{12 a^4} \] Input:
Integrate[1/(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^2),x]
Output:
((a*(11*a^2 + 15*a*b*x^2 + 6*b^2*x^4))/(a + b*x^2)^3 + 12*Log[x] - 6*Log[a + b*x^2])/(12*a^4)
Time = 0.38 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01, 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.
\(\displaystyle \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle b^4 \int \frac {1}{b^4 x \left (b x^2+a\right )^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1}{x \left (a+b x^2\right )^4}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^2 \left (b x^2+a\right )^4}dx^2\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {b}{a^4 \left (b x^2+a\right )}-\frac {b}{a^3 \left (b x^2+a\right )^2}-\frac {b}{a^2 \left (b x^2+a\right )^3}-\frac {b}{a \left (b x^2+a\right )^4}+\frac {1}{a^4 x^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {\log \left (a+b x^2\right )}{a^4}+\frac {\log \left (x^2\right )}{a^4}+\frac {1}{a^3 \left (a+b x^2\right )}+\frac {1}{2 a^2 \left (a+b x^2\right )^2}+\frac {1}{3 a \left (a+b x^2\right )^3}\right )\) |
Input:
Int[1/(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^2),x]
Output:
(1/(3*a*(a + b*x^2)^3) + 1/(2*a^2*(a + b*x^2)^2) + 1/(a^3*(a + b*x^2)) + L og[x^2]/a^4 - 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 = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90
method | result | size |
norman | \(\frac {-\frac {3 b \,x^{2}}{2 a^{2}}-\frac {9 b^{2} x^{4}}{4 a^{3}}-\frac {11 b^{3} x^{6}}{12 a^{4}}}{\left (b \,x^{2}+a \right )^{3}}+\frac {\ln \left (x \right )}{a^{4}}-\frac {\ln \left (b \,x^{2}+a \right )}{2 a^{4}}\) | \(63\) |
default | \(-\frac {b \left (\frac {\ln \left (b \,x^{2}+a \right )}{b}-\frac {a^{3}}{3 b \left (b \,x^{2}+a \right )^{3}}-\frac {a^{2}}{2 b \left (b \,x^{2}+a \right )^{2}}-\frac {a}{b \left (b \,x^{2}+a \right )}\right )}{2 a^{4}}+\frac {\ln \left (x \right )}{a^{4}}\) | \(76\) |
risch | \(\frac {\frac {b^{2} x^{4}}{2 a^{3}}+\frac {5 b \,x^{2}}{4 a^{2}}+\frac {11}{12 a}}{\left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}+\frac {\ln \left (x \right )}{a^{4}}-\frac {\ln \left (b \,x^{2}+a \right )}{2 a^{4}}\) | \(77\) |
parallelrisch | \(\frac {12 \ln \left (x \right ) x^{6} b^{3}-6 \ln \left (b \,x^{2}+a \right ) x^{6} b^{3}-11 b^{3} x^{6}+36 \ln \left (x \right ) x^{4} a \,b^{2}-18 \ln \left (b \,x^{2}+a \right ) x^{4} a \,b^{2}-27 b^{2} x^{4} a +36 \ln \left (x \right ) x^{2} a^{2} b -18 \ln \left (b \,x^{2}+a \right ) x^{2} a^{2} b -18 a^{2} b \,x^{2}+12 a^{3} \ln \left (x \right )-6 \ln \left (b \,x^{2}+a \right ) a^{3}}{12 a^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \left (b \,x^{2}+a \right )}\) | \(164\) |
Input:
int(1/x/(b^2*x^4+2*a*b*x^2+a^2)^2,x,method=_RETURNVERBOSE)
Output:
(-3/2*b/a^2*x^2-9/4*b^2/a^3*x^4-11/12*b^3/a^4*x^6)/(b*x^2+a)^3+ln(x)/a^4-1 /2*ln(b*x^2+a)/a^4
Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (62) = 124\).
Time = 0.07 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.91 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {6 \, a b^{2} x^{4} + 15 \, a^{2} b x^{2} + 11 \, a^{3} - 6 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \log \left (b x^{2} + a\right ) + 12 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \log \left (x\right )}{12 \, {\left (a^{4} b^{3} x^{6} + 3 \, a^{5} b^{2} x^{4} + 3 \, a^{6} b x^{2} + a^{7}\right )}} \] Input:
integrate(1/x/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")
Output:
1/12*(6*a*b^2*x^4 + 15*a^2*b*x^2 + 11*a^3 - 6*(b^3*x^6 + 3*a*b^2*x^4 + 3*a ^2*b*x^2 + a^3)*log(b*x^2 + a) + 12*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*log(x))/(a^4*b^3*x^6 + 3*a^5*b^2*x^4 + 3*a^6*b*x^2 + a^7)
Time = 0.24 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {11 a^{2} + 15 a b x^{2} + 6 b^{2} x^{4}}{12 a^{6} + 36 a^{5} b x^{2} + 36 a^{4} b^{2} x^{4} + 12 a^{3} b^{3} x^{6}} + \frac {\log {\left (x \right )}}{a^{4}} - \frac {\log {\left (\frac {a}{b} + x^{2} \right )}}{2 a^{4}} \] Input:
integrate(1/x/(b**2*x**4+2*a*b*x**2+a**2)**2,x)
Output:
(11*a**2 + 15*a*b*x**2 + 6*b**2*x**4)/(12*a**6 + 36*a**5*b*x**2 + 36*a**4* b**2*x**4 + 12*a**3*b**3*x**6) + log(x)/a**4 - log(a/b + x**2)/(2*a**4)
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {6 \, b^{2} x^{4} + 15 \, a b x^{2} + 11 \, a^{2}}{12 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )}} - \frac {\log \left (b x^{2} + a\right )}{2 \, a^{4}} + \frac {\log \left (x^{2}\right )}{2 \, a^{4}} \] Input:
integrate(1/x/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")
Output:
1/12*(6*b^2*x^4 + 15*a*b*x^2 + 11*a^2)/(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3*a^ 5*b*x^2 + a^6) - 1/2*log(b*x^2 + a)/a^4 + 1/2*log(x^2)/a^4
Time = 0.12 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\log \left (x^{2}\right )}{2 \, a^{4}} - \frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{4}} + \frac {11 \, b^{3} x^{6} + 39 \, a b^{2} x^{4} + 48 \, a^{2} b x^{2} + 22 \, a^{3}}{12 \, {\left (b x^{2} + a\right )}^{3} a^{4}} \] Input:
integrate(1/x/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")
Output:
1/2*log(x^2)/a^4 - 1/2*log(abs(b*x^2 + a))/a^4 + 1/12*(11*b^3*x^6 + 39*a*b ^2*x^4 + 48*a^2*b*x^2 + 22*a^3)/((b*x^2 + a)^3*a^4)
Time = 0.07 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\ln \left (x\right )}{a^4}+\frac {\frac {11}{12\,a}+\frac {5\,b\,x^2}{4\,a^2}+\frac {b^2\,x^4}{2\,a^3}}{a^3+3\,a^2\,b\,x^2+3\,a\,b^2\,x^4+b^3\,x^6}-\frac {\ln \left (b\,x^2+a\right )}{2\,a^4} \] Input:
int(1/(x*(a^2 + b^2*x^4 + 2*a*b*x^2)^2),x)
Output:
log(x)/a^4 + (11/(12*a) + (5*b*x^2)/(4*a^2) + (b^2*x^4)/(2*a^3))/(a^3 + b^ 3*x^6 + 3*a^2*b*x^2 + 3*a*b^2*x^4) - log(a + b*x^2)/(2*a^4)
Time = 0.16 (sec) , antiderivative size = 161, normalized size of antiderivative = 2.30 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {-6 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3}-18 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b \,x^{2}-18 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} x^{4}-6 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{3} x^{6}+12 \,\mathrm {log}\left (x \right ) a^{3}+36 \,\mathrm {log}\left (x \right ) a^{2} b \,x^{2}+36 \,\mathrm {log}\left (x \right ) a \,b^{2} x^{4}+12 \,\mathrm {log}\left (x \right ) b^{3} x^{6}+9 a^{3}+9 a^{2} b \,x^{2}-2 b^{3} x^{6}}{12 a^{4} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:
int(1/x/(b^2*x^4+2*a*b*x^2+a^2)^2,x)
Output:
( - 6*log(a + b*x**2)*a**3 - 18*log(a + b*x**2)*a**2*b*x**2 - 18*log(a + b *x**2)*a*b**2*x**4 - 6*log(a + b*x**2)*b**3*x**6 + 12*log(x)*a**3 + 36*log (x)*a**2*b*x**2 + 36*log(x)*a*b**2*x**4 + 12*log(x)*b**3*x**6 + 9*a**3 + 9 *a**2*b*x**2 - 2*b**3*x**6)/(12*a**4*(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x**6))