Integrand size = 24, antiderivative size = 84 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {1}{2 a^4 x^2}-\frac {b}{6 a^2 \left (a+b x^2\right )^3}-\frac {b}{2 a^3 \left (a+b x^2\right )^2}-\frac {3 b}{2 a^4 \left (a+b x^2\right )}-\frac {4 b \log (x)}{a^5}+\frac {2 b \log \left (a+b x^2\right )}{a^5} \] Output:
-1/2/a^4/x^2-1/6*b/a^2/(b*x^2+a)^3-1/2*b/a^3/(b*x^2+a)^2-3/2*b/a^4/(b*x^2+ a)-4*b*ln(x)/a^5+2*b*ln(b*x^2+a)/a^5
Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {\frac {a \left (3 a^3+22 a^2 b x^2+30 a b^2 x^4+12 b^3 x^6\right )}{x^2 \left (a+b x^2\right )^3}+24 b \log (x)-12 b \log \left (a+b x^2\right )}{6 a^5} \] Input:
Integrate[1/(x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^2),x]
Output:
-1/6*((a*(3*a^3 + 22*a^2*b*x^2 + 30*a*b^2*x^4 + 12*b^3*x^6))/(x^2*(a + b*x ^2)^3) + 24*b*Log[x] - 12*b*Log[a + b*x^2])/a^5
Time = 0.41 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, 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.
\(\displaystyle \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle b^4 \int \frac {1}{b^4 x^3 \left (b x^2+a\right )^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {1}{x^3 \left (a+b x^2\right )^4}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (b x^2+a\right )^4}dx^2\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{2} \int \left (\frac {4 b^2}{a^5 \left (b x^2+a\right )}+\frac {3 b^2}{a^4 \left (b x^2+a\right )^2}+\frac {2 b^2}{a^3 \left (b x^2+a\right )^3}+\frac {b^2}{a^2 \left (b x^2+a\right )^4}-\frac {4 b}{a^5 x^2}+\frac {1}{a^4 x^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {4 b \log \left (x^2\right )}{a^5}+\frac {4 b \log \left (a+b x^2\right )}{a^5}-\frac {3 b}{a^4 \left (a+b x^2\right )}-\frac {1}{a^4 x^2}-\frac {b}{a^3 \left (a+b x^2\right )^2}-\frac {b}{3 a^2 \left (a+b x^2\right )^3}\right )\) |
Input:
Int[1/(x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^2),x]
Output:
(-(1/(a^4*x^2)) - b/(3*a^2*(a + b*x^2)^3) - b/(a^3*(a + b*x^2)^2) - (3*b)/ (a^4*(a + b*x^2)) - (4*b*Log[x^2])/a^5 + (4*b*Log[a + b*x^2])/a^5)/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 = 0.13 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.90
method | result | size |
norman | \(\frac {-\frac {1}{2 a}+\frac {6 b^{2} x^{4}}{a^{3}}+\frac {9 b^{3} x^{6}}{a^{4}}+\frac {11 b^{4} x^{8}}{3 a^{5}}}{x^{2} \left (b \,x^{2}+a \right )^{3}}-\frac {4 b \ln \left (x \right )}{a^{5}}+\frac {2 b \ln \left (b \,x^{2}+a \right )}{a^{5}}\) | \(76\) |
default | \(\frac {b^{2} \left (\frac {4 \ln \left (b \,x^{2}+a \right )}{b}-\frac {a^{3}}{3 b \left (b \,x^{2}+a \right )^{3}}-\frac {a^{2}}{b \left (b \,x^{2}+a \right )^{2}}-\frac {3 a}{b \left (b \,x^{2}+a \right )}\right )}{2 a^{5}}-\frac {1}{2 a^{4} x^{2}}-\frac {4 b \ln \left (x \right )}{a^{5}}\) | \(89\) |
risch | \(\frac {-\frac {2 b^{3} x^{6}}{a^{4}}-\frac {5 b^{2} x^{4}}{a^{3}}-\frac {11 b \,x^{2}}{3 a^{2}}-\frac {1}{2 a}}{x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \left (b \,x^{2}+a \right )}-\frac {4 b \ln \left (x \right )}{a^{5}}+\frac {2 b \ln \left (-b \,x^{2}-a \right )}{a^{5}}\) | \(97\) |
parallelrisch | \(-\frac {24 b^{4} \ln \left (x \right ) x^{8}-12 \ln \left (b \,x^{2}+a \right ) x^{8} b^{4}-22 b^{4} x^{8}+72 b^{3} a \ln \left (x \right ) x^{6}-36 \ln \left (b \,x^{2}+a \right ) x^{6} a \,b^{3}-54 a \,b^{3} x^{6}+72 a^{2} b^{2} \ln \left (x \right ) x^{4}-36 \ln \left (b \,x^{2}+a \right ) x^{4} a^{2} b^{2}-36 a^{2} b^{2} x^{4}+24 a^{3} b \ln \left (x \right ) x^{2}-12 \ln \left (b \,x^{2}+a \right ) x^{2} a^{3} b +3 a^{4}}{6 a^{5} x^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \left (b \,x^{2}+a \right )}\) | \(186\) |
Input:
int(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^2,x,method=_RETURNVERBOSE)
Output:
(-1/2/a+6*b^2/a^3*x^4+9*b^3/a^4*x^6+11/3*b^4/a^5*x^8)/x^2/(b*x^2+a)^3-4*b* ln(x)/a^5+2*b*ln(b*x^2+a)/a^5
Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (76) = 152\).
Time = 0.07 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.94 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {12 \, a b^{3} x^{6} + 30 \, a^{2} b^{2} x^{4} + 22 \, a^{3} b x^{2} + 3 \, a^{4} - 12 \, {\left (b^{4} x^{8} + 3 \, a b^{3} x^{6} + 3 \, a^{2} b^{2} x^{4} + a^{3} b x^{2}\right )} \log \left (b x^{2} + a\right ) + 24 \, {\left (b^{4} x^{8} + 3 \, a b^{3} x^{6} + 3 \, a^{2} b^{2} x^{4} + a^{3} b x^{2}\right )} \log \left (x\right )}{6 \, {\left (a^{5} b^{3} x^{8} + 3 \, a^{6} b^{2} x^{6} + 3 \, a^{7} b x^{4} + a^{8} x^{2}\right )}} \] Input:
integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")
Output:
-1/6*(12*a*b^3*x^6 + 30*a^2*b^2*x^4 + 22*a^3*b*x^2 + 3*a^4 - 12*(b^4*x^8 + 3*a*b^3*x^6 + 3*a^2*b^2*x^4 + a^3*b*x^2)*log(b*x^2 + a) + 24*(b^4*x^8 + 3 *a*b^3*x^6 + 3*a^2*b^2*x^4 + a^3*b*x^2)*log(x))/(a^5*b^3*x^8 + 3*a^6*b^2*x ^6 + 3*a^7*b*x^4 + a^8*x^2)
Time = 0.29 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {- 3 a^{3} - 22 a^{2} b x^{2} - 30 a b^{2} x^{4} - 12 b^{3} x^{6}}{6 a^{7} x^{2} + 18 a^{6} b x^{4} + 18 a^{5} b^{2} x^{6} + 6 a^{4} b^{3} x^{8}} - \frac {4 b \log {\left (x \right )}}{a^{5}} + \frac {2 b \log {\left (\frac {a}{b} + x^{2} \right )}}{a^{5}} \] Input:
integrate(1/x**3/(b**2*x**4+2*a*b*x**2+a**2)**2,x)
Output:
(-3*a**3 - 22*a**2*b*x**2 - 30*a*b**2*x**4 - 12*b**3*x**6)/(6*a**7*x**2 + 18*a**6*b*x**4 + 18*a**5*b**2*x**6 + 6*a**4*b**3*x**8) - 4*b*log(x)/a**5 + 2*b*log(a/b + x**2)/a**5
Time = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.18 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {12 \, b^{3} x^{6} + 30 \, a b^{2} x^{4} + 22 \, a^{2} b x^{2} + 3 \, a^{3}}{6 \, {\left (a^{4} b^{3} x^{8} + 3 \, a^{5} b^{2} x^{6} + 3 \, a^{6} b x^{4} + a^{7} x^{2}\right )}} + \frac {2 \, b \log \left (b x^{2} + a\right )}{a^{5}} - \frac {2 \, b \log \left (x^{2}\right )}{a^{5}} \] Input:
integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")
Output:
-1/6*(12*b^3*x^6 + 30*a*b^2*x^4 + 22*a^2*b*x^2 + 3*a^3)/(a^4*b^3*x^8 + 3*a ^5*b^2*x^6 + 3*a^6*b*x^4 + a^7*x^2) + 2*b*log(b*x^2 + a)/a^5 - 2*b*log(x^2 )/a^5
Time = 0.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {2 \, b \log \left (x^{2}\right )}{a^{5}} + \frac {2 \, b \log \left ({\left | b x^{2} + a \right |}\right )}{a^{5}} + \frac {4 \, b x^{2} - a}{2 \, a^{5} x^{2}} - \frac {22 \, b^{4} x^{6} + 75 \, a b^{3} x^{4} + 87 \, a^{2} b^{2} x^{2} + 35 \, a^{3} b}{6 \, {\left (b x^{2} + a\right )}^{3} a^{5}} \] Input:
integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")
Output:
-2*b*log(x^2)/a^5 + 2*b*log(abs(b*x^2 + a))/a^5 + 1/2*(4*b*x^2 - a)/(a^5*x ^2) - 1/6*(22*b^4*x^6 + 75*a*b^3*x^4 + 87*a^2*b^2*x^2 + 35*a^3*b)/((b*x^2 + a)^3*a^5)
Time = 17.48 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {2\,b\,\ln \left (b\,x^2+a\right )}{a^5}-\frac {\frac {1}{2\,a}+\frac {11\,b\,x^2}{3\,a^2}+\frac {5\,b^2\,x^4}{a^3}+\frac {2\,b^3\,x^6}{a^4}}{a^3\,x^2+3\,a^2\,b\,x^4+3\,a\,b^2\,x^6+b^3\,x^8}-\frac {4\,b\,\ln \left (x\right )}{a^5} \] Input:
int(1/(x^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^2),x)
Output:
(2*b*log(a + b*x^2))/a^5 - (1/(2*a) + (11*b*x^2)/(3*a^2) + (5*b^2*x^4)/a^3 + (2*b^3*x^6)/a^4)/(a^3*x^2 + b^3*x^8 + 3*a^2*b*x^4 + 3*a*b^2*x^6) - (4*b *log(x))/a^5
Time = 0.17 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.23 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {12 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3} b \,x^{2}+36 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b^{2} x^{4}+36 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{3} x^{6}+12 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{4} x^{8}-24 \,\mathrm {log}\left (x \right ) a^{3} b \,x^{2}-72 \,\mathrm {log}\left (x \right ) a^{2} b^{2} x^{4}-72 \,\mathrm {log}\left (x \right ) a \,b^{3} x^{6}-24 \,\mathrm {log}\left (x \right ) b^{4} x^{8}-3 a^{4}-18 a^{3} b \,x^{2}-18 a^{2} b^{2} x^{4}+4 b^{4} x^{8}}{6 a^{5} x^{2} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:
int(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^2,x)
Output:
(12*log(a + b*x**2)*a**3*b*x**2 + 36*log(a + b*x**2)*a**2*b**2*x**4 + 36*l og(a + b*x**2)*a*b**3*x**6 + 12*log(a + b*x**2)*b**4*x**8 - 24*log(x)*a**3 *b*x**2 - 72*log(x)*a**2*b**2*x**4 - 72*log(x)*a*b**3*x**6 - 24*log(x)*b** 4*x**8 - 3*a**4 - 18*a**3*b*x**2 - 18*a**2*b**2*x**4 + 4*b**4*x**8)/(6*a** 5*x**2*(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x**6))