Integrand size = 17, antiderivative size = 95 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^3} \, dx=-\frac {1}{6 b^3 x^6}+\frac {3 c}{4 b^4 x^4}-\frac {3 c^2}{b^5 x^2}-\frac {c^3}{4 b^4 \left (b+c x^2\right )^2}-\frac {2 c^3}{b^5 \left (b+c x^2\right )}-\frac {10 c^3 \log (x)}{b^6}+\frac {5 c^3 \log \left (b+c x^2\right )}{b^6} \] Output:
-1/6/b^3/x^6+3/4*c/b^4/x^4-3*c^2/b^5/x^2-1/4*c^3/b^4/(c*x^2+b)^2-2*c^3/b^5 /(c*x^2+b)-10*c^3*ln(x)/b^6+5*c^3*ln(c*x^2+b)/b^6
Time = 0.05 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^3} \, dx=-\frac {\frac {b \left (2 b^4-5 b^3 c x^2+20 b^2 c^2 x^4+90 b c^3 x^6+60 c^4 x^8\right )}{x^6 \left (b+c x^2\right )^2}+120 c^3 \log (x)-60 c^3 \log \left (b+c x^2\right )}{12 b^6} \] Input:
Integrate[1/(x*(b*x^2 + c*x^4)^3),x]
Output:
-1/12*((b*(2*b^4 - 5*b^3*c*x^2 + 20*b^2*c^2*x^4 + 90*b*c^3*x^6 + 60*c^4*x^ 8))/(x^6*(b + c*x^2)^2) + 120*c^3*Log[x] - 60*c^3*Log[b + c*x^2])/b^6
Time = 0.42 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {9, 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.
\(\displaystyle \int \frac {1}{x \left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {1}{x^7 \left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^8 \left (c x^2+b\right )^3}dx^2\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{2} \int \left (\frac {10 c^4}{b^6 \left (c x^2+b\right )}+\frac {4 c^4}{b^5 \left (c x^2+b\right )^2}+\frac {c^4}{b^4 \left (c x^2+b\right )^3}-\frac {10 c^3}{b^6 x^2}+\frac {6 c^2}{b^5 x^4}-\frac {3 c}{b^4 x^6}+\frac {1}{b^3 x^8}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {10 c^3 \log \left (x^2\right )}{b^6}+\frac {10 c^3 \log \left (b+c x^2\right )}{b^6}-\frac {4 c^3}{b^5 \left (b+c x^2\right )}-\frac {6 c^2}{b^5 x^2}-\frac {c^3}{2 b^4 \left (b+c x^2\right )^2}+\frac {3 c}{2 b^4 x^4}-\frac {1}{3 b^3 x^6}\right )\) |
Input:
Int[1/(x*(b*x^2 + c*x^4)^3),x]
Output:
(-1/3*1/(b^3*x^6) + (3*c)/(2*b^4*x^4) - (6*c^2)/(b^5*x^2) - c^3/(2*b^4*(b + c*x^2)^2) - (4*c^3)/(b^5*(b + c*x^2)) - (10*c^3*Log[x^2])/b^6 + (10*c^3* Log[b + c*x^2])/b^6)/2
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, 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]
Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.94
method | result | size |
norman | \(\frac {-\frac {1}{6 b}+\frac {5 c \,x^{2}}{12 b^{2}}-\frac {5 c^{2} x^{4}}{3 b^{3}}+\frac {10 c^{4} x^{8}}{b^{5}}+\frac {15 c^{5} x^{10}}{2 b^{6}}}{x^{6} \left (c \,x^{2}+b \right )^{2}}-\frac {10 c^{3} \ln \left (x \right )}{b^{6}}+\frac {5 c^{3} \ln \left (c \,x^{2}+b \right )}{b^{6}}\) | \(89\) |
risch | \(\frac {-\frac {5 c^{4} x^{8}}{b^{5}}-\frac {15 c^{3} x^{6}}{2 b^{4}}-\frac {5 c^{2} x^{4}}{3 b^{3}}+\frac {5 c \,x^{2}}{12 b^{2}}-\frac {1}{6 b}}{x^{6} \left (c \,x^{2}+b \right )^{2}}-\frac {10 c^{3} \ln \left (x \right )}{b^{6}}+\frac {5 c^{3} \ln \left (-c \,x^{2}-b \right )}{b^{6}}\) | \(92\) |
default | \(-\frac {1}{6 b^{3} x^{6}}+\frac {3 c}{4 b^{4} x^{4}}-\frac {3 c^{2}}{b^{5} x^{2}}-\frac {10 c^{3} \ln \left (x \right )}{b^{6}}+\frac {c^{4} \left (-\frac {b^{2}}{2 c \left (c \,x^{2}+b \right )^{2}}+\frac {10 \ln \left (c \,x^{2}+b \right )}{c}-\frac {4 b}{c \left (c \,x^{2}+b \right )}\right )}{2 b^{6}}\) | \(94\) |
parallelrisch | \(-\frac {120 \ln \left (x \right ) x^{10} c^{5}-60 \ln \left (c \,x^{2}+b \right ) x^{10} c^{5}-90 c^{5} x^{10}+240 \ln \left (x \right ) x^{8} b \,c^{4}-120 \ln \left (c \,x^{2}+b \right ) x^{8} b \,c^{4}-120 c^{4} x^{8} b +120 \ln \left (x \right ) x^{6} b^{2} c^{3}-60 \ln \left (c \,x^{2}+b \right ) x^{6} b^{2} c^{3}+20 c^{2} x^{4} b^{3}-5 x^{2} c \,b^{4}+2 b^{5}}{12 b^{6} x^{6} \left (c \,x^{2}+b \right )^{2}}\) | \(147\) |
Input:
int(1/x/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
(-1/6/b+5/12*c/b^2*x^2-5/3*c^2/b^3*x^4+10*c^4/b^5*x^8+15/2*c^5/b^6*x^10)/x ^6/(c*x^2+b)^2-10*c^3*ln(x)/b^6+5*c^3*ln(c*x^2+b)/b^6
Time = 0.18 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.53 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^3} \, dx=-\frac {60 \, b c^{4} x^{8} + 90 \, b^{2} c^{3} x^{6} + 20 \, b^{3} c^{2} x^{4} - 5 \, b^{4} c x^{2} + 2 \, b^{5} - 60 \, {\left (c^{5} x^{10} + 2 \, b c^{4} x^{8} + b^{2} c^{3} x^{6}\right )} \log \left (c x^{2} + b\right ) + 120 \, {\left (c^{5} x^{10} + 2 \, b c^{4} x^{8} + b^{2} c^{3} x^{6}\right )} \log \left (x\right )}{12 \, {\left (b^{6} c^{2} x^{10} + 2 \, b^{7} c x^{8} + b^{8} x^{6}\right )}} \] Input:
integrate(1/x/(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
-1/12*(60*b*c^4*x^8 + 90*b^2*c^3*x^6 + 20*b^3*c^2*x^4 - 5*b^4*c*x^2 + 2*b^ 5 - 60*(c^5*x^10 + 2*b*c^4*x^8 + b^2*c^3*x^6)*log(c*x^2 + b) + 120*(c^5*x^ 10 + 2*b*c^4*x^8 + b^2*c^3*x^6)*log(x))/(b^6*c^2*x^10 + 2*b^7*c*x^8 + b^8* x^6)
Time = 0.28 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^3} \, dx=\frac {- 2 b^{4} + 5 b^{3} c x^{2} - 20 b^{2} c^{2} x^{4} - 90 b c^{3} x^{6} - 60 c^{4} x^{8}}{12 b^{7} x^{6} + 24 b^{6} c x^{8} + 12 b^{5} c^{2} x^{10}} - \frac {10 c^{3} \log {\left (x \right )}}{b^{6}} + \frac {5 c^{3} \log {\left (\frac {b}{c} + x^{2} \right )}}{b^{6}} \] Input:
integrate(1/x/(c*x**4+b*x**2)**3,x)
Output:
(-2*b**4 + 5*b**3*c*x**2 - 20*b**2*c**2*x**4 - 90*b*c**3*x**6 - 60*c**4*x* *8)/(12*b**7*x**6 + 24*b**6*c*x**8 + 12*b**5*c**2*x**10) - 10*c**3*log(x)/ b**6 + 5*c**3*log(b/c + x**2)/b**6
Time = 0.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^3} \, dx=-\frac {60 \, c^{4} x^{8} + 90 \, b c^{3} x^{6} + 20 \, b^{2} c^{2} x^{4} - 5 \, b^{3} c x^{2} + 2 \, b^{4}}{12 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )}} + \frac {5 \, c^{3} \log \left (c x^{2} + b\right )}{b^{6}} - \frac {5 \, c^{3} \log \left (x^{2}\right )}{b^{6}} \] Input:
integrate(1/x/(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
-1/12*(60*c^4*x^8 + 90*b*c^3*x^6 + 20*b^2*c^2*x^4 - 5*b^3*c*x^2 + 2*b^4)/( b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6) + 5*c^3*log(c*x^2 + b)/b^6 - 5*c^3*l og(x^2)/b^6
Time = 0.14 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^3} \, dx=-\frac {5 \, c^{3} \log \left (x^{2}\right )}{b^{6}} + \frac {5 \, c^{3} \log \left ({\left | c x^{2} + b \right |}\right )}{b^{6}} - \frac {30 \, c^{5} x^{4} + 68 \, b c^{4} x^{2} + 39 \, b^{2} c^{3}}{4 \, {\left (c x^{2} + b\right )}^{2} b^{6}} + \frac {110 \, c^{3} x^{6} - 36 \, b c^{2} x^{4} + 9 \, b^{2} c x^{2} - 2 \, b^{3}}{12 \, b^{6} x^{6}} \] Input:
integrate(1/x/(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
-5*c^3*log(x^2)/b^6 + 5*c^3*log(abs(c*x^2 + b))/b^6 - 1/4*(30*c^5*x^4 + 68 *b*c^4*x^2 + 39*b^2*c^3)/((c*x^2 + b)^2*b^6) + 1/12*(110*c^3*x^6 - 36*b*c^ 2*x^4 + 9*b^2*c*x^2 - 2*b^3)/(b^6*x^6)
Time = 17.48 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^3} \, dx=\frac {5\,c^3\,\ln \left (c\,x^2+b\right )}{b^6}-\frac {\frac {1}{6\,b}-\frac {5\,c\,x^2}{12\,b^2}+\frac {5\,c^2\,x^4}{3\,b^3}+\frac {15\,c^3\,x^6}{2\,b^4}+\frac {5\,c^4\,x^8}{b^5}}{b^2\,x^6+2\,b\,c\,x^8+c^2\,x^{10}}-\frac {10\,c^3\,\ln \left (x\right )}{b^6} \] Input:
int(1/(x*(b*x^2 + c*x^4)^3),x)
Output:
(5*c^3*log(b + c*x^2))/b^6 - (1/(6*b) - (5*c*x^2)/(12*b^2) + (5*c^2*x^4)/( 3*b^3) + (15*c^3*x^6)/(2*b^4) + (5*c^4*x^8)/b^5)/(b^2*x^6 + c^2*x^10 + 2*b *c*x^8) - (10*c^3*log(x))/b^6
Time = 0.17 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.67 \[ \int \frac {1}{x \left (b x^2+c x^4\right )^3} \, dx=\frac {60 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{2} c^{3} x^{6}+120 \,\mathrm {log}\left (c \,x^{2}+b \right ) b \,c^{4} x^{8}+60 \,\mathrm {log}\left (c \,x^{2}+b \right ) c^{5} x^{10}-120 \,\mathrm {log}\left (x \right ) b^{2} c^{3} x^{6}-240 \,\mathrm {log}\left (x \right ) b \,c^{4} x^{8}-120 \,\mathrm {log}\left (x \right ) c^{5} x^{10}-2 b^{5}+5 b^{4} c \,x^{2}-20 b^{3} c^{2} x^{4}-60 b^{2} c^{3} x^{6}+30 c^{5} x^{10}}{12 b^{6} x^{6} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:
int(1/x/(c*x^4+b*x^2)^3,x)
Output:
(60*log(b + c*x**2)*b**2*c**3*x**6 + 120*log(b + c*x**2)*b*c**4*x**8 + 60* log(b + c*x**2)*c**5*x**10 - 120*log(x)*b**2*c**3*x**6 - 240*log(x)*b*c**4 *x**8 - 120*log(x)*c**5*x**10 - 2*b**5 + 5*b**4*c*x**2 - 20*b**3*c**2*x**4 - 60*b**2*c**3*x**6 + 30*c**5*x**10)/(12*b**6*x**6*(b**2 + 2*b*c*x**2 + c **2*x**4))