Integrand size = 15, antiderivative size = 86 \[ \int \frac {x}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {1}{4 b^3 x^4}+\frac {3 c}{2 b^4 x^2}+\frac {c^2}{4 b^3 \left (b+c x^2\right )^2}+\frac {3 c^2}{2 b^4 \left (b+c x^2\right )}+\frac {6 c^2 \log (x)}{b^5}-\frac {3 c^2 \log \left (b+c x^2\right )}{b^5} \] Output:
-1/4/b^3/x^4+3/2*c/b^4/x^2+1/4*c^2/b^3/(c*x^2+b)^2+3/2*c^2/b^4/(c*x^2+b)+6 *c^2*ln(x)/b^5-3*c^2*ln(c*x^2+b)/b^5
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \frac {x}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {b \left (-b^3+4 b^2 c x^2+18 b c^2 x^4+12 c^3 x^6\right )}{x^4 \left (b+c x^2\right )^2}+24 c^2 \log (x)-12 c^2 \log \left (b+c x^2\right )}{4 b^5} \] Input:
Integrate[x/(b*x^2 + c*x^4)^3,x]
Output:
((b*(-b^3 + 4*b^2*c*x^2 + 18*b*c^2*x^4 + 12*c^3*x^6))/(x^4*(b + c*x^2)^2) + 24*c^2*Log[x] - 12*c^2*Log[b + c*x^2])/(4*b^5)
Time = 0.39 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, 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.
\(\displaystyle \int \frac {x}{\left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {1}{x^5 \left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^6 \left (c x^2+b\right )^3}dx^2\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {6 c^3}{b^5 \left (c x^2+b\right )}-\frac {3 c^3}{b^4 \left (c x^2+b\right )^2}-\frac {c^3}{b^3 \left (c x^2+b\right )^3}+\frac {6 c^2}{b^5 x^2}-\frac {3 c}{b^4 x^4}+\frac {1}{b^3 x^6}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {6 c^2 \log \left (x^2\right )}{b^5}-\frac {6 c^2 \log \left (b+c x^2\right )}{b^5}+\frac {3 c^2}{b^4 \left (b+c x^2\right )}+\frac {3 c}{b^4 x^2}+\frac {c^2}{2 b^3 \left (b+c x^2\right )^2}-\frac {1}{2 b^3 x^4}\right )\) |
Input:
Int[x/(b*x^2 + c*x^4)^3,x]
Output:
(-1/2*1/(b^3*x^4) + (3*c)/(b^4*x^2) + c^2/(2*b^3*(b + c*x^2)^2) + (3*c^2)/ (b^4*(b + c*x^2)) + (6*c^2*Log[x^2])/b^5 - (6*c^2*Log[b + c*x^2])/b^5)/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 = 77, normalized size of antiderivative = 0.90
method | result | size |
risch | \(\frac {\frac {3 c^{3} x^{6}}{b^{4}}+\frac {9 c^{2} x^{4}}{2 b^{3}}+\frac {c \,x^{2}}{b^{2}}-\frac {1}{4 b}}{x^{4} \left (c \,x^{2}+b \right )^{2}}+\frac {6 c^{2} \ln \left (x \right )}{b^{5}}-\frac {3 c^{2} \ln \left (c \,x^{2}+b \right )}{b^{5}}\) | \(77\) |
norman | \(\frac {\frac {c \,x^{3}}{b^{2}}-\frac {x}{4 b}-\frac {6 c^{3} x^{7}}{b^{4}}-\frac {9 c^{4} x^{9}}{2 b^{5}}}{x^{5} \left (c \,x^{2}+b \right )^{2}}+\frac {6 c^{2} \ln \left (x \right )}{b^{5}}-\frac {3 c^{2} \ln \left (c \,x^{2}+b \right )}{b^{5}}\) | \(78\) |
default | \(-\frac {1}{4 b^{3} x^{4}}+\frac {3 c}{2 b^{4} x^{2}}+\frac {6 c^{2} \ln \left (x \right )}{b^{5}}-\frac {c^{3} \left (-\frac {b^{2}}{2 c \left (c \,x^{2}+b \right )^{2}}+\frac {6 \ln \left (c \,x^{2}+b \right )}{c}-\frac {3 b}{c \left (c \,x^{2}+b \right )}\right )}{2 b^{5}}\) | \(83\) |
parallelrisch | \(\frac {24 \ln \left (x \right ) x^{8} c^{4}-12 \ln \left (c \,x^{2}+b \right ) x^{8} c^{4}-18 c^{4} x^{8}+48 \ln \left (x \right ) x^{6} b \,c^{3}-24 \ln \left (c \,x^{2}+b \right ) x^{6} b \,c^{3}-24 b \,x^{6} c^{3}+24 \ln \left (x \right ) x^{4} b^{2} c^{2}-12 \ln \left (c \,x^{2}+b \right ) x^{4} b^{2} c^{2}+4 x^{2} b^{3} c -b^{4}}{4 b^{5} x^{4} \left (c \,x^{2}+b \right )^{2}}\) | \(136\) |
Input:
int(x/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
(3*c^3/b^4*x^6+9/2*c^2/b^3*x^4+c/b^2*x^2-1/4/b)/x^4/(c*x^2+b)^2+6*c^2*ln(x )/b^5-3*c^2*ln(c*x^2+b)/b^5
Time = 0.18 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.56 \[ \int \frac {x}{\left (b x^2+c x^4\right )^3} \, dx=\frac {12 \, b c^{3} x^{6} + 18 \, b^{2} c^{2} x^{4} + 4 \, b^{3} c x^{2} - b^{4} - 12 \, {\left (c^{4} x^{8} + 2 \, b c^{3} x^{6} + b^{2} c^{2} x^{4}\right )} \log \left (c x^{2} + b\right ) + 24 \, {\left (c^{4} x^{8} + 2 \, b c^{3} x^{6} + b^{2} c^{2} x^{4}\right )} \log \left (x\right )}{4 \, {\left (b^{5} c^{2} x^{8} + 2 \, b^{6} c x^{6} + b^{7} x^{4}\right )}} \] Input:
integrate(x/(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
1/4*(12*b*c^3*x^6 + 18*b^2*c^2*x^4 + 4*b^3*c*x^2 - b^4 - 12*(c^4*x^8 + 2*b *c^3*x^6 + b^2*c^2*x^4)*log(c*x^2 + b) + 24*(c^4*x^8 + 2*b*c^3*x^6 + b^2*c ^2*x^4)*log(x))/(b^5*c^2*x^8 + 2*b^6*c*x^6 + b^7*x^4)
Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.05 \[ \int \frac {x}{\left (b x^2+c x^4\right )^3} \, dx=\frac {- b^{3} + 4 b^{2} c x^{2} + 18 b c^{2} x^{4} + 12 c^{3} x^{6}}{4 b^{6} x^{4} + 8 b^{5} c x^{6} + 4 b^{4} c^{2} x^{8}} + \frac {6 c^{2} \log {\left (x \right )}}{b^{5}} - \frac {3 c^{2} \log {\left (\frac {b}{c} + x^{2} \right )}}{b^{5}} \] Input:
integrate(x/(c*x**4+b*x**2)**3,x)
Output:
(-b**3 + 4*b**2*c*x**2 + 18*b*c**2*x**4 + 12*c**3*x**6)/(4*b**6*x**4 + 8*b **5*c*x**6 + 4*b**4*c**2*x**8) + 6*c**2*log(x)/b**5 - 3*c**2*log(b/c + x** 2)/b**5
Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.07 \[ \int \frac {x}{\left (b x^2+c x^4\right )^3} \, dx=\frac {12 \, c^{3} x^{6} + 18 \, b c^{2} x^{4} + 4 \, b^{2} c x^{2} - b^{3}}{4 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )}} - \frac {3 \, c^{2} \log \left (c x^{2} + b\right )}{b^{5}} + \frac {3 \, c^{2} \log \left (x^{2}\right )}{b^{5}} \] Input:
integrate(x/(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
1/4*(12*c^3*x^6 + 18*b*c^2*x^4 + 4*b^2*c*x^2 - b^3)/(b^4*c^2*x^8 + 2*b^5*c *x^6 + b^6*x^4) - 3*c^2*log(c*x^2 + b)/b^5 + 3*c^2*log(x^2)/b^5
Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92 \[ \int \frac {x}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {3 \, c^{2} \log \left ({\left | c x^{2} + b \right |}\right )}{b^{5}} + \frac {6 \, c^{2} \log \left ({\left | x \right |}\right )}{b^{5}} + \frac {12 \, c^{3} x^{6} + 18 \, b c^{2} x^{4} + 4 \, b^{2} c x^{2} - b^{3}}{4 \, {\left (c x^{4} + b x^{2}\right )}^{2} b^{4}} \] Input:
integrate(x/(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
-3*c^2*log(abs(c*x^2 + b))/b^5 + 6*c^2*log(abs(x))/b^5 + 1/4*(12*c^3*x^6 + 18*b*c^2*x^4 + 4*b^2*c*x^2 - b^3)/((c*x^4 + b*x^2)^2*b^4)
Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02 \[ \int \frac {x}{\left (b x^2+c x^4\right )^3} \, dx=\frac {\frac {c\,x^2}{b^2}-\frac {1}{4\,b}+\frac {9\,c^2\,x^4}{2\,b^3}+\frac {3\,c^3\,x^6}{b^4}}{b^2\,x^4+2\,b\,c\,x^6+c^2\,x^8}-\frac {3\,c^2\,\ln \left (c\,x^2+b\right )}{b^5}+\frac {6\,c^2\,\ln \left (x\right )}{b^5} \] Input:
int(x/(b*x^2 + c*x^4)^3,x)
Output:
((c*x^2)/b^2 - 1/(4*b) + (9*c^2*x^4)/(2*b^3) + (3*c^3*x^6)/b^4)/(b^2*x^4 + c^2*x^8 + 2*b*c*x^6) - (3*c^2*log(b + c*x^2))/b^5 + (6*c^2*log(x))/b^5
Time = 0.17 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.72 \[ \int \frac {x}{\left (b x^2+c x^4\right )^3} \, dx=\frac {-12 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{2} c^{2} x^{4}-24 \,\mathrm {log}\left (c \,x^{2}+b \right ) b \,c^{3} x^{6}-12 \,\mathrm {log}\left (c \,x^{2}+b \right ) c^{4} x^{8}+24 \,\mathrm {log}\left (x \right ) b^{2} c^{2} x^{4}+48 \,\mathrm {log}\left (x \right ) b \,c^{3} x^{6}+24 \,\mathrm {log}\left (x \right ) c^{4} x^{8}-b^{4}+4 b^{3} c \,x^{2}+12 b^{2} c^{2} x^{4}-6 c^{4} x^{8}}{4 b^{5} x^{4} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:
int(x/(c*x^4+b*x^2)^3,x)
Output:
( - 12*log(b + c*x**2)*b**2*c**2*x**4 - 24*log(b + c*x**2)*b*c**3*x**6 - 1 2*log(b + c*x**2)*c**4*x**8 + 24*log(x)*b**2*c**2*x**4 + 48*log(x)*b*c**3* x**6 + 24*log(x)*c**4*x**8 - b**4 + 4*b**3*c*x**2 + 12*b**2*c**2*x**4 - 6* c**4*x**8)/(4*b**5*x**4*(b**2 + 2*b*c*x**2 + c**2*x**4))