Integrand size = 17, antiderivative size = 67 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {1}{2 b^3 x^2}-\frac {c}{4 b^2 \left (b+c x^2\right )^2}-\frac {c}{b^3 \left (b+c x^2\right )}-\frac {3 c \log (x)}{b^4}+\frac {3 c \log \left (b+c x^2\right )}{2 b^4} \] Output:
-1/2/b^3/x^2-1/4*c/b^2/(c*x^2+b)^2-c/b^3/(c*x^2+b)-3*c*ln(x)/b^4+3/2*c*ln( c*x^2+b)/b^4
Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {\frac {b \left (2 b^2+9 b c x^2+6 c^2 x^4\right )}{x^2 \left (b+c x^2\right )^2}+12 c \log (x)-6 c \log \left (b+c x^2\right )}{4 b^4} \] Input:
Integrate[x^3/(b*x^2 + c*x^4)^3,x]
Output:
-1/4*((b*(2*b^2 + 9*b*c*x^2 + 6*c^2*x^4))/(x^2*(b + c*x^2)^2) + 12*c*Log[x ] - 6*c*Log[b + c*x^2])/b^4
Time = 0.34 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.03, 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.
\(\displaystyle \int \frac {x^3}{\left (b x^2+c x^4\right )^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {1}{x^3 \left (b+c x^2\right )^3}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (c x^2+b\right )^3}dx^2\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {1}{2} \int \left (\frac {3 c^2}{b^4 \left (c x^2+b\right )}+\frac {2 c^2}{b^3 \left (c x^2+b\right )^2}+\frac {c^2}{b^2 \left (c x^2+b\right )^3}-\frac {3 c}{b^4 x^2}+\frac {1}{b^3 x^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {3 c \log \left (x^2\right )}{b^4}+\frac {3 c \log \left (b+c x^2\right )}{b^4}-\frac {2 c}{b^3 \left (b+c x^2\right )}-\frac {1}{b^3 x^2}-\frac {c}{2 b^2 \left (b+c x^2\right )^2}\right )\) |
Input:
Int[x^3/(b*x^2 + c*x^4)^3,x]
Output:
(-(1/(b^3*x^2)) - c/(2*b^2*(b + c*x^2)^2) - (2*c)/(b^3*(b + c*x^2)) - (3*c *Log[x^2])/b^4 + (3*c*Log[b + c*x^2])/b^4)/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 = 0.12 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99
method | result | size |
risch | \(\frac {-\frac {3 c^{2} x^{4}}{2 b^{3}}-\frac {9 c \,x^{2}}{4 b^{2}}-\frac {1}{2 b}}{x^{2} \left (c \,x^{2}+b \right )^{2}}-\frac {3 c \ln \left (x \right )}{b^{4}}+\frac {3 c \ln \left (-c \,x^{2}-b \right )}{2 b^{4}}\) | \(66\) |
norman | \(\frac {\frac {3 c^{2} x^{7}}{b^{3}}-\frac {x^{3}}{2 b}+\frac {9 c^{3} x^{9}}{4 b^{4}}}{x^{5} \left (c \,x^{2}+b \right )^{2}}-\frac {3 c \ln \left (x \right )}{b^{4}}+\frac {3 c \ln \left (c \,x^{2}+b \right )}{2 b^{4}}\) | \(68\) |
default | \(-\frac {1}{2 b^{3} x^{2}}-\frac {3 c \ln \left (x \right )}{b^{4}}+\frac {c^{2} \left (-\frac {b^{2}}{2 c \left (c \,x^{2}+b \right )^{2}}+\frac {3 \ln \left (c \,x^{2}+b \right )}{c}-\frac {2 b}{c \left (c \,x^{2}+b \right )}\right )}{2 b^{4}}\) | \(72\) |
parallelrisch | \(-\frac {12 c^{3} \ln \left (x \right ) x^{6}-6 \ln \left (c \,x^{2}+b \right ) x^{6} c^{3}-9 c^{3} x^{6}+24 b \,c^{2} \ln \left (x \right ) x^{4}-12 \ln \left (c \,x^{2}+b \right ) x^{4} b \,c^{2}-12 b \,c^{2} x^{4}+12 b^{2} c \ln \left (x \right ) x^{2}-6 \ln \left (c \,x^{2}+b \right ) x^{2} b^{2} c +2 b^{3}}{4 b^{4} x^{2} \left (c \,x^{2}+b \right )^{2}}\) | \(123\) |
Input:
int(x^3/(c*x^4+b*x^2)^3,x,method=_RETURNVERBOSE)
Output:
(-3/2*c^2/b^3*x^4-9/4*c/b^2*x^2-1/2/b)/x^2/(c*x^2+b)^2-3*c*ln(x)/b^4+3/2/b ^4*c*ln(-c*x^2-b)
Time = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.78 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {6 \, b c^{2} x^{4} + 9 \, b^{2} c x^{2} + 2 \, b^{3} - 6 \, {\left (c^{3} x^{6} + 2 \, b c^{2} x^{4} + b^{2} c x^{2}\right )} \log \left (c x^{2} + b\right ) + 12 \, {\left (c^{3} x^{6} + 2 \, b c^{2} x^{4} + b^{2} c x^{2}\right )} \log \left (x\right )}{4 \, {\left (b^{4} c^{2} x^{6} + 2 \, b^{5} c x^{4} + b^{6} x^{2}\right )}} \] Input:
integrate(x^3/(c*x^4+b*x^2)^3,x, algorithm="fricas")
Output:
-1/4*(6*b*c^2*x^4 + 9*b^2*c*x^2 + 2*b^3 - 6*(c^3*x^6 + 2*b*c^2*x^4 + b^2*c *x^2)*log(c*x^2 + b) + 12*(c^3*x^6 + 2*b*c^2*x^4 + b^2*c*x^2)*log(x))/(b^4 *c^2*x^6 + 2*b^5*c*x^4 + b^6*x^2)
Time = 0.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.19 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^3} \, dx=\frac {- 2 b^{2} - 9 b c x^{2} - 6 c^{2} x^{4}}{4 b^{5} x^{2} + 8 b^{4} c x^{4} + 4 b^{3} c^{2} x^{6}} - \frac {3 c \log {\left (x \right )}}{b^{4}} + \frac {3 c \log {\left (\frac {b}{c} + x^{2} \right )}}{2 b^{4}} \] Input:
integrate(x**3/(c*x**4+b*x**2)**3,x)
Output:
(-2*b**2 - 9*b*c*x**2 - 6*c**2*x**4)/(4*b**5*x**2 + 8*b**4*c*x**4 + 4*b**3 *c**2*x**6) - 3*c*log(x)/b**4 + 3*c*log(b/c + x**2)/(2*b**4)
Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.15 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^3} \, dx=-\frac {6 \, c^{2} x^{4} + 9 \, b c x^{2} + 2 \, b^{2}}{4 \, {\left (b^{3} c^{2} x^{6} + 2 \, b^{4} c x^{4} + b^{5} x^{2}\right )}} + \frac {3 \, c \log \left (c x^{2} + b\right )}{2 \, b^{4}} - \frac {3 \, c \log \left (x^{2}\right )}{2 \, b^{4}} \] Input:
integrate(x^3/(c*x^4+b*x^2)^3,x, algorithm="maxima")
Output:
-1/4*(6*c^2*x^4 + 9*b*c*x^2 + 2*b^2)/(b^3*c^2*x^6 + 2*b^4*c*x^4 + b^5*x^2) + 3/2*c*log(c*x^2 + b)/b^4 - 3/2*c*log(x^2)/b^4
Time = 0.15 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3 \, c \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, b^{4}} - \frac {3 \, c \log \left ({\left | x \right |}\right )}{b^{4}} - \frac {6 \, b c^{2} x^{4} + 9 \, b^{2} c x^{2} + 2 \, b^{3}}{4 \, {\left (c x^{2} + b\right )}^{2} b^{4} x^{2}} \] Input:
integrate(x^3/(c*x^4+b*x^2)^3,x, algorithm="giac")
Output:
3/2*c*log(abs(c*x^2 + b))/b^4 - 3*c*log(abs(x))/b^4 - 1/4*(6*b*c^2*x^4 + 9 *b^2*c*x^2 + 2*b^3)/((c*x^2 + b)^2*b^4*x^2)
Time = 17.64 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^3} \, dx=\frac {3\,c\,\ln \left (c\,x^2+b\right )}{2\,b^4}-\frac {\frac {1}{2\,b}+\frac {9\,c\,x^2}{4\,b^2}+\frac {3\,c^2\,x^4}{2\,b^3}}{b^2\,x^2+2\,b\,c\,x^4+c^2\,x^6}-\frac {3\,c\,\ln \left (x\right )}{b^4} \] Input:
int(x^3/(b*x^2 + c*x^4)^3,x)
Output:
(3*c*log(b + c*x^2))/(2*b^4) - (1/(2*b) + (9*c*x^2)/(4*b^2) + (3*c^2*x^4)/ (2*b^3))/(b^2*x^2 + c^2*x^6 + 2*b*c*x^4) - (3*c*log(x))/b^4
Time = 0.17 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.99 \[ \int \frac {x^3}{\left (b x^2+c x^4\right )^3} \, dx=\frac {6 \,\mathrm {log}\left (c \,x^{2}+b \right ) b^{2} c \,x^{2}+12 \,\mathrm {log}\left (c \,x^{2}+b \right ) b \,c^{2} x^{4}+6 \,\mathrm {log}\left (c \,x^{2}+b \right ) c^{3} x^{6}-12 \,\mathrm {log}\left (x \right ) b^{2} c \,x^{2}-24 \,\mathrm {log}\left (x \right ) b \,c^{2} x^{4}-12 \,\mathrm {log}\left (x \right ) c^{3} x^{6}-2 b^{3}-6 b^{2} c \,x^{2}+3 c^{3} x^{6}}{4 b^{4} x^{2} \left (c^{2} x^{4}+2 b c \,x^{2}+b^{2}\right )} \] Input:
int(x^3/(c*x^4+b*x^2)^3,x)
Output:
(6*log(b + c*x**2)*b**2*c*x**2 + 12*log(b + c*x**2)*b*c**2*x**4 + 6*log(b + c*x**2)*c**3*x**6 - 12*log(x)*b**2*c*x**2 - 24*log(x)*b*c**2*x**4 - 12*l og(x)*c**3*x**6 - 2*b**3 - 6*b**2*c*x**2 + 3*c**3*x**6)/(4*b**4*x**2*(b**2 + 2*b*c*x**2 + c**2*x**4))