Integrand size = 17, antiderivative size = 84 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=-\frac {1}{4 a^2 x^4}+\frac {2 b}{3 a^3 x^3}-\frac {3 b^2}{2 a^4 x^2}+\frac {4 b^3}{a^5 x}+\frac {b^4}{a^5 (a+b x)}+\frac {5 b^4 \log (x)}{a^6}-\frac {5 b^4 \log (a+b x)}{a^6} \] Output:
-1/4/a^2/x^4+2/3*b/a^3/x^3-3/2*b^2/a^4/x^2+4*b^3/a^5/x+b^4/a^5/(b*x+a)+5*b ^4*ln(x)/a^6-5*b^4*ln(b*x+a)/a^6
Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=\frac {\frac {a \left (-3 a^4+5 a^3 b x-10 a^2 b^2 x^2+30 a b^3 x^3+60 b^4 x^4\right )}{x^4 (a+b x)}+60 b^4 \log (x)-60 b^4 \log (a+b x)}{12 a^6} \] Input:
Integrate[1/(x*(a*x^2 + b*x^3)^2),x]
Output:
((a*(-3*a^4 + 5*a^3*b*x - 10*a^2*b^2*x^2 + 30*a*b^3*x^3 + 60*b^4*x^4))/(x^ 4*(a + b*x)) + 60*b^4*Log[x] - 60*b^4*Log[a + b*x])/(12*a^6)
Time = 0.37 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {9, 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.
\(\displaystyle \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {1}{x^5 (a+b x)^2}dx\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \int \left (-\frac {5 b^5}{a^6 (a+b x)}+\frac {5 b^4}{a^6 x}-\frac {b^5}{a^5 (a+b x)^2}-\frac {4 b^3}{a^5 x^2}+\frac {3 b^2}{a^4 x^3}-\frac {2 b}{a^3 x^4}+\frac {1}{a^2 x^5}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 b^4 \log (x)}{a^6}-\frac {5 b^4 \log (a+b x)}{a^6}+\frac {b^4}{a^5 (a+b x)}+\frac {4 b^3}{a^5 x}-\frac {3 b^2}{2 a^4 x^2}+\frac {2 b}{3 a^3 x^3}-\frac {1}{4 a^2 x^4}\) |
Input:
Int[1/(x*(a*x^2 + b*x^3)^2),x]
Output:
-1/4*1/(a^2*x^4) + (2*b)/(3*a^3*x^3) - (3*b^2)/(2*a^4*x^2) + (4*b^3)/(a^5* x) + b^4/(a^5*(a + b*x)) + (5*b^4*Log[x])/a^6 - (5*b^4*Log[a + b*x])/a^6
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])
Time = 0.38 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94
method | result | size |
default | \(-\frac {1}{4 a^{2} x^{4}}+\frac {2 b}{3 a^{3} x^{3}}-\frac {3 b^{2}}{2 a^{4} x^{2}}+\frac {4 b^{3}}{a^{5} x}+\frac {b^{4}}{a^{5} \left (b x +a \right )}+\frac {5 b^{4} \ln \left (x \right )}{a^{6}}-\frac {5 b^{4} \ln \left (b x +a \right )}{a^{6}}\) | \(79\) |
norman | \(\frac {-\frac {5 b^{5} x^{5}}{a^{6}}-\frac {1}{4 a}+\frac {5 b x}{12 a^{2}}-\frac {5 b^{2} x^{2}}{6 a^{3}}+\frac {5 b^{3} x^{3}}{2 a^{4}}}{x^{4} \left (b x +a \right )}+\frac {5 b^{4} \ln \left (x \right )}{a^{6}}-\frac {5 b^{4} \ln \left (b x +a \right )}{a^{6}}\) | \(83\) |
risch | \(\frac {\frac {5 b^{4} x^{4}}{a^{5}}+\frac {5 b^{3} x^{3}}{2 a^{4}}-\frac {5 b^{2} x^{2}}{6 a^{3}}+\frac {5 b x}{12 a^{2}}-\frac {1}{4 a}}{x^{4} \left (b x +a \right )}+\frac {5 b^{4} \ln \left (-x \right )}{a^{6}}-\frac {5 b^{4} \ln \left (b x +a \right )}{a^{6}}\) | \(85\) |
parallelrisch | \(\frac {60 \ln \left (x \right ) x^{5} b^{5}-60 \ln \left (b x +a \right ) x^{5} b^{5}+60 \ln \left (x \right ) x^{4} a \,b^{4}-60 \ln \left (b x +a \right ) x^{4} a \,b^{4}-60 b^{5} x^{5}+30 a^{2} b^{3} x^{3}-10 a^{3} b^{2} x^{2}+5 a^{4} b x -3 a^{5}}{12 a^{6} \left (b x +a \right ) x^{4}}\) | \(109\) |
Input:
int(1/x/(b*x^3+a*x^2)^2,x,method=_RETURNVERBOSE)
Output:
-1/4/a^2/x^4+2/3*b/a^3/x^3-3/2*b^2/a^4/x^2+4*b^3/a^5/x+b^4/a^5/(b*x+a)+5*b ^4*ln(x)/a^6-5*b^4*ln(b*x+a)/a^6
Time = 0.08 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.29 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=\frac {60 \, a b^{4} x^{4} + 30 \, a^{2} b^{3} x^{3} - 10 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b x - 3 \, a^{5} - 60 \, {\left (b^{5} x^{5} + a b^{4} x^{4}\right )} \log \left (b x + a\right ) + 60 \, {\left (b^{5} x^{5} + a b^{4} x^{4}\right )} \log \left (x\right )}{12 \, {\left (a^{6} b x^{5} + a^{7} x^{4}\right )}} \] Input:
integrate(1/x/(b*x^3+a*x^2)^2,x, algorithm="fricas")
Output:
1/12*(60*a*b^4*x^4 + 30*a^2*b^3*x^3 - 10*a^3*b^2*x^2 + 5*a^4*b*x - 3*a^5 - 60*(b^5*x^5 + a*b^4*x^4)*log(b*x + a) + 60*(b^5*x^5 + a*b^4*x^4)*log(x))/ (a^6*b*x^5 + a^7*x^4)
Time = 0.17 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=\frac {- 3 a^{4} + 5 a^{3} b x - 10 a^{2} b^{2} x^{2} + 30 a b^{3} x^{3} + 60 b^{4} x^{4}}{12 a^{6} x^{4} + 12 a^{5} b x^{5}} + \frac {5 b^{4} \left (\log {\left (x \right )} - \log {\left (\frac {a}{b} + x \right )}\right )}{a^{6}} \] Input:
integrate(1/x/(b*x**3+a*x**2)**2,x)
Output:
(-3*a**4 + 5*a**3*b*x - 10*a**2*b**2*x**2 + 30*a*b**3*x**3 + 60*b**4*x**4) /(12*a**6*x**4 + 12*a**5*b*x**5) + 5*b**4*(log(x) - log(a/b + x))/a**6
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=\frac {60 \, b^{4} x^{4} + 30 \, a b^{3} x^{3} - 10 \, a^{2} b^{2} x^{2} + 5 \, a^{3} b x - 3 \, a^{4}}{12 \, {\left (a^{5} b x^{5} + a^{6} x^{4}\right )}} - \frac {5 \, b^{4} \log \left (b x + a\right )}{a^{6}} + \frac {5 \, b^{4} \log \left (x\right )}{a^{6}} \] Input:
integrate(1/x/(b*x^3+a*x^2)^2,x, algorithm="maxima")
Output:
1/12*(60*b^4*x^4 + 30*a*b^3*x^3 - 10*a^2*b^2*x^2 + 5*a^3*b*x - 3*a^4)/(a^5 *b*x^5 + a^6*x^4) - 5*b^4*log(b*x + a)/a^6 + 5*b^4*log(x)/a^6
Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=-\frac {5 \, b^{4} \log \left ({\left | b x + a \right |}\right )}{a^{6}} + \frac {5 \, b^{4} \log \left ({\left | x \right |}\right )}{a^{6}} + \frac {60 \, a b^{4} x^{4} + 30 \, a^{2} b^{3} x^{3} - 10 \, a^{3} b^{2} x^{2} + 5 \, a^{4} b x - 3 \, a^{5}}{12 \, {\left (b x + a\right )} a^{6} x^{4}} \] Input:
integrate(1/x/(b*x^3+a*x^2)^2,x, algorithm="giac")
Output:
-5*b^4*log(abs(b*x + a))/a^6 + 5*b^4*log(abs(x))/a^6 + 1/12*(60*a*b^4*x^4 + 30*a^2*b^3*x^3 - 10*a^3*b^2*x^2 + 5*a^4*b*x - 3*a^5)/((b*x + a)*a^6*x^4)
Time = 8.65 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=\frac {\frac {5\,b^3\,x^3}{2\,a^4}-\frac {5\,b^2\,x^2}{6\,a^3}-\frac {1}{4\,a}+\frac {5\,b^4\,x^4}{a^5}+\frac {5\,b\,x}{12\,a^2}}{b\,x^5+a\,x^4}-\frac {10\,b^4\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^6} \] Input:
int(1/(x*(a*x^2 + b*x^3)^2),x)
Output:
((5*b^3*x^3)/(2*a^4) - (5*b^2*x^2)/(6*a^3) - 1/(4*a) + (5*b^4*x^4)/a^5 + ( 5*b*x)/(12*a^2))/(a*x^4 + b*x^5) - (10*b^4*atanh((2*b*x)/a + 1))/a^6
Time = 0.22 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.29 \[ \int \frac {1}{x \left (a x^2+b x^3\right )^2} \, dx=\frac {-60 \,\mathrm {log}\left (b x +a \right ) a \,b^{4} x^{4}-60 \,\mathrm {log}\left (b x +a \right ) b^{5} x^{5}+60 \,\mathrm {log}\left (x \right ) a \,b^{4} x^{4}+60 \,\mathrm {log}\left (x \right ) b^{5} x^{5}-3 a^{5}+5 a^{4} b x -10 a^{3} b^{2} x^{2}+30 a^{2} b^{3} x^{3}-60 b^{5} x^{5}}{12 a^{6} x^{4} \left (b x +a \right )} \] Input:
int(1/x/(b*x^3+a*x^2)^2,x)
Output:
( - 60*log(a + b*x)*a*b**4*x**4 - 60*log(a + b*x)*b**5*x**5 + 60*log(x)*a* b**4*x**4 + 60*log(x)*b**5*x**5 - 3*a**5 + 5*a**4*b*x - 10*a**3*b**2*x**2 + 30*a**2*b**3*x**3 - 60*b**5*x**5)/(12*a**6*x**4*(a + b*x))