Integrand size = 22, antiderivative size = 81 \[ \int \frac {1}{x (a+b x)^3 (a c-b c x)^3} \, dx=\frac {1}{4 a^2 c^3 \left (a^2-b^2 x^2\right )^2}+\frac {1}{2 a^4 c^3 \left (a^2-b^2 x^2\right )}+\frac {\log (x)}{a^6 c^3}-\frac {\log \left (a^2-b^2 x^2\right )}{2 a^6 c^3} \] Output:
1/4/a^2/c^3/(-b^2*x^2+a^2)^2+1/2/a^4/c^3/(-b^2*x^2+a^2)+ln(x)/a^6/c^3-1/2* ln(-b^2*x^2+a^2)/a^6/c^3
Time = 0.02 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.42 \[ \int \frac {1}{x (a+b x)^3 (a c-b c x)^3} \, dx=\frac {1}{16 a^4 c^3 (-a+b x)^2}-\frac {5}{16 a^5 c^3 (-a+b x)}+\frac {1}{16 a^4 c^3 (a+b x)^2}+\frac {5}{16 a^5 c^3 (a+b x)}+\frac {\log (x)}{a^6 c^3}-\frac {\log (a-b x)}{2 a^6 c^3}-\frac {\log (a+b x)}{2 a^6 c^3} \] Input:
Integrate[1/(x*(a + b*x)^3*(a*c - b*c*x)^3),x]
Output:
1/(16*a^4*c^3*(-a + b*x)^2) - 5/(16*a^5*c^3*(-a + b*x)) + 1/(16*a^4*c^3*(a + b*x)^2) + 5/(16*a^5*c^3*(a + b*x)) + Log[x]/(a^6*c^3) - Log[a - b*x]/(2 *a^6*c^3) - Log[a + b*x]/(2*a^6*c^3)
Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {82, 243, 27, 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.
\(\displaystyle \int \frac {1}{x (a+b x)^3 (a c-b c x)^3} \, dx\) |
\(\Big \downarrow \) 82 |
\(\displaystyle \int \frac {1}{x \left (a^2 c-b^2 c x^2\right )^3}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{c^3 x^2 \left (a^2-b^2 x^2\right )^3}dx^2\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{x^2 \left (a^2-b^2 x^2\right )^3}dx^2}{2 c^3}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {b^2}{a^6 \left (a^2-b^2 x^2\right )}+\frac {b^2}{a^4 \left (a^2-b^2 x^2\right )^2}+\frac {b^2}{a^2 \left (a^2-b^2 x^2\right )^3}+\frac {1}{a^6 x^2}\right )dx^2}{2 c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\log \left (x^2\right )}{a^6}+\frac {1}{2 a^2 \left (a^2-b^2 x^2\right )^2}-\frac {\log \left (a^2-b^2 x^2\right )}{a^6}+\frac {1}{a^4 \left (a^2-b^2 x^2\right )}}{2 c^3}\) |
Input:
Int[1/(x*(a + b*x)^3*(a*c - b*c*x)^3),x]
Output:
(1/(2*a^2*(a^2 - b^2*x^2)^2) + 1/(a^4*(a^2 - b^2*x^2)) + Log[x^2]/a^6 - Lo g[a^2 - b^2*x^2]/a^6)/(2*c^3)
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[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) )^(p_.), x_] :> Int[(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && IntegerQ[m]
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 = 0.21 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\frac {-\frac {b^{2} x^{2}}{2 a^{4}}+\frac {3}{4 a^{2}}}{\left (b x +a \right )^{2} c^{3} \left (-b x +a \right )^{2}}+\frac {\ln \left (x \right )}{a^{6} c^{3}}-\frac {\ln \left (-b^{2} x^{2}+a^{2}\right )}{2 a^{6} c^{3}}\) | \(68\) |
default | \(\frac {\frac {\ln \left (x \right )}{a^{6}}-\frac {\ln \left (b x +a \right )}{2 a^{6}}+\frac {5}{16 a^{5} \left (b x +a \right )}+\frac {1}{16 a^{4} \left (b x +a \right )^{2}}-\frac {\ln \left (-b x +a \right )}{2 a^{6}}+\frac {5}{16 a^{5} \left (-b x +a \right )}+\frac {1}{16 a^{4} \left (-b x +a \right )^{2}}}{c^{3}}\) | \(85\) |
norman | \(\frac {\frac {b^{2} x^{2}}{a^{4} c}-\frac {3 b^{4} x^{4}}{4 a^{6} c}}{\left (b x +a \right )^{2} c^{2} \left (-b x +a \right )^{2}}+\frac {\ln \left (x \right )}{a^{6} c^{3}}-\frac {\ln \left (-b x +a \right )}{2 a^{6} c^{3}}-\frac {\ln \left (b x +a \right )}{2 a^{6} c^{3}}\) | \(87\) |
parallelrisch | \(\frac {4 b^{4} \ln \left (x \right ) x^{4}-2 \ln \left (b x -a \right ) x^{4} b^{4}-2 \ln \left (b x +a \right ) x^{4} b^{4}-3 b^{4} x^{4}-8 a^{2} b^{2} \ln \left (x \right ) x^{2}+4 \ln \left (b x -a \right ) x^{2} a^{2} b^{2}+4 \ln \left (b x +a \right ) x^{2} a^{2} b^{2}+4 a^{2} b^{2} x^{2}+4 a^{4} \ln \left (x \right )-2 \ln \left (b x -a \right ) a^{4}-2 \ln \left (b x +a \right ) a^{4}}{4 a^{6} c^{3} \left (b x +a \right )^{2} \left (b x -a \right )^{2}}\) | \(165\) |
Input:
int(1/x/(b*x+a)^3/(-b*c*x+a*c)^3,x,method=_RETURNVERBOSE)
Output:
(-1/2/a^4*b^2*x^2+3/4/a^2)/(b*x+a)^2/c^3/(-b*x+a)^2+ln(x)/a^6/c^3-1/2*ln(- b^2*x^2+a^2)/a^6/c^3
Time = 0.07 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.48 \[ \int \frac {1}{x (a+b x)^3 (a c-b c x)^3} \, dx=-\frac {2 \, a^{2} b^{2} x^{2} - 3 \, a^{4} + 2 \, {\left (b^{4} x^{4} - 2 \, a^{2} b^{2} x^{2} + a^{4}\right )} \log \left (b^{2} x^{2} - a^{2}\right ) - 4 \, {\left (b^{4} x^{4} - 2 \, a^{2} b^{2} x^{2} + a^{4}\right )} \log \left (x\right )}{4 \, {\left (a^{6} b^{4} c^{3} x^{4} - 2 \, a^{8} b^{2} c^{3} x^{2} + a^{10} c^{3}\right )}} \] Input:
integrate(1/x/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="fricas")
Output:
-1/4*(2*a^2*b^2*x^2 - 3*a^4 + 2*(b^4*x^4 - 2*a^2*b^2*x^2 + a^4)*log(b^2*x^ 2 - a^2) - 4*(b^4*x^4 - 2*a^2*b^2*x^2 + a^4)*log(x))/(a^6*b^4*c^3*x^4 - 2* a^8*b^2*c^3*x^2 + a^10*c^3)
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x (a+b x)^3 (a c-b c x)^3} \, dx=- \frac {- 3 a^{2} + 2 b^{2} x^{2}}{4 a^{8} c^{3} - 8 a^{6} b^{2} c^{3} x^{2} + 4 a^{4} b^{4} c^{3} x^{4}} + \frac {\log {\left (x \right )}}{a^{6} c^{3}} - \frac {\log {\left (- \frac {a^{2}}{b^{2}} + x^{2} \right )}}{2 a^{6} c^{3}} \] Input:
integrate(1/x/(b*x+a)**3/(-b*c*x+a*c)**3,x)
Output:
-(-3*a**2 + 2*b**2*x**2)/(4*a**8*c**3 - 8*a**6*b**2*c**3*x**2 + 4*a**4*b** 4*c**3*x**4) + log(x)/(a**6*c**3) - log(-a**2/b**2 + x**2)/(2*a**6*c**3)
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x (a+b x)^3 (a c-b c x)^3} \, dx=-\frac {2 \, b^{2} x^{2} - 3 \, a^{2}}{4 \, {\left (a^{4} b^{4} c^{3} x^{4} - 2 \, a^{6} b^{2} c^{3} x^{2} + a^{8} c^{3}\right )}} - \frac {\log \left (b x + a\right )}{2 \, a^{6} c^{3}} - \frac {\log \left (b x - a\right )}{2 \, a^{6} c^{3}} + \frac {\log \left (x\right )}{a^{6} c^{3}} \] Input:
integrate(1/x/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="maxima")
Output:
-1/4*(2*b^2*x^2 - 3*a^2)/(a^4*b^4*c^3*x^4 - 2*a^6*b^2*c^3*x^2 + a^8*c^3) - 1/2*log(b*x + a)/(a^6*c^3) - 1/2*log(b*x - a)/(a^6*c^3) + log(x)/(a^6*c^3 )
Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x (a+b x)^3 (a c-b c x)^3} \, dx=\frac {\log \left (x^{2}\right )}{2 \, a^{6} c^{3}} - \frac {\log \left ({\left | b^{2} x^{2} - a^{2} \right |}\right )}{2 \, a^{6} c^{3}} + \frac {3 \, b^{4} x^{4} - 8 \, a^{2} b^{2} x^{2} + 6 \, a^{4}}{4 \, {\left (b^{2} x^{2} - a^{2}\right )}^{2} a^{6} c^{3}} \] Input:
integrate(1/x/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="giac")
Output:
1/2*log(x^2)/(a^6*c^3) - 1/2*log(abs(b^2*x^2 - a^2))/(a^6*c^3) + 1/4*(3*b^ 4*x^4 - 8*a^2*b^2*x^2 + 6*a^4)/((b^2*x^2 - a^2)^2*a^6*c^3)
Time = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x (a+b x)^3 (a c-b c x)^3} \, dx=\frac {\frac {3}{4\,a^2}-\frac {b^2\,x^2}{2\,a^4}}{a^4\,c^3-2\,a^2\,b^2\,c^3\,x^2+b^4\,c^3\,x^4}-\frac {\ln \left (a^2-b^2\,x^2\right )}{2\,a^6\,c^3}+\frac {\ln \left (x\right )}{a^6\,c^3} \] Input:
int(1/(x*(a*c - b*c*x)^3*(a + b*x)^3),x)
Output:
(3/(4*a^2) - (b^2*x^2)/(2*a^4))/(a^4*c^3 + b^4*c^3*x^4 - 2*a^2*b^2*c^3*x^2 ) - log(a^2 - b^2*x^2)/(2*a^6*c^3) + log(x)/(a^6*c^3)
Time = 0.16 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.12 \[ \int \frac {1}{x (a+b x)^3 (a c-b c x)^3} \, dx=\frac {-2 \,\mathrm {log}\left (-b x -a \right ) a^{4}+4 \,\mathrm {log}\left (-b x -a \right ) a^{2} b^{2} x^{2}-2 \,\mathrm {log}\left (-b x -a \right ) b^{4} x^{4}-2 \,\mathrm {log}\left (-b x +a \right ) a^{4}+4 \,\mathrm {log}\left (-b x +a \right ) a^{2} b^{2} x^{2}-2 \,\mathrm {log}\left (-b x +a \right ) b^{4} x^{4}+4 \,\mathrm {log}\left (x \right ) a^{4}-8 \,\mathrm {log}\left (x \right ) a^{2} b^{2} x^{2}+4 \,\mathrm {log}\left (x \right ) b^{4} x^{4}+2 a^{4}-b^{4} x^{4}}{4 a^{6} c^{3} \left (b^{4} x^{4}-2 a^{2} b^{2} x^{2}+a^{4}\right )} \] Input:
int(1/x/(b*x+a)^3/(-b*c*x+a*c)^3,x)
Output:
( - 2*log( - a - b*x)*a**4 + 4*log( - a - b*x)*a**2*b**2*x**2 - 2*log( - a - b*x)*b**4*x**4 - 2*log(a - b*x)*a**4 + 4*log(a - b*x)*a**2*b**2*x**2 - 2*log(a - b*x)*b**4*x**4 + 4*log(x)*a**4 - 8*log(x)*a**2*b**2*x**2 + 4*log (x)*b**4*x**4 + 2*a**4 - b**4*x**4)/(4*a**6*c**3*(a**4 - 2*a**2*b**2*x**2 + b**4*x**4))