Integrand size = 22, antiderivative size = 104 \[ \int \frac {1}{x^3 (a+b x)^3 (a c-b c x)^3} \, dx=-\frac {1}{2 a^6 c^3 x^2}+\frac {b^2}{4 a^4 c^3 \left (a^2-b^2 x^2\right )^2}+\frac {b^2}{a^6 c^3 \left (a^2-b^2 x^2\right )}+\frac {3 b^2 \log (x)}{a^8 c^3}-\frac {3 b^2 \log \left (a^2-b^2 x^2\right )}{2 a^8 c^3} \] Output:
-1/2/a^6/c^3/x^2+1/4*b^2/a^4/c^3/(-b^2*x^2+a^2)^2+b^2/a^6/c^3/(-b^2*x^2+a^ 2)+3*b^2*ln(x)/a^8/c^3-3/2*b^2*ln(-b^2*x^2+a^2)/a^8/c^3
Time = 0.04 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.44 \[ \int \frac {1}{x^3 (a+b x)^3 (a c-b c x)^3} \, dx=-\frac {1}{2 a^6 c^3 x^2}+\frac {b^2}{16 a^6 c^3 (-a+b x)^2}-\frac {9 b^2}{16 a^7 c^3 (-a+b x)}+\frac {b^2}{16 a^6 c^3 (a+b x)^2}+\frac {9 b^2}{16 a^7 c^3 (a+b x)}+\frac {3 b^2 \log (x)}{a^8 c^3}-\frac {3 b^2 \log (a-b x)}{2 a^8 c^3}-\frac {3 b^2 \log (a+b x)}{2 a^8 c^3} \] Input:
Integrate[1/(x^3*(a + b*x)^3*(a*c - b*c*x)^3),x]
Output:
-1/2*1/(a^6*c^3*x^2) + b^2/(16*a^6*c^3*(-a + b*x)^2) - (9*b^2)/(16*a^7*c^3 *(-a + b*x)) + b^2/(16*a^6*c^3*(a + b*x)^2) + (9*b^2)/(16*a^7*c^3*(a + b*x )) + (3*b^2*Log[x])/(a^8*c^3) - (3*b^2*Log[a - b*x])/(2*a^8*c^3) - (3*b^2* Log[a + b*x])/(2*a^8*c^3)
Time = 0.23 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.91, 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.
\(\displaystyle \int \frac {1}{x^3 (a+b x)^3 (a c-b c x)^3} \, dx\) |
\(\Big \downarrow \) 82 |
\(\displaystyle \int \frac {1}{x^3 \left (a^2 c-b^2 c x^2\right )^3}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {1}{c^3 x^4 \left (a^2-b^2 x^2\right )^3}dx^2\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{x^4 \left (a^2-b^2 x^2\right )^3}dx^2}{2 c^3}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\int \left (\frac {3 b^4}{a^8 \left (a^2-b^2 x^2\right )}+\frac {2 b^4}{a^6 \left (a^2-b^2 x^2\right )^2}+\frac {b^4}{a^4 \left (a^2-b^2 x^2\right )^3}+\frac {3 b^2}{a^8 x^2}+\frac {1}{a^6 x^4}\right )dx^2}{2 c^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {3 b^2 \log \left (x^2\right )}{a^8}-\frac {1}{a^6 x^2}-\frac {3 b^2 \log \left (a^2-b^2 x^2\right )}{a^8}+\frac {2 b^2}{a^6 \left (a^2-b^2 x^2\right )}+\frac {b^2}{2 a^4 \left (a^2-b^2 x^2\right )^2}}{2 c^3}\) |
Input:
Int[1/(x^3*(a + b*x)^3*(a*c - b*c*x)^3),x]
Output:
(-(1/(a^6*x^2)) + b^2/(2*a^4*(a^2 - b^2*x^2)^2) + (2*b^2)/(a^6*(a^2 - b^2* x^2)) + (3*b^2*Log[x^2])/a^8 - (3*b^2*Log[a^2 - b^2*x^2])/a^8)/(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 = 0.22 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {-\frac {3 b^{4} x^{4}}{2 a^{6}}+\frac {9 b^{2} x^{2}}{4 a^{4}}-\frac {1}{2 a^{2}}}{x^{2} \left (b x +a \right )^{2} c^{3} \left (-b x +a \right )^{2}}+\frac {3 b^{2} \ln \left (x \right )}{a^{8} c^{3}}-\frac {3 b^{2} \ln \left (-b^{2} x^{2}+a^{2}\right )}{2 a^{8} c^{3}}\) | \(89\) |
norman | \(\frac {-\frac {1}{2 a^{2} c}+\frac {3 b^{4} x^{4}}{a^{6} c}-\frac {9 b^{6} x^{6}}{4 a^{8} c}}{x^{2} \left (b x +a \right )^{2} c^{2} \left (-b x +a \right )^{2}}+\frac {3 b^{2} \ln \left (x \right )}{a^{8} c^{3}}-\frac {3 b^{2} \ln \left (-b x +a \right )}{2 a^{8} c^{3}}-\frac {3 b^{2} \ln \left (b x +a \right )}{2 a^{8} c^{3}}\) | \(109\) |
default | \(\frac {-\frac {1}{2 a^{6} x^{2}}+\frac {3 b^{2} \ln \left (x \right )}{a^{8}}+\frac {9 b^{2}}{16 a^{7} \left (b x +a \right )}+\frac {b^{2}}{16 a^{6} \left (b x +a \right )^{2}}-\frac {3 b^{2} \ln \left (b x +a \right )}{2 a^{8}}+\frac {9 b^{2}}{16 a^{7} \left (-b x +a \right )}+\frac {b^{2}}{16 a^{6} \left (-b x +a \right )^{2}}-\frac {3 b^{2} \ln \left (-b x +a \right )}{2 a^{8}}}{c^{3}}\) | \(115\) |
parallelrisch | \(\frac {12 \ln \left (x \right ) x^{6} b^{6}-6 \ln \left (b x -a \right ) x^{6} b^{6}-6 \ln \left (b x +a \right ) x^{6} b^{6}-9 b^{6} x^{6}-24 \ln \left (x \right ) x^{4} a^{2} b^{4}+12 \ln \left (b x -a \right ) x^{4} a^{2} b^{4}+12 \ln \left (b x +a \right ) x^{4} a^{2} b^{4}+12 a^{2} x^{4} b^{4}+12 \ln \left (x \right ) x^{2} a^{4} b^{2}-6 \ln \left (b x -a \right ) x^{2} a^{4} b^{2}-6 \ln \left (b x +a \right ) x^{2} a^{4} b^{2}-2 a^{6}}{4 a^{8} c^{3} x^{2} \left (b x +a \right )^{2} \left (b x -a \right )^{2}}\) | \(191\) |
Input:
int(1/x^3/(b*x+a)^3/(-b*c*x+a*c)^3,x,method=_RETURNVERBOSE)
Output:
(-3/2/a^6*b^4*x^4+9/4/a^4*b^2*x^2-1/2/a^2)/x^2/(b*x+a)^2/c^3/(-b*x+a)^2+3* b^2*ln(x)/a^8/c^3-3/2*b^2*ln(-b^2*x^2+a^2)/a^8/c^3
Time = 0.07 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.42 \[ \int \frac {1}{x^3 (a+b x)^3 (a c-b c x)^3} \, dx=-\frac {6 \, a^{2} b^{4} x^{4} - 9 \, a^{4} b^{2} x^{2} + 2 \, a^{6} + 6 \, {\left (b^{6} x^{6} - 2 \, a^{2} b^{4} x^{4} + a^{4} b^{2} x^{2}\right )} \log \left (b^{2} x^{2} - a^{2}\right ) - 12 \, {\left (b^{6} x^{6} - 2 \, a^{2} b^{4} x^{4} + a^{4} b^{2} x^{2}\right )} \log \left (x\right )}{4 \, {\left (a^{8} b^{4} c^{3} x^{6} - 2 \, a^{10} b^{2} c^{3} x^{4} + a^{12} c^{3} x^{2}\right )}} \] Input:
integrate(1/x^3/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="fricas")
Output:
-1/4*(6*a^2*b^4*x^4 - 9*a^4*b^2*x^2 + 2*a^6 + 6*(b^6*x^6 - 2*a^2*b^4*x^4 + a^4*b^2*x^2)*log(b^2*x^2 - a^2) - 12*(b^6*x^6 - 2*a^2*b^4*x^4 + a^4*b^2*x ^2)*log(x))/(a^8*b^4*c^3*x^6 - 2*a^10*b^2*c^3*x^4 + a^12*c^3*x^2)
Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^3 (a+b x)^3 (a c-b c x)^3} \, dx=- \frac {2 a^{4} - 9 a^{2} b^{2} x^{2} + 6 b^{4} x^{4}}{4 a^{10} c^{3} x^{2} - 8 a^{8} b^{2} c^{3} x^{4} + 4 a^{6} b^{4} c^{3} x^{6}} + \frac {3 b^{2} \log {\left (x \right )}}{a^{8} c^{3}} - \frac {3 b^{2} \log {\left (- \frac {a^{2}}{b^{2}} + x^{2} \right )}}{2 a^{8} c^{3}} \] Input:
integrate(1/x**3/(b*x+a)**3/(-b*c*x+a*c)**3,x)
Output:
-(2*a**4 - 9*a**2*b**2*x**2 + 6*b**4*x**4)/(4*a**10*c**3*x**2 - 8*a**8*b** 2*c**3*x**4 + 4*a**6*b**4*c**3*x**6) + 3*b**2*log(x)/(a**8*c**3) - 3*b**2* log(-a**2/b**2 + x**2)/(2*a**8*c**3)
Time = 0.03 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^3 (a+b x)^3 (a c-b c x)^3} \, dx=-\frac {6 \, b^{4} x^{4} - 9 \, a^{2} b^{2} x^{2} + 2 \, a^{4}}{4 \, {\left (a^{6} b^{4} c^{3} x^{6} - 2 \, a^{8} b^{2} c^{3} x^{4} + a^{10} c^{3} x^{2}\right )}} - \frac {3 \, b^{2} \log \left (b x + a\right )}{2 \, a^{8} c^{3}} - \frac {3 \, b^{2} \log \left (b x - a\right )}{2 \, a^{8} c^{3}} + \frac {3 \, b^{2} \log \left (x\right )}{a^{8} c^{3}} \] Input:
integrate(1/x^3/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="maxima")
Output:
-1/4*(6*b^4*x^4 - 9*a^2*b^2*x^2 + 2*a^4)/(a^6*b^4*c^3*x^6 - 2*a^8*b^2*c^3* x^4 + a^10*c^3*x^2) - 3/2*b^2*log(b*x + a)/(a^8*c^3) - 3/2*b^2*log(b*x - a )/(a^8*c^3) + 3*b^2*log(x)/(a^8*c^3)
Time = 0.12 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^3 (a+b x)^3 (a c-b c x)^3} \, dx=\frac {3 \, b^{2} \log \left (x^{2}\right )}{2 \, a^{8} c^{3}} - \frac {3 \, b^{2} \log \left ({\left | b^{2} x^{2} - a^{2} \right |}\right )}{2 \, a^{8} c^{3}} + \frac {9 \, b^{6} x^{4} - 22 \, a^{2} b^{4} x^{2} + 14 \, a^{4} b^{2}}{4 \, {\left (b^{2} x^{2} - a^{2}\right )}^{2} a^{8} c^{3}} - \frac {3 \, b^{2} x^{2} + a^{2}}{2 \, a^{8} c^{3} x^{2}} \] Input:
integrate(1/x^3/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="giac")
Output:
3/2*b^2*log(x^2)/(a^8*c^3) - 3/2*b^2*log(abs(b^2*x^2 - a^2))/(a^8*c^3) + 1 /4*(9*b^6*x^4 - 22*a^2*b^4*x^2 + 14*a^4*b^2)/((b^2*x^2 - a^2)^2*a^8*c^3) - 1/2*(3*b^2*x^2 + a^2)/(a^8*c^3*x^2)
Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^3 (a+b x)^3 (a c-b c x)^3} \, dx=\frac {3\,b^2\,\ln \left (x\right )}{a^8\,c^3}-\frac {3\,b^2\,\ln \left (a^2-b^2\,x^2\right )}{2\,a^8\,c^3}-\frac {\frac {1}{2\,a^2}-\frac {9\,b^2\,x^2}{4\,a^4}+\frac {3\,b^4\,x^4}{2\,a^6}}{a^4\,c^3\,x^2-2\,a^2\,b^2\,c^3\,x^4+b^4\,c^3\,x^6} \] Input:
int(1/(x^3*(a*c - b*c*x)^3*(a + b*x)^3),x)
Output:
(3*b^2*log(x))/(a^8*c^3) - (3*b^2*log(a^2 - b^2*x^2))/(2*a^8*c^3) - (1/(2* a^2) - (9*b^2*x^2)/(4*a^4) + (3*b^4*x^4)/(2*a^6))/(a^4*c^3*x^2 + b^4*c^3*x ^6 - 2*a^2*b^2*c^3*x^4)
Time = 0.16 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.96 \[ \int \frac {1}{x^3 (a+b x)^3 (a c-b c x)^3} \, dx=\frac {-6 \,\mathrm {log}\left (-b x -a \right ) a^{4} b^{2} x^{2}+12 \,\mathrm {log}\left (-b x -a \right ) a^{2} b^{4} x^{4}-6 \,\mathrm {log}\left (-b x -a \right ) b^{6} x^{6}-6 \,\mathrm {log}\left (-b x +a \right ) a^{4} b^{2} x^{2}+12 \,\mathrm {log}\left (-b x +a \right ) a^{2} b^{4} x^{4}-6 \,\mathrm {log}\left (-b x +a \right ) b^{6} x^{6}+12 \,\mathrm {log}\left (x \right ) a^{4} b^{2} x^{2}-24 \,\mathrm {log}\left (x \right ) a^{2} b^{4} x^{4}+12 \,\mathrm {log}\left (x \right ) b^{6} x^{6}-2 a^{6}+6 a^{4} b^{2} x^{2}-3 b^{6} x^{6}}{4 a^{8} c^{3} x^{2} \left (b^{4} x^{4}-2 a^{2} b^{2} x^{2}+a^{4}\right )} \] Input:
int(1/x^3/(b*x+a)^3/(-b*c*x+a*c)^3,x)
Output:
( - 6*log( - a - b*x)*a**4*b**2*x**2 + 12*log( - a - b*x)*a**2*b**4*x**4 - 6*log( - a - b*x)*b**6*x**6 - 6*log(a - b*x)*a**4*b**2*x**2 + 12*log(a - b*x)*a**2*b**4*x**4 - 6*log(a - b*x)*b**6*x**6 + 12*log(x)*a**4*b**2*x**2 - 24*log(x)*a**2*b**4*x**4 + 12*log(x)*b**6*x**6 - 2*a**6 + 6*a**4*b**2*x* *2 - 3*b**6*x**6)/(4*a**8*c**3*x**2*(a**4 - 2*a**2*b**2*x**2 + b**4*x**4))