Integrand size = 22, antiderivative size = 86 \[ \int \frac {1}{x^2 (a+b x)^3 (a c-b c x)^3} \, dx=-\frac {1}{a^6 c^3 x}+\frac {b^2 x}{4 a^4 c^3 \left (a^2-b^2 x^2\right )^2}+\frac {7 b^2 x}{8 a^6 c^3 \left (a^2-b^2 x^2\right )}+\frac {15 b \text {arctanh}\left (\frac {b x}{a}\right )}{8 a^7 c^3} \] Output:
-1/a^6/c^3/x+1/4*b^2*x/a^4/c^3/(-b^2*x^2+a^2)^2+7/8*b^2*x/a^6/c^3/(-b^2*x^ 2+a^2)+15/8*b*arctanh(b*x/a)/a^7/c^3
Time = 0.02 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x^2 (a+b x)^3 (a c-b c x)^3} \, dx=\frac {-16 a^5+50 a^3 b^2 x^2-30 a b^4 x^4-15 b x \left (a^2-b^2 x^2\right )^2 \log (a-b x)+15 b x \left (a^2-b^2 x^2\right )^2 \log (a+b x)}{16 a^7 c^3 x (a-b x)^2 (a+b x)^2} \] Input:
Integrate[1/(x^2*(a + b*x)^3*(a*c - b*c*x)^3),x]
Output:
(-16*a^5 + 50*a^3*b^2*x^2 - 30*a*b^4*x^4 - 15*b*x*(a^2 - b^2*x^2)^2*Log[a - b*x] + 15*b*x*(a^2 - b^2*x^2)^2*Log[a + b*x])/(16*a^7*c^3*x*(a - b*x)^2* (a + b*x)^2)
Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {82, 253, 27, 253, 264, 221}
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.
\(\displaystyle \int \frac {1}{x^2 (a+b x)^3 (a c-b c x)^3} \, dx\) |
\(\Big \downarrow \) 82 |
\(\displaystyle \int \frac {1}{x^2 \left (a^2 c-b^2 c x^2\right )^3}dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {5 \int \frac {1}{c^2 x^2 \left (a^2-b^2 x^2\right )^2}dx}{4 a^2 c}+\frac {1}{4 a^2 c^3 x \left (a^2-b^2 x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5 \int \frac {1}{x^2 \left (a^2-b^2 x^2\right )^2}dx}{4 a^2 c^3}+\frac {1}{4 a^2 c^3 x \left (a^2-b^2 x^2\right )^2}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {1}{x^2 \left (a^2-b^2 x^2\right )}dx}{2 a^2}+\frac {1}{2 a^2 x \left (a^2-b^2 x^2\right )}\right )}{4 a^2 c^3}+\frac {1}{4 a^2 c^3 x \left (a^2-b^2 x^2\right )^2}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {b^2 \int \frac {1}{a^2-b^2 x^2}dx}{a^2}-\frac {1}{a^2 x}\right )}{2 a^2}+\frac {1}{2 a^2 x \left (a^2-b^2 x^2\right )}\right )}{4 a^2 c^3}+\frac {1}{4 a^2 c^3 x \left (a^2-b^2 x^2\right )^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {1}{4 a^2 c^3 x \left (a^2-b^2 x^2\right )^2}+\frac {5 \left (\frac {1}{2 a^2 x \left (a^2-b^2 x^2\right )}+\frac {3 \left (\frac {b \text {arctanh}\left (\frac {b x}{a}\right )}{a^3}-\frac {1}{a^2 x}\right )}{2 a^2}\right )}{4 a^2 c^3}\) |
Input:
Int[1/(x^2*(a + b*x)^3*(a*c - b*c*x)^3),x]
Output:
1/(4*a^2*c^3*x*(a^2 - b^2*x^2)^2) + (5*(1/(2*a^2*x*(a^2 - b^2*x^2)) + (3*( -(1/(a^2*x)) + (b*ArcTanh[(b*x)/a])/a^3))/(2*a^2)))/(4*a^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_.)*((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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.98
method | result | size |
risch | \(\frac {-\frac {15 b^{4} x^{4}}{8 a^{6}}+\frac {25 b^{2} x^{2}}{8 a^{4}}-\frac {1}{a^{2}}}{x \left (b x +a \right )^{2} c^{3} \left (-b x +a \right )^{2}}-\frac {15 b \ln \left (b x -a \right )}{16 a^{7} c^{3}}+\frac {15 b \ln \left (b x +a \right )}{16 a^{7} c^{3}}\) | \(84\) |
norman | \(\frac {-\frac {1}{a^{2} c}+\frac {25 b^{2} x^{2}}{8 a^{4} c}-\frac {15 b^{4} x^{4}}{8 a^{6} c}}{x \left (b x +a \right )^{2} c^{2} \left (-b x +a \right )^{2}}-\frac {15 b \ln \left (-b x +a \right )}{16 a^{7} c^{3}}+\frac {15 b \ln \left (b x +a \right )}{16 a^{7} c^{3}}\) | \(92\) |
default | \(\frac {-\frac {1}{a^{6} x}+\frac {15 b \ln \left (b x +a \right )}{16 a^{7}}-\frac {7 b}{16 a^{6} \left (b x +a \right )}-\frac {b}{16 a^{5} \left (b x +a \right )^{2}}-\frac {15 b \ln \left (-b x +a \right )}{16 a^{7}}+\frac {7 b}{16 a^{6} \left (-b x +a \right )}+\frac {b}{16 a^{5} \left (-b x +a \right )^{2}}}{c^{3}}\) | \(93\) |
parallelrisch | \(\frac {-15 \ln \left (b x -a \right ) x^{5} b^{9}+15 \ln \left (b x +a \right ) x^{5} b^{9}+30 \ln \left (b x -a \right ) x^{3} a^{2} b^{7}-30 \ln \left (b x +a \right ) x^{3} a^{2} b^{7}-30 x^{4} a \,b^{8}-15 \ln \left (b x -a \right ) x \,a^{4} b^{5}+15 \ln \left (b x +a \right ) x \,a^{4} b^{5}+50 x^{2} a^{3} b^{6}-16 a^{5} b^{4}}{16 a^{7} b^{4} c^{3} x \left (b x +a \right )^{2} \left (b x -a \right )^{2}}\) | \(158\) |
Input:
int(1/x^2/(b*x+a)^3/(-b*c*x+a*c)^3,x,method=_RETURNVERBOSE)
Output:
(-15/8/a^6*b^4*x^4+25/8/a^4*b^2*x^2-1/a^2)/x/(b*x+a)^2/c^3/(-b*x+a)^2-15/1 6*b/a^7/c^3*ln(b*x-a)+15/16*b/a^7/c^3*ln(b*x+a)
Time = 0.07 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.56 \[ \int \frac {1}{x^2 (a+b x)^3 (a c-b c x)^3} \, dx=-\frac {30 \, a b^{4} x^{4} - 50 \, a^{3} b^{2} x^{2} + 16 \, a^{5} - 15 \, {\left (b^{5} x^{5} - 2 \, a^{2} b^{3} x^{3} + a^{4} b x\right )} \log \left (b x + a\right ) + 15 \, {\left (b^{5} x^{5} - 2 \, a^{2} b^{3} x^{3} + a^{4} b x\right )} \log \left (b x - a\right )}{16 \, {\left (a^{7} b^{4} c^{3} x^{5} - 2 \, a^{9} b^{2} c^{3} x^{3} + a^{11} c^{3} x\right )}} \] Input:
integrate(1/x^2/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="fricas")
Output:
-1/16*(30*a*b^4*x^4 - 50*a^3*b^2*x^2 + 16*a^5 - 15*(b^5*x^5 - 2*a^2*b^3*x^ 3 + a^4*b*x)*log(b*x + a) + 15*(b^5*x^5 - 2*a^2*b^3*x^3 + a^4*b*x)*log(b*x - a))/(a^7*b^4*c^3*x^5 - 2*a^9*b^2*c^3*x^3 + a^11*c^3*x)
Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.10 \[ \int \frac {1}{x^2 (a+b x)^3 (a c-b c x)^3} \, dx=- \frac {8 a^{4} - 25 a^{2} b^{2} x^{2} + 15 b^{4} x^{4}}{8 a^{10} c^{3} x - 16 a^{8} b^{2} c^{3} x^{3} + 8 a^{6} b^{4} c^{3} x^{5}} - \frac {b \left (\frac {15 \log {\left (- \frac {a}{b} + x \right )}}{16} - \frac {15 \log {\left (\frac {a}{b} + x \right )}}{16}\right )}{a^{7} c^{3}} \] Input:
integrate(1/x**2/(b*x+a)**3/(-b*c*x+a*c)**3,x)
Output:
-(8*a**4 - 25*a**2*b**2*x**2 + 15*b**4*x**4)/(8*a**10*c**3*x - 16*a**8*b** 2*c**3*x**3 + 8*a**6*b**4*c**3*x**5) - b*(15*log(-a/b + x)/16 - 15*log(a/b + x)/16)/(a**7*c**3)
Time = 0.03 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x^2 (a+b x)^3 (a c-b c x)^3} \, dx=-\frac {15 \, b^{4} x^{4} - 25 \, a^{2} b^{2} x^{2} + 8 \, a^{4}}{8 \, {\left (a^{6} b^{4} c^{3} x^{5} - 2 \, a^{8} b^{2} c^{3} x^{3} + a^{10} c^{3} x\right )}} + \frac {15 \, b \log \left (b x + a\right )}{16 \, a^{7} c^{3}} - \frac {15 \, b \log \left (b x - a\right )}{16 \, a^{7} c^{3}} \] Input:
integrate(1/x^2/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="maxima")
Output:
-1/8*(15*b^4*x^4 - 25*a^2*b^2*x^2 + 8*a^4)/(a^6*b^4*c^3*x^5 - 2*a^8*b^2*c^ 3*x^3 + a^10*c^3*x) + 15/16*b*log(b*x + a)/(a^7*c^3) - 15/16*b*log(b*x - a )/(a^7*c^3)
Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^2 (a+b x)^3 (a c-b c x)^3} \, dx=\frac {15 \, b \log \left ({\left | b x + a \right |}\right )}{16 \, a^{7} c^{3}} - \frac {15 \, b \log \left ({\left | b x - a \right |}\right )}{16 \, a^{7} c^{3}} - \frac {7 \, b^{4} x^{3} - 9 \, a^{2} b^{2} x}{8 \, {\left (b^{2} x^{2} - a^{2}\right )}^{2} a^{6} c^{3}} - \frac {1}{a^{6} c^{3} x} \] Input:
integrate(1/x^2/(b*x+a)^3/(-b*c*x+a*c)^3,x, algorithm="giac")
Output:
15/16*b*log(abs(b*x + a))/(a^7*c^3) - 15/16*b*log(abs(b*x - a))/(a^7*c^3) - 1/8*(7*b^4*x^3 - 9*a^2*b^2*x)/((b^2*x^2 - a^2)^2*a^6*c^3) - 1/(a^6*c^3*x )
Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^2 (a+b x)^3 (a c-b c x)^3} \, dx=\frac {15\,b\,\mathrm {atanh}\left (\frac {b\,x}{a}\right )}{8\,a^7\,c^3}-\frac {\frac {1}{a^2}-\frac {25\,b^2\,x^2}{8\,a^4}+\frac {15\,b^4\,x^4}{8\,a^6}}{a^4\,c^3\,x-2\,a^2\,b^2\,c^3\,x^3+b^4\,c^3\,x^5} \] Input:
int(1/(x^2*(a*c - b*c*x)^3*(a + b*x)^3),x)
Output:
(15*b*atanh((b*x)/a))/(8*a^7*c^3) - (1/a^2 - (25*b^2*x^2)/(8*a^4) + (15*b^ 4*x^4)/(8*a^6))/(a^4*c^3*x + b^4*c^3*x^5 - 2*a^2*b^2*c^3*x^3)
Time = 0.16 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.87 \[ \int \frac {1}{x^2 (a+b x)^3 (a c-b c x)^3} \, dx=\frac {15 \,\mathrm {log}\left (-b x -a \right ) a^{4} b x -30 \,\mathrm {log}\left (-b x -a \right ) a^{2} b^{3} x^{3}+15 \,\mathrm {log}\left (-b x -a \right ) b^{5} x^{5}-15 \,\mathrm {log}\left (-b x +a \right ) a^{4} b x +30 \,\mathrm {log}\left (-b x +a \right ) a^{2} b^{3} x^{3}-15 \,\mathrm {log}\left (-b x +a \right ) b^{5} x^{5}-16 a^{5}+50 a^{3} b^{2} x^{2}-30 a \,b^{4} x^{4}}{16 a^{7} c^{3} x \left (b^{4} x^{4}-2 a^{2} b^{2} x^{2}+a^{4}\right )} \] Input:
int(1/x^2/(b*x+a)^3/(-b*c*x+a*c)^3,x)
Output:
(15*log( - a - b*x)*a**4*b*x - 30*log( - a - b*x)*a**2*b**3*x**3 + 15*log( - a - b*x)*b**5*x**5 - 15*log(a - b*x)*a**4*b*x + 30*log(a - b*x)*a**2*b* *3*x**3 - 15*log(a - b*x)*b**5*x**5 - 16*a**5 + 50*a**3*b**2*x**2 - 30*a*b **4*x**4)/(16*a**7*c**3*x*(a**4 - 2*a**2*b**2*x**2 + b**4*x**4))