Integrand size = 13, antiderivative size = 72 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^6} \, dx=\frac {a^4}{2 b^5 \left (a+\frac {b}{x}\right )^2}-\frac {4 a^3}{b^5 \left (a+\frac {b}{x}\right )}-\frac {1}{2 b^3 x^2}+\frac {3 a}{b^4 x}-\frac {6 a^2 \log \left (a+\frac {b}{x}\right )}{b^5} \] Output:
1/2*a^4/b^5/(a+b/x)^2-4*a^3/b^5/(a+b/x)-1/2/b^3/x^2+3*a/b^4/x-6*a^2*ln(a+b /x)/b^5
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^6} \, dx=\frac {\frac {b \left (-b^3+4 a b^2 x+18 a^2 b x^2+12 a^3 x^3\right )}{x^2 (b+a x)^2}+12 a^2 \log (x)-12 a^2 \log (b+a x)}{2 b^5} \] Input:
Integrate[1/((a + b/x)^3*x^6),x]
Output:
((b*(-b^3 + 4*a*b^2*x + 18*a^2*b*x^2 + 12*a^3*x^3))/(x^2*(b + a*x)^2) + 12 *a^2*Log[x] - 12*a^2*Log[b + a*x])/(2*b^5)
Time = 0.34 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {795, 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.
\(\displaystyle \int \frac {1}{x^6 \left (a+\frac {b}{x}\right )^3} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {1}{x^3 (a x+b)^3}dx\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \int \left (-\frac {6 a^3}{b^5 (a x+b)}-\frac {3 a^3}{b^4 (a x+b)^2}-\frac {a^3}{b^3 (a x+b)^3}+\frac {6 a^2}{b^5 x}-\frac {3 a}{b^4 x^2}+\frac {1}{b^3 x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {6 a^2 \log (x)}{b^5}-\frac {6 a^2 \log (a x+b)}{b^5}+\frac {3 a^2}{b^4 (a x+b)}+\frac {a^2}{2 b^3 (a x+b)^2}+\frac {3 a}{b^4 x}-\frac {1}{2 b^3 x^2}\) |
Input:
Int[1/((a + b/x)^3*x^6),x]
Output:
-1/2*1/(b^3*x^2) + (3*a)/(b^4*x) + a^2/(2*b^3*(b + a*x)^2) + (3*a^2)/(b^4* (b + a*x)) + (6*a^2*Log[x])/b^5 - (6*a^2*Log[b + a*x])/b^5
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_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* (b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01
method | result | size |
default | \(-\frac {1}{2 b^{3} x^{2}}+\frac {3 a}{b^{4} x}+\frac {6 a^{2} \ln \left (x \right )}{b^{5}}-\frac {6 a^{2} \ln \left (a x +b \right )}{b^{5}}+\frac {3 a^{2}}{b^{4} \left (a x +b \right )}+\frac {a^{2}}{2 b^{3} \left (a x +b \right )^{2}}\) | \(73\) |
risch | \(\frac {\frac {6 a^{3} x^{3}}{b^{4}}+\frac {9 a^{2} x^{2}}{b^{3}}+\frac {2 a x}{b^{2}}-\frac {1}{2 b}}{x^{2} \left (a x +b \right )^{2}}+\frac {6 a^{2} \ln \left (-x \right )}{b^{5}}-\frac {6 a^{2} \ln \left (a x +b \right )}{b^{5}}\) | \(74\) |
norman | \(\frac {-\frac {x^{3}}{2 b}+\frac {2 a \,x^{4}}{b^{2}}-\frac {12 a^{3} x^{6}}{b^{4}}-\frac {9 a^{4} x^{7}}{b^{5}}}{x^{5} \left (a x +b \right )^{2}}+\frac {6 a^{2} \ln \left (x \right )}{b^{5}}-\frac {6 a^{2} \ln \left (a x +b \right )}{b^{5}}\) | \(77\) |
parallelrisch | \(\frac {12 \ln \left (x \right ) x^{4} a^{6}-12 \ln \left (a x +b \right ) x^{4} a^{6}+24 \ln \left (x \right ) x^{3} a^{5} b -24 \ln \left (a x +b \right ) x^{3} a^{5} b +12 \ln \left (x \right ) x^{2} a^{4} b^{2}-12 \ln \left (a x +b \right ) x^{2} a^{4} b^{2}+12 a^{5} b \,x^{3}+18 a^{4} x^{2} b^{2}+4 x \,a^{3} b^{3}-a^{2} b^{4}}{2 a^{2} b^{5} x^{2} \left (a x +b \right )^{2}}\) | \(137\) |
Input:
int(1/(a+b/x)^3/x^6,x,method=_RETURNVERBOSE)
Output:
-1/2/b^3/x^2+3*a/b^4/x+6/b^5*a^2*ln(x)-6/b^5*a^2*ln(a*x+b)+3*a^2/b^4/(a*x+ b)+1/2*a^2/b^3/(a*x+b)^2
Time = 0.09 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.81 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^6} \, dx=\frac {12 \, a^{3} b x^{3} + 18 \, a^{2} b^{2} x^{2} + 4 \, a b^{3} x - b^{4} - 12 \, {\left (a^{4} x^{4} + 2 \, a^{3} b x^{3} + a^{2} b^{2} x^{2}\right )} \log \left (a x + b\right ) + 12 \, {\left (a^{4} x^{4} + 2 \, a^{3} b x^{3} + a^{2} b^{2} x^{2}\right )} \log \left (x\right )}{2 \, {\left (a^{2} b^{5} x^{4} + 2 \, a b^{6} x^{3} + b^{7} x^{2}\right )}} \] Input:
integrate(1/(a+b/x)^3/x^6,x, algorithm="fricas")
Output:
1/2*(12*a^3*b*x^3 + 18*a^2*b^2*x^2 + 4*a*b^3*x - b^4 - 12*(a^4*x^4 + 2*a^3 *b*x^3 + a^2*b^2*x^2)*log(a*x + b) + 12*(a^4*x^4 + 2*a^3*b*x^3 + a^2*b^2*x ^2)*log(x))/(a^2*b^5*x^4 + 2*a*b^6*x^3 + b^7*x^2)
Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^6} \, dx=\frac {6 a^{2} \left (\log {\left (x \right )} - \log {\left (x + \frac {b}{a} \right )}\right )}{b^{5}} + \frac {12 a^{3} x^{3} + 18 a^{2} b x^{2} + 4 a b^{2} x - b^{3}}{2 a^{2} b^{4} x^{4} + 4 a b^{5} x^{3} + 2 b^{6} x^{2}} \] Input:
integrate(1/(a+b/x)**3/x**6,x)
Output:
6*a**2*(log(x) - log(x + b/a))/b**5 + (12*a**3*x**3 + 18*a**2*b*x**2 + 4*a *b**2*x - b**3)/(2*a**2*b**4*x**4 + 4*a*b**5*x**3 + 2*b**6*x**2)
Time = 0.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^6} \, dx=\frac {12 \, a^{3} x^{3} + 18 \, a^{2} b x^{2} + 4 \, a b^{2} x - b^{3}}{2 \, {\left (a^{2} b^{4} x^{4} + 2 \, a b^{5} x^{3} + b^{6} x^{2}\right )}} - \frac {6 \, a^{2} \log \left (a x + b\right )}{b^{5}} + \frac {6 \, a^{2} \log \left (x\right )}{b^{5}} \] Input:
integrate(1/(a+b/x)^3/x^6,x, algorithm="maxima")
Output:
1/2*(12*a^3*x^3 + 18*a^2*b*x^2 + 4*a*b^2*x - b^3)/(a^2*b^4*x^4 + 2*a*b^5*x ^3 + b^6*x^2) - 6*a^2*log(a*x + b)/b^5 + 6*a^2*log(x)/b^5
Time = 0.12 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^6} \, dx=-\frac {6 \, a^{2} \log \left ({\left | a x + b \right |}\right )}{b^{5}} + \frac {6 \, a^{2} \log \left ({\left | x \right |}\right )}{b^{5}} + \frac {12 \, a^{3} x^{3} + 18 \, a^{2} b x^{2} + 4 \, a b^{2} x - b^{3}}{2 \, {\left (a x^{2} + b x\right )}^{2} b^{4}} \] Input:
integrate(1/(a+b/x)^3/x^6,x, algorithm="giac")
Output:
-6*a^2*log(abs(a*x + b))/b^5 + 6*a^2*log(abs(x))/b^5 + 1/2*(12*a^3*x^3 + 1 8*a^2*b*x^2 + 4*a*b^2*x - b^3)/((a*x^2 + b*x)^2*b^4)
Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^6} \, dx=\frac {\frac {9\,a^2\,x^2}{b^3}-\frac {1}{2\,b}+\frac {6\,a^3\,x^3}{b^4}+\frac {2\,a\,x}{b^2}}{a^2\,x^4+2\,a\,b\,x^3+b^2\,x^2}-\frac {12\,a^2\,\mathrm {atanh}\left (\frac {2\,a\,x}{b}+1\right )}{b^5} \] Input:
int(1/(x^6*(a + b/x)^3),x)
Output:
((9*a^2*x^2)/b^3 - 1/(2*b) + (6*a^3*x^3)/b^4 + (2*a*x)/b^2)/(a^2*x^4 + b^2 *x^2 + 2*a*b*x^3) - (12*a^2*atanh((2*a*x)/b + 1))/b^5
Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.92 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^3 x^6} \, dx=\frac {-12 \,\mathrm {log}\left (a x +b \right ) a^{4} x^{4}-24 \,\mathrm {log}\left (a x +b \right ) a^{3} b \,x^{3}-12 \,\mathrm {log}\left (a x +b \right ) a^{2} b^{2} x^{2}+12 \,\mathrm {log}\left (x \right ) a^{4} x^{4}+24 \,\mathrm {log}\left (x \right ) a^{3} b \,x^{3}+12 \,\mathrm {log}\left (x \right ) a^{2} b^{2} x^{2}-6 a^{4} x^{4}+12 a^{2} b^{2} x^{2}+4 a \,b^{3} x -b^{4}}{2 b^{5} x^{2} \left (a^{2} x^{2}+2 a b x +b^{2}\right )} \] Input:
int(1/(a+b/x)^3/x^6,x)
Output:
( - 12*log(a*x + b)*a**4*x**4 - 24*log(a*x + b)*a**3*b*x**3 - 12*log(a*x + b)*a**2*b**2*x**2 + 12*log(x)*a**4*x**4 + 24*log(x)*a**3*b*x**3 + 12*log( x)*a**2*b**2*x**2 - 6*a**4*x**4 + 12*a**2*b**2*x**2 + 4*a*b**3*x - b**4)/( 2*b**5*x**2*(a**2*x**2 + 2*a*b*x + b**2))