Integrand size = 56, antiderivative size = 23 \[ \int \frac {18+2 x+18 \log \left (\frac {5}{x}\right )+\left (162-126 x-34 x^2-2 x^3\right ) \log ^2\left (\frac {5}{x}\right )}{\left (81+18 x+x^2\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=-((-2+x) x)+\frac {2 x}{(9+x) \log \left (\frac {5}{x}\right )} \] Output:
2*x/(x+9)/ln(5/x)-x*(-2+x)
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {18+2 x+18 \log \left (\frac {5}{x}\right )+\left (162-126 x-34 x^2-2 x^3\right ) \log ^2\left (\frac {5}{x}\right )}{\left (81+18 x+x^2\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=2 \left (x-\frac {x^2}{2}+\frac {x}{(9+x) \log \left (\frac {5}{x}\right )}\right ) \] Input:
Integrate[(18 + 2*x + 18*Log[5/x] + (162 - 126*x - 34*x^2 - 2*x^3)*Log[5/x ]^2)/((81 + 18*x + x^2)*Log[5/x]^2),x]
Output:
2*(x - x^2/2 + x/((9 + x)*Log[5/x]))
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.
\(\displaystyle \int \frac {\left (-2 x^3-34 x^2-126 x+162\right ) \log ^2\left (\frac {5}{x}\right )+2 x+18 \log \left (\frac {5}{x}\right )+18}{\left (x^2+18 x+81\right ) \log ^2\left (\frac {5}{x}\right )} \, dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (-2 x^3-34 x^2-126 x+162\right ) \log ^2\left (\frac {5}{x}\right )+2 x+18 \log \left (\frac {5}{x}\right )+18}{(x+9)^2 \log ^2\left (\frac {5}{x}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-2 (x-1)+\frac {2}{(x+9) \log ^2\left (\frac {5}{x}\right )}+\frac {18}{(x+9)^2 \log \left (\frac {5}{x}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int \frac {1}{(x+9) \log ^2\left (\frac {5}{x}\right )}dx+18 \int \frac {1}{(x+9)^2 \log \left (\frac {5}{x}\right )}dx-(1-x)^2\) |
Input:
Int[(18 + 2*x + 18*Log[5/x] + (162 - 126*x - 34*x^2 - 2*x^3)*Log[5/x]^2)/( (81 + 18*x + x^2)*Log[5/x]^2),x]
Output:
$Aborted
Time = 0.58 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13
method | result | size |
risch | \(-x^{2}+2 x +\frac {2 x}{\left (x +9\right ) \ln \left (\frac {5}{x}\right )}\) | \(26\) |
norman | \(\frac {-162 \ln \left (\frac {5}{x}\right )+2 x -x^{3} \ln \left (\frac {5}{x}\right )-7 \ln \left (\frac {5}{x}\right ) x^{2}}{\left (x +9\right ) \ln \left (\frac {5}{x}\right )}\) | \(49\) |
parallelrisch | \(\frac {-162 \ln \left (\frac {5}{x}\right )+2 x -x^{3} \ln \left (\frac {5}{x}\right )-7 \ln \left (\frac {5}{x}\right ) x^{2}}{\left (x +9\right ) \ln \left (\frac {5}{x}\right )}\) | \(49\) |
derivativedivides | \(-\frac {\left (-\frac {450 \ln \left (\frac {5}{x}\right )}{x^{2}}+25 \ln \left (\frac {5}{x}\right )-\frac {50}{x^{2}}+\frac {175 \ln \left (\frac {5}{x}\right )}{x}\right ) x^{2}}{5 \left (5+\frac {45}{x}\right ) \ln \left (\frac {5}{x}\right )}\) | \(59\) |
default | \(-\frac {\left (-\frac {450 \ln \left (\frac {5}{x}\right )}{x^{2}}+25 \ln \left (\frac {5}{x}\right )-\frac {50}{x^{2}}+\frac {175 \ln \left (\frac {5}{x}\right )}{x}\right ) x^{2}}{5 \left (5+\frac {45}{x}\right ) \ln \left (\frac {5}{x}\right )}\) | \(59\) |
Input:
int(((-2*x^3-34*x^2-126*x+162)*ln(5/x)^2+18*ln(5/x)+2*x+18)/(x^2+18*x+81)/ ln(5/x)^2,x,method=_RETURNVERBOSE)
Output:
-x^2+2*x+2*x/(x+9)/ln(5/x)
Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {18+2 x+18 \log \left (\frac {5}{x}\right )+\left (162-126 x-34 x^2-2 x^3\right ) \log ^2\left (\frac {5}{x}\right )}{\left (81+18 x+x^2\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=-\frac {{\left (x^{3} + 7 \, x^{2} - 18 \, x\right )} \log \left (\frac {5}{x}\right ) - 2 \, x}{{\left (x + 9\right )} \log \left (\frac {5}{x}\right )} \] Input:
integrate(((-2*x^3-34*x^2-126*x+162)*log(5/x)^2+18*log(5/x)+2*x+18)/(x^2+1 8*x+81)/log(5/x)^2,x, algorithm="fricas")
Output:
-((x^3 + 7*x^2 - 18*x)*log(5/x) - 2*x)/((x + 9)*log(5/x))
Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {18+2 x+18 \log \left (\frac {5}{x}\right )+\left (162-126 x-34 x^2-2 x^3\right ) \log ^2\left (\frac {5}{x}\right )}{\left (81+18 x+x^2\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=- x^{2} + 2 x + \frac {2 x}{\left (x + 9\right ) \log {\left (\frac {5}{x} \right )}} \] Input:
integrate(((-2*x**3-34*x**2-126*x+162)*ln(5/x)**2+18*ln(5/x)+2*x+18)/(x**2 +18*x+81)/ln(5/x)**2,x)
Output:
-x**2 + 2*x + 2*x/((x + 9)*log(5/x))
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (23) = 46\).
Time = 0.15 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.57 \[ \int \frac {18+2 x+18 \log \left (\frac {5}{x}\right )+\left (162-126 x-34 x^2-2 x^3\right ) \log ^2\left (\frac {5}{x}\right )}{\left (81+18 x+x^2\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=-\frac {x^{3} \log \left (5\right ) + 7 \, x^{2} \log \left (5\right ) - 2 \, x {\left (9 \, \log \left (5\right ) + 1\right )} - {\left (x^{3} + 7 \, x^{2} - 18 \, x\right )} \log \left (x\right )}{x \log \left (5\right ) - {\left (x + 9\right )} \log \left (x\right ) + 9 \, \log \left (5\right )} \] Input:
integrate(((-2*x^3-34*x^2-126*x+162)*log(5/x)^2+18*log(5/x)+2*x+18)/(x^2+1 8*x+81)/log(5/x)^2,x, algorithm="maxima")
Output:
-(x^3*log(5) + 7*x^2*log(5) - 2*x*(9*log(5) + 1) - (x^3 + 7*x^2 - 18*x)*lo g(x))/(x*log(5) - (x + 9)*log(x) + 9*log(5))
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {18+2 x+18 \log \left (\frac {5}{x}\right )+\left (162-126 x-34 x^2-2 x^3\right ) \log ^2\left (\frac {5}{x}\right )}{\left (81+18 x+x^2\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=x^{2} {\left (\frac {2}{x} - 1\right )} + \frac {2}{\frac {9 \, \log \left (\frac {5}{x}\right )}{x} + \log \left (\frac {5}{x}\right )} \] Input:
integrate(((-2*x^3-34*x^2-126*x+162)*log(5/x)^2+18*log(5/x)+2*x+18)/(x^2+1 8*x+81)/log(5/x)^2,x, algorithm="giac")
Output:
x^2*(2/x - 1) + 2/(9*log(5/x)/x + log(5/x))
Time = 2.42 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.96 \[ \int \frac {18+2 x+18 \log \left (\frac {5}{x}\right )+\left (162-126 x-34 x^2-2 x^3\right ) \log ^2\left (\frac {5}{x}\right )}{\left (81+18 x+x^2\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=\frac {18\,x}{x+9}-\frac {7\,x^2}{x+9}-\frac {x^3}{x+9}+\frac {2\,x}{\ln \left (\frac {5}{x}\right )\,\left (x+9\right )} \] Input:
int((2*x + 18*log(5/x) - log(5/x)^2*(126*x + 34*x^2 + 2*x^3 - 162) + 18)/( log(5/x)^2*(18*x + x^2 + 81)),x)
Output:
(18*x)/(x + 9) - (7*x^2)/(x + 9) - x^3/(x + 9) + (2*x)/(log(5/x)*(x + 9))
Time = 0.20 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.96 \[ \int \frac {18+2 x+18 \log \left (\frac {5}{x}\right )+\left (162-126 x-34 x^2-2 x^3\right ) \log ^2\left (\frac {5}{x}\right )}{\left (81+18 x+x^2\right ) \log ^2\left (\frac {5}{x}\right )} \, dx=\frac {x \left (-\mathrm {log}\left (\frac {5}{x}\right ) x^{2}-7 \,\mathrm {log}\left (\frac {5}{x}\right ) x +18 \,\mathrm {log}\left (\frac {5}{x}\right )+2\right )}{\mathrm {log}\left (\frac {5}{x}\right ) \left (x +9\right )} \] Input:
int(((-2*x^3-34*x^2-126*x+162)*log(5/x)^2+18*log(5/x)+2*x+18)/(x^2+18*x+81 )/log(5/x)^2,x)
Output:
(x*( - log(5/x)*x**2 - 7*log(5/x)*x + 18*log(5/x) + 2))/(log(5/x)*(x + 9))