\(\int \frac {-135+750 x-81 x^2+477 x^3-150 x^4+(-810 x^2+2655 x^3-1125 x^4+(-162 x^2+531 x^3-225 x^4) \log (\frac {1}{9} (-9 x+25 x^2))) \log (5+\log (\frac {1}{9} (-9 x+25 x^2)))}{-45 x+125 x^2+(-9 x+25 x^2) \log (\frac {1}{9} (-9 x+25 x^2))} \, dx\) [1873]

Optimal result
Mathematica [A] (verified)
Rubi [F]
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [A] (verification not implemented)
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 121, antiderivative size = 29 \[ \int \frac {-135+750 x-81 x^2+477 x^3-150 x^4+\left (-810 x^2+2655 x^3-1125 x^4+\left (-162 x^2+531 x^3-225 x^4\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right ) \log \left (5+\log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right )}{-45 x+125 x^2+\left (-9 x+25 x^2\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )} \, dx=-3+3 \left (5-(-3+x) x^2\right ) \log \left (5+\log \left (-x+\frac {25 x^2}{9}\right )\right ) \] Output:

3*ln(ln(25/9*x^2-x)+5)*(5-x^2*(-3+x))-3
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-135+750 x-81 x^2+477 x^3-150 x^4+\left (-810 x^2+2655 x^3-1125 x^4+\left (-162 x^2+531 x^3-225 x^4\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right ) \log \left (5+\log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right )}{-45 x+125 x^2+\left (-9 x+25 x^2\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )} \, dx=-3 \left (-5+(-3+x) x^2\right ) \log \left (5+\log \left (\frac {1}{9} x (-9+25 x)\right )\right ) \] Input:

Integrate[(-135 + 750*x - 81*x^2 + 477*x^3 - 150*x^4 + (-810*x^2 + 2655*x^ 
3 - 1125*x^4 + (-162*x^2 + 531*x^3 - 225*x^4)*Log[(-9*x + 25*x^2)/9])*Log[ 
5 + Log[(-9*x + 25*x^2)/9]])/(-45*x + 125*x^2 + (-9*x + 25*x^2)*Log[(-9*x 
+ 25*x^2)/9]),x]
 

Output:

-3*(-5 + (-3 + x)*x^2)*Log[5 + Log[(x*(-9 + 25*x))/9]]
 

Rubi [F]

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.

\(\displaystyle \int \frac {-150 x^4+477 x^3-81 x^2+\left (-1125 x^4+2655 x^3-810 x^2+\left (-225 x^4+531 x^3-162 x^2\right ) \log \left (\frac {1}{9} \left (25 x^2-9 x\right )\right )\right ) \log \left (\log \left (\frac {1}{9} \left (25 x^2-9 x\right )\right )+5\right )+750 x-135}{125 x^2+\left (25 x^2-9 x\right ) \log \left (\frac {1}{9} \left (25 x^2-9 x\right )\right )-45 x} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {150 x^4-477 x^3+81 x^2-\left (-1125 x^4+2655 x^3-810 x^2+\left (-225 x^4+531 x^3-162 x^2\right ) \log \left (\frac {1}{9} \left (25 x^2-9 x\right )\right )\right ) \log \left (\log \left (\frac {1}{9} \left (25 x^2-9 x\right )\right )+5\right )-750 x+135}{(9-25 x) x \left (\log \left (\frac {1}{9} x (25 x-9)\right )+5\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (-\frac {150 x^3}{(25 x-9) \left (\log \left (\frac {1}{9} x (25 x-9)\right )+5\right )}+\frac {477 x^2}{(25 x-9) \left (\log \left (\frac {1}{9} x (25 x-9)\right )+5\right )}-9 (x-2) x \log \left (\log \left (\frac {1}{9} x (25 x-9)\right )+5\right )-\frac {81 x}{(25 x-9) \left (\log \left (\frac {1}{9} x (25 x-9)\right )+5\right )}+\frac {750}{(25 x-9) \left (\log \left (\frac {1}{9} x (25 x-9)\right )+5\right )}-\frac {135}{(25 x-9) x \left (\log \left (\frac {1}{9} x (25 x-9)\right )+5\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -6 \int \frac {x^2}{\log \left (\frac {1}{9} x (25 x-9)\right )+5}dx-9 \int x^2 \log \left (\log \left (\frac {1}{9} x (25 x-9)\right )+5\right )dx+\frac {1782}{625} \int \frac {1}{\log \left (\frac {1}{9} x (25 x-9)\right )+5}dx+15 \int \frac {1}{x \left (\log \left (\frac {1}{9} x (25 x-9)\right )+5\right )}dx+\frac {423}{25} \int \frac {x}{\log \left (\frac {1}{9} x (25 x-9)\right )+5}dx+\frac {250413}{625} \int \frac {1}{(25 x-9) \left (\log \left (\frac {1}{9} x (25 x-9)\right )+5\right )}dx+18 \int x \log \left (\log \left (\frac {1}{9} x (25 x-9)\right )+5\right )dx\)

Input:

Int[(-135 + 750*x - 81*x^2 + 477*x^3 - 150*x^4 + (-810*x^2 + 2655*x^3 - 11 
25*x^4 + (-162*x^2 + 531*x^3 - 225*x^4)*Log[(-9*x + 25*x^2)/9])*Log[5 + Lo 
g[(-9*x + 25*x^2)/9]])/(-45*x + 125*x^2 + (-9*x + 25*x^2)*Log[(-9*x + 25*x 
^2)/9]),x]
 

Output:

$Aborted
 
Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.83

method result size
parallelrisch \(-3 \ln \left (\ln \left (\frac {25}{9} x^{2}-x \right )+5\right ) x^{3}+9 \ln \left (\ln \left (\frac {25}{9} x^{2}-x \right )+5\right ) x^{2}+15 \ln \left (\ln \left (\frac {25}{9} x^{2}-x \right )+5\right )\) \(53\)

Input:

int((((-225*x^4+531*x^3-162*x^2)*ln(25/9*x^2-x)-1125*x^4+2655*x^3-810*x^2) 
*ln(ln(25/9*x^2-x)+5)-150*x^4+477*x^3-81*x^2+750*x-135)/((25*x^2-9*x)*ln(2 
5/9*x^2-x)+125*x^2-45*x),x,method=_RETURNVERBOSE)
 

Output:

-3*ln(ln(25/9*x^2-x)+5)*x^3+9*ln(ln(25/9*x^2-x)+5)*x^2+15*ln(ln(25/9*x^2-x 
)+5)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {-135+750 x-81 x^2+477 x^3-150 x^4+\left (-810 x^2+2655 x^3-1125 x^4+\left (-162 x^2+531 x^3-225 x^4\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right ) \log \left (5+\log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right )}{-45 x+125 x^2+\left (-9 x+25 x^2\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )} \, dx=-3 \, {\left (x^{3} - 3 \, x^{2} - 5\right )} \log \left (\log \left (\frac {25}{9} \, x^{2} - x\right ) + 5\right ) \] Input:

integrate((((-225*x^4+531*x^3-162*x^2)*log(25/9*x^2-x)-1125*x^4+2655*x^3-8 
10*x^2)*log(log(25/9*x^2-x)+5)-150*x^4+477*x^3-81*x^2+750*x-135)/((25*x^2- 
9*x)*log(25/9*x^2-x)+125*x^2-45*x),x, algorithm="fricas")
 

Output:

-3*(x^3 - 3*x^2 - 5)*log(log(25/9*x^2 - x) + 5)
 

Sympy [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.45 \[ \int \frac {-135+750 x-81 x^2+477 x^3-150 x^4+\left (-810 x^2+2655 x^3-1125 x^4+\left (-162 x^2+531 x^3-225 x^4\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right ) \log \left (5+\log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right )}{-45 x+125 x^2+\left (-9 x+25 x^2\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )} \, dx=\left (- 3 x^{3} + 9 x^{2} - \frac {54189}{312500}\right ) \log {\left (\log {\left (\frac {25 x^{2}}{9} - x \right )} + 5 \right )} + \frac {4741689 \log {\left (\log {\left (\frac {25 x^{2}}{9} - x \right )} + 5 \right )}}{312500} \] Input:

integrate((((-225*x**4+531*x**3-162*x**2)*ln(25/9*x**2-x)-1125*x**4+2655*x 
**3-810*x**2)*ln(ln(25/9*x**2-x)+5)-150*x**4+477*x**3-81*x**2+750*x-135)/( 
(25*x**2-9*x)*ln(25/9*x**2-x)+125*x**2-45*x),x)
 

Output:

(-3*x**3 + 9*x**2 - 54189/312500)*log(log(25*x**2/9 - x) + 5) + 4741689*lo 
g(log(25*x**2/9 - x) + 5)/312500
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-135+750 x-81 x^2+477 x^3-150 x^4+\left (-810 x^2+2655 x^3-1125 x^4+\left (-162 x^2+531 x^3-225 x^4\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right ) \log \left (5+\log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right )}{-45 x+125 x^2+\left (-9 x+25 x^2\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )} \, dx=-3 \, {\left (x^{3} - 3 \, x^{2} - 5\right )} \log \left (-2 \, \log \left (3\right ) + \log \left (25 \, x - 9\right ) + \log \left (x\right ) + 5\right ) \] Input:

integrate((((-225*x^4+531*x^3-162*x^2)*log(25/9*x^2-x)-1125*x^4+2655*x^3-8 
10*x^2)*log(log(25/9*x^2-x)+5)-150*x^4+477*x^3-81*x^2+750*x-135)/((25*x^2- 
9*x)*log(25/9*x^2-x)+125*x^2-45*x),x, algorithm="maxima")
 

Output:

-3*(x^3 - 3*x^2 - 5)*log(-2*log(3) + log(25*x - 9) + log(x) + 5)
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {-135+750 x-81 x^2+477 x^3-150 x^4+\left (-810 x^2+2655 x^3-1125 x^4+\left (-162 x^2+531 x^3-225 x^4\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right ) \log \left (5+\log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right )}{-45 x+125 x^2+\left (-9 x+25 x^2\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )} \, dx=-3 \, {\left (x^{3} - 3 \, x^{2}\right )} \log \left (\log \left (\frac {25}{9} \, x^{2} - x\right ) + 5\right ) + 15 \, \log \left (-2 \, \log \left (3\right ) + \log \left (25 \, x^{2} - 9 \, x\right ) + 5\right ) \] Input:

integrate((((-225*x^4+531*x^3-162*x^2)*log(25/9*x^2-x)-1125*x^4+2655*x^3-8 
10*x^2)*log(log(25/9*x^2-x)+5)-150*x^4+477*x^3-81*x^2+750*x-135)/((25*x^2- 
9*x)*log(25/9*x^2-x)+125*x^2-45*x),x, algorithm="giac")
 

Output:

-3*(x^3 - 3*x^2)*log(log(25/9*x^2 - x) + 5) + 15*log(-2*log(3) + log(25*x^ 
2 - 9*x) + 5)
 

Mupad [B] (verification not implemented)

Time = 3.98 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-135+750 x-81 x^2+477 x^3-150 x^4+\left (-810 x^2+2655 x^3-1125 x^4+\left (-162 x^2+531 x^3-225 x^4\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right ) \log \left (5+\log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right )}{-45 x+125 x^2+\left (-9 x+25 x^2\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )} \, dx=3\,\ln \left (\ln \left (\frac {25\,x^2}{9}-x\right )+5\right )\,\left (-x^3+3\,x^2+5\right ) \] Input:

int((log(log((25*x^2)/9 - x) + 5)*(log((25*x^2)/9 - x)*(162*x^2 - 531*x^3 
+ 225*x^4) + 810*x^2 - 2655*x^3 + 1125*x^4) - 750*x + 81*x^2 - 477*x^3 + 1 
50*x^4 + 135)/(45*x + log((25*x^2)/9 - x)*(9*x - 25*x^2) - 125*x^2),x)
 

Output:

3*log(log((25*x^2)/9 - x) + 5)*(3*x^2 - x^3 + 5)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {-135+750 x-81 x^2+477 x^3-150 x^4+\left (-810 x^2+2655 x^3-1125 x^4+\left (-162 x^2+531 x^3-225 x^4\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right ) \log \left (5+\log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )\right )}{-45 x+125 x^2+\left (-9 x+25 x^2\right ) \log \left (\frac {1}{9} \left (-9 x+25 x^2\right )\right )} \, dx=3 \,\mathrm {log}\left (\mathrm {log}\left (\frac {25}{9} x^{2}-x \right )+5\right ) \left (-x^{3}+3 x^{2}+5\right ) \] Input:

int((((-225*x^4+531*x^3-162*x^2)*log(25/9*x^2-x)-1125*x^4+2655*x^3-810*x^2 
)*log(log(25/9*x^2-x)+5)-150*x^4+477*x^3-81*x^2+750*x-135)/((25*x^2-9*x)*l 
og(25/9*x^2-x)+125*x^2-45*x),x)
 

Output:

3*log(log((25*x**2 - 9*x)/9) + 5)*( - x**3 + 3*x**2 + 5)