\(\int \frac {25-25 x^2-25 \log (-3 x)+\log (\frac {1}{4} (4+e^5)) (-3-x+4 x^2+(3+2 x) \log (-3 x))}{\log (\frac {1}{4} (4+e^5)) (x^2-2 x^3+x^4+(2 x-2 x^2) \log (-3 x)+\log ^2(-3 x))} \, dx\) [2186]

Optimal result

Integrand size = 91, antiderivative size = 33 \[ \int \frac {25-25 x^2-25 \log (-3 x)+\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-3-x+4 x^2+(3+2 x) \log (-3 x)\right )}{\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (x^2-2 x^3+x^4+\left (2 x-2 x^2\right ) \log (-3 x)+\log ^2(-3 x)\right )} \, dx=\frac {3+x-\frac {25}{\log \left (1+\frac {e^5}{4}\right )}}{1-x+\frac {\log (-3 x)}{x}} \] Output:

(x+3-25/ln(1+1/4*exp(5)))/(ln(-3*x)/x-x+1)
 

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.61 \[ \int \frac {25-25 x^2-25 \log (-3 x)+\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-3-x+4 x^2+(3+2 x) \log (-3 x)\right )}{\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (x^2-2 x^3+x^4+\left (2 x-2 x^2\right ) \log (-3 x)+\log ^2(-3 x)\right )} \, dx=\frac {x \left (-25+3 \log \left (\frac {1}{4} \left (4+e^5\right )\right )+x \log \left (\frac {1}{4} \left (4+e^5\right )\right )\right )}{\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (x-x^2+\log (-3 x)\right )} \] Input:

Integrate[(25 - 25*x^2 - 25*Log[-3*x] + Log[(4 + E^5)/4]*(-3 - x + 4*x^2 + 
 (3 + 2*x)*Log[-3*x]))/(Log[(4 + E^5)/4]*(x^2 - 2*x^3 + x^4 + (2*x - 2*x^2 
)*Log[-3*x] + Log[-3*x]^2)),x]
 

Output:

(x*(-25 + 3*Log[(4 + E^5)/4] + x*Log[(4 + E^5)/4]))/(Log[(4 + E^5)/4]*(x - 
 x^2 + Log[-3*x]))
 

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.

\(\displaystyle \int \frac {-25 x^2+\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (4 x^2-x+(2 x+3) \log (-3 x)-3\right )-25 \log (-3 x)+25}{\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (x^4-2 x^3+x^2+\left (2 x-2 x^2\right ) \log (-3 x)+\log ^2(-3 x)\right )} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {-25 x^2-25 \log (-3 x)-\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-4 x^2+x-(2 x+3) \log (-3 x)+3\right )+25}{x^4-2 x^3+x^2+\log ^2(-3 x)+2 \left (x-x^2\right ) \log (-3 x)}dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}\)

\(\Big \downarrow \) 7239

\(\displaystyle \frac {\int \frac {-25 x^2-25 \log (-3 x)-\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-4 x^2+x-(2 x+3) \log (-3 x)+3\right )+25}{\left (-x^2+x+\log (-3 x)\right )^2}dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {\int \left (\frac {-2 \log \left (\frac {1}{4} \left (4+e^5\right )\right ) x-3 \log \left (\frac {1}{4} \left (4+e^5\right )\right )+25}{x^2-x-\log (-3 x)}+\frac {\left (2 x^2-x-1\right ) \left (\log \left (\frac {1}{4} \left (4+e^5\right )\right ) x+3 \log \left (\frac {1}{4} \left (4+e^5\right )\right )-25\right )}{\left (x^2-x-\log (-3 x)\right )^2}\right )dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (25+\log (64)-3 \log \left (4+e^5\right )\right ) \int \frac {1}{\left (x^2-x-\log (-3 x)\right )^2}dx+\left (25-4 \log \left (\frac {1}{4} \left (4+e^5\right )\right )\right ) \int \frac {x}{\left (x^2-x-\log (-3 x)\right )^2}dx-5 \left (10-\log \left (\frac {1}{4} \left (4+e^5\right )\right )\right ) \int \frac {x^2}{\left (x^2-x-\log (-3 x)\right )^2}dx+\left (25+\log (64)-3 \log \left (4+e^5\right )\right ) \int \frac {1}{x^2-x-\log (-3 x)}dx-2 \log \left (\frac {1}{4} \left (4+e^5\right )\right ) \int \frac {x}{x^2-x-\log (-3 x)}dx+2 \log \left (\frac {1}{4} \left (4+e^5\right )\right ) \int \frac {x^3}{\left (x^2-x-\log (-3 x)\right )^2}dx}{\log \left (\frac {1}{4} \left (4+e^5\right )\right )}\)

Input:

Int[(25 - 25*x^2 - 25*Log[-3*x] + Log[(4 + E^5)/4]*(-3 - x + 4*x^2 + (3 + 
2*x)*Log[-3*x]))/(Log[(4 + E^5)/4]*(x^2 - 2*x^3 + x^4 + (2*x - 2*x^2)*Log[ 
-3*x] + Log[-3*x]^2)),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.58

method	result	size
parallelrisch	\(\frac {-\ln \left (1+\frac {{\mathrm e}^{5}}{4}\right ) x^{2}-3 \ln \left (1+\frac {{\mathrm e}^{5}}{4}\right ) x +25 x}{\ln \left (1+\frac {{\mathrm e}^{5}}{4}\right ) \left (x^{2}-x -\ln \left (-3 x \right )\right )}\)	\(52\)
norman	\(\frac {-\ln \left (-3 x \right )-\frac {\left (8 \ln \left (2\right )-4 \ln \left (4+{\mathrm e}^{5}\right )+25\right ) x}{2 \ln \left (2\right )-\ln \left (4+{\mathrm e}^{5}\right )}}{x^{2}-x -\ln \left (-3 x \right )}\)	\(54\)
risch	\(\frac {\left (2 x \ln \left (2\right )-\ln \left (4+{\mathrm e}^{5}\right ) x +6 \ln \left (2\right )-3 \ln \left (4+{\mathrm e}^{5}\right )+25\right ) x}{\left (-2 \ln \left (2\right )+\ln \left (4+{\mathrm e}^{5}\right )\right ) \left (x^{2}-x -\ln \left (-3 x \right )\right )}\)	\(56\)
default	\(\frac {-\frac {225 x}{-9 x^{2}+9 \ln \left (-3 x \right )+9 x}-\frac {6 \ln \left (2\right ) \left (3 x^{2}+9 x \right )}{-9 x^{2}+9 \ln \left (-3 x \right )+9 x}+\frac {3 \ln \left (4+{\mathrm e}^{5}\right ) \left (3 x^{2}+9 x \right )}{-9 x^{2}+9 \ln \left (-3 x \right )+9 x}}{\ln \left (1+\frac {{\mathrm e}^{5}}{4}\right )}\)	\(95\)
derivativedivides	\(-\frac {3 \left (\frac {75 x}{-9 x^{2}+9 \ln \left (-3 x \right )+9 x}+\frac {2 \ln \left (2\right ) \left (3 x^{2}+9 x \right )}{-9 x^{2}+9 \ln \left (-3 x \right )+9 x}-\frac {\ln \left (4+{\mathrm e}^{5}\right ) \left (3 x^{2}+9 x \right )}{-9 x^{2}+9 \ln \left (-3 x \right )+9 x}\right )}{\ln \left (1+\frac {{\mathrm e}^{5}}{4}\right )}\)	\(96\)

Input:

int((((3+2*x)*ln(-3*x)+4*x^2-x-3)*ln(1+1/4*exp(5))-25*ln(-3*x)-25*x^2+25)/ 
(ln(-3*x)^2+(-2*x^2+2*x)*ln(-3*x)+x^4-2*x^3+x^2)/ln(1+1/4*exp(5)),x,method 
=_RETURNVERBOSE)
 

Output:

1/ln(1+1/4*exp(5))*(-ln(1+1/4*exp(5))*x^2-3*ln(1+1/4*exp(5))*x+25*x)/(x^2- 
x-ln(-3*x))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36 \[ \int \frac {25-25 x^2-25 \log (-3 x)+\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-3-x+4 x^2+(3+2 x) \log (-3 x)\right )}{\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (x^2-2 x^3+x^4+\left (2 x-2 x^2\right ) \log (-3 x)+\log ^2(-3 x)\right )} \, dx=-\frac {{\left (x^{2} + 3 \, x\right )} \log \left (\frac {1}{4} \, e^{5} + 1\right ) - 25 \, x}{{\left (x^{2} - x - \log \left (-3 \, x\right )\right )} \log \left (\frac {1}{4} \, e^{5} + 1\right )} \] Input:

integrate((((3+2*x)*log(-3*x)+4*x^2-x-3)*log(1+1/4*exp(5))-25*log(-3*x)-25 
*x^2+25)/(log(-3*x)^2+(-2*x^2+2*x)*log(-3*x)+x^4-2*x^3+x^2)/log(1+1/4*exp( 
5)),x, algorithm="fricas")
 

Output:

-((x^2 + 3*x)*log(1/4*e^5 + 1) - 25*x)/((x^2 - x - log(-3*x))*log(1/4*e^5 
+ 1))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (24) = 48\).

Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.91 \[ \int \frac {25-25 x^2-25 \log (-3 x)+\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-3-x+4 x^2+(3+2 x) \log (-3 x)\right )}{\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (x^2-2 x^3+x^4+\left (2 x-2 x^2\right ) \log (-3 x)+\log ^2(-3 x)\right )} \, dx=\frac {x^{2} \log {\left (1 + \frac {e^{5}}{4} \right )} - 25 x + 3 x \log {\left (1 + \frac {e^{5}}{4} \right )}}{- x^{2} \log {\left (1 + \frac {e^{5}}{4} \right )} + x \log {\left (1 + \frac {e^{5}}{4} \right )} + \log {\left (- 3 x \right )} \log {\left (1 + \frac {e^{5}}{4} \right )}} \] Input:

integrate((((3+2*x)*ln(-3*x)+4*x**2-x-3)*ln(1+1/4*exp(5))-25*ln(-3*x)-25*x 
**2+25)/(ln(-3*x)**2+(-2*x**2+2*x)*ln(-3*x)+x**4-2*x**3+x**2)/ln(1+1/4*exp 
(5)),x)
 

Output:

(x**2*log(1 + exp(5)/4) - 25*x + 3*x*log(1 + exp(5)/4))/(-x**2*log(1 + exp 
(5)/4) + x*log(1 + exp(5)/4) + log(-3*x)*log(1 + exp(5)/4))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (30) = 60\).

Time = 0.14 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.85 \[ \int \frac {25-25 x^2-25 \log (-3 x)+\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-3-x+4 x^2+(3+2 x) \log (-3 x)\right )}{\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (x^2-2 x^3+x^4+\left (2 x-2 x^2\right ) \log (-3 x)+\log ^2(-3 x)\right )} \, dx=\frac {x^{2} {\left (2 \, \log \left (2\right ) - \log \left (e^{5} + 4\right )\right )} + x {\left (6 \, \log \left (2\right ) - 3 \, \log \left (e^{5} + 4\right ) + 25\right )}}{{\left (x^{2} - x - \log \left (3\right ) - \log \left (-x\right )\right )} \log \left (\frac {1}{4} \, e^{5} + 1\right )} \] Input:

integrate((((3+2*x)*log(-3*x)+4*x^2-x-3)*log(1+1/4*exp(5))-25*log(-3*x)-25 
*x^2+25)/(log(-3*x)^2+(-2*x^2+2*x)*log(-3*x)+x^4-2*x^3+x^2)/log(1+1/4*exp( 
5)),x, algorithm="maxima")
 

Output:

(x^2*(2*log(2) - log(e^5 + 4)) + x*(6*log(2) - 3*log(e^5 + 4) + 25))/((x^2 
 - x - log(3) - log(-x))*log(1/4*e^5 + 1))
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.79 \[ \int \frac {25-25 x^2-25 \log (-3 x)+\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-3-x+4 x^2+(3+2 x) \log (-3 x)\right )}{\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (x^2-2 x^3+x^4+\left (2 x-2 x^2\right ) \log (-3 x)+\log ^2(-3 x)\right )} \, dx=\frac {2 \, x^{2} \log \left (2\right ) - x^{2} \log \left (e^{5} + 4\right ) + 6 \, x \log \left (2\right ) - 3 \, x \log \left (e^{5} + 4\right ) + 25 \, x}{{\left (x^{2} - x - \log \left (-3 \, x\right )\right )} \log \left (\frac {1}{4} \, e^{5} + 1\right )} \] Input:

integrate((((3+2*x)*log(-3*x)+4*x^2-x-3)*log(1+1/4*exp(5))-25*log(-3*x)-25 
*x^2+25)/(log(-3*x)^2+(-2*x^2+2*x)*log(-3*x)+x^4-2*x^3+x^2)/log(1+1/4*exp( 
5)),x, algorithm="giac")
 

Output:

(2*x^2*log(2) - x^2*log(e^5 + 4) + 6*x*log(2) - 3*x*log(e^5 + 4) + 25*x)/( 
(x^2 - x - log(-3*x))*log(1/4*e^5 + 1))
 

Mupad [B] (verification not implemented)

Time = 5.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.33 \[ \int \frac {25-25 x^2-25 \log (-3 x)+\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-3-x+4 x^2+(3+2 x) \log (-3 x)\right )}{\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (x^2-2 x^3+x^4+\left (2 x-2 x^2\right ) \log (-3 x)+\log ^2(-3 x)\right )} \, dx=\frac {x\,\left (3\,\ln \left (\frac {{\mathrm {e}}^5}{4}+1\right )+x\,\ln \left (\frac {{\mathrm {e}}^5}{4}+1\right )-25\right )}{\ln \left (\frac {{\mathrm {e}}^5}{4}+1\right )\,\left (x+\ln \left (-3\,x\right )-x^2\right )} \] Input:

int(-(25*log(-3*x) + 25*x^2 + log(exp(5)/4 + 1)*(x - 4*x^2 - log(-3*x)*(2* 
x + 3) + 3) - 25)/(log(exp(5)/4 + 1)*(log(-3*x)*(2*x - 2*x^2) + log(-3*x)^ 
2 + x^2 - 2*x^3 + x^4)),x)
 

Output:

(x*(3*log(exp(5)/4 + 1) + x*log(exp(5)/4 + 1) - 25))/(log(exp(5)/4 + 1)*(x 
 + log(-3*x) - x^2))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.58 \[ \int \frac {25-25 x^2-25 \log (-3 x)+\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (-3-x+4 x^2+(3+2 x) \log (-3 x)\right )}{\log \left (\frac {1}{4} \left (4+e^5\right )\right ) \left (x^2-2 x^3+x^4+\left (2 x-2 x^2\right ) \log (-3 x)+\log ^2(-3 x)\right )} \, dx=\frac {\mathrm {log}\left (-3 x \right ) \mathrm {log}\left (\frac {e^{5}}{4}+1\right )+4 \,\mathrm {log}\left (\frac {e^{5}}{4}+1\right ) x -25 x}{\mathrm {log}\left (\frac {e^{5}}{4}+1\right ) \left (\mathrm {log}\left (-3 x \right )-x^{2}+x \right )} \] Input:

int((((3+2*x)*log(-3*x)+4*x^2-x-3)*log(1+1/4*exp(5))-25*log(-3*x)-25*x^2+2 
5)/(log(-3*x)^2+(-2*x^2+2*x)*log(-3*x)+x^4-2*x^3+x^2)/log(1+1/4*exp(5)),x)
 

Output:

(log( - 3*x)*log((e**5 + 4)/4) + 4*log((e**5 + 4)/4)*x - 25*x)/(log((e**5 
+ 4)/4)*(log( - 3*x) - x**2 + x))