\(\int \frac {e^x (5-9 x+5 x^2)+(30 x^2-30 x^3) \log (5)+(e^x (-1+2 x-x^2)+(-6 x^2+6 x^3) \log (5)) \log (1-x)+(e^x (-5+10 x-5 x^2)+e^x (1-2 x+x^2) \log (1-x)) \log (5-\log (1-x))}{30 x^2-30 x^3+(-6 x^2+6 x^3) \log (1-x)} \, dx\) [84]

Optimal result

Integrand size = 142, antiderivative size = 28 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=x \log (5)+\frac {e^x (-1+\log (5-\log (1-x)))}{6 x} \] Output:

x*ln(5)+1/6*(ln(-ln(1-x)+5)-1)*exp(x)/x
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=\frac {1}{6} \left (-\frac {e^x}{x}+6 x \log (5)+\frac {e^x \log (5-\log (1-x))}{x}\right ) \] Input:

Integrate[(E^x*(5 - 9*x + 5*x^2) + (30*x^2 - 30*x^3)*Log[5] + (E^x*(-1 + 2 
*x - x^2) + (-6*x^2 + 6*x^3)*Log[5])*Log[1 - x] + (E^x*(-5 + 10*x - 5*x^2) 
 + E^x*(1 - 2*x + x^2)*Log[1 - x])*Log[5 - Log[1 - x]])/(30*x^2 - 30*x^3 + 
 (-6*x^2 + 6*x^3)*Log[1 - x]),x]
 

Output:

(-(E^x/x) + 6*x*Log[5] + (E^x*Log[5 - Log[1 - x]])/x)/6
 

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.

\(\displaystyle \int \frac {e^x \left (5 x^2-9 x+5\right )+\left (e^x \left (-5 x^2+10 x-5\right )+e^x \left (x^2-2 x+1\right ) \log (1-x)\right ) \log (5-\log (1-x))+\left (e^x \left (-x^2+2 x-1\right )+\left (6 x^3-6 x^2\right ) \log (5)\right ) \log (1-x)+\left (30 x^2-30 x^3\right ) \log (5)}{-30 x^3+30 x^2+\left (6 x^3-6 x^2\right ) \log (1-x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^x \left (5 x^2-9 x+5\right )+\left (e^x \left (-5 x^2+10 x-5\right )+e^x \left (x^2-2 x+1\right ) \log (1-x)\right ) \log (5-\log (1-x))+\left (e^x \left (-x^2+2 x-1\right )+\left (6 x^3-6 x^2\right ) \log (5)\right ) \log (1-x)+\left (30 x^2-30 x^3\right ) \log (5)}{6 (1-x) x^2 (5-\log (1-x))}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \int \frac {e^x \left (5 x^2-9 x+5\right )-\left (e^x \left (x^2-2 x+1\right )+6 \left (x^2-x^3\right ) \log (5)\right ) \log (1-x)-\left (5 e^x \left (x^2-2 x+1\right )-e^x \left (x^2-2 x+1\right ) \log (1-x)\right ) \log (5-\log (1-x))+30 \left (x^2-x^3\right ) \log (5)}{(1-x) x^2 (5-\log (1-x))}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{6} \int \left (\frac {e^x \left (-\log (1-x) x^2+\log (1-x) \log (5-\log (1-x)) x^2-5 \log (5-\log (1-x)) x^2+5 x^2+2 \log (1-x) x-2 \log (1-x) \log (5-\log (1-x)) x+10 \log (5-\log (1-x)) x-9 x-\log (1-x)+\log (1-x) \log (5-\log (1-x))-5 \log (5-\log (1-x))+5\right )}{(x-1) x^2 (\log (1-x)-5)}+6 \log (5)\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \frac {1}{6} \int \left (\frac {e^x \left (-\log (1-x) x^2+\log (1-x) \log (5-\log (1-x)) x^2-5 \log (5-\log (1-x)) x^2+5 x^2+2 \log (1-x) x-2 \log (1-x) \log (5-\log (1-x)) x+10 \log (5-\log (1-x)) x-9 x-\log (1-x)+\log (1-x) \log (5-\log (1-x))-5 \log (5-\log (1-x))+5\right )}{(x-1) x^2 (\log (1-x)-5)}+6 \log (5)\right )dx\)

Input:

Int[(E^x*(5 - 9*x + 5*x^2) + (30*x^2 - 30*x^3)*Log[5] + (E^x*(-1 + 2*x - x 
^2) + (-6*x^2 + 6*x^3)*Log[5])*Log[1 - x] + (E^x*(-5 + 10*x - 5*x^2) + E^x 
*(1 - 2*x + x^2)*Log[1 - x])*Log[5 - Log[1 - x]])/(30*x^2 - 30*x^3 + (-6*x 
^2 + 6*x^3)*Log[1 - x]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 4.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32

method	result	size
risch	\(\frac {{\mathrm e}^{x} \ln \left (-\ln \left (1-x \right )+5\right )}{6 x}+\frac {6 x^{2} \ln \left (5\right )-{\mathrm e}^{x}}{6 x}\)	\(37\)
parallelrisch	\(\frac {6 x^{2} \ln \left (5\right )+12 x \ln \left (5\right )+{\mathrm e}^{x} \ln \left (-\ln \left (1-x \right )+5\right )-{\mathrm e}^{x}}{6 x}\)	\(37\)

Input:

int((((x^2-2*x+1)*exp(x)*ln(1-x)+(-5*x^2+10*x-5)*exp(x))*ln(-ln(1-x)+5)+(( 
-x^2+2*x-1)*exp(x)+(6*x^3-6*x^2)*ln(5))*ln(1-x)+(5*x^2-9*x+5)*exp(x)+(-30* 
x^3+30*x^2)*ln(5))/((6*x^3-6*x^2)*ln(1-x)-30*x^3+30*x^2),x,method=_RETURNV 
ERBOSE)
 

Output:

1/6/x*exp(x)*ln(-ln(1-x)+5)+1/6*(6*x^2*ln(5)-exp(x))/x
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=\frac {6 \, x^{2} \log \left (5\right ) + e^{x} \log \left (-\log \left (-x + 1\right ) + 5\right ) - e^{x}}{6 \, x} \] Input:

integrate((((x^2-2*x+1)*exp(x)*log(1-x)+(-5*x^2+10*x-5)*exp(x))*log(-log(1 
-x)+5)+((-x^2+2*x-1)*exp(x)+(6*x^3-6*x^2)*log(5))*log(1-x)+(5*x^2-9*x+5)*e 
xp(x)+(-30*x^3+30*x^2)*log(5))/((6*x^3-6*x^2)*log(1-x)-30*x^3+30*x^2),x, a 
lgorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=x \log {\left (5 \right )} + \frac {\left (\log {\left (5 - \log {\left (1 - x \right )} \right )} - 1\right ) e^{x}}{6 x} \] Input:

integrate((((x**2-2*x+1)*exp(x)*ln(1-x)+(-5*x**2+10*x-5)*exp(x))*ln(-ln(1- 
x)+5)+((-x**2+2*x-1)*exp(x)+(6*x**3-6*x**2)*ln(5))*ln(1-x)+(5*x**2-9*x+5)* 
exp(x)+(-30*x**3+30*x**2)*ln(5))/((6*x**3-6*x**2)*ln(1-x)-30*x**3+30*x**2) 
,x)
 

Output:

x*log(5) + (log(5 - log(1 - x)) - 1)*exp(x)/(6*x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (25) = 50\).

Time = 0.14 (sec) , antiderivative size = 106, normalized size of antiderivative = 3.79 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=-\log \left (5\right ) \log \left (-x + 1\right ) \log \left (\log \left (-x + 1\right ) - 5\right ) + {\left ({\left (\log \left (-x + 1\right ) - 5\right )} \log \left (\log \left (-x + 1\right ) - 5\right ) - \log \left (-x + 1\right ) + 5\right )} \log \left (5\right ) + 5 \, \log \left (5\right ) \log \left (\log \left (-x + 1\right ) - 5\right ) + \frac {6 \, x^{2} \log \left (5\right ) + 6 \, x \log \left (5\right ) \log \left (-x + 1\right ) + e^{x} \log \left (-\log \left (-x + 1\right ) + 5\right ) - e^{x}}{6 \, x} \] Input:

integrate((((x^2-2*x+1)*exp(x)*log(1-x)+(-5*x^2+10*x-5)*exp(x))*log(-log(1 
-x)+5)+((-x^2+2*x-1)*exp(x)+(6*x^3-6*x^2)*log(5))*log(1-x)+(5*x^2-9*x+5)*e 
xp(x)+(-30*x^3+30*x^2)*log(5))/((6*x^3-6*x^2)*log(1-x)-30*x^3+30*x^2),x, a 
lgorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=\frac {6 \, x^{2} \log \left (5\right ) + e^{x} \log \left (-\log \left (-x + 1\right ) + 5\right ) - e^{x}}{6 \, x} \] Input:

integrate((((x^2-2*x+1)*exp(x)*log(1-x)+(-5*x^2+10*x-5)*exp(x))*log(-log(1 
-x)+5)+((-x^2+2*x-1)*exp(x)+(6*x^3-6*x^2)*log(5))*log(1-x)+(5*x^2-9*x+5)*e 
xp(x)+(-30*x^3+30*x^2)*log(5))/((6*x^3-6*x^2)*log(1-x)-30*x^3+30*x^2),x, a 
lgorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=\int \frac {\ln \left (5-\ln \left (1-x\right )\right )\,\left ({\mathrm {e}}^x\,\left (5\,x^2-10\,x+5\right )-{\mathrm {e}}^x\,\ln \left (1-x\right )\,\left (x^2-2\,x+1\right )\right )+\ln \left (1-x\right )\,\left ({\mathrm {e}}^x\,\left (x^2-2\,x+1\right )+\ln \left (5\right )\,\left (6\,x^2-6\,x^3\right )\right )-\ln \left (5\right )\,\left (30\,x^2-30\,x^3\right )-{\mathrm {e}}^x\,\left (5\,x^2-9\,x+5\right )}{\ln \left (1-x\right )\,\left (6\,x^2-6\,x^3\right )-30\,x^2+30\,x^3} \,d x \] Input:

int((log(5 - log(1 - x))*(exp(x)*(5*x^2 - 10*x + 5) - exp(x)*log(1 - x)*(x 
^2 - 2*x + 1)) + log(1 - x)*(exp(x)*(x^2 - 2*x + 1) + log(5)*(6*x^2 - 6*x^ 
3)) - log(5)*(30*x^2 - 30*x^3) - exp(x)*(5*x^2 - 9*x + 5))/(log(1 - x)*(6* 
x^2 - 6*x^3) - 30*x^2 + 30*x^3),x)
 

Output:

int((log(5 - log(1 - x))*(exp(x)*(5*x^2 - 10*x + 5) - exp(x)*log(1 - x)*(x 
^2 - 2*x + 1)) + log(1 - x)*(exp(x)*(x^2 - 2*x + 1) + log(5)*(6*x^2 - 6*x^ 
3)) - log(5)*(30*x^2 - 30*x^3) - exp(x)*(5*x^2 - 9*x + 5))/(log(1 - x)*(6* 
x^2 - 6*x^3) - 30*x^2 + 30*x^3), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^x \left (5-9 x+5 x^2\right )+\left (30 x^2-30 x^3\right ) \log (5)+\left (e^x \left (-1+2 x-x^2\right )+\left (-6 x^2+6 x^3\right ) \log (5)\right ) \log (1-x)+\left (e^x \left (-5+10 x-5 x^2\right )+e^x \left (1-2 x+x^2\right ) \log (1-x)\right ) \log (5-\log (1-x))}{30 x^2-30 x^3+\left (-6 x^2+6 x^3\right ) \log (1-x)} \, dx=\frac {e^{x} \mathrm {log}\left (-\mathrm {log}\left (1-x \right )+5\right )-e^{x}+6 \,\mathrm {log}\left (5\right ) x^{2}}{6 x} \] Input:

int((((x^2-2*x+1)*exp(x)*log(1-x)+(-5*x^2+10*x-5)*exp(x))*log(-log(1-x)+5) 
+((-x^2+2*x-1)*exp(x)+(6*x^3-6*x^2)*log(5))*log(1-x)+(5*x^2-9*x+5)*exp(x)+ 
(-30*x^3+30*x^2)*log(5))/((6*x^3-6*x^2)*log(1-x)-30*x^3+30*x^2),x)
 

Output:

(e**x*log( - log( - x + 1) + 5) - e**x + 6*log(5)*x**2)/(6*x)