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
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
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
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
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
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)
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
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
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)
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)