Integrand size = 71, antiderivative size = 20 \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=\log \left (-95 \left (-2+\frac {2 e^x}{x (\log (5)+\log (x))}\right )\right ) \] Output:
ln(-95*exp(x)/x/(1/2*ln(5)+1/2*ln(x))+190)
\[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=\int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx \] Input:
Integrate[(E^x*(1 + (1 - x)*Log[5]) + E^x*(1 - x)*Log[x])/(-(E^x*x*Log[5]) + x^2*Log[5]^2 + (-(E^x*x) + 2*x^2*Log[5])*Log[x] + x^2*Log[x]^2),x]
Output:
Integrate[(E^x*(1 + (1 - x)*Log[5]) + E^x*(1 - x)*Log[x])/(-(E^x*x*Log[5]) + x^2*Log[5]^2 + (-(E^x*x) + 2*x^2*Log[5])*Log[x] + x^2*Log[x]^2), 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.
\(\displaystyle \int \frac {e^x ((1-x) \log (5)+1)+e^x (1-x) \log (x)}{x^2 \log ^2(x)+x^2 \log ^2(5)+\left (2 x^2 \log (5)-e^x x\right ) \log (x)-e^x x \log (5)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^x (x \log (x)+x \log (5)-\log (5 x)-1)}{x \left (e^x+x (-\log (x))-x \log (5)\right ) \log (5 x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {e^x \log (x)}{\left (-e^x+x \log (x)+x \log (5)\right ) \log (5 x)}+\frac {e^x}{x \left (-e^x+x \log (x)+x \log (5)\right )}+\frac {e^x}{x \left (-e^x+x \log (x)+x \log (5)\right ) \log (5 x)}-\frac {e^x \log (5)}{\left (-e^x+x \log (x)+x \log (5)\right ) \log (5 x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {e^x}{x \left (\log (x) x+\log (5) x-e^x\right )}dx-\log (5) \int \frac {e^x}{\left (\log (x) x+\log (5) x-e^x\right ) \log (5 x)}dx+\int \frac {e^x}{x \left (\log (x) x+\log (5) x-e^x\right ) \log (5 x)}dx-\int \frac {e^x \log (x)}{\left (\log (x) x+\log (5) x-e^x\right ) \log (5 x)}dx\) |
Input:
Int[(E^x*(1 + (1 - x)*Log[5]) + E^x*(1 - x)*Log[x])/(-(E^x*x*Log[5]) + x^2 *Log[5]^2 + (-(E^x*x) + 2*x^2*Log[5])*Log[x] + x^2*Log[x]^2),x]
Output:
$Aborted
Time = 0.63 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35
method | result | size |
risch | \(\ln \left (\ln \left (x \right )+\frac {x \ln \left (5\right )-{\mathrm e}^{x}}{x}\right )-\ln \left (\ln \left (5\right )+\ln \left (x \right )\right )\) | \(27\) |
norman | \(-\ln \left (x \right )-\ln \left (\ln \left (5\right )+\ln \left (x \right )\right )+\ln \left (x \ln \left (5\right )+x \ln \left (x \right )-{\mathrm e}^{x}\right )\) | \(28\) |
parallelrisch | \(-\ln \left (x \right )-\ln \left (\ln \left (5\right )+\ln \left (x \right )\right )+\ln \left (x \ln \left (5\right )+x \ln \left (x \right )-{\mathrm e}^{x}\right )\) | \(28\) |
Input:
int(((1-x)*exp(x)*ln(x)+((1-x)*ln(5)+1)*exp(x))/(x^2*ln(x)^2+(-exp(x)*x+2* x^2*ln(5))*ln(x)-x*exp(x)*ln(5)+x^2*ln(5)^2),x,method=_RETURNVERBOSE)
Output:
ln(ln(x)+1/x*(x*ln(5)-exp(x)))-ln(ln(5)+ln(x))
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=\log \left (\frac {x \log \left (5\right ) + x \log \left (x\right ) - e^{x}}{x}\right ) - \log \left (\log \left (5\right ) + \log \left (x\right )\right ) \] Input:
integrate(((1-x)*exp(x)*log(x)+((1-x)*log(5)+1)*exp(x))/(x^2*log(x)^2+(-ex p(x)*x+2*x^2*log(5))*log(x)-x*exp(x)*log(5)+x^2*log(5)^2),x, algorithm="fr icas")
Output:
log((x*log(5) + x*log(x) - e^x)/x) - log(log(5) + log(x))
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=- \log {\left (x \right )} - \log {\left (\log {\left (x \right )} + \log {\left (5 \right )} \right )} + \log {\left (- x \log {\left (x \right )} - x \log {\left (5 \right )} + e^{x} \right )} \] Input:
integrate(((1-x)*exp(x)*ln(x)+((1-x)*ln(5)+1)*exp(x))/(x**2*ln(x)**2+(-exp (x)*x+2*x**2*ln(5))*ln(x)-x*exp(x)*ln(5)+x**2*ln(5)**2),x)
Output:
-log(x) - log(log(x) + log(5)) + log(-x*log(x) - x*log(5) + exp(x))
Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=\log \left (-x \log \left (5\right ) - x \log \left (x\right ) + e^{x}\right ) - \log \left (x\right ) - \log \left (\log \left (5\right ) + \log \left (x\right )\right ) \] Input:
integrate(((1-x)*exp(x)*log(x)+((1-x)*log(5)+1)*exp(x))/(x^2*log(x)^2+(-ex p(x)*x+2*x^2*log(5))*log(x)-x*exp(x)*log(5)+x^2*log(5)^2),x, algorithm="ma xima")
Output:
log(-x*log(5) - x*log(x) + e^x) - log(x) - log(log(5) + log(x))
Time = 0.12 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=\log \left (-x \log \left (5\right ) - x \log \left (x\right ) + e^{x}\right ) - \log \left (x\right ) - \log \left (\log \left (5\right ) + \log \left (x\right )\right ) \] Input:
integrate(((1-x)*exp(x)*log(x)+((1-x)*log(5)+1)*exp(x))/(x^2*log(x)^2+(-ex p(x)*x+2*x^2*log(5))*log(x)-x*exp(x)*log(5)+x^2*log(5)^2),x, algorithm="gi ac")
Output:
log(-x*log(5) - x*log(x) + e^x) - log(x) - log(log(5) + log(x))
Timed out. \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=-\int \frac {{\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x-1\right )-1\right )+{\mathrm {e}}^x\,\ln \left (x\right )\,\left (x-1\right )}{x^2\,{\ln \left (5\right )}^2+x^2\,{\ln \left (x\right )}^2+\ln \left (x\right )\,\left (2\,x^2\,\ln \left (5\right )-x\,{\mathrm {e}}^x\right )-x\,{\mathrm {e}}^x\,\ln \left (5\right )} \,d x \] Input:
int(-(exp(x)*(log(5)*(x - 1) - 1) + exp(x)*log(x)*(x - 1))/(x^2*log(5)^2 + x^2*log(x)^2 + log(x)*(2*x^2*log(5) - x*exp(x)) - x*exp(x)*log(5)),x)
Output:
-int((exp(x)*(log(5)*(x - 1) - 1) + exp(x)*log(x)*(x - 1))/(x^2*log(5)^2 + x^2*log(x)^2 + log(x)*(2*x^2*log(5) - x*exp(x)) - x*exp(x)*log(5)), x)
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {e^x (1+(1-x) \log (5))+e^x (1-x) \log (x)}{-e^x x \log (5)+x^2 \log ^2(5)+\left (-e^x x+2 x^2 \log (5)\right ) \log (x)+x^2 \log ^2(x)} \, dx=\mathrm {log}\left (e^{x}-\mathrm {log}\left (x \right ) x -\mathrm {log}\left (5\right ) x \right )-\mathrm {log}\left (\mathrm {log}\left (x \right )+\mathrm {log}\left (5\right )\right )-\mathrm {log}\left (x \right ) \] Input:
int(((1-x)*exp(x)*log(x)+((1-x)*log(5)+1)*exp(x))/(x^2*log(x)^2+(-exp(x)*x +2*x^2*log(5))*log(x)-x*exp(x)*log(5)+x^2*log(5)^2),x)
Output:
log(e**x - log(x)*x - log(5)*x) - log(log(x) + log(5)) - log(x)