Integrand size = 82, antiderivative size = 27 \[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=-e^{-4+e^{2 x}} x \log (x \log (5))+\log \left (-e^x+\log (x)\right ) \]
Time = 2.88 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=-e^{-4+e^{2 x}} x \log (x \log (5))+\log \left (e^x-\log (x)\right ) \]
Integrate[(1 - E^x*x + E^(-4 + E^(2*x))*(E^x*x - x*Log[x] + (E^x*x + 2*E^( 3*x)*x^2 + (-x - 2*E^(2*x)*x^2)*Log[x])*Log[x*Log[5]]))/(-(E^x*x) + x*Log[ x]),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.
\(\displaystyle \int \frac {e^{e^{2 x}-4} \left (\left (2 e^{3 x} x^2+\left (-2 e^{2 x} x^2-x\right ) \log (x)+e^x x\right ) \log (x \log (5))+e^x x-x \log (x)\right )-e^x x+1}{x \log (x)-e^x x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x \log (x)-1}{x \left (e^x-\log (x)\right )}-2 e^{2 x+e^{2 x}-4} x \log (x \log (5))+\frac {-e^{e^{2 x}}-e^{e^{2 x}} \log (x \log (5))+e^4}{e^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\int \frac {\operatorname {ExpIntegralEi}\left (e^{2 x}\right )}{x}dx}{2 e^4}+2 \int \frac {\int e^{2 x+e^{2 x}-4} xdx}{x}dx-2 \log (x \log (5)) \int e^{2 x+e^{2 x}-4} xdx-\int \frac {1}{x \left (e^x-\log (x)\right )}dx+\int \frac {\log (x)}{e^x-\log (x)}dx-\frac {\operatorname {ExpIntegralEi}\left (e^{2 x}\right )}{2 e^4}-\frac {\operatorname {ExpIntegralEi}\left (e^{2 x}\right ) \log (x \log (5))}{2 e^4}+x\) |
Int[(1 - E^x*x + E^(-4 + E^(2*x))*(E^x*x - x*Log[x] + (E^x*x + 2*E^(3*x)*x ^2 + (-x - 2*E^(2*x)*x^2)*Log[x])*Log[x*Log[5]]))/(-(E^x*x) + x*Log[x]),x]
3.24.15.3.1 Defintions of rubi rules used
Time = 8.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
parallelrisch | \(\ln \left (\ln \left (x \right )-{\mathrm e}^{x}\right )-{\mathrm e}^{{\mathrm e}^{2 x}-4} x \ln \left (x \ln \left (5\right )\right )\) | \(25\) |
risch | \(\ln \left (\ln \left (x \right )-{\mathrm e}^{x}\right )+\left (-x \ln \left (\ln \left (5\right )\right )-x \ln \left (x \right )\right ) {\mathrm e}^{{\mathrm e}^{2 x}-4}\) | \(30\) |
int(((((-2*exp(2*x)*x^2-x)*ln(x)+2*x^2*exp(x)*exp(2*x)+exp(x)*x)*ln(x*ln(5 ))-x*ln(x)+exp(x)*x)*exp(exp(2*x)-4)-exp(x)*x+1)/(x*ln(x)-exp(x)*x),x,meth od=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=-{\left (x \log \left (x\right ) + x \log \left (\log \left (5\right )\right )\right )} e^{\left (e^{\left (2 \, x\right )} - 4\right )} + \log \left (-e^{x} + \log \left (x\right )\right ) \]
integrate(((((-2*exp(2*x)*x^2-x)*log(x)+2*x^2*exp(x)*exp(2*x)+exp(x)*x)*lo g(x*log(5))-x*log(x)+exp(x)*x)*exp(exp(2*x)-4)-exp(x)*x+1)/(x*log(x)-exp(x )*x),x, algorithm=\
Time = 134.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=\left (- x \log {\left (x \right )} - x \log {\left (\log {\left (5 \right )} \right )}\right ) e^{e^{2 x} - 4} + \log {\left (e^{x} - \log {\left (x \right )} \right )} \]
integrate(((((-2*exp(2*x)*x**2-x)*ln(x)+2*x**2*exp(x)*exp(2*x)+exp(x)*x)*l n(x*ln(5))-x*ln(x)+exp(x)*x)*exp(exp(2*x)-4)-exp(x)*x+1)/(x*ln(x)-exp(x)*x ),x)
Time = 0.31 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=-{\left (x \log \left (x\right ) + x \log \left (\log \left (5\right )\right )\right )} e^{\left (e^{\left (2 \, x\right )} - 4\right )} + \log \left (e^{x} - \log \left (x\right )\right ) \]
integrate(((((-2*exp(2*x)*x^2-x)*log(x)+2*x^2*exp(x)*exp(2*x)+exp(x)*x)*lo g(x*log(5))-x*log(x)+exp(x)*x)*exp(exp(2*x)-4)-exp(x)*x+1)/(x*log(x)-exp(x )*x),x, algorithm=\
\[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=\int { \frac {x e^{x} - {\left (x e^{x} + {\left (2 \, x^{2} e^{\left (3 \, x\right )} + x e^{x} - {\left (2 \, x^{2} e^{\left (2 \, x\right )} + x\right )} \log \left (x\right )\right )} \log \left (x \log \left (5\right )\right ) - x \log \left (x\right )\right )} e^{\left (e^{\left (2 \, x\right )} - 4\right )} - 1}{x e^{x} - x \log \left (x\right )} \,d x } \]
integrate(((((-2*exp(2*x)*x^2-x)*log(x)+2*x^2*exp(x)*exp(2*x)+exp(x)*x)*lo g(x*log(5))-x*log(x)+exp(x)*x)*exp(exp(2*x)-4)-exp(x)*x+1)/(x*log(x)-exp(x )*x),x, algorithm=\
integrate((x*e^x - (x*e^x + (2*x^2*e^(3*x) + x*e^x - (2*x^2*e^(2*x) + x)*l og(x))*log(x*log(5)) - x*log(x))*e^(e^(2*x) - 4) - 1)/(x*e^x - x*log(x)), x)
Time = 9.36 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {1-e^x x+e^{-4+e^{2 x}} \left (e^x x-x \log (x)+\left (e^x x+2 e^{3 x} x^2+\left (-x-2 e^{2 x} x^2\right ) \log (x)\right ) \log (x \log (5))\right )}{-e^x x+x \log (x)} \, dx=\ln \left (\ln \left (x\right )-{\mathrm {e}}^x\right )-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}-4}\,\left (x\,\ln \left (\ln \left (5\right )\right )+x\,\ln \left (x\right )\right ) \]