Integrand size = 45, antiderivative size = 22 \[ \int \frac {e^{-x} \left (2+\left (-1+e^x\right ) \log (x)-x \log ^2(x)-x \log (x) \log \left (\frac {100 \log ^2(x)}{x^2}\right )\right )}{x \log (x)} \, dx=\log (x)+e^{-x} \left (\log (x)+\log \left (\frac {100 \log ^2(x)}{x^2}\right )\right ) \] Output:
ln(x)+(ln(100*ln(x)^2/x^2)+ln(x))/exp(x)
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-x} \left (2+\left (-1+e^x\right ) \log (x)-x \log ^2(x)-x \log (x) \log \left (\frac {100 \log ^2(x)}{x^2}\right )\right )}{x \log (x)} \, dx=\log (x)+e^{-x} \log (x)+e^{-x} \log \left (\frac {100 \log ^2(x)}{x^2}\right ) \] Input:
Integrate[(2 + (-1 + E^x)*Log[x] - x*Log[x]^2 - x*Log[x]*Log[(100*Log[x]^2 )/x^2])/(E^x*x*Log[x]),x]
Output:
Log[x] + Log[x]/E^x + Log[(100*Log[x]^2)/x^2]/E^x
Time = 0.66 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.68, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {e^{-x} \left (-x \log \left (\frac {100 \log ^2(x)}{x^2}\right ) \log (x)-x \log ^2(x)+\left (e^x-1\right ) \log (x)+2\right )}{x \log (x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{-x} \left (-x \log \left (\frac {100 \log ^2(x)}{x^2}\right ) \log (x)-x \log ^2(x)-\log (x)+2\right )}{x \log (x)}+\frac {1}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{-x} \left (x \log \left (\frac {100 \log ^2(x)}{x^2}\right ) \log (x)+x \log ^2(x)\right )}{x \log (x)}+\log (x)\) |
Input:
Int[(2 + (-1 + E^x)*Log[x] - x*Log[x]^2 - x*Log[x]*Log[(100*Log[x]^2)/x^2] )/(E^x*x*Log[x]),x]
Output:
Log[x] + (x*Log[x]^2 + x*Log[x]*Log[(100*Log[x]^2)/x^2])/(E^x*x*Log[x])
Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36
method | result | size |
parallelrisch | \(-\left (-{\mathrm e}^{x} \ln \left (x \right )-\ln \left (x \right )-\ln \left (\frac {100 \ln \left (x \right )^{2}}{x^{2}}\right )\right ) {\mathrm e}^{-x}\) | \(30\) |
risch | \(2 \,{\mathrm e}^{-x} \ln \left (\ln \left (x \right )\right )+\frac {\left (i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-i \pi \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{3}+i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-i \pi \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{2}}{x^{2}}\right )^{3}-i \pi \operatorname {csgn}\left (i \ln \left (x \right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )^{2}+i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{2}}{x^{2}}\right )^{2}+i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{2}}{x^{2}}\right )^{2}-i \pi \,\operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )^{2}}{x^{2}}\right )-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+2 \,{\mathrm e}^{x} \ln \left (x \right )+4 \ln \left (2\right )+4 \ln \left (5\right )-2 \ln \left (x \right )\right ) {\mathrm e}^{-x}}{2}\) | \(235\) |
Input:
int((-x*ln(x)*ln(100*ln(x)^2/x^2)-x*ln(x)^2+(exp(x)-1)*ln(x)+2)/x/exp(x)/l n(x),x,method=_RETURNVERBOSE)
Output:
-(-exp(x)*ln(x)-ln(x)-ln(100*ln(x)^2/x^2))/exp(x)
Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-x} \left (2+\left (-1+e^x\right ) \log (x)-x \log ^2(x)-x \log (x) \log \left (\frac {100 \log ^2(x)}{x^2}\right )\right )}{x \log (x)} \, dx={\left ({\left (e^{x} + 1\right )} \log \left (x\right ) + \log \left (\frac {100 \, \log \left (x\right )^{2}}{x^{2}}\right )\right )} e^{\left (-x\right )} \] Input:
integrate((-x*log(x)*log(100*log(x)^2/x^2)-x*log(x)^2+(-1+exp(x))*log(x)+2 )/x/exp(x)/log(x),x, algorithm="fricas")
Output:
((e^x + 1)*log(x) + log(100*log(x)^2/x^2))*e^(-x)
Time = 4.89 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-x} \left (2+\left (-1+e^x\right ) \log (x)-x \log ^2(x)-x \log (x) \log \left (\frac {100 \log ^2(x)}{x^2}\right )\right )}{x \log (x)} \, dx=\left (\log {\left (x \right )} + \log {\left (\frac {100 \log {\left (x \right )}^{2}}{x^{2}} \right )}\right ) e^{- x} + \log {\left (x \right )} \] Input:
integrate((-x*ln(x)*ln(100*ln(x)**2/x**2)-x*ln(x)**2+(-1+exp(x))*ln(x)+2)/ x/exp(x)/ln(x),x)
Output:
(log(x) + log(100*log(x)**2/x**2))*exp(-x) + log(x)
\[ \int \frac {e^{-x} \left (2+\left (-1+e^x\right ) \log (x)-x \log ^2(x)-x \log (x) \log \left (\frac {100 \log ^2(x)}{x^2}\right )\right )}{x \log (x)} \, dx=\int { -\frac {{\left (x \log \left (x\right )^{2} + x \log \left (x\right ) \log \left (\frac {100 \, \log \left (x\right )^{2}}{x^{2}}\right ) - {\left (e^{x} - 1\right )} \log \left (x\right ) - 2\right )} e^{\left (-x\right )}}{x \log \left (x\right )} \,d x } \] Input:
integrate((-x*log(x)*log(100*log(x)^2/x^2)-x*log(x)^2+(-1+exp(x))*log(x)+2 )/x/exp(x)/log(x),x, algorithm="maxima")
Output:
-(log(x) - 2*log(log(x)))*e^(-x) - Ei(-x) - integrate((2*x*(log(5) + log(2 )) - 1)*e^(-x)/x, x) + log(x)
Time = 0.12 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-x} \left (2+\left (-1+e^x\right ) \log (x)-x \log ^2(x)-x \log (x) \log \left (\frac {100 \log ^2(x)}{x^2}\right )\right )}{x \log (x)} \, dx=e^{\left (-x\right )} \log \left (100 \, \log \left (x\right )^{2}\right ) - e^{\left (-x\right )} \log \left (x\right ) + \log \left (x\right ) \] Input:
integrate((-x*log(x)*log(100*log(x)^2/x^2)-x*log(x)^2+(-1+exp(x))*log(x)+2 )/x/exp(x)/log(x),x, algorithm="giac")
Output:
e^(-x)*log(100*log(x)^2) - e^(-x)*log(x) + log(x)
Timed out. \[ \int \frac {e^{-x} \left (2+\left (-1+e^x\right ) \log (x)-x \log ^2(x)-x \log (x) \log \left (\frac {100 \log ^2(x)}{x^2}\right )\right )}{x \log (x)} \, dx=\int -\frac {{\mathrm {e}}^{-x}\,\left (x\,{\ln \left (x\right )}^2-\ln \left (x\right )\,\left ({\mathrm {e}}^x-1\right )+x\,\ln \left (x\right )\,\ln \left (\frac {100\,{\ln \left (x\right )}^2}{x^2}\right )-2\right )}{x\,\ln \left (x\right )} \,d x \] Input:
int(-(exp(-x)*(x*log(x)^2 - log(x)*(exp(x) - 1) + x*log(x)*log((100*log(x) ^2)/x^2) - 2))/(x*log(x)),x)
Output:
int(-(exp(-x)*(x*log(x)^2 - log(x)*(exp(x) - 1) + x*log(x)*log((100*log(x) ^2)/x^2) - 2))/(x*log(x)), x)
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-x} \left (2+\left (-1+e^x\right ) \log (x)-x \log ^2(x)-x \log (x) \log \left (\frac {100 \log ^2(x)}{x^2}\right )\right )}{x \log (x)} \, dx=\frac {e^{x} \mathrm {log}\left (x \right )+\mathrm {log}\left (\frac {100 \mathrm {log}\left (x \right )^{2}}{x^{2}}\right )+\mathrm {log}\left (x \right )}{e^{x}} \] Input:
int((-x*log(x)*log(100*log(x)^2/x^2)-x*log(x)^2+(-1+exp(x))*log(x)+2)/x/ex p(x)/log(x),x)
Output:
(e**x*log(x) + log((100*log(x)**2)/x**2) + log(x))/e**x