Integrand size = 59, antiderivative size = 19 \[ \int \frac {e^{-x} \left (5 e^x \log (x)+e^{\frac {1}{5} e^{-x} \left (-20 \log (2)+e^x \log (\log (x))\right )} \left (e^x+20 x \log (2) \log (x)\right )\right )}{5 x \log (x)} \, dx=2^{-4 e^{-x}} \sqrt [5]{\log (x)}+\log (x) \]
Time = 0.45 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {e^{-x} \left (5 e^x \log (x)+e^{\frac {1}{5} e^{-x} \left (-20 \log (2)+e^x \log (\log (x))\right )} \left (e^x+20 x \log (2) \log (x)\right )\right )}{5 x \log (x)} \, dx=\frac {1}{5} \left (\frac {2^{1-4 e^{-x}} \log (32) \sqrt [5]{\log (x)}}{\log (4)}+5 \log (x)\right ) \]
Integrate[(5*E^x*Log[x] + E^((-20*Log[2] + E^x*Log[Log[x]])/(5*E^x))*(E^x + 20*x*Log[2]*Log[x]))/(5*E^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.
\(\displaystyle \int \frac {e^{-x} \left (5 e^x \log (x)+e^{\frac {1}{5} e^{-x} \left (e^x \log (\log (x))-20 \log (2)\right )} \left (e^x+20 x \log (2) \log (x)\right )\right )}{5 x \log (x)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {e^{-x} \left (5 e^x \log (x)+2^{-4 e^{-x}} \left (20 x \log (2) \log (x)+e^x\right ) \sqrt [5]{\log (x)}\right )}{x \log (x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (\frac {2^{-4 e^{-x}} \left (5\ 2^{4 e^{-x}} \log ^{\frac {4}{5}}(x)+1\right )}{x \log ^{\frac {4}{5}}(x)}+4^{e^{-x} \left (-2+e^x\right )} e^{-x} \log (32) \sqrt [5]{\log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\int \frac {2^{-4 e^{-x}}}{x \log ^{\frac {4}{5}}(x)}dx+\log (32) \int 4^{e^{-x} \left (-2+e^x\right )} e^{-x} \sqrt [5]{\log (x)}dx+5 \log (x)\right )\) |
Int[(5*E^x*Log[x] + E^((-20*Log[2] + E^x*Log[Log[x]])/(5*E^x))*(E^x + 20*x *Log[2]*Log[x]))/(5*E^x*x*Log[x]),x]
3.4.24.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 7.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\ln \left (x \right )+\ln \left (x \right )^{\frac {1}{5}} \left (\frac {1}{16}\right )^{{\mathrm e}^{-x}}\) | \(15\) |
parallelrisch | \(\ln \left (x \right )+{\mathrm e}^{\frac {\left ({\mathrm e}^{x} \ln \left (\ln \left (x \right )\right )-20 \ln \left (2\right )\right ) {\mathrm e}^{-x}}{5}}\) | \(22\) |
int(1/5*((20*x*ln(2)*ln(x)+exp(x))*exp(1/5*(exp(x)*ln(ln(x))-20*ln(2))/exp (x))+5*exp(x)*ln(x))/x/exp(x)/ln(x),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-x} \left (5 e^x \log (x)+e^{\frac {1}{5} e^{-x} \left (-20 \log (2)+e^x \log (\log (x))\right )} \left (e^x+20 x \log (2) \log (x)\right )\right )}{5 x \log (x)} \, dx=e^{\left (\frac {1}{5} \, {\left (e^{x} \log \left (\log \left (x\right )\right ) - 20 \, \log \left (2\right )\right )} e^{\left (-x\right )}\right )} + \log \left (x\right ) \]
integrate(1/5*((20*x*log(2)*log(x)+exp(x))*exp(1/5*(exp(x)*log(log(x))-20* log(2))/exp(x))+5*exp(x)*log(x))/x/exp(x)/log(x),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-x} \left (5 e^x \log (x)+e^{\frac {1}{5} e^{-x} \left (-20 \log (2)+e^x \log (\log (x))\right )} \left (e^x+20 x \log (2) \log (x)\right )\right )}{5 x \log (x)} \, dx=e^{\left (\frac {e^{x} \log {\left (\log {\left (x \right )} \right )}}{5} - 4 \log {\left (2 \right )}\right ) e^{- x}} + \log {\left (x \right )} \]
integrate(1/5*((20*x*ln(2)*ln(x)+exp(x))*exp(1/5*(exp(x)*ln(ln(x))-20*ln(2 ))/exp(x))+5*exp(x)*ln(x))/x/exp(x)/ln(x),x)
\[ \int \frac {e^{-x} \left (5 e^x \log (x)+e^{\frac {1}{5} e^{-x} \left (-20 \log (2)+e^x \log (\log (x))\right )} \left (e^x+20 x \log (2) \log (x)\right )\right )}{5 x \log (x)} \, dx=\int { \frac {{\left ({\left (20 \, x \log \left (2\right ) \log \left (x\right ) + e^{x}\right )} e^{\left (\frac {1}{5} \, {\left (e^{x} \log \left (\log \left (x\right )\right ) - 20 \, \log \left (2\right )\right )} e^{\left (-x\right )}\right )} + 5 \, e^{x} \log \left (x\right )\right )} e^{\left (-x\right )}}{5 \, x \log \left (x\right )} \,d x } \]
integrate(1/5*((20*x*log(2)*log(x)+exp(x))*exp(1/5*(exp(x)*log(log(x))-20* log(2))/exp(x))+5*exp(x)*log(x))/x/exp(x)/log(x),x, algorithm=\
1/5*integrate((20*x*log(2)*log(x) + e^x)*e^(-4*e^(-x)*log(2) - x)/(x*log(x )^(4/5)), x) + log(x)
\[ \int \frac {e^{-x} \left (5 e^x \log (x)+e^{\frac {1}{5} e^{-x} \left (-20 \log (2)+e^x \log (\log (x))\right )} \left (e^x+20 x \log (2) \log (x)\right )\right )}{5 x \log (x)} \, dx=\int { \frac {{\left ({\left (20 \, x \log \left (2\right ) \log \left (x\right ) + e^{x}\right )} e^{\left (\frac {1}{5} \, {\left (e^{x} \log \left (\log \left (x\right )\right ) - 20 \, \log \left (2\right )\right )} e^{\left (-x\right )}\right )} + 5 \, e^{x} \log \left (x\right )\right )} e^{\left (-x\right )}}{5 \, x \log \left (x\right )} \,d x } \]
integrate(1/5*((20*x*log(2)*log(x)+exp(x))*exp(1/5*(exp(x)*log(log(x))-20* log(2))/exp(x))+5*exp(x)*log(x))/x/exp(x)/log(x),x, algorithm=\
integrate(1/5*((20*x*log(2)*log(x) + e^x)*e^(1/5*(e^x*log(log(x)) - 20*log (2))*e^(-x)) + 5*e^x*log(x))*e^(-x)/(x*log(x)), x)
Time = 7.72 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {e^{-x} \left (5 e^x \log (x)+e^{\frac {1}{5} e^{-x} \left (-20 \log (2)+e^x \log (\log (x))\right )} \left (e^x+20 x \log (2) \log (x)\right )\right )}{5 x \log (x)} \, dx=\ln \left (x\right )+\frac {{\ln \left (x\right )}^{1/5}}{2^{4\,{\mathrm {e}}^{-x}}} \]