Integrand size = 95, antiderivative size = 24 \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\log \left (e^{1+e^{-x} x}+\log \left (\log \left (125 e^2 \log ^2(x)\right )\right )\right ) \]
\[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx \]
Integrate[(2*E^x + E^(1 + x/E^x)*(x - x^2)*Log[x]*Log[125*E^2*Log[x]^2])/( E^(1 + x + x/E^x)*x*Log[x]*Log[125*E^2*Log[x]^2] + E^x*x*Log[x]*Log[125*E^ 2*Log[x]^2]*Log[Log[125*E^2*Log[x]^2]]),x]
Integrate[(2*E^x + E^(1 + x/E^x)*(x - x^2)*Log[x]*Log[125*E^2*Log[x]^2])/( E^(1 + x + x/E^x)*x*Log[x]*Log[125*E^2*Log[x]^2] + E^x*x*Log[x]*Log[125*E^ 2*Log[x]^2]*Log[Log[125*E^2*Log[x]^2]]), x]
Time = 1.55 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {7292, 7235}
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.
| \(\displaystyle \int \frac {e^{e^{-x} x+1} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )+2 e^x}{e^{e^{-x} x+x+1} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right ) \log \left (125 e^2 \log ^2(x)\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{-x} \left (e^{e^{-x} x+1} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )+2 e^x\right )}{x \log (x) \left (\log \left (\log ^2(x)\right )+2 \left (1+\frac {3 \log (5)}{2}\right )\right ) \left (e^{e^{-x} x+1}+\log \left (\log \left (125 \log ^2(x)\right )+2\right )\right )}dx\) |
| \(\Big \downarrow \) 7235 |
| \(\displaystyle \log \left (e^{e^{-x} x+1}+\log \left (\log \left (125 \log ^2(x)\right )+2\right )\right )\) |
Int[(2*E^x + E^(1 + x/E^x)*(x - x^2)*Log[x]*Log[125*E^2*Log[x]^2])/(E^(1 + x + x/E^x)*x*Log[x]*Log[125*E^2*Log[x]^2] + E^x*x*Log[x]*Log[125*E^2*Log[ x]^2]*Log[Log[125*E^2*Log[x]^2]]),x]
3.26.93.3.1 Defintions of rubi rules used
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.70 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.29
\[\ln \left ({\mathrm e}^{1+x \,{\mathrm e}^{-x}}+\ln \left (3 \ln \left (5\right )+2+2 \ln \left (\ln \left (x \right )\right )-\frac {i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right ) {\left (-\operatorname {csgn}\left (i \ln \left (x \right )^{2}\right )+\operatorname {csgn}\left (i \ln \left (x \right )\right )\right )}^{2}}{2}\right )\right )\]
int(((-x^2+x)*exp(1)*ln(x)*exp(x/exp(x))*ln(125*exp(2)*ln(x)^2)+2*exp(x))/ (x*exp(x)*ln(x)*ln(125*exp(2)*ln(x)^2)*ln(ln(125*exp(2)*ln(x)^2))+x*exp(1) *exp(x)*ln(x)*exp(x/exp(x))*ln(125*exp(2)*ln(x)^2)),x)
ln(exp(1+x*exp(-x))+ln(3*ln(5)+2+2*ln(ln(x))-1/2*I*Pi*csgn(I*ln(x)^2)*(-cs gn(I*ln(x)^2)+csgn(I*ln(x)))^2))
Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\log \left ({\left (e^{x} \log \left (\log \left (125 \, e^{2} \log \left (x\right )^{2}\right )\right ) + e^{\left ({\left ({\left (x + 1\right )} e^{x} + x\right )} e^{\left (-x\right )}\right )}\right )} e^{\left (-x\right )}\right ) \]
integrate(((-x^2+x)*exp(1)*log(x)*exp(x/exp(x))*log(125*exp(2)*log(x)^2)+2 *exp(x))/(x*exp(x)*log(x)*log(125*exp(2)*log(x)^2)*log(log(125*exp(2)*log( x)^2))+x*exp(1)*exp(x)*log(x)*exp(x/exp(x))*log(125*exp(2)*log(x)^2)),x, a lgorithm=\
Time = 1.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\log {\left (e^{x e^{- x}} + \frac {\log {\left (\log {\left (125 e^{2} \log {\left (x \right )}^{2} \right )} \right )}}{e} \right )} \]
integrate(((-x**2+x)*exp(1)*ln(x)*exp(x/exp(x))*ln(125*exp(2)*ln(x)**2)+2* exp(x))/(x*exp(x)*ln(x)*ln(125*exp(2)*ln(x)**2)*ln(ln(125*exp(2)*ln(x)**2) )+x*exp(1)*exp(x)*ln(x)*exp(x/exp(x))*ln(125*exp(2)*ln(x)**2)),x)
Time = 0.33 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\log \left (e^{\left (x e^{\left (-x\right )} + 1\right )} + \log \left (3 \, \log \left (5\right ) + 2 \, \log \left (\log \left (x\right )\right ) + 2\right )\right ) \]
integrate(((-x^2+x)*exp(1)*log(x)*exp(x/exp(x))*log(125*exp(2)*log(x)^2)+2 *exp(x))/(x*exp(x)*log(x)*log(125*exp(2)*log(x)^2)*log(log(125*exp(2)*log( x)^2))+x*exp(1)*exp(x)*log(x)*exp(x/exp(x))*log(125*exp(2)*log(x)^2)),x, a lgorithm=\
Exception generated. \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\text {Exception raised: TypeError} \]
integrate(((-x^2+x)*exp(1)*log(x)*exp(x/exp(x))*log(125*exp(2)*log(x)^2)+2 *exp(x))/(x*exp(x)*log(x)*log(125*exp(2)*log(x)^2)*log(log(125*exp(2)*log( x)^2))+x*exp(1)*exp(x)*log(x)*exp(x/exp(x))*log(125*exp(2)*log(x)^2)),x, a lgorithm=\
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:ln of unsigned or minus infinity Er ror: Bad Argument Value
Time = 12.43 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {2 e^x+e^{1+e^{-x} x} \left (x-x^2\right ) \log (x) \log \left (125 e^2 \log ^2(x)\right )}{e^{1+x+e^{-x} x} x \log (x) \log \left (125 e^2 \log ^2(x)\right )+e^x x \log (x) \log \left (125 e^2 \log ^2(x)\right ) \log \left (\log \left (125 e^2 \log ^2(x)\right )\right )} \, dx=\ln \left ({\mathrm {e}}^{x\,{\mathrm {e}}^{-x}+1}+\ln \left (\ln \left (125\,{\mathrm {e}}^2\,{\ln \left (x\right )}^2\right )\right )\right ) \]