Integrand size = 95, antiderivative size = 29 \[ \int \frac {e^x (3-3 x) \log (3)+\left (-2 e^x \log (3)-2 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right )+\left (3 e^x \log (3)+3 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right ) \log \left (\frac {3}{\log \left (\frac {e^x+x}{x}\right )}\right )}{\left (e^x+x\right ) \log \left (\frac {e^x+x}{x}\right )} \, dx=\log (3) \left (x+3 \left (3-x+x \log \left (\frac {3}{\log \left (\frac {e^x+x}{x}\right )}\right )\right )\right ) \]
Time = 0.68 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {e^x (3-3 x) \log (3)+\left (-2 e^x \log (3)-2 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right )+\left (3 e^x \log (3)+3 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right ) \log \left (\frac {3}{\log \left (\frac {e^x+x}{x}\right )}\right )}{\left (e^x+x\right ) \log \left (\frac {e^x+x}{x}\right )} \, dx=\log (3) \left (-2 x+3 x \log \left (\frac {3}{\log \left (\frac {e^x+x}{x}\right )}\right )\right ) \]
Integrate[(E^x*(3 - 3*x)*Log[3] + (-2*E^x*Log[3] - 2*x*Log[3])*Log[(E^x + x)/x] + (3*E^x*Log[3] + 3*x*Log[3])*Log[(E^x + x)/x]*Log[3/Log[(E^x + x)/x ]])/((E^x + x)*Log[(E^x + x)/x]),x]
Time = 1.35 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {7239, 27, 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.
\(\displaystyle \int \frac {e^x (3-3 x) \log (3)+\left (-2 x \log (3)-2 e^x \log (3)\right ) \log \left (\frac {x+e^x}{x}\right )+\left (3 x \log (3)+3 e^x \log (3)\right ) \log \left (\frac {x+e^x}{x}\right ) \log \left (\frac {3}{\log \left (\frac {x+e^x}{x}\right )}\right )}{\left (x+e^x\right ) \log \left (\frac {x+e^x}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \log (3) \left (-\frac {3 e^x (x-1)}{\left (x+e^x\right ) \log \left (\frac {x+e^x}{x}\right )}+3 \log \left (\frac {3}{\log \left (\frac {x+e^x}{x}\right )}\right )-2\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (3) \int \left (\frac {3 e^x (1-x)}{\left (x+e^x\right ) \log \left (\frac {x+e^x}{x}\right )}+3 \log \left (\frac {3}{\log \left (\frac {x+e^x}{x}\right )}\right )-2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log (3) \left (3 x \log \left (\frac {3}{\log \left (\frac {x+e^x}{x}\right )}\right )-2 x\right )\) |
Int[(E^x*(3 - 3*x)*Log[3] + (-2*E^x*Log[3] - 2*x*Log[3])*Log[(E^x + x)/x] + (3*E^x*Log[3] + 3*x*Log[3])*Log[(E^x + x)/x]*Log[3/Log[(E^x + x)/x]])/(( E^x + x)*Log[(E^x + x)/x]),x]
3.24.69.3.1 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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \(3 \ln \left (\frac {3}{\ln \left (\frac {{\mathrm e}^{x}+x}{x}\right )}\right ) \ln \left (3\right ) x -2 x \ln \left (3\right )\) | \(26\) |
risch | \(\text {Expression too large to display}\) | \(797\) |
int(((3*ln(3)*exp(x)+3*x*ln(3))*ln(1/x*(exp(x)+x))*ln(3/ln(1/x*(exp(x)+x)) )+(-2*ln(3)*exp(x)-2*x*ln(3))*ln(1/x*(exp(x)+x))+(-3*x+3)*ln(3)*exp(x))/(e xp(x)+x)/ln(1/x*(exp(x)+x)),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {e^x (3-3 x) \log (3)+\left (-2 e^x \log (3)-2 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right )+\left (3 e^x \log (3)+3 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right ) \log \left (\frac {3}{\log \left (\frac {e^x+x}{x}\right )}\right )}{\left (e^x+x\right ) \log \left (\frac {e^x+x}{x}\right )} \, dx=3 \, x \log \left (3\right ) \log \left (\frac {3}{\log \left (\frac {x + e^{x}}{x}\right )}\right ) - 2 \, x \log \left (3\right ) \]
integrate(((3*log(3)*exp(x)+3*x*log(3))*log(1/x*(exp(x)+x))*log(3/log(1/x* (exp(x)+x)))+(-2*log(3)*exp(x)-2*x*log(3))*log(1/x*(exp(x)+x))+(-3*x+3)*lo g(3)*exp(x))/(exp(x)+x)/log(1/x*(exp(x)+x)),x, algorithm=\
Timed out. \[ \int \frac {e^x (3-3 x) \log (3)+\left (-2 e^x \log (3)-2 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right )+\left (3 e^x \log (3)+3 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right ) \log \left (\frac {3}{\log \left (\frac {e^x+x}{x}\right )}\right )}{\left (e^x+x\right ) \log \left (\frac {e^x+x}{x}\right )} \, dx=\text {Timed out} \]
integrate(((3*ln(3)*exp(x)+3*x*ln(3))*ln(1/x*(exp(x)+x))*ln(3/ln(1/x*(exp( x)+x)))+(-2*ln(3)*exp(x)-2*x*ln(3))*ln(1/x*(exp(x)+x))+(-3*x+3)*ln(3)*exp( x))/(exp(x)+x)/ln(1/x*(exp(x)+x)),x)
Time = 0.33 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {e^x (3-3 x) \log (3)+\left (-2 e^x \log (3)-2 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right )+\left (3 e^x \log (3)+3 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right ) \log \left (\frac {3}{\log \left (\frac {e^x+x}{x}\right )}\right )}{\left (e^x+x\right ) \log \left (\frac {e^x+x}{x}\right )} \, dx=-3 \, x \log \left (3\right ) \log \left (\log \left (x + e^{x}\right ) - \log \left (x\right )\right ) + {\left (3 \, \log \left (3\right )^{2} - 2 \, \log \left (3\right )\right )} x \]
integrate(((3*log(3)*exp(x)+3*x*log(3))*log(1/x*(exp(x)+x))*log(3/log(1/x* (exp(x)+x)))+(-2*log(3)*exp(x)-2*x*log(3))*log(1/x*(exp(x)+x))+(-3*x+3)*lo g(3)*exp(x))/(exp(x)+x)/log(1/x*(exp(x)+x)),x, algorithm=\
\[ \int \frac {e^x (3-3 x) \log (3)+\left (-2 e^x \log (3)-2 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right )+\left (3 e^x \log (3)+3 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right ) \log \left (\frac {3}{\log \left (\frac {e^x+x}{x}\right )}\right )}{\left (e^x+x\right ) \log \left (\frac {e^x+x}{x}\right )} \, dx=\int { -\frac {3 \, {\left (x - 1\right )} e^{x} \log \left (3\right ) - 3 \, {\left (x \log \left (3\right ) + e^{x} \log \left (3\right )\right )} \log \left (\frac {x + e^{x}}{x}\right ) \log \left (\frac {3}{\log \left (\frac {x + e^{x}}{x}\right )}\right ) + 2 \, {\left (x \log \left (3\right ) + e^{x} \log \left (3\right )\right )} \log \left (\frac {x + e^{x}}{x}\right )}{{\left (x + e^{x}\right )} \log \left (\frac {x + e^{x}}{x}\right )} \,d x } \]
integrate(((3*log(3)*exp(x)+3*x*log(3))*log(1/x*(exp(x)+x))*log(3/log(1/x* (exp(x)+x)))+(-2*log(3)*exp(x)-2*x*log(3))*log(1/x*(exp(x)+x))+(-3*x+3)*lo g(3)*exp(x))/(exp(x)+x)/log(1/x*(exp(x)+x)),x, algorithm=\
integrate(-(3*(x - 1)*e^x*log(3) - 3*(x*log(3) + e^x*log(3))*log((x + e^x) /x)*log(3/log((x + e^x)/x)) + 2*(x*log(3) + e^x*log(3))*log((x + e^x)/x))/ ((x + e^x)*log((x + e^x)/x)), x)
Time = 12.41 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {e^x (3-3 x) \log (3)+\left (-2 e^x \log (3)-2 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right )+\left (3 e^x \log (3)+3 x \log (3)\right ) \log \left (\frac {e^x+x}{x}\right ) \log \left (\frac {3}{\log \left (\frac {e^x+x}{x}\right )}\right )}{\left (e^x+x\right ) \log \left (\frac {e^x+x}{x}\right )} \, dx=3\,x\,\ln \left (3\right )\,\ln \left (\frac {3}{\ln \left (\frac {x+{\mathrm {e}}^x}{x}\right )}\right )-2\,x\,\ln \left (3\right ) \]