Integrand size = 218, antiderivative size = 23 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5 \log (x) \left (5+\log \left (-5 e^{-\frac {2-x}{e^9}}+x\right )\right ) \]
Time = 0.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5 \log (x) \left (5+\log \left (-5 e^{\frac {-2+x}{e^9}}+x\right )\right ) \]
Integrate[(Log[x]*(5 + Log[(-5 + E^((2 - x)/E^9)*x)/E^((2 - x)/E^9)])*(-12 5*E^9 + 25*E^(9 + (2 - x)/E^9)*x + (-25*x + 5*E^(9 + (2 - x)/E^9)*x)*Log[x ] + (-25*E^9 + 5*E^(9 + (2 - x)/E^9)*x)*Log[(-5 + E^((2 - x)/E^9)*x)/E^((2 - x)/E^9)]))/((-25*E^9*x + 5*E^(9 + (2 - x)/E^9)*x^2)*Log[x] + (-5*E^9*x + E^(9 + (2 - x)/E^9)*x^2)*Log[x]*Log[(-5 + E^((2 - x)/E^9)*x)/E^((2 - x)/ E^9)]),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.
| \(\displaystyle \int \frac {\log (x) \left (\log \left (e^{-\frac {2-x}{e^9}} \left (e^{\frac {2-x}{e^9}} x-5\right )\right )+5\right ) \left (25 e^{\frac {2-x}{e^9}+9} x+\left (5 e^{\frac {2-x}{e^9}+9} x-25 x\right ) \log (x)+\left (5 e^{\frac {2-x}{e^9}+9} x-25 e^9\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (e^{\frac {2-x}{e^9}} x-5\right )\right )-125 e^9\right )}{\left (5 e^{\frac {2-x}{e^9}+9} x^2-25 e^9 x\right ) \log (x)+\left (e^{\frac {2-x}{e^9}+9} x^2-5 e^9 x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (e^{\frac {2-x}{e^9}} x-5\right )\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {5 e^9 \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right ) \left (\log \left (x-5 e^{\frac {x-2}{e^9}}\right )+5\right )-5 \left (e^{9+\frac {2}{e^9}}-5 e^{\frac {x}{e^9}}\right ) x \log (x)}{e^9 x \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right )}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int -\frac {5 \left (\left (e^{9+\frac {2}{e^9}}-5 e^{\frac {x}{e^9}}\right ) x \log (x)-e^9 \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right ) \left (\log \left (x-5 e^{-\frac {2-x}{e^9}}\right )+5\right )\right )}{x \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right )}dx}{e^9}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {5 \int \frac {\left (e^{9+\frac {2}{e^9}}-5 e^{\frac {x}{e^9}}\right ) x \log (x)-e^9 \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right ) \left (\log \left (x-5 e^{-\frac {2-x}{e^9}}\right )+5\right )}{x \left (5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x\right )}dx}{e^9}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {5 \int \left (\frac {-x \log (x)-e^9 \log \left (x-5 e^{\frac {x-2}{e^9}}\right )-5 e^9}{x}-\frac {e^{\frac {2}{e^9}} \left (e^9-x\right ) \log (x)}{e^{\frac {2}{e^9}} x-5 e^{\frac {x}{e^9}}}\right )dx}{e^9}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {5 \left (-e^{9+\frac {2}{e^9}} \int \frac {\int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x}dx}{x}dx-e^{\frac {2}{e^9}} \int \frac {\int \frac {x}{e^{\frac {2}{e^9}} x-5 e^{\frac {x}{e^9}}}dx}{x}dx+e^{9+\frac {2}{e^9}} \log (x) \int \frac {1}{5 e^{\frac {x}{e^9}}-e^{\frac {2}{e^9}} x}dx+e^{\frac {2}{e^9}} \log (x) \int \frac {x}{e^{\frac {2}{e^9}} x-5 e^{\frac {x}{e^9}}}dx-e^9 \int \frac {\log \left (x-5 e^{\frac {x-2}{e^9}}\right )}{x}dx+x+x (-\log (x))-5 e^9 \log (x)\right )}{e^9}\) |
Int[(Log[x]*(5 + Log[(-5 + E^((2 - x)/E^9)*x)/E^((2 - x)/E^9)])*(-125*E^9 + 25*E^(9 + (2 - x)/E^9)*x + (-25*x + 5*E^(9 + (2 - x)/E^9)*x)*Log[x] + (- 25*E^9 + 5*E^(9 + (2 - x)/E^9)*x)*Log[(-5 + E^((2 - x)/E^9)*x)/E^((2 - x)/ E^9)]))/((-25*E^9*x + 5*E^(9 + (2 - x)/E^9)*x^2)*Log[x] + (-5*E^9*x + E^(9 + (2 - x)/E^9)*x^2)*Log[x]*Log[(-5 + E^((2 - x)/E^9)*x)/E^((2 - x)/E^9)]) ,x]
3.2.34.3.1 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]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.41 (sec) , antiderivative size = 212, normalized size of antiderivative = 9.22
\[-5 \ln \left (x \right ) \ln \left ({\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}\right )+5 \ln \left (x \right ) \ln \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )-\frac {5 i \pi \ln \left (x \right ) \operatorname {csgn}\left (i \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (-2+x \right ) {\mathrm e}^{-9}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (-2+x \right ) {\mathrm e}^{-9}} \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )\right )}{2}+\frac {5 i \pi \ln \left (x \right ) \operatorname {csgn}\left (i \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (-2+x \right ) {\mathrm e}^{-9}} \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )\right )^{2}}{2}+\frac {5 i \pi \ln \left (x \right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (-2+x \right ) {\mathrm e}^{-9}}\right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (-2+x \right ) {\mathrm e}^{-9}} \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )\right )^{2}}{2}-\frac {5 i \pi \ln \left (x \right ) \operatorname {csgn}\left (i {\mathrm e}^{\left (-2+x \right ) {\mathrm e}^{-9}} \left (x \,{\mathrm e}^{-\left (-2+x \right ) {\mathrm e}^{-9}}-5\right )\right )^{3}}{2}+25 \ln \left (x \right )\]
int(((5*x*exp(9)*exp((2-x)/exp(9))-25*exp(9))*ln((x*exp((2-x)/exp(9))-5)/e xp((2-x)/exp(9)))+(5*x*exp(9)*exp((2-x)/exp(9))-25*x)*ln(x)+25*x*exp(9)*ex p((2-x)/exp(9))-125*exp(9))*exp(ln(ln((x*exp((2-x)/exp(9))-5)/exp((2-x)/ex p(9)))+5)+ln(ln(x)))/((x^2*exp(9)*exp((2-x)/exp(9))-5*x*exp(9))*ln(x)*ln(( x*exp((2-x)/exp(9))-5)/exp((2-x)/exp(9)))+(5*x^2*exp(9)*exp((2-x)/exp(9))- 25*x*exp(9))*ln(x)),x)
-5*ln(x)*ln(exp(-(-2+x)*exp(-9)))+5*ln(x)*ln(x*exp(-(-2+x)*exp(-9))-5)-5/2 *I*Pi*ln(x)*csgn(I*(x*exp(-(-2+x)*exp(-9))-5))*csgn(I*exp((-2+x)*exp(-9))) *csgn(I*exp((-2+x)*exp(-9))*(x*exp(-(-2+x)*exp(-9))-5))+5/2*I*Pi*ln(x)*csg n(I*(x*exp(-(-2+x)*exp(-9))-5))*csgn(I*exp((-2+x)*exp(-9))*(x*exp(-(-2+x)* exp(-9))-5))^2+5/2*I*Pi*ln(x)*csgn(I*exp((-2+x)*exp(-9)))*csgn(I*exp((-2+x )*exp(-9))*(x*exp(-(-2+x)*exp(-9))-5))^2-5/2*I*Pi*ln(x)*csgn(I*exp((-2+x)* exp(-9))*(x*exp(-(-2+x)*exp(-9))-5))^3+25*ln(x)
Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5 \, \log \left ({\left (x e^{\left (-{\left (x - 9 \, e^{9} - 2\right )} e^{\left (-9\right )}\right )} - 5 \, e^{9}\right )} e^{\left ({\left (x - 9 \, e^{9} - 2\right )} e^{\left (-9\right )}\right )}\right ) \log \left (x\right ) + 25 \, \log \left (x\right ) \]
integrate(((5*x*exp(9)*exp((2-x)/exp(9))-25*exp(9))*log((x*exp((2-x)/exp(9 ))-5)/exp((2-x)/exp(9)))+(5*x*exp(9)*exp((2-x)/exp(9))-25*x)*log(x)+25*x*e xp(9)*exp((2-x)/exp(9))-125*exp(9))*exp(log(log((x*exp((2-x)/exp(9))-5)/ex p((2-x)/exp(9)))+5)+log(log(x)))/((x^2*exp(9)*exp((2-x)/exp(9))-5*x*exp(9) )*log(x)*log((x*exp((2-x)/exp(9))-5)/exp((2-x)/exp(9)))+(5*x^2*exp(9)*exp( (2-x)/exp(9))-25*x*exp(9))*log(x)),x, algorithm=\
5*log((x*e^(-(x - 9*e^9 - 2)*e^(-9)) - 5*e^9)*e^((x - 9*e^9 - 2)*e^(-9)))* log(x) + 25*log(x)
Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5 \log {\left (x \right )} \log {\left (\left (x e^{\frac {2 - x}{e^{9}}} - 5\right ) e^{- \frac {2 - x}{e^{9}}} \right )} + 25 \log {\left (x \right )} \]
integrate(((5*x*exp(9)*exp((2-x)/exp(9))-25*exp(9))*ln((x*exp((2-x)/exp(9) )-5)/exp((2-x)/exp(9)))+(5*x*exp(9)*exp((2-x)/exp(9))-25*x)*ln(x)+25*x*exp (9)*exp((2-x)/exp(9))-125*exp(9))*exp(ln(ln((x*exp((2-x)/exp(9))-5)/exp((2 -x)/exp(9)))+5)+ln(ln(x)))/((x**2*exp(9)*exp((2-x)/exp(9))-5*x*exp(9))*ln( x)*ln((x*exp((2-x)/exp(9))-5)/exp((2-x)/exp(9)))+(5*x**2*exp(9)*exp((2-x)/ exp(9))-25*x*exp(9))*ln(x)),x)
Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5 \, {\left (e^{9} \log \left (x e^{\left (2 \, e^{\left (-9\right )}\right )} - 5 \, e^{\left (x e^{\left (-9\right )}\right )}\right ) \log \left (x\right ) + {\left (5 \, e^{9} - 2\right )} \log \left (x\right )\right )} e^{\left (-9\right )} \]
integrate(((5*x*exp(9)*exp((2-x)/exp(9))-25*exp(9))*log((x*exp((2-x)/exp(9 ))-5)/exp((2-x)/exp(9)))+(5*x*exp(9)*exp((2-x)/exp(9))-25*x)*log(x)+25*x*e xp(9)*exp((2-x)/exp(9))-125*exp(9))*exp(log(log((x*exp((2-x)/exp(9))-5)/ex p((2-x)/exp(9)))+5)+log(log(x)))/((x^2*exp(9)*exp((2-x)/exp(9))-5*x*exp(9) )*log(x)*log((x*exp((2-x)/exp(9))-5)/exp((2-x)/exp(9)))+(5*x^2*exp(9)*exp( (2-x)/exp(9))-25*x*exp(9))*log(x)),x, algorithm=\
Time = 0.47 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5 \, {\left (e^{9} \log \left (x e^{\left (2 \, e^{\left (-9\right )}\right )} - 5 \, e^{\left (x e^{\left (-9\right )}\right )}\right ) \log \left (x\right ) + 5 \, e^{9} \log \left (x\right ) - 2 \, \log \left (x\right )\right )} e^{\left (-9\right )} \]
integrate(((5*x*exp(9)*exp((2-x)/exp(9))-25*exp(9))*log((x*exp((2-x)/exp(9 ))-5)/exp((2-x)/exp(9)))+(5*x*exp(9)*exp((2-x)/exp(9))-25*x)*log(x)+25*x*e xp(9)*exp((2-x)/exp(9))-125*exp(9))*exp(log(log((x*exp((2-x)/exp(9))-5)/ex p((2-x)/exp(9)))+5)+log(log(x)))/((x^2*exp(9)*exp((2-x)/exp(9))-5*x*exp(9) )*log(x)*log((x*exp((2-x)/exp(9))-5)/exp((2-x)/exp(9)))+(5*x^2*exp(9)*exp( (2-x)/exp(9))-25*x*exp(9))*log(x)),x, algorithm=\
Time = 12.12 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {\log (x) \left (5+\log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right ) \left (-125 e^9+25 e^{9+\frac {2-x}{e^9}} x+\left (-25 x+5 e^{9+\frac {2-x}{e^9}} x\right ) \log (x)+\left (-25 e^9+5 e^{9+\frac {2-x}{e^9}} x\right ) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )\right )}{\left (-25 e^9 x+5 e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x)+\left (-5 e^9 x+e^{9+\frac {2-x}{e^9}} x^2\right ) \log (x) \log \left (e^{-\frac {2-x}{e^9}} \left (-5+e^{\frac {2-x}{e^9}} x\right )\right )} \, dx=5\,\ln \left (x\right )\,\left (\ln \left (x-5\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{-9}}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-9}}\right )+5\right ) \]
int((exp(log(log(x)) + log(log(exp(exp(-9)*(x - 2))*(x*exp(-exp(-9)*(x - 2 )) - 5)) + 5))*(125*exp(9) + log(x)*(25*x - 5*x*exp(-exp(-9)*(x - 2))*exp( 9)) + log(exp(exp(-9)*(x - 2))*(x*exp(-exp(-9)*(x - 2)) - 5))*(25*exp(9) - 5*x*exp(-exp(-9)*(x - 2))*exp(9)) - 25*x*exp(-exp(-9)*(x - 2))*exp(9)))/( log(x)*(25*x*exp(9) - 5*x^2*exp(-exp(-9)*(x - 2))*exp(9)) + log(exp(exp(-9 )*(x - 2))*(x*exp(-exp(-9)*(x - 2)) - 5))*log(x)*(5*x*exp(9) - x^2*exp(-ex p(-9)*(x - 2))*exp(9))),x)