Integrand size = 75, antiderivative size = 23 \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=e^{-x} \left (5+\frac {e}{2}+x\right ) \log \left (\log \left (5-e^x\right )\right ) \]
Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\frac {1}{2} e^{-x} (10+e+2 x) \log \left (\log \left (5-e^x\right )\right ) \]
Integrate[(E^x*(10 + E + 2*x) + (40 + 5*E + E^x*(-8 - E - 2*x) + 10*x)*Log [5 - E^x]*Log[Log[5 - E^x]])/((-10*E^x + 2*E^(2*x))*Log[5 - E^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.
\(\displaystyle \int \frac {e^x (2 x+e+10)+\left (e^x (-2 x-e-8)+10 x+5 e+40\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (2 e^{2 x}-10 e^x\right ) \log \left (5-e^x\right )} \, dx\) |
\(\Big \downarrow \) 2721 |
\(\displaystyle \int \frac {e^{-x} \left (e^x (2 x+e+10)+\left (e^x (-2 x-e-8)+10 x+5 e+40\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )\right )}{\left (2 e^x-10\right ) \log \left (5-e^x\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-x} \left (-e^x (2 x+e+10)-\left (e^x (-2 x-e-8)+10 x+5 e+40\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )\right )}{2 \left (5-e^x\right ) \log \left (5-e^x\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {e^{-x} \left (e^x (2 x+e+10)+\left (10 x-e^x (2 x+e+8)+5 (8+e)\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )\right )}{\left (5-e^x\right ) \log \left (5-e^x\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {e^{-x} \left (e^x (2 x+e+10)+\left (10 x-e^x (2 x+e+8)+5 (8+e)\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )\right )}{\left (5-e^x\right ) \log \left (5-e^x\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {e^{-x} \left (2 \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right ) x-2 x+8 \left (1+\frac {e}{8}\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )-10 \left (1+\frac {e}{10}\right )\right )}{\log \left (5-e^x\right )}-\frac {5 e^{-x} (2 x+e+10)}{\left (-5+e^x\right ) \log \left (5-e^x\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{5} (10+e) \text {Subst}\left (\int \frac {1}{x \log (5-x)}dx,x,e^x\right )+2 \int \frac {e^{-x} x}{\log \left (5-e^x\right )}dx+10 \int \frac {e^{-x} x}{\left (-5+e^x\right ) \log \left (5-e^x\right )}dx-(8+e) \int e^{-x} \log \left (\log \left (5-e^x\right )\right )dx-2 \int e^{-x} x \log \left (\log \left (5-e^x\right )\right )dx+\frac {1}{5} (10+e) \log \left (\log \left (5-e^x\right )\right )\right )\) |
Int[(E^x*(10 + E + 2*x) + (40 + 5*E + E^x*(-8 - E - 2*x) + 10*x)*Log[5 - E ^x]*Log[Log[5 - E^x]])/((-10*E^x + 2*E^(2*x))*Log[5 - E^x]),x]
3.7.84.3.1 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_.)*((a_.)*(F_)^(v_) + (b_.)*(F_)^(w_))^(n_), x_Symbol] :> Int[u*F^(n *v)*(a + b*F^ExpandToSum[w - v, x])^n, x] /; FreeQ[{F, a, b, n}, x] && ILtQ [n, 0] && LinearQ[{v, w}, x]
Time = 4.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\frac {\left ({\mathrm e}+2 x +10\right ) {\mathrm e}^{-x} \ln \left (\ln \left (5-{\mathrm e}^{x}\right )\right )}{2}\) | \(22\) |
parallelrisch | \(\frac {\left (\ln \left (\ln \left (5-{\mathrm e}^{x}\right )\right ) {\mathrm e}+2 \ln \left (\ln \left (5-{\mathrm e}^{x}\right )\right ) x +10 \ln \left (\ln \left (5-{\mathrm e}^{x}\right )\right )\right ) {\mathrm e}^{-x}}{2}\) | \(40\) |
int((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*ln(5-exp(x))*ln(ln(5-exp(x) ))+(exp(1)+2*x+10)*exp(x))/(2*exp(x)^2-10*exp(x))/ln(5-exp(x)),x,method=_R ETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\frac {1}{2} \, {\left (2 \, x + e + 10\right )} e^{\left (-x\right )} \log \left (\log \left (-e^{x} + 5\right )\right ) \]
integrate((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*log(5-exp(x))*log(log (5-exp(x)))+(exp(1)+2*x+10)*exp(x))/(2*exp(x)^2-10*exp(x))/log(5-exp(x)),x , algorithm=\
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\frac {\left (2 x + e + 10\right ) e^{- x} \log {\left (\log {\left (5 - e^{x} \right )} \right )}}{2} \]
integrate((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*ln(5-exp(x))*ln(ln(5- exp(x)))+(exp(1)+2*x+10)*exp(x))/(2*exp(x)**2-10*exp(x))/ln(5-exp(x)),x)
Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\frac {1}{2} \, {\left (2 \, x + e + 10\right )} e^{\left (-x\right )} \log \left (\log \left (-e^{x} + 5\right )\right ) \]
integrate((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*log(5-exp(x))*log(log (5-exp(x)))+(exp(1)+2*x+10)*exp(x))/(2*exp(x)^2-10*exp(x))/log(5-exp(x)),x , algorithm=\
\[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\int { -\frac {{\left ({\left (2 \, x + e + 8\right )} e^{x} - 10 \, x - 5 \, e - 40\right )} \log \left (-e^{x} + 5\right ) \log \left (\log \left (-e^{x} + 5\right )\right ) - {\left (2 \, x + e + 10\right )} e^{x}}{2 \, {\left (e^{\left (2 \, x\right )} - 5 \, e^{x}\right )} \log \left (-e^{x} + 5\right )} \,d x } \]
integrate((((-exp(1)-2*x-8)*exp(x)+5*exp(1)+10*x+40)*log(5-exp(x))*log(log (5-exp(x)))+(exp(1)+2*x+10)*exp(x))/(2*exp(x)^2-10*exp(x))/log(5-exp(x)),x , algorithm=\
integrate(-1/2*(((2*x + e + 8)*e^x - 10*x - 5*e - 40)*log(-e^x + 5)*log(lo g(-e^x + 5)) - (2*x + e + 10)*e^x)/((e^(2*x) - 5*e^x)*log(-e^x + 5)), x)
Timed out. \[ \int \frac {e^x (10+e+2 x)+\left (40+5 e+e^x (-8-e-2 x)+10 x\right ) \log \left (5-e^x\right ) \log \left (\log \left (5-e^x\right )\right )}{\left (-10 e^x+2 e^{2 x}\right ) \log \left (5-e^x\right )} \, dx=\int \frac {{\mathrm {e}}^x\,\left (2\,x+\mathrm {e}+10\right )+\ln \left (5-{\mathrm {e}}^x\right )\,\ln \left (\ln \left (5-{\mathrm {e}}^x\right )\right )\,\left (10\,x+5\,\mathrm {e}-{\mathrm {e}}^x\,\left (2\,x+\mathrm {e}+8\right )+40\right )}{\ln \left (5-{\mathrm {e}}^x\right )\,\left (2\,{\mathrm {e}}^{2\,x}-10\,{\mathrm {e}}^x\right )} \,d x \]
int((exp(x)*(2*x + exp(1) + 10) + log(5 - exp(x))*log(log(5 - exp(x)))*(10 *x + 5*exp(1) - exp(x)*(2*x + exp(1) + 8) + 40))/(log(5 - exp(x))*(2*exp(2 *x) - 10*exp(x))),x)