Integrand size = 178, antiderivative size = 25 \[ \int \frac {e^{-5+x} \left (2+e^x\right ) \left (80+14 x-16 x^2+2 x^3+e^x \left (80+14 x-16 x^2+2 x^3\right )\right )+\left (-92+40 x-4 x^2+e^x \left (-46+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (46-20 x+2 x^2+e^x \left (23-10 x+x^2\right )\right )\right ) \log \left (-2+e^{-5+x} \left (2+e^x\right )\right )}{-100+40 x-4 x^2+e^x \left (-50+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (50-20 x+2 x^2+e^x \left (25-10 x+x^2\right )\right )} \, dx=\left (2+\frac {2}{-5+x}+x\right ) \log \left (-2+e^{-5+x} \left (2+e^x\right )\right ) \]
Time = 0.89 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {e^{-5+x} \left (2+e^x\right ) \left (80+14 x-16 x^2+2 x^3+e^x \left (80+14 x-16 x^2+2 x^3\right )\right )+\left (-92+40 x-4 x^2+e^x \left (-46+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (46-20 x+2 x^2+e^x \left (23-10 x+x^2\right )\right )\right ) \log \left (-2+e^{-5+x} \left (2+e^x\right )\right )}{-100+40 x-4 x^2+e^x \left (-50+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (50-20 x+2 x^2+e^x \left (25-10 x+x^2\right )\right )} \, dx=2 \log \left (-2 e^5+2 e^x+e^{2 x}\right )+\frac {\left (2-5 x+x^2\right ) \log \left (-2+e^{-5+x} \left (2+e^x\right )\right )}{-5+x} \]
Integrate[(E^(-5 + x)*(2 + E^x)*(80 + 14*x - 16*x^2 + 2*x^3 + E^x*(80 + 14 *x - 16*x^2 + 2*x^3)) + (-92 + 40*x - 4*x^2 + E^x*(-46 + 20*x - 2*x^2) + E ^(-5 + x)*(2 + E^x)*(46 - 20*x + 2*x^2 + E^x*(23 - 10*x + x^2)))*Log[-2 + E^(-5 + x)*(2 + E^x)])/(-100 + 40*x - 4*x^2 + E^x*(-50 + 20*x - 2*x^2) + E ^(-5 + x)*(2 + E^x)*(50 - 20*x + 2*x^2 + E^x*(25 - 10*x + x^2))),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.
| \(\displaystyle \int \frac {\left (-4 x^2+e^x \left (-2 x^2+20 x-46\right )+e^{x-5} \left (e^x+2\right ) \left (2 x^2+e^x \left (x^2-10 x+23\right )-20 x+46\right )+40 x-92\right ) \log \left (e^{x-5} \left (e^x+2\right )-2\right )+e^{x-5} \left (e^x+2\right ) \left (2 x^3-16 x^2+e^x \left (2 x^3-16 x^2+14 x+80\right )+14 x+80\right )}{-4 x^2+e^x \left (-2 x^2+20 x-50\right )+e^{x-5} \left (e^x+2\right ) \left (2 x^2+e^x \left (x^2-10 x+25\right )-20 x+50\right )+40 x-100} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {\left (x^2-10 x+23\right ) \log \left (e^{x-5} \left (e^x+2\right )-2\right )+\frac {2 e^x \left (e^x+1\right ) \left (x^3-8 x^2+7 x+40\right )}{2 e^x+e^{2 x}-2 e^5}}{(5-x)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 x^3-16 x^2+x^2 \log \left (e^{x-5} \left (e^x+2\right )-2\right )+14 x-10 x \log \left (e^{x-5} \left (e^x+2\right )-2\right )+23 \log \left (e^{x-5} \left (e^x+2\right )-2\right )+80}{(x-5)^2}-\frac {2 \left (e^x-2 e^5\right ) \left (x^2-3 x-8\right )}{\left (2 e^x+e^{2 x}-2 e^5\right ) (x-5)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -8 e^5 \int \frac {1}{\left (2 e^5-2 e^x-e^{2 x}\right ) (x-5)}dx-8 \int \frac {e^x}{\left (-2 e^5+2 e^x+e^{2 x}\right ) (x-5)}dx-4 \int \frac {e^{2 x}}{\left (-2 e^5+2 e^x+e^{2 x}\right ) (x-5)}dx-2 \int \frac {e^{2 x} x}{-2 e^5+2 e^x+e^{2 x}}dx-\frac {\operatorname {PolyLog}\left (2,-\frac {e^x}{1-\sqrt {1+2 e^5}}\right )}{\sqrt {1+2 e^5}}+\operatorname {PolyLog}\left (2,-\frac {e^x}{1-\sqrt {1+2 e^5}}\right )+\frac {\operatorname {PolyLog}\left (2,-\frac {e^x}{1+\sqrt {1+2 e^5}}\right )}{\sqrt {1+2 e^5}}+\operatorname {PolyLog}\left (2,-\frac {e^x}{1+\sqrt {1+2 e^5}}\right )+2 \log \left (-2 e^x-e^{2 x}+2 e^5\right )+x \log \left (e^{x-5} \left (e^x+2\right )-2\right )-\frac {2 \log \left (e^{x-5} \left (e^x+2\right )-2\right )}{5-x}-\frac {x \log \left (\frac {e^x}{1-\sqrt {1+2 e^5}}+1\right )}{\sqrt {1+2 e^5}}+x \log \left (\frac {e^x}{1-\sqrt {1+2 e^5}}+1\right )+\frac {x \log \left (\frac {e^x}{1+\sqrt {1+2 e^5}}+1\right )}{\sqrt {1+2 e^5}}+x \log \left (\frac {e^x}{1+\sqrt {1+2 e^5}}+1\right )+4 \log (5-x)\) |
Int[(E^(-5 + x)*(2 + E^x)*(80 + 14*x - 16*x^2 + 2*x^3 + E^x*(80 + 14*x - 1 6*x^2 + 2*x^3)) + (-92 + 40*x - 4*x^2 + E^x*(-46 + 20*x - 2*x^2) + E^(-5 + x)*(2 + E^x)*(46 - 20*x + 2*x^2 + E^x*(23 - 10*x + x^2)))*Log[-2 + E^(-5 + x)*(2 + E^x)])/(-100 + 40*x - 4*x^2 + E^x*(-50 + 20*x - 2*x^2) + E^(-5 + x)*(2 + E^x)*(50 - 20*x + 2*x^2 + E^x*(25 - 10*x + x^2))),x]
3.24.53.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 16.46 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88
| method | result | size |
| risch | \(\frac {\left (x^{2}-5 x +2\right ) \ln \left ({\mathrm e}^{-5+2 x}+2 \,{\mathrm e}^{-5+x}-2\right )}{-5+x}+2 \ln \left ({\mathrm e}^{2 x}+2 \,{\mathrm e}^{x}-2 \,{\mathrm e}^{5}\right )\) | \(47\) |
| parallelrisch | \(\frac {4 \ln \left ({\mathrm e}^{\ln \left ({\mathrm e}^{x}+2\right )+x -5}-2\right ) x^{2}-12 \ln \left ({\mathrm e}^{\ln \left ({\mathrm e}^{x}+2\right )+x -5}-2\right ) x -32 \ln \left ({\mathrm e}^{\ln \left ({\mathrm e}^{x}+2\right )+x -5}-2\right )}{4 x -20}\) | \(55\) |
int(((((x^2-10*x+23)*exp(x)+2*x^2-20*x+46)*exp(ln(exp(x)+2)+x-5)+(-2*x^2+2 0*x-46)*exp(x)-4*x^2+40*x-92)*ln(exp(ln(exp(x)+2)+x-5)-2)+((2*x^3-16*x^2+1 4*x+80)*exp(x)+2*x^3-16*x^2+14*x+80)*exp(ln(exp(x)+2)+x-5))/(((x^2-10*x+25 )*exp(x)+2*x^2-20*x+50)*exp(ln(exp(x)+2)+x-5)+(-2*x^2+20*x-50)*exp(x)-4*x^ 2+40*x-100),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36 \[ \int \frac {e^{-5+x} \left (2+e^x\right ) \left (80+14 x-16 x^2+2 x^3+e^x \left (80+14 x-16 x^2+2 x^3\right )\right )+\left (-92+40 x-4 x^2+e^x \left (-46+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (46-20 x+2 x^2+e^x \left (23-10 x+x^2\right )\right )\right ) \log \left (-2+e^{-5+x} \left (2+e^x\right )\right )}{-100+40 x-4 x^2+e^x \left (-50+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (50-20 x+2 x^2+e^x \left (25-10 x+x^2\right )\right )} \, dx=\frac {{\left (x^{2} - 3 \, x - 8\right )} \log \left (-{\left (2 \, e^{5} - e^{\left (2 \, x\right )} - 2 \, e^{x}\right )} e^{\left (-5\right )}\right )}{x - 5} \]
integrate(((((x^2-10*x+23)*exp(x)+2*x^2-20*x+46)*exp(log(exp(x)+2)+x-5)+(- 2*x^2+20*x-46)*exp(x)-4*x^2+40*x-92)*log(exp(log(exp(x)+2)+x-5)-2)+((2*x^3 -16*x^2+14*x+80)*exp(x)+2*x^3-16*x^2+14*x+80)*exp(log(exp(x)+2)+x-5))/(((x ^2-10*x+25)*exp(x)+2*x^2-20*x+50)*exp(log(exp(x)+2)+x-5)+(-2*x^2+20*x-50)* exp(x)-4*x^2+40*x-100),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (20) = 40\).
Time = 0.33 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.76 \[ \int \frac {e^{-5+x} \left (2+e^x\right ) \left (80+14 x-16 x^2+2 x^3+e^x \left (80+14 x-16 x^2+2 x^3\right )\right )+\left (-92+40 x-4 x^2+e^x \left (-46+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (46-20 x+2 x^2+e^x \left (23-10 x+x^2\right )\right )\right ) \log \left (-2+e^{-5+x} \left (2+e^x\right )\right )}{-100+40 x-4 x^2+e^x \left (-50+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (50-20 x+2 x^2+e^x \left (25-10 x+x^2\right )\right )} \, dx=2 \log {\left (e^{2 x} + 2 e^{x} - 2 e^{5} \right )} + \frac {\left (x^{2} - 5 x + 2\right ) \log {\left (\frac {\left (e^{x} + 2\right ) e^{x}}{e^{5}} - 2 \right )}}{x - 5} \]
integrate(((((x**2-10*x+23)*exp(x)+2*x**2-20*x+46)*exp(ln(exp(x)+2)+x-5)+( -2*x**2+20*x-46)*exp(x)-4*x**2+40*x-92)*ln(exp(ln(exp(x)+2)+x-5)-2)+((2*x* *3-16*x**2+14*x+80)*exp(x)+2*x**3-16*x**2+14*x+80)*exp(ln(exp(x)+2)+x-5))/ (((x**2-10*x+25)*exp(x)+2*x**2-20*x+50)*exp(ln(exp(x)+2)+x-5)+(-2*x**2+20* x-50)*exp(x)-4*x**2+40*x-100),x)
2*log(exp(2*x) + 2*exp(x) - 2*exp(5)) + (x**2 - 5*x + 2)*log((exp(x) + 2)* exp(-5)*exp(x) - 2)/(x - 5)
Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^{-5+x} \left (2+e^x\right ) \left (80+14 x-16 x^2+2 x^3+e^x \left (80+14 x-16 x^2+2 x^3\right )\right )+\left (-92+40 x-4 x^2+e^x \left (-46+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (46-20 x+2 x^2+e^x \left (23-10 x+x^2\right )\right )\right ) \log \left (-2+e^{-5+x} \left (2+e^x\right )\right )}{-100+40 x-4 x^2+e^x \left (-50+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (50-20 x+2 x^2+e^x \left (25-10 x+x^2\right )\right )} \, dx=-\frac {5 \, x^{2} - {\left (x^{2} - 3 \, x - 8\right )} \log \left (-2 \, e^{5} + e^{\left (2 \, x\right )} + 2 \, e^{x}\right ) - 25 \, x + 10}{x - 5} \]
integrate(((((x^2-10*x+23)*exp(x)+2*x^2-20*x+46)*exp(log(exp(x)+2)+x-5)+(- 2*x^2+20*x-46)*exp(x)-4*x^2+40*x-92)*log(exp(log(exp(x)+2)+x-5)-2)+((2*x^3 -16*x^2+14*x+80)*exp(x)+2*x^3-16*x^2+14*x+80)*exp(log(exp(x)+2)+x-5))/(((x ^2-10*x+25)*exp(x)+2*x^2-20*x+50)*exp(log(exp(x)+2)+x-5)+(-2*x^2+20*x-50)* exp(x)-4*x^2+40*x-100),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (23) = 46\).
Time = 5.89 (sec) , antiderivative size = 104, normalized size of antiderivative = 4.16 \[ \int \frac {e^{-5+x} \left (2+e^x\right ) \left (80+14 x-16 x^2+2 x^3+e^x \left (80+14 x-16 x^2+2 x^3\right )\right )+\left (-92+40 x-4 x^2+e^x \left (-46+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (46-20 x+2 x^2+e^x \left (23-10 x+x^2\right )\right )\right ) \log \left (-2+e^{-5+x} \left (2+e^x\right )\right )}{-100+40 x-4 x^2+e^x \left (-50+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (50-20 x+2 x^2+e^x \left (25-10 x+x^2\right )\right )} \, dx=\frac {x^{2} \log \left (-2 \, e^{5} + e^{\left (2 \, x\right )} + 2 \, e^{x}\right ) - 5 \, x^{2} + 2 \, x \log \left (2 \, e^{5} - e^{\left (2 \, x\right )} - 2 \, e^{x}\right ) - 5 \, x \log \left (-2 \, e^{5} + e^{\left (2 \, x\right )} + 2 \, e^{x}\right ) + 25 \, x - 10 \, \log \left (2 \, e^{5} - e^{\left (2 \, x\right )} - 2 \, e^{x}\right ) + 2 \, \log \left (-2 \, e^{5} + e^{\left (2 \, x\right )} + 2 \, e^{x}\right ) - 10}{x - 5} \]
integrate(((((x^2-10*x+23)*exp(x)+2*x^2-20*x+46)*exp(log(exp(x)+2)+x-5)+(- 2*x^2+20*x-46)*exp(x)-4*x^2+40*x-92)*log(exp(log(exp(x)+2)+x-5)-2)+((2*x^3 -16*x^2+14*x+80)*exp(x)+2*x^3-16*x^2+14*x+80)*exp(log(exp(x)+2)+x-5))/(((x ^2-10*x+25)*exp(x)+2*x^2-20*x+50)*exp(log(exp(x)+2)+x-5)+(-2*x^2+20*x-50)* exp(x)-4*x^2+40*x-100),x, algorithm=\
(x^2*log(-2*e^5 + e^(2*x) + 2*e^x) - 5*x^2 + 2*x*log(2*e^5 - e^(2*x) - 2*e ^x) - 5*x*log(-2*e^5 + e^(2*x) + 2*e^x) + 25*x - 10*log(2*e^5 - e^(2*x) - 2*e^x) + 2*log(-2*e^5 + e^(2*x) + 2*e^x) - 10)/(x - 5)
Time = 13.19 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.44 \[ \int \frac {e^{-5+x} \left (2+e^x\right ) \left (80+14 x-16 x^2+2 x^3+e^x \left (80+14 x-16 x^2+2 x^3\right )\right )+\left (-92+40 x-4 x^2+e^x \left (-46+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (46-20 x+2 x^2+e^x \left (23-10 x+x^2\right )\right )\right ) \log \left (-2+e^{-5+x} \left (2+e^x\right )\right )}{-100+40 x-4 x^2+e^x \left (-50+20 x-2 x^2\right )+e^{-5+x} \left (2+e^x\right ) \left (50-20 x+2 x^2+e^x \left (25-10 x+x^2\right )\right )} \, dx=-3\,\ln \left ({\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^5+2\,{\mathrm {e}}^x\right )-\ln \left ({\mathrm {e}}^{-5}\,{\mathrm {e}}^x\,\left ({\mathrm {e}}^x+2\right )-2\right )\,\left (\frac {10\,x-2\,x^2}{x-5}+\frac {x^2-10\,x+23}{x-5}\right ) \]
int((log(exp(x + log(exp(x) + 2) - 5) - 2)*(exp(x)*(2*x^2 - 20*x + 46) - 4 0*x + 4*x^2 - exp(x + log(exp(x) + 2) - 5)*(exp(x)*(x^2 - 10*x + 23) - 20* x + 2*x^2 + 46) + 92) - exp(x + log(exp(x) + 2) - 5)*(14*x - 16*x^2 + 2*x^ 3 + exp(x)*(14*x - 16*x^2 + 2*x^3 + 80) + 80))/(exp(x)*(2*x^2 - 20*x + 50) - 40*x + 4*x^2 - exp(x + log(exp(x) + 2) - 5)*(exp(x)*(x^2 - 10*x + 25) - 20*x + 2*x^2 + 50) + 100),x)