3.8.55 \(\int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)+(e^x+x) \log (x)}{e^x+x}} (e^{2 x} (2+2 x) \log (3)+e^x (2 x+4 x^2) \log (3)+(-2 x+2 x^2+2 x^3) \log (3))}{3 e^{2 x} x+6 e^x x^2+3 x^3} \, dx\) [755]

3.8.55.1 Optimal result

Integrand size = 104, antiderivative size = 21 \[ \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)+\left (e^x+x\right ) \log (x)}{e^x+x}} \left (e^{2 x} (2+2 x) \log (3)+e^x \left (2 x+4 x^2\right ) \log (3)+\left (-2 x+2 x^2+2 x^3\right ) \log (3)\right )}{3 e^{2 x} x+6 e^x x^2+3 x^3} \, dx=4+\frac {2}{3} e^{5+x+\frac {1}{e^x+x}} x \log (3) \]

output

4+2/3*exp(4/(4*exp(x)+4*x)+x+ln(x)+5)*ln(3)
 

3.8.55.2 Mathematica [A] (verified)

Time = 5.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)+\left (e^x+x\right ) \log (x)}{e^x+x}} \left (e^{2 x} (2+2 x) \log (3)+e^x \left (2 x+4 x^2\right ) \log (3)+\left (-2 x+2 x^2+2 x^3\right ) \log (3)\right )}{3 e^{2 x} x+6 e^x x^2+3 x^3} \, dx=\frac {2}{3} e^{5+x+\frac {1}{e^x+x}} x \log (3) \]

input

Integrate[(E^((1 + 5*x + x^2 + E^x*(5 + x) + (E^x + x)*Log[x])/(E^x + x))* 
(E^(2*x)*(2 + 2*x)*Log[3] + E^x*(2*x + 4*x^2)*Log[3] + (-2*x + 2*x^2 + 2*x 
^3)*Log[3]))/(3*E^(2*x)*x + 6*E^x*x^2 + 3*x^3),x]
 

output

(2*E^(5 + x + (E^x + x)^(-1))*x*Log[3])/3
 

3.8.55.3 Rubi [F]

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.

\(\displaystyle \int \frac {\left (e^x \left (4 x^2+2 x\right ) \log (3)+\left (2 x^3+2 x^2-2 x\right ) \log (3)+e^{2 x} (2 x+2) \log (3)\right ) \exp \left (\frac {x^2+5 x+e^x (x+5)+\left (x+e^x\right ) \log (x)+1}{x+e^x}\right )}{3 x^3+6 e^x x^2+3 e^{2 x} x} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {2 \left (x^3+2 e^x x^2+x^2+e^x x+e^{2 x} x-x+e^{2 x}\right ) \log (3) \exp \left (\frac {x^2+5 x+e^x (x+5)+\left (x+e^x\right ) \log (x)+1}{x+e^x}\right )}{3 x \left (x+e^x\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{3} \log (3) \int \frac {e^{\frac {x^2+5 x+e^x (x+5)+1}{x+e^x}} \left (x^3+2 e^x x^2+x^2+e^x x+e^{2 x} x-x+e^{2 x}\right )}{\left (x+e^x\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {2}{3} \log (3) \int \left (e^{\frac {x^2+5 x+e^x (x+5)+1}{x+e^x}} x-\frac {e^{\frac {x^2+5 x+e^x (x+5)+1}{x+e^x}} x}{x+e^x}+\frac {e^{\frac {x^2+5 x+e^x (x+5)+1}{x+e^x}} (x-1) x}{\left (x+e^x\right )^2}+e^{\frac {x^2+5 x+e^x (x+5)+1}{x+e^x}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2}{3} \log (3) \left (\int e^{\frac {x^2+5 x+e^x (x+5)+1}{x+e^x}}dx+\int e^{\frac {x^2+5 x+e^x (x+5)+1}{x+e^x}} xdx-\int \frac {e^{\frac {x^2+5 x+e^x (x+5)+1}{x+e^x}} x}{\left (x+e^x\right )^2}dx+\int \frac {e^{\frac {x^2+5 x+e^x (x+5)+1}{x+e^x}} x^2}{\left (x+e^x\right )^2}dx-\int \frac {e^{\frac {x^2+5 x+e^x (x+5)+1}{x+e^x}} x}{x+e^x}dx\right )\)

input

Int[(E^((1 + 5*x + x^2 + E^x*(5 + x) + (E^x + x)*Log[x])/(E^x + x))*(E^(2* 
x)*(2 + 2*x)*Log[3] + E^x*(2*x + 4*x^2)*Log[3] + (-2*x + 2*x^2 + 2*x^3)*Lo 
g[3]))/(3*E^(2*x)*x + 6*E^x*x^2 + 3*x^3),x]
 

output

$Aborted
 

3.8.55.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.8.55.4 Maple [A] (verified)

Time = 4.77 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62

method	result	size
parallelrisch	\(\frac {2 \ln \left (3\right ) {\mathrm e}^{\frac {\left ({\mathrm e}^{x}+x \right ) \ln \left (x \right )+\left (5+x \right ) {\mathrm e}^{x}+x^{2}+5 x +1}{{\mathrm e}^{x}+x}}}{3}\)	\(34\)
risch	\(\frac {2 \ln \left (3\right ) {\mathrm e}^{\frac {{\mathrm e}^{x} \ln \left (x \right )+x \ln \left (x \right )+{\mathrm e}^{x} x +x^{2}+5 \,{\mathrm e}^{x}+5 x +1}{{\mathrm e}^{x}+x}}}{3}\)	\(38\)

input

int(((2+2*x)*ln(3)*exp(x)^2+(4*x^2+2*x)*ln(3)*exp(x)+(2*x^3+2*x^2-2*x)*ln( 
3))*exp(((exp(x)+x)*ln(x)+(5+x)*exp(x)+x^2+5*x+1)/(exp(x)+x))/(3*x*exp(x)^ 
2+6*exp(x)*x^2+3*x^3),x,method=_RETURNVERBOSE)
 

output

2/3*ln(3)*exp(((exp(x)+x)*ln(x)+(5+x)*exp(x)+x^2+5*x+1)/(exp(x)+x))
 

3.8.55.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)+\left (e^x+x\right ) \log (x)}{e^x+x}} \left (e^{2 x} (2+2 x) \log (3)+e^x \left (2 x+4 x^2\right ) \log (3)+\left (-2 x+2 x^2+2 x^3\right ) \log (3)\right )}{3 e^{2 x} x+6 e^x x^2+3 x^3} \, dx=\frac {2}{3} \, e^{\left (\frac {x^{2} + {\left (x + 5\right )} e^{x} + {\left (x + e^{x}\right )} \log \left (x\right ) + 5 \, x + 1}{x + e^{x}}\right )} \log \left (3\right ) \]

input

integrate(((2+2*x)*log(3)*exp(x)^2+(4*x^2+2*x)*log(3)*exp(x)+(2*x^3+2*x^2- 
2*x)*log(3))*exp(((exp(x)+x)*log(x)+(5+x)*exp(x)+x^2+5*x+1)/(exp(x)+x))/(3 
*x*exp(x)^2+6*exp(x)*x^2+3*x^3),x, algorithm=\
 

output

2/3*e^((x^2 + (x + 5)*e^x + (x + e^x)*log(x) + 5*x + 1)/(x + e^x))*log(3)
 

3.8.55.6 Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.71 \[ \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)+\left (e^x+x\right ) \log (x)}{e^x+x}} \left (e^{2 x} (2+2 x) \log (3)+e^x \left (2 x+4 x^2\right ) \log (3)+\left (-2 x+2 x^2+2 x^3\right ) \log (3)\right )}{3 e^{2 x} x+6 e^x x^2+3 x^3} \, dx=\frac {2 e^{\frac {x^{2} + 5 x + \left (x + 5\right ) e^{x} + \left (x + e^{x}\right ) \log {\left (x \right )} + 1}{x + e^{x}}} \log {\left (3 \right )}}{3} \]

input

integrate(((2+2*x)*ln(3)*exp(x)**2+(4*x**2+2*x)*ln(3)*exp(x)+(2*x**3+2*x** 
2-2*x)*ln(3))*exp(((exp(x)+x)*ln(x)+(5+x)*exp(x)+x**2+5*x+1)/(exp(x)+x))/( 
3*x*exp(x)**2+6*exp(x)*x**2+3*x**3),x)
 

output

2*exp((x**2 + 5*x + (x + 5)*exp(x) + (x + exp(x))*log(x) + 1)/(x + exp(x)) 
)*log(3)/3
 

3.8.55.7 Maxima [F]

\[ \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)+\left (e^x+x\right ) \log (x)}{e^x+x}} \left (e^{2 x} (2+2 x) \log (3)+e^x \left (2 x+4 x^2\right ) \log (3)+\left (-2 x+2 x^2+2 x^3\right ) \log (3)\right )}{3 e^{2 x} x+6 e^x x^2+3 x^3} \, dx=\int { \frac {2 \, {\left ({\left (x + 1\right )} e^{\left (2 \, x\right )} \log \left (3\right ) + {\left (2 \, x^{2} + x\right )} e^{x} \log \left (3\right ) + {\left (x^{3} + x^{2} - x\right )} \log \left (3\right )\right )} e^{\left (\frac {x^{2} + {\left (x + 5\right )} e^{x} + {\left (x + e^{x}\right )} \log \left (x\right ) + 5 \, x + 1}{x + e^{x}}\right )}}{3 \, {\left (x^{3} + 2 \, x^{2} e^{x} + x e^{\left (2 \, x\right )}\right )}} \,d x } \]

input

integrate(((2+2*x)*log(3)*exp(x)^2+(4*x^2+2*x)*log(3)*exp(x)+(2*x^3+2*x^2- 
2*x)*log(3))*exp(((exp(x)+x)*log(x)+(5+x)*exp(x)+x^2+5*x+1)/(exp(x)+x))/(3 
*x*exp(x)^2+6*exp(x)*x^2+3*x^3),x, algorithm=\
 

output

2/3*integrate(((x + 1)*e^(2*x)*log(3) + (2*x^2 + x)*e^x*log(3) + (x^3 + x^ 
2 - x)*log(3))*e^((x^2 + (x + 5)*e^x + (x + e^x)*log(x) + 5*x + 1)/(x + e^ 
x))/(x^3 + 2*x^2*e^x + x*e^(2*x)), x)
 

3.8.55.8 Giac [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)+\left (e^x+x\right ) \log (x)}{e^x+x}} \left (e^{2 x} (2+2 x) \log (3)+e^x \left (2 x+4 x^2\right ) \log (3)+\left (-2 x+2 x^2+2 x^3\right ) \log (3)\right )}{3 e^{2 x} x+6 e^x x^2+3 x^3} \, dx=\frac {2}{3} \, x e^{\left (\frac {x^{2} + x e^{x} + 5 \, x + 5 \, e^{x} + 1}{x + e^{x}}\right )} \log \left (3\right ) \]

input

integrate(((2+2*x)*log(3)*exp(x)^2+(4*x^2+2*x)*log(3)*exp(x)+(2*x^3+2*x^2- 
2*x)*log(3))*exp(((exp(x)+x)*log(x)+(5+x)*exp(x)+x^2+5*x+1)/(exp(x)+x))/(3 
*x*exp(x)^2+6*exp(x)*x^2+3*x^3),x, algorithm=\
 

output

2/3*x*e^((x^2 + x*e^x + 5*x + 5*e^x + 1)/(x + e^x))*log(3)
 

3.8.55.9 Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)+\left (e^x+x\right ) \log (x)}{e^x+x}} \left (e^{2 x} (2+2 x) \log (3)+e^x \left (2 x+4 x^2\right ) \log (3)+\left (-2 x+2 x^2+2 x^3\right ) \log (3)\right )}{3 e^{2 x} x+6 e^x x^2+3 x^3} \, dx=\int \frac {{\mathrm {e}}^{\frac {5\,x+{\mathrm {e}}^x\,\left (x+5\right )+x^2+\ln \left (x\right )\,\left (x+{\mathrm {e}}^x\right )+1}{x+{\mathrm {e}}^x}}\,\left (\ln \left (3\right )\,\left (2\,x^3+2\,x^2-2\,x\right )+{\mathrm {e}}^x\,\ln \left (3\right )\,\left (4\,x^2+2\,x\right )+{\mathrm {e}}^{2\,x}\,\ln \left (3\right )\,\left (2\,x+2\right )\right )}{3\,x\,{\mathrm {e}}^{2\,x}+6\,x^2\,{\mathrm {e}}^x+3\,x^3} \,d x \]

input

int((exp((5*x + exp(x)*(x + 5) + x^2 + log(x)*(x + exp(x)) + 1)/(x + exp(x 
)))*(log(3)*(2*x^2 - 2*x + 2*x^3) + exp(x)*log(3)*(2*x + 4*x^2) + exp(2*x) 
*log(3)*(2*x + 2)))/(3*x*exp(2*x) + 6*x^2*exp(x) + 3*x^3),x)
 

output

int((exp((5*x + exp(x)*(x + 5) + x^2 + log(x)*(x + exp(x)) + 1)/(x + exp(x 
)))*(log(3)*(2*x^2 - 2*x + 2*x^3) + exp(x)*log(3)*(2*x + 4*x^2) + exp(2*x) 
*log(3)*(2*x + 2)))/(3*x*exp(2*x) + 6*x^2*exp(x) + 3*x^3), x)