\(\int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)+(e^x+x) \log (x)}{e^x+x}} (e^{2 x} (2+2 x) \log (3)+e^x (2 x+4 x^2) \log (3)+(-2 x+2 x^2+2 x^3) \log (3))}{3 e^{2 x} x+6 e^x x^2+3 x^3} \, dx\) [6755]

Optimal result

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

[Out]

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

Rubi [F]

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

[In]

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)*Log[3]))/(3*E^(2*x)*x + 6*E^x*x^2 + 3*x^3),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} \left (e^{2 x} (1+x)+e^x x (1+2 x)+x \left (-1+x+x^2\right )\right ) \log (3)}{3 \left (e^x+x\right )^2} \, dx \\ & = \frac {1}{3} (2 \log (3)) \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} \left (e^{2 x} (1+x)+e^x x (1+2 x)+x \left (-1+x+x^2\right )\right )}{\left (e^x+x\right )^2} \, dx \\ & = \frac {1}{3} (2 \log (3)) \int \left (e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}}+e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} x+\frac {e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} (-1+x) x}{\left (e^x+x\right )^2}-\frac {e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} x}{e^x+x}\right ) \, dx \\ & = \frac {1}{3} (2 \log (3)) \int e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} \, dx+\frac {1}{3} (2 \log (3)) \int e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} x \, dx+\frac {1}{3} (2 \log (3)) \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} (-1+x) x}{\left (e^x+x\right )^2} \, dx-\frac {1}{3} (2 \log (3)) \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} x}{e^x+x} \, dx \\ & = \frac {1}{3} (2 \log (3)) \int e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} \, dx+\frac {1}{3} (2 \log (3)) \int e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} x \, dx-\frac {1}{3} (2 \log (3)) \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} x}{e^x+x} \, dx+\frac {1}{3} (2 \log (3)) \int \left (-\frac {e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} x}{\left (e^x+x\right )^2}+\frac {e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} x^2}{\left (e^x+x\right )^2}\right ) \, dx \\ & = \frac {1}{3} (2 \log (3)) \int e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} \, dx+\frac {1}{3} (2 \log (3)) \int e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} x \, dx-\frac {1}{3} (2 \log (3)) \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} x}{\left (e^x+x\right )^2} \, dx+\frac {1}{3} (2 \log (3)) \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} x^2}{\left (e^x+x\right )^2} \, dx-\frac {1}{3} (2 \log (3)) \int \frac {e^{\frac {1+5 x+x^2+e^x (5+x)}{e^x+x}} x}{e^x+x} \, dx \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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]

[Out]

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

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	\(\frac {2 \ln \left (3\right ) {\mathrm e}^{\frac {{\mathrm e}^{x} \ln \left (x \right )+x \ln \left (x \right )+{\mathrm e}^{x} x +x^{2}+5 \,{\mathrm e}^{x}+5 x +1}{{\mathrm e}^{x}+x}}}{3}\)	\(38\)

[In]

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)

[Out]

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

Fricas [A] (verification not implemented)

none

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

[In]

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="fricas")

[Out]

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

Sympy [A] (verification not implemented)

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

[In]

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)

[Out]

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

Maxima [F]

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

[In]

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="maxima")

[Out]

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)

Giac [A] (verification not implemented)

none

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

[In]

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="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

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)

[Out]

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)