\(\int \frac {e^{2+e^{\frac {e^x+2 x}{x}}+x+\frac {e^x+2 x}{x}} (-3+3 x) \log (2)}{x^2} \, dx\) [8553]

Optimal result

Integrand size = 40, antiderivative size = 21 \[ \int \frac {e^{2+e^{\frac {e^x+2 x}{x}}+x+\frac {e^x+2 x}{x}} (-3+3 x) \log (2)}{x^2} \, dx=3 e^{2+e^{\frac {e^x+2 x}{x}}} \log (2) \]

[Out]

3*ln(2)*x*exp(exp((exp(x)+2*x)/x)-ln(x)+2)

Rubi [F]

\[ \int \frac {e^{2+e^{\frac {e^x+2 x}{x}}+x+\frac {e^x+2 x}{x}} (-3+3 x) \log (2)}{x^2} \, dx=\int \frac {e^{2+e^{\frac {e^x+2 x}{x}}+x+\frac {e^x+2 x}{x}} (-3+3 x) \log (2)}{x^2} \, dx \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {e^{2+e^{\frac {e^x+2 x}{x}}+x+\frac {e^x+2 x}{x}} (-3+3 x) \log (2)}{x^2} \, dx=3 e^{2+e^{2+\frac {e^x}{x}}} \log (2) \]

[In]

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

[Out]

3*E^(2 + E^(2 + E^x/x))*Log[2]

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90

method	result	size
risch	\(3 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{x}+2 x}{x}}+2}\)	\(19\)
parallelrisch	\(3 \ln \left (2\right ) x \,{\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{x}+2 x}{x}}-\ln \left (x \right )+2}\)	\(24\)

[In]

int((-3+3*x)*ln(2)*exp(x)*exp((exp(x)+2*x)/x)*exp(exp((exp(x)+2*x)/x)-ln(x)+2)/x,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (23) = 46\).

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.48 \[ \int \frac {e^{2+e^{\frac {e^x+2 x}{x}}+x+\frac {e^x+2 x}{x}} (-3+3 x) \log (2)}{x^2} \, dx=3 \, x e^{\left (-x + \frac {x^{2} + x e^{\left (\frac {2 \, x + e^{x}}{x}\right )} - x \log \left (x\right ) + 4 \, x + e^{x}}{x} - \frac {2 \, x + e^{x}}{x}\right )} \log \left (2\right ) \]

[In]

integrate((-3+3*x)*log(2)*exp(x)*exp((exp(x)+2*x)/x)*exp(exp((exp(x)+2*x)/x)-log(x)+2)/x,x, algorithm="fricas"
)

[Out]

3*x*e^(-x + (x^2 + x*e^((2*x + e^x)/x) - x*log(x) + 4*x + e^x)/x - (2*x + e^x)/x)*log(2)

Sympy [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2+e^{\frac {e^x+2 x}{x}}+x+\frac {e^x+2 x}{x}} (-3+3 x) \log (2)}{x^2} \, dx=3 e^{e^{\frac {2 x + e^{x}}{x}} + 2} \log {\left (2 \right )} \]

[In]

integrate((-3+3*x)*ln(2)*exp(x)*exp((exp(x)+2*x)/x)*exp(exp((exp(x)+2*x)/x)-ln(x)+2)/x,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2+e^{\frac {e^x+2 x}{x}}+x+\frac {e^x+2 x}{x}} (-3+3 x) \log (2)}{x^2} \, dx=3 \, e^{\left (e^{\left (\frac {e^{x}}{x} + 2\right )} + 2\right )} \log \left (2\right ) \]

[In]

integrate((-3+3*x)*log(2)*exp(x)*exp((exp(x)+2*x)/x)*exp(exp((exp(x)+2*x)/x)-log(x)+2)/x,x, algorithm="maxima"
)

[Out]

3*e^(e^(e^x/x + 2) + 2)*log(2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \frac {e^{2+e^{\frac {e^x+2 x}{x}}+x+\frac {e^x+2 x}{x}} (-3+3 x) \log (2)}{x^2} \, dx=3 \, e^{\left (e^{\left (\frac {e^{x}}{x} + 2\right )} + 2\right )} \log \left (2\right ) \]

[In]

integrate((-3+3*x)*log(2)*exp(x)*exp((exp(x)+2*x)/x)*exp(exp((exp(x)+2*x)/x)-log(x)+2)/x,x, algorithm="giac")

[Out]

3*e^(e^(e^x/x + 2) + 2)*log(2)

Mupad [B] (verification not implemented)

Time = 13.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^{2+e^{\frac {e^x+2 x}{x}}+x+\frac {e^x+2 x}{x}} (-3+3 x) \log (2)}{x^2} \, dx=3\,{\mathrm {e}}^2\,{\mathrm {e}}^{{\mathrm {e}}^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x}}}\,\ln \left (2\right ) \]

[In]

int((exp((2*x + exp(x))/x)*exp(exp((2*x + exp(x))/x) - log(x) + 2)*exp(x)*log(2)*(3*x - 3))/x,x)

[Out]

3*exp(2)*exp(exp(2)*exp(exp(x)/x))*log(2)