\(\int \frac {1+e^{e^x+x} (5 x-e^{2 x} x+x^2+e^{9+x} (-2 x-e^x x)+e^x (2 x+x^2))}{x} \, dx\) [9147]

Optimal result

Integrand size = 56, antiderivative size = 26 \[ \int \frac {1+e^{e^x+x} \left (5 x-e^{2 x} x+x^2+e^{9+x} \left (-2 x-e^x x\right )+e^x \left (2 x+x^2\right )\right )}{x} \, dx=e^{e^x+x} \left (4-e^x-e^{9+x}+x\right )+\log (x) \]

[Out]

exp(exp(x)+x)*(x-exp(x)-exp(x+9)+4)+ln(x)

Rubi [F]

\[ \int \frac {1+e^{e^x+x} \left (5 x-e^{2 x} x+x^2+e^{9+x} \left (-2 x-e^x x\right )+e^x \left (2 x+x^2\right )\right )}{x} \, dx=\int \frac {1+e^{e^x+x} \left (5 x-e^{2 x} x+x^2+e^{9+x} \left (-2 x-e^x x\right )+e^x \left (2 x+x^2\right )\right )}{x} \, dx \]

[In]

Int[(1 + E^(E^x + x)*(5*x - E^(2*x)*x + x^2 + E^(9 + x)*(-2*x - E^x*x) + E^x*(2*x + x^2)))/x,x]

[Out]

5*E^E^x - 2*E^E^x*(1 - E^9) + 2*E^(E^x + x)*(1 - E^9) - 2*E^E^x*(1 + E^9) + 2*E^(E^x + x)*(1 + E^9) - E^(E^x +
2*x)*(1 + E^9) + Log[x] + Defer[Int][E^(E^x + x)*x, x] + Defer[Int][E^(E^x + 2*x)*x, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-e^{e^x+3 x} \left (1+e^9\right )-e^{e^x+2 x} \left (-2+2 e^9-x\right )+\frac {1}{x}+e^{e^x+x} (5+x)\right ) \, dx \\ & = \log (x)+\left (-1-e^9\right ) \int e^{e^x+3 x} \, dx-\int e^{e^x+2 x} \left (-2+2 e^9-x\right ) \, dx+\int e^{e^x+x} (5+x) \, dx \\ & = \log (x)+\left (-1-e^9\right ) \text {Subst}\left (\int e^x x^2 \, dx,x,e^x\right )+\int \left (5 e^{e^x+x}+e^{e^x+x} x\right ) \, dx-\int \left (-2 e^{e^x+2 x} \left (1-e^9\right )-e^{e^x+2 x} x\right ) \, dx \\ & = -e^{e^x+2 x} \left (1+e^9\right )+\log (x)+5 \int e^{e^x+x} \, dx+\left (2 \left (1-e^9\right )\right ) \int e^{e^x+2 x} \, dx+\left (2 \left (1+e^9\right )\right ) \text {Subst}\left (\int e^x x \, dx,x,e^x\right )+\int e^{e^x+x} x \, dx+\int e^{e^x+2 x} x \, dx \\ & = 2 e^{e^x+x} \left (1+e^9\right )-e^{e^x+2 x} \left (1+e^9\right )+\log (x)+5 \text {Subst}\left (\int e^x \, dx,x,e^x\right )+\left (2 \left (1-e^9\right )\right ) \text {Subst}\left (\int e^x x \, dx,x,e^x\right )-\left (2 \left (1+e^9\right )\right ) \text {Subst}\left (\int e^x \, dx,x,e^x\right )+\int e^{e^x+x} x \, dx+\int e^{e^x+2 x} x \, dx \\ & = 5 e^{e^x}+2 e^{e^x+x} \left (1-e^9\right )-2 e^{e^x} \left (1+e^9\right )+2 e^{e^x+x} \left (1+e^9\right )-e^{e^x+2 x} \left (1+e^9\right )+\log (x)-\left (2 \left (1-e^9\right )\right ) \text {Subst}\left (\int e^x \, dx,x,e^x\right )+\int e^{e^x+x} x \, dx+\int e^{e^x+2 x} x \, dx \\ & = 5 e^{e^x}-2 e^{e^x} \left (1-e^9\right )+2 e^{e^x+x} \left (1-e^9\right )-2 e^{e^x} \left (1+e^9\right )+2 e^{e^x+x} \left (1+e^9\right )-e^{e^x+2 x} \left (1+e^9\right )+\log (x)+\int e^{e^x+x} x \, dx+\int e^{e^x+2 x} x \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 3.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {1+e^{e^x+x} \left (5 x-e^{2 x} x+x^2+e^{9+x} \left (-2 x-e^x x\right )+e^x \left (2 x+x^2\right )\right )}{x} \, dx=e^{e^x} \left (e^{2 x} \left (-1-e^9\right )+e^x (4+x)\right )+\log (x) \]

[In]

Integrate[(1 + E^(E^x + x)*(5*x - E^(2*x)*x + x^2 + E^(9 + x)*(-2*x - E^x*x) + E^x*(2*x + x^2)))/x,x]

[Out]

E^E^x*(E^(2*x)*(-1 - E^9) + E^x*(4 + x)) + Log[x]

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88

method	result	size
risch	\({\mathrm e}^{{\mathrm e}^{x}+x} \left (x -{\mathrm e}^{x}-{\mathrm e}^{x +9}+4\right )+\ln \left (x \right )\)	\(23\)
norman	\({\mathrm e}^{{\mathrm e}^{x}+x} x +\left (-{\mathrm e}^{9}-1\right ) {\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}+x}+4 \,{\mathrm e}^{{\mathrm e}^{x}+x}+\ln \left (x \right )\)	\(32\)
parallelrisch	\(-{\mathrm e}^{{\mathrm e}^{x}+x} {\mathrm e}^{x}-{\mathrm e}^{{\mathrm e}^{x}+x} {\mathrm e}^{x +9}+{\mathrm e}^{{\mathrm e}^{x}+x} x +\ln \left (x \right )+4 \,{\mathrm e}^{{\mathrm e}^{x}+x}\)	\(38\)

[In]

int((((-exp(x)*x-2*x)*exp(x+9)-x*exp(x)^2+(x^2+2*x)*exp(x)+x^2+5*x)*exp(exp(x)+x)+1)/x,x,method=_RETURNVERBOSE
)

[Out]

exp(exp(x)+x)*(x-exp(x)-exp(x+9)+4)+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.54 \[ \int \frac {1+e^{e^x+x} \left (5 x-e^{2 x} x+x^2+e^{9+x} \left (-2 x-e^x x\right )+e^x \left (2 x+x^2\right )\right )}{x} \, dx={\left ({\left ({\left (x + 4\right )} e^{9} - {\left (e^{9} + 1\right )} e^{\left (x + 9\right )}\right )} e^{\left ({\left (x e^{9} + e^{\left (x + 9\right )}\right )} e^{\left (-9\right )}\right )} + e^{9} \log \left (x\right )\right )} e^{\left (-9\right )} \]

[In]

integrate((((-exp(x)*x-2*x)*exp(x+9)-x*exp(x)^2+(x^2+2*x)*exp(x)+x^2+5*x)*exp(exp(x)+x)+1)/x,x, algorithm="fri
cas")

[Out]

(((x + 4)*e^9 - (e^9 + 1)*e^(x + 9))*e^((x*e^9 + e^(x + 9))*e^(-9)) + e^9*log(x))*e^(-9)

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {1+e^{e^x+x} \left (5 x-e^{2 x} x+x^2+e^{9+x} \left (-2 x-e^x x\right )+e^x \left (2 x+x^2\right )\right )}{x} \, dx=\left (x - e^{9} e^{x} - e^{x} + 4\right ) e^{x + e^{x}} + \log {\left (x \right )} \]

[In]

integrate((((-exp(x)*x-2*x)*exp(x+9)-x*exp(x)**2+(x**2+2*x)*exp(x)+x**2+5*x)*exp(exp(x)+x)+1)/x,x)

[Out]

(x - exp(9)*exp(x) - exp(x) + 4)*exp(x + exp(x)) + log(x)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {1+e^{e^x+x} \left (5 x-e^{2 x} x+x^2+e^{9+x} \left (-2 x-e^x x\right )+e^x \left (2 x+x^2\right )\right )}{x} \, dx=-{\left ({\left (e^{9} + 1\right )} e^{\left (2 \, x\right )} - {\left (x + 4\right )} e^{x} + 5\right )} e^{\left (e^{x}\right )} + 5 \, e^{\left (e^{x}\right )} + \log \left (x\right ) \]

[In]

integrate((((-exp(x)*x-2*x)*exp(x+9)-x*exp(x)^2+(x^2+2*x)*exp(x)+x^2+5*x)*exp(exp(x)+x)+1)/x,x, algorithm="max
ima")

[Out]

-((e^9 + 1)*e^(2*x) - (x + 4)*e^x + 5)*e^(e^x) + 5*e^(e^x) + log(x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (22) = 44\).

Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92 \[ \int \frac {1+e^{e^x+x} \left (5 x-e^{2 x} x+x^2+e^{9+x} \left (-2 x-e^x x\right )+e^x \left (2 x+x^2\right )\right )}{x} \, dx={\left (x e^{\left (4 \, x + e^{x}\right )} + e^{\left (3 \, x\right )} \log \left (x\right ) - e^{\left (5 \, x + e^{x} + 9\right )} - e^{\left (5 \, x + e^{x}\right )} + 4 \, e^{\left (4 \, x + e^{x}\right )}\right )} e^{\left (-3 \, x\right )} \]

[In]

integrate((((-exp(x)*x-2*x)*exp(x+9)-x*exp(x)^2+(x^2+2*x)*exp(x)+x^2+5*x)*exp(exp(x)+x)+1)/x,x, algorithm="gia
c")

[Out]

(x*e^(4*x + e^x) + e^(3*x)*log(x) - e^(5*x + e^x + 9) - e^(5*x + e^x) + 4*e^(4*x + e^x))*e^(-3*x)

Mupad [B] (verification not implemented)

Time = 12.74 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {1+e^{e^x+x} \left (5 x-e^{2 x} x+x^2+e^{9+x} \left (-2 x-e^x x\right )+e^x \left (2 x+x^2\right )\right )}{x} \, dx=4\,{\mathrm {e}}^{x+{\mathrm {e}}^x}+\ln \left (x\right )-{\mathrm {e}}^{2\,x+{\mathrm {e}}^x}\,\left ({\mathrm {e}}^9+1\right )+x\,{\mathrm {e}}^{x+{\mathrm {e}}^x} \]

[In]

int((exp(x + exp(x))*(5*x - x*exp(2*x) + exp(x)*(2*x + x^2) - exp(x + 9)*(2*x + x*exp(x)) + x^2) + 1)/x,x)

[Out]

4*exp(x + exp(x)) + log(x) - exp(2*x + exp(x))*(exp(9) + 1) + x*exp(x + exp(x))