\(\int \frac {3}{4} e^{e^x-x} (16-16 x+16 e^x x+\frac {4}{3} e^{-e^x+x} (20+8 x)) \, dx\) [6483]

Optimal result

Integrand size = 42, antiderivative size = 22 \[ \int \frac {3}{4} e^{e^x-x} \left (16-16 x+16 e^x x+\frac {4}{3} e^{-e^x+x} (20+8 x)\right ) \, dx=-5+4 \left (3+x \left (5+3 e^{e^x-x}+x\right )\right ) \]

[Out]

7+4*x*(5+x+4/exp(-ln(3/4*exp(exp(x)))+x))

Rubi [F]

\[ \int \frac {3}{4} e^{e^x-x} \left (16-16 x+16 e^x x+\frac {4}{3} e^{-e^x+x} (20+8 x)\right ) \, dx=\int \frac {3}{4} e^{e^x-x} \left (16-16 x+16 e^x x+\frac {4}{3} e^{-e^x+x} (20+8 x)\right ) \, dx \]

[In]

Int[(3*E^(E^x - x)*(16 - 16*x + 16*E^x*x + (4*E^(-E^x + x)*(20 + 8*x))/3))/4,x]

[Out]

-12*E^(E^x - x) + (5 + 2*x)^2 + 12*ExpIntegralEi[E^x] + 12*Defer[Int][E^E^x*x, x] - 12*Defer[Int][E^(E^x - x)*
x, x]

Rubi steps \begin{align*} \text {integral}& = \frac {3}{4} \int e^{e^x-x} \left (16-16 x+16 e^x x+\frac {4}{3} e^{-e^x+x} (20+8 x)\right ) \, dx \\ & = \frac {3}{4} \int \left (16 e^{e^x-x}+16 e^{e^x} x-16 e^{e^x-x} x+\frac {16}{3} (5+2 x)\right ) \, dx \\ & = (5+2 x)^2+12 \int e^{e^x-x} \, dx+12 \int e^{e^x} x \, dx-12 \int e^{e^x-x} x \, dx \\ & = (5+2 x)^2+12 \int e^{e^x} x \, dx-12 \int e^{e^x-x} x \, dx+12 \text {Subst}\left (\int \frac {e^x}{x^2} \, dx,x,e^x\right ) \\ & = -12 e^{e^x-x}+(5+2 x)^2+12 \int e^{e^x} x \, dx-12 \int e^{e^x-x} x \, dx+12 \text {Subst}\left (\int \frac {e^x}{x} \, dx,x,e^x\right ) \\ & = -12 e^{e^x-x}+(5+2 x)^2+12 \text {Ei}\left (e^x\right )+12 \int e^{e^x} x \, dx-12 \int e^{e^x-x} x \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {3}{4} e^{e^x-x} \left (16-16 x+16 e^x x+\frac {4}{3} e^{-e^x+x} (20+8 x)\right ) \, dx=4 x \left (5+3 e^{e^x-x}+x\right ) \]

[In]

Integrate[(3*E^(E^x - x)*(16 - 16*x + 16*E^x*x + (4*E^(-E^x + x)*(20 + 8*x))/3))/4,x]

[Out]

4*x*(5 + 3*E^(E^x - x) + x)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73

method	result	size
norman	\(\left (12 x \,{\mathrm e}^{2 \,{\mathrm e}^{x}}+20 x \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}+4 x^{2} {\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x}}\right ) {\mathrm e}^{-x} {\mathrm e}^{-{\mathrm e}^{x}}\)	\(38\)
default	\(\frac {3 \left (16 x +4 x^{2} {\mathrm e}^{-\ln \left (\frac {3 \,{\mathrm e}^{{\mathrm e}^{x}}}{4}\right )+x}+20 \,{\mathrm e}^{-\ln \left (\frac {3 \,{\mathrm e}^{{\mathrm e}^{x}}}{4}\right )+x} x \right ) {\mathrm e}^{{\mathrm e}^{x}-x}}{4}\)	\(49\)
parallelrisch	\(\frac {3 \left (8 x^{2} {\mathrm e}^{-\ln \left (\frac {3 \,{\mathrm e}^{{\mathrm e}^{x}}}{4}\right )+x}+40 \,{\mathrm e}^{-\ln \left (\frac {3 \,{\mathrm e}^{{\mathrm e}^{x}}}{4}\right )+x} x +32 x \right ) {\mathrm e}^{{\mathrm e}^{x}-x}}{8}\)	\(50\)

[In]

int(((8*x+20)*exp(-ln(3/4*exp(exp(x)))+x)+16*exp(x)*x-16*x+16)/exp(-ln(3/4*exp(exp(x)))+x),x,method=_RETURNVER
BOSE)

[Out]

(12*x*exp(exp(x))^2+20*x*exp(x)*exp(exp(x))+4*x^2*exp(x)*exp(exp(x)))/exp(x)/exp(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {3}{4} e^{e^x-x} \left (16-16 x+16 e^x x+\frac {4}{3} e^{-e^x+x} (20+8 x)\right ) \, dx=4 \, x^{2} + 16 \, x e^{\left (-x + e^{x} + \log \left (\frac {3}{4}\right )\right )} + 20 \, x \]

[In]

integrate(((8*x+20)*exp(-log(3/4*exp(exp(x)))+x)+16*exp(x)*x-16*x+16)/exp(-log(3/4*exp(exp(x)))+x),x, algorith
m="fricas")

[Out]

4*x^2 + 16*x*e^(-x + e^x + log(3/4)) + 20*x

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {3}{4} e^{e^x-x} \left (16-16 x+16 e^x x+\frac {4}{3} e^{-e^x+x} (20+8 x)\right ) \, dx=4 x^{2} + 20 x + 12 x e^{- x} e^{e^{x}} \]

[In]

integrate(((8*x+20)*exp(-ln(3/4*exp(exp(x)))+x)+16*exp(x)*x-16*x+16)/exp(-ln(3/4*exp(exp(x)))+x),x)

[Out]

4*x**2 + 20*x + 12*x*exp(-x)*exp(exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {3}{4} e^{e^x-x} \left (16-16 x+16 e^x x+\frac {4}{3} e^{-e^x+x} (20+8 x)\right ) \, dx=4 \, x^{2} + 12 \, x e^{\left (-x + e^{x}\right )} + 20 \, x \]

[In]

integrate(((8*x+20)*exp(-log(3/4*exp(exp(x)))+x)+16*exp(x)*x-16*x+16)/exp(-log(3/4*exp(exp(x)))+x),x, algorith
m="maxima")

[Out]

4*x^2 + 12*x*e^(-x + e^x) + 20*x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {3}{4} e^{e^x-x} \left (16-16 x+16 e^x x+\frac {4}{3} e^{-e^x+x} (20+8 x)\right ) \, dx=4 \, x^{2} + 12 \, x e^{\left (-x + e^{x}\right )} + 20 \, x \]

[In]

integrate(((8*x+20)*exp(-log(3/4*exp(exp(x)))+x)+16*exp(x)*x-16*x+16)/exp(-log(3/4*exp(exp(x)))+x),x, algorith
m="giac")

[Out]

4*x^2 + 12*x*e^(-x + e^x) + 20*x

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {3}{4} e^{e^x-x} \left (16-16 x+16 e^x x+\frac {4}{3} e^{-e^x+x} (20+8 x)\right ) \, dx=4\,x\,\left (x+3\,{\mathrm {e}}^{{\mathrm {e}}^x-x}+5\right ) \]

[In]

int(exp(log((3*exp(exp(x)))/4) - x)*(exp(x - log((3*exp(exp(x)))/4))*(8*x + 20) - 16*x + 16*x*exp(x) + 16),x)

[Out]

4*x*(x + 3*exp(exp(x) - x) + 5)