\(\int \frac {1}{4} (e^{e^x} (-1+e^x (-64-x))+80 x) \, dx\) [1315]

Optimal result

Integrand size = 25, antiderivative size = 20 \[ \int \frac {1}{4} \left (e^{e^x} \left (-1+e^x (-64-x)\right )+80 x\right ) \, dx=-e^{e^x} \left (16+\frac {x}{4}\right )+10 x^2 \]

[Out]

10*x^2-exp(exp(x))*(1/4*x+16)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {12, 2326} \[ \int \frac {1}{4} \left (e^{e^x} \left (-1+e^x (-64-x)\right )+80 x\right ) \, dx=10 x^2-\frac {1}{4} e^{e^x} (x+64) \]

[In]

Int[(E^E^x*(-1 + E^x*(-64 - x)) + 80*x)/4,x]

[Out]

10*x^2 - (E^E^x*(64 + x))/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \left (e^{e^x} \left (-1+e^x (-64-x)\right )+80 x\right ) \, dx \\ & = 10 x^2+\frac {1}{4} \int e^{e^x} \left (-1+e^x (-64-x)\right ) \, dx \\ & = 10 x^2-\frac {1}{4} e^{e^x} (64+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {1}{4} \left (e^{e^x} \left (-1+e^x (-64-x)\right )+80 x\right ) \, dx=10 x^2-\frac {1}{4} e^{e^x} (64+x) \]

[In]

Integrate[(E^E^x*(-1 + E^x*(-64 - x)) + 80*x)/4,x]

[Out]

10*x^2 - (E^E^x*(64 + x))/4

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {\left (-x -64\right ) {\mathrm e}^{{\mathrm e}^{x}}}{4}+10 x^{2}\)	\(17\)
default	\(-\frac {x \,{\mathrm e}^{{\mathrm e}^{x}}}{4}-16 \,{\mathrm e}^{{\mathrm e}^{x}}+10 x^{2}\)	\(18\)
norman	\(-\frac {x \,{\mathrm e}^{{\mathrm e}^{x}}}{4}-16 \,{\mathrm e}^{{\mathrm e}^{x}}+10 x^{2}\)	\(18\)
parallelrisch	\(-\frac {x \,{\mathrm e}^{{\mathrm e}^{x}}}{4}-16 \,{\mathrm e}^{{\mathrm e}^{x}}+10 x^{2}\)	\(18\)

[In]

int(1/4*((-x-64)*exp(x)-1)*exp(exp(x))+20*x,x,method=_RETURNVERBOSE)

[Out]

1/4*(-x-64)*exp(exp(x))+10*x^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {1}{4} \left (e^{e^x} \left (-1+e^x (-64-x)\right )+80 x\right ) \, dx=10 \, x^{2} - \frac {1}{4} \, {\left (x + 64\right )} e^{\left (e^{x}\right )} \]

[In]

integrate(1/4*((-x-64)*exp(x)-1)*exp(exp(x))+20*x,x, algorithm="fricas")

[Out]

10*x^2 - 1/4*(x + 64)*e^(e^x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {1}{4} \left (e^{e^x} \left (-1+e^x (-64-x)\right )+80 x\right ) \, dx=10 x^{2} + \frac {\left (- x - 64\right ) e^{e^{x}}}{4} \]

[In]

integrate(1/4*((-x-64)*exp(x)-1)*exp(exp(x))+20*x,x)

[Out]

10*x**2 + (-x - 64)*exp(exp(x))/4

Maxima [F]

\[ \int \frac {1}{4} \left (e^{e^x} \left (-1+e^x (-64-x)\right )+80 x\right ) \, dx=\int { -\frac {1}{4} \, {\left ({\left (x + 64\right )} e^{x} + 1\right )} e^{\left (e^{x}\right )} + 20 \, x \,d x } \]

[In]

integrate(1/4*((-x-64)*exp(x)-1)*exp(exp(x))+20*x,x, algorithm="maxima")

[Out]

10*x^2 - 1/4*x*e^(e^x) - 1/4*Ei(e^x) - 16*e^(e^x) + 1/4*integrate(e^(e^x), x)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {1}{4} \left (e^{e^x} \left (-1+e^x (-64-x)\right )+80 x\right ) \, dx=10 \, x^{2} - \frac {1}{4} \, {\left (x e^{\left (x + e^{x}\right )} + 64 \, e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \]

[In]

integrate(1/4*((-x-64)*exp(x)-1)*exp(exp(x))+20*x,x, algorithm="giac")

[Out]

10*x^2 - 1/4*(x*e^(x + e^x) + 64*e^(x + e^x))*e^(-x)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {1}{4} \left (e^{e^x} \left (-1+e^x (-64-x)\right )+80 x\right ) \, dx=10\,x^2-\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^x}}{4}-16\,{\mathrm {e}}^{{\mathrm {e}}^x} \]

[In]

int(20*x - (exp(exp(x))*(exp(x)*(x + 64) + 1))/4,x)

[Out]

10*x^2 - (x*exp(exp(x)))/4 - 16*exp(exp(x))