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

Optimal. Leaf size=20 \[ -e^{e^x} \left (16+\frac {x}{4}\right )+10 x^2 \]

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Rubi [A]  time = 0.02, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {12, 2288} \begin {gather*} 10 x^2-\frac {1}{4} e^{e^x} (x+64) \end {gather*}

Antiderivative was successfully verified.

[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 2288

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.90 \begin {gather*} 10 x^2-\frac {1}{4} e^{e^x} (64+x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

10*x^2 - (E^E^x*(64 + x))/4

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fricas [A]  time = 0.59, size = 14, normalized size = 0.70 \begin {gather*} 10 \, x^{2} - \frac {1}{4} \, {\left (x + 64\right )} e^{\left (e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.53, size = 27, normalized size = 1.35 \begin {gather*} 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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.03, size = 17, normalized size = 0.85

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

Verification of antiderivative is not currently implemented for this CAS.

[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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 10 \, x^{2} - \frac {1}{4} \, x e^{\left (e^{x}\right )} - \frac {1}{4} \, {\rm Ei}\left (e^{x}\right ) - 16 \, e^{\left (e^{x}\right )} + \frac {1}{4} \, \int e^{\left (e^{x}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.06, size = 17, normalized size = 0.85 \begin {gather*} 10\,x^2-\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^x}}{4}-16\,{\mathrm {e}}^{{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.14, size = 15, normalized size = 0.75 \begin {gather*} 10 x^{2} + \frac {\left (- x - 64\right ) e^{e^{x}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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