3.103.76 \(\int (16 e^{2 e^x+x}+e^{e^x+x} (64-16 e-8 \log (9))) \, dx\)

Optimal. Leaf size=21 \[ 2 \left (2 \left (4-e+e^{e^x}\right )-\log (9)\right )^2 \]

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Rubi [A]  time = 0.02, antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 5, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2282, 2194} \begin {gather*} 8 e^{2 e^x}+8 e^{e^x} (8-2 e-\log (9)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[16*E^(2*E^x + x) + E^(E^x + x)*(64 - 16*E - 8*Log[9]),x]

[Out]

8*E^(2*E^x) + 8*E^E^x*(8 - 2*E - Log[9])

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=16 \int e^{2 e^x+x} \, dx+(8 (8-2 e-\log (9))) \int e^{e^x+x} \, dx\\ &=16 \operatorname {Subst}\left (\int e^{2 x} \, dx,x,e^x\right )+(8 (8-2 e-\log (9))) \operatorname {Subst}\left (\int e^x \, dx,x,e^x\right )\\ &=8 e^{2 e^x}+8 e^{e^x} (8-2 e-\log (9))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 1.00 \begin {gather*} 8 e^{e^x} \left (8-2 e+e^{e^x}-\log (9)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[16*E^(2*E^x + x) + E^(E^x + x)*(64 - 16*E - 8*Log[9]),x]

[Out]

8*E^E^x*(8 - 2*E + E^E^x - Log[9])

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fricas [B]  time = 0.77, size = 33, normalized size = 1.57 \begin {gather*} -8 \, {\left (2 \, {\left (e + \log \relax (3) - 4\right )} e^{\left (2 \, x + e^{x}\right )} - e^{\left (2 \, x + 2 \, e^{x}\right )}\right )} e^{\left (-2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*exp(x)*exp(exp(x))^2+(-16*log(3)-16*exp(1)+64)*exp(x)*exp(exp(x)),x, algorithm="fricas")

[Out]

-8*(2*(e + log(3) - 4)*e^(2*x + e^x) - e^(2*x + 2*e^x))*e^(-2*x)

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giac [A]  time = 0.18, size = 19, normalized size = 0.90 \begin {gather*} -16 \, {\left (e + \log \relax (3) - 4\right )} e^{\left (e^{x}\right )} + 8 \, e^{\left (2 \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*exp(x)*exp(exp(x))^2+(-16*log(3)-16*exp(1)+64)*exp(x)*exp(exp(x)),x, algorithm="giac")

[Out]

-16*(e + log(3) - 4)*e^(e^x) + 8*e^(2*e^x)

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maple [A]  time = 0.05, size = 23, normalized size = 1.10

method	result	size
norman	\(\left (-16 \ln \relax (3)-16 \,{\mathrm e}+64\right ) {\mathrm e}^{{\mathrm e}^{x}}+8 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}\)	\(23\)
derivativedivides	\(-16 \,{\mathrm e} \,{\mathrm e}^{{\mathrm e}^{x}}-16 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \relax (3)+64 \,{\mathrm e}^{{\mathrm e}^{x}}+8 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}\)	\(28\)
default	\(-16 \,{\mathrm e} \,{\mathrm e}^{{\mathrm e}^{x}}-16 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \relax (3)+64 \,{\mathrm e}^{{\mathrm e}^{x}}+8 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}\)	\(28\)
risch	\(-16 \,{\mathrm e} \,{\mathrm e}^{{\mathrm e}^{x}}-16 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \relax (3)+64 \,{\mathrm e}^{{\mathrm e}^{x}}+8 \,{\mathrm e}^{2 \,{\mathrm e}^{x}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(16*exp(x)*exp(exp(x))^2+(-16*ln(3)-16*exp(1)+64)*exp(x)*exp(exp(x)),x,method=_RETURNVERBOSE)

[Out]

(-16*ln(3)-16*exp(1)+64)*exp(exp(x))+8*exp(exp(x))^2

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maxima [A]  time = 0.35, size = 19, normalized size = 0.90 \begin {gather*} -16 \, {\left (e + \log \relax (3) - 4\right )} e^{\left (e^{x}\right )} + 8 \, e^{\left (2 \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*exp(x)*exp(exp(x))^2+(-16*log(3)-16*exp(1)+64)*exp(x)*exp(exp(x)),x, algorithm="maxima")

[Out]

-16*(e + log(3) - 4)*e^(e^x) + 8*e^(2*e^x)

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mupad [B]  time = 0.18, size = 18, normalized size = 0.86 \begin {gather*} 8\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^{{\mathrm {e}}^x}-2\,\mathrm {e}-2\,\ln \relax (3)+8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(16*exp(2*exp(x))*exp(x) - exp(exp(x))*exp(x)*(16*exp(1) + 16*log(3) - 64),x)

[Out]

8*exp(exp(x))*(exp(exp(x)) - 2*exp(1) - 2*log(3) + 8)

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sympy [A]  time = 0.28, size = 24, normalized size = 1.14 \begin {gather*} 8 e^{2 e^{x}} + \left (- 16 e - 16 \log {\relax (3 )} + 64\right ) e^{e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*exp(x)*exp(exp(x))**2+(-16*ln(3)-16*exp(1)+64)*exp(x)*exp(exp(x)),x)

[Out]

8*exp(2*exp(x)) + (-16*E - 16*log(3) + 64)*exp(exp(x))

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