3.40.30 \(\int \frac {16 e^{e^x+x}}{18 e^{e^x}+9 \log (3)} \, dx\)

Optimal. Leaf size=15 \[ \frac {8}{9} \log \left (2 e^{e^x}+\log (3)\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {12, 2282, 2246, 31} \begin {gather*} \frac {8}{9} \log \left (2 e^{e^x}+\log (3)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16*E^(E^x + x))/(18*E^E^x + 9*Log[3]),x]

[Out]

(8*Log[2*E^E^x + Log[3]])/9

Rule 12

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2246

Int[((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)*((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.),
x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int[(a + b*x)^p, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b,
c, d, e, n, p}, 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 \frac {e^{e^x+x}}{18 e^{e^x}+9 \log (3)} \, dx\\ &=16 \operatorname {Subst}\left (\int \frac {e^x}{9 \left (2 e^x+\log (3)\right )} \, dx,x,e^x\right )\\ &=\frac {16}{9} \operatorname {Subst}\left (\int \frac {e^x}{2 e^x+\log (3)} \, dx,x,e^x\right )\\ &=\frac {16}{9} \operatorname {Subst}\left (\int \frac {1}{2 x+\log (3)} \, dx,x,e^{e^x}\right )\\ &=\frac {8}{9} \log \left (2 e^{e^x}+\log (3)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 15, normalized size = 1.00 \begin {gather*} \frac {8}{9} \log \left (2 e^{e^x}+\log (3)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*E^(E^x + x))/(18*E^E^x + 9*Log[3]),x]

[Out]

(8*Log[2*E^E^x + Log[3]])/9

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fricas [A]  time = 0.63, size = 20, normalized size = 1.33 \begin {gather*} -\frac {8}{9} \, x + \frac {8}{9} \, \log \left (e^{x} \log \relax (3) + 2 \, e^{\left (x + e^{x}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*exp(x)*exp(exp(x))/(18*exp(exp(x))+9*log(3)),x, algorithm="fricas")

[Out]

-8/9*x + 8/9*log(e^x*log(3) + 2*e^(x + e^x))

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giac [A]  time = 0.15, size = 11, normalized size = 0.73 \begin {gather*} \frac {8}{9} \, \log \left (2 \, e^{\left (e^{x}\right )} + \log \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*exp(x)*exp(exp(x))/(18*exp(exp(x))+9*log(3)),x, algorithm="giac")

[Out]

8/9*log(2*e^(e^x) + log(3))

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maple [A]  time = 0.03, size = 12, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {8 \ln \left (2 \,{\mathrm e}^{{\mathrm e}^{x}}+\ln \relax (3)\right )}{9}\)	\(12\)
default	\(\frac {8 \ln \left (2 \,{\mathrm e}^{{\mathrm e}^{x}}+\ln \relax (3)\right )}{9}\)	\(12\)
risch	\(\frac {8 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}+\frac {\ln \relax (3)}{2}\right )}{9}\)	\(12\)
norman	\(\frac {8 \ln \left (18 \,{\mathrm e}^{{\mathrm e}^{x}}+9 \ln \relax (3)\right )}{9}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(16*exp(x)*exp(exp(x))/(18*exp(exp(x))+9*ln(3)),x,method=_RETURNVERBOSE)

[Out]

8/9*ln(2*exp(exp(x))+ln(3))

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maxima [A]  time = 0.34, size = 11, normalized size = 0.73 \begin {gather*} \frac {8}{9} \, \log \left (2 \, e^{\left (e^{x}\right )} + \log \relax (3)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*exp(x)*exp(exp(x))/(18*exp(exp(x))+9*log(3)),x, algorithm="maxima")

[Out]

8/9*log(2*e^(e^x) + log(3))

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mupad [B]  time = 0.09, size = 11, normalized size = 0.73 \begin {gather*} \frac {8\,\ln \left (2\,{\mathrm {e}}^{{\mathrm {e}}^x}+\ln \relax (3)\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*exp(exp(x))*exp(x))/(18*exp(exp(x)) + 9*log(3)),x)

[Out]

(8*log(2*exp(exp(x)) + log(3)))/9

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sympy [A]  time = 0.12, size = 14, normalized size = 0.93 \begin {gather*} \frac {8 \log {\left (e^{e^{x}} + \frac {\log {\relax (3 )}}{2} \right )}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16*exp(x)*exp(exp(x))/(18*exp(exp(x))+9*ln(3)),x)

[Out]

8*log(exp(exp(x)) + log(3)/2)/9

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