3.69.60 \(\int \frac {e^{\frac {x}{3 e^2}}}{(12 e^2+3 e^{2+\frac {x}{3 e^2}}) \log (4+e^{\frac {x}{3 e^2}})} \, dx\)

Optimal. Leaf size=18 \[ e^2+\log \left (\log \left (4+e^{\frac {x}{3 e^2}}\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 14, normalized size of antiderivative = 0.78, number of steps used = 5, number of rules used = 5, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {2282, 12, 2390, 2302, 29} \begin {gather*} \log \left (\log \left (e^{\frac {x}{3 e^2}}+4\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(x/(3*E^2))/((12*E^2 + 3*E^(2 + x/(3*E^2)))*Log[4 + E^(x/(3*E^2))]),x]

[Out]

Log[Log[4 + E^(x/(3*E^2))]]

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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]]]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\left (3 e^2\right ) \operatorname {Subst}\left (\int \frac {1}{3 e^2 (4+x) \log (4+x)} \, dx,x,e^{\frac {x}{3 e^2}}\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{(4+x) \log (4+x)} \, dx,x,e^{\frac {x}{3 e^2}}\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,4+e^{\frac {x}{3 e^2}}\right )\\ &=\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (4+e^{\frac {x}{3 e^2}}\right )\right )\\ &=\log \left (\log \left (4+e^{\frac {x}{3 e^2}}\right )\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 14, normalized size = 0.78 \begin {gather*} \log \left (\log \left (4+e^{\frac {x}{3 e^2}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(x/(3*E^2))/((12*E^2 + 3*E^(2 + x/(3*E^2)))*Log[4 + E^(x/(3*E^2))]),x]

[Out]

Log[Log[4 + E^(x/(3*E^2))]]

________________________________________________________________________________________

fricas [A]  time = 0.59, size = 21, normalized size = 1.17 \begin {gather*} \log \left (\log \left ({\left (4 \, e^{2} + e^{\left (\frac {1}{3} \, {\left (x + 6 \, e^{2}\right )} e^{\left (-2\right )}\right )}\right )} e^{\left (-2\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(1/3*x/exp(2))/(3*exp(2)*exp(1/3*x/exp(2))+12*exp(2))/log(exp(1/3*x/exp(2))+4),x, algorithm="fric
as")

[Out]

log(log((4*e^2 + e^(1/3*(x + 6*e^2)*e^(-2)))*e^(-2)))

________________________________________________________________________________________

giac [A]  time = 0.32, size = 11, normalized size = 0.61 \begin {gather*} \log \left ({\left | \log \left (e^{\left (\frac {1}{3} \, x e^{\left (-2\right )}\right )} + 4\right ) \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(1/3*x/exp(2))/(3*exp(2)*exp(1/3*x/exp(2))+12*exp(2))/log(exp(1/3*x/exp(2))+4),x, algorithm="giac
")

[Out]

log(abs(log(e^(1/3*x*e^(-2)) + 4)))

________________________________________________________________________________________

maple [A]  time = 0.05, size = 11, normalized size = 0.61

method	result	size
risch	\(\ln \left (\ln \left ({\mathrm e}^{\frac {x \,{\mathrm e}^{-2}}{3}}+4\right )\right )\)	\(11\)
derivativedivides	\(\ln \left (\ln \left ({\mathrm e}^{\frac {x \,{\mathrm e}^{-2}}{3}}+4\right )\right )\)	\(13\)
default	\(\ln \left (\ln \left ({\mathrm e}^{\frac {x \,{\mathrm e}^{-2}}{3}}+4\right )\right )\)	\(13\)
norman	\(\ln \left (\ln \left ({\mathrm e}^{\frac {x \,{\mathrm e}^{-2}}{3}}+4\right )\right )\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(1/3*x/exp(2))/(3*exp(2)*exp(1/3*x/exp(2))+12*exp(2))/ln(exp(1/3*x/exp(2))+4),x,method=_RETURNVERBOSE)

[Out]

ln(ln(exp(1/3*x*exp(-2))+4))

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 10, normalized size = 0.56 \begin {gather*} \log \left (\log \left (e^{\left (\frac {1}{3} \, x e^{\left (-2\right )}\right )} + 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(1/3*x/exp(2))/(3*exp(2)*exp(1/3*x/exp(2))+12*exp(2))/log(exp(1/3*x/exp(2))+4),x, algorithm="maxi
ma")

[Out]

log(log(e^(1/3*x*e^(-2)) + 4))

________________________________________________________________________________________

mupad [B]  time = 4.23, size = 10, normalized size = 0.56 \begin {gather*} \ln \left (\ln \left ({\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-2}}{3}}+4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp((x*exp(-2))/3)/(log(exp((x*exp(-2))/3) + 4)*(12*exp(2) + 3*exp(2)*exp((x*exp(-2))/3))),x)

[Out]

log(log(exp((x*exp(-2))/3) + 4))

________________________________________________________________________________________

sympy [A]  time = 0.19, size = 12, normalized size = 0.67 \begin {gather*} \log {\left (\log {\left (e^{\frac {x}{3 e^{2}}} + 4 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(1/3*x/exp(2))/(3*exp(2)*exp(1/3*x/exp(2))+12*exp(2))/ln(exp(1/3*x/exp(2))+4),x)

[Out]

log(log(exp(x*exp(-2)/3) + 4))

________________________________________________________________________________________