\(\int \frac {e^{\frac {x}{3 e^2}}}{(12 e^2+3 e^{2+\frac {x}{3 e^2}}) \log (4+e^{\frac {x}{3 e^2}})} \, dx\) [6860]

Optimal result

Integrand size = 48, antiderivative size = 18 \[ \int \frac {e^{\frac {x}{3 e^2}}}{\left (12 e^2+3 e^{2+\frac {x}{3 e^2}}\right ) \log \left (4+e^{\frac {x}{3 e^2}}\right )} \, dx=e^2+\log \left (\log \left (4+e^{\frac {x}{3 e^2}}\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.104, Rules used = {2320, 12, 2437, 2339, 29} \[ \int \frac {e^{\frac {x}{3 e^2}}}{\left (12 e^2+3 e^{2+\frac {x}{3 e^2}}\right ) \log \left (4+e^{\frac {x}{3 e^2}}\right )} \, dx=\log \left (\log \left (e^{\frac {x}{3 e^2}}+4\right )\right ) \]

[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 2320

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 2339

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 2437

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {e^{\frac {x}{3 e^2}}}{\left (12 e^2+3 e^{2+\frac {x}{3 e^2}}\right ) \log \left (4+e^{\frac {x}{3 e^2}}\right )} \, dx=\log \left (\log \left (4+e^{\frac {x}{3 e^2}}\right )\right ) \]

[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))]]

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61

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

[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*exp(-2)*x)+4))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.17 \[ \int \frac {e^{\frac {x}{3 e^2}}}{\left (12 e^2+3 e^{2+\frac {x}{3 e^2}}\right ) \log \left (4+e^{\frac {x}{3 e^2}}\right )} \, dx=\log \left (\log \left ({\left (4 \, e^{2} + e^{\left (\frac {1}{3} \, {\left (x + 6 \, e^{2}\right )} e^{\left (-2\right )}\right )}\right )} e^{\left (-2\right )}\right )\right ) \]

[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)))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.67 \[ \int \frac {e^{\frac {x}{3 e^2}}}{\left (12 e^2+3 e^{2+\frac {x}{3 e^2}}\right ) \log \left (4+e^{\frac {x}{3 e^2}}\right )} \, dx=\log {\left (\log {\left (e^{\frac {x}{3 e^{2}}} + 4 \right )} \right )} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {e^{\frac {x}{3 e^2}}}{\left (12 e^2+3 e^{2+\frac {x}{3 e^2}}\right ) \log \left (4+e^{\frac {x}{3 e^2}}\right )} \, dx=\log \left (\log \left (e^{\left (\frac {1}{3} \, x e^{\left (-2\right )}\right )} + 4\right )\right ) \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {e^{\frac {x}{3 e^2}}}{\left (12 e^2+3 e^{2+\frac {x}{3 e^2}}\right ) \log \left (4+e^{\frac {x}{3 e^2}}\right )} \, dx=\log \left (\log \left (e^{\left (\frac {1}{3} \, x e^{\left (-2\right )}\right )} + 4\right )\right ) \]

[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(log(e^(1/3*x*e^(-2)) + 4))

Mupad [B] (verification not implemented)

Time = 12.14 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.56 \[ \int \frac {e^{\frac {x}{3 e^2}}}{\left (12 e^2+3 e^{2+\frac {x}{3 e^2}}\right ) \log \left (4+e^{\frac {x}{3 e^2}}\right )} \, dx=\ln \left (\ln \left ({\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{-2}}{3}}+4\right )\right ) \]

[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))