\(\int 4\ 3^{-1-e^4+x} e^{3^{-1-e^4+x} (4+2\ 3^{1+e^4-x})} \log (3) \, dx\) [609]

Optimal result

Integrand size = 41, antiderivative size = 16 \[ \int 4\ 3^{-1-e^4+x} e^{3^{-1-e^4+x} \left (4+2\ 3^{1+e^4-x}\right )} \log (3) \, dx=e^{2+4\ 3^{-1-e^4+x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {12, 2320, 2225} \[ \int 4\ 3^{-1-e^4+x} e^{3^{-1-e^4+x} \left (4+2\ 3^{1+e^4-x}\right )} \log (3) \, dx=e^{4\ 3^{x-e^4-1}+2} \]

[In]

Int[4*3^(-1 - E^4 + x)*E^(3^(-1 - E^4 + x)*(4 + 2*3^(1 + E^4 - x)))*Log[3],x]

[Out]

E^(2 + 4*3^(-1 - E^4 + x))

Rule 12

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

Rule 2225

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 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]]]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int 4\ 3^{-1-e^4+x} e^{3^{-1-e^4+x} \left (4+2\ 3^{1+e^4-x}\right )} \log (3) \, dx=e^{2+4\ 3^{-1-e^4+x}} \]

[In]

Integrate[4*3^(-1 - E^4 + x)*E^(3^(-1 - E^4 + x)*(4 + 2*3^(1 + E^4 - x)))*Log[3],x]

[Out]

E^(2 + 4*3^(-1 - E^4 + x))

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.62

method	result	size
risch	\({\mathrm e}^{\frac {2 \left (3^{{\mathrm e}^{4}-x +1}+2\right ) 3^{x -{\mathrm e}^{4}}}{3}}\)	\(26\)
parallelrisch	\({\mathrm e}^{2 \left ({\mathrm e}^{\left ({\mathrm e}^{4}-x +1\right ) \ln \left (3\right )}+2\right ) {\mathrm e}^{-\left ({\mathrm e}^{4}-x +1\right ) \ln \left (3\right )}}\)	\(30\)
derivativedivides	\({\mathrm e}^{\left (2 \,{\mathrm e}^{\left ({\mathrm e}^{4}-x +1\right ) \ln \left (3\right )}+4\right ) {\mathrm e}^{-\left ({\mathrm e}^{4}-x +1\right ) \ln \left (3\right )}}\)	\(31\)
default	\({\mathrm e}^{\left (2 \,{\mathrm e}^{\left ({\mathrm e}^{4}-x +1\right ) \ln \left (3\right )}+4\right ) {\mathrm e}^{-\left ({\mathrm e}^{4}-x +1\right ) \ln \left (3\right )}}\)	\(31\)
norman	\({\mathrm e}^{\left (2 \,{\mathrm e}^{\left ({\mathrm e}^{4}-x +1\right ) \ln \left (3\right )}+4\right ) {\mathrm e}^{-\left ({\mathrm e}^{4}-x +1\right ) \ln \left (3\right )}}\)	\(31\)

[In]

int(4*ln(3)*exp((2*exp((exp(4)-x+1)*ln(3))+4)/exp((exp(4)-x+1)*ln(3)))/exp((exp(4)-x+1)*ln(3)),x,method=_RETUR
NVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int 4\ 3^{-1-e^4+x} e^{3^{-1-e^4+x} \left (4+2\ 3^{1+e^4-x}\right )} \log (3) \, dx=e^{\left (\frac {2 \, {\left (3^{-x + e^{4} + 1} + 2\right )}}{3^{-x + e^{4} + 1}}\right )} \]

[In]

integrate(4*log(3)*exp((2*exp((exp(4)-x+1)*log(3))+4)/exp((exp(4)-x+1)*log(3)))/exp((exp(4)-x+1)*log(3)),x, al
gorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.69 \[ \int 4\ 3^{-1-e^4+x} e^{3^{-1-e^4+x} \left (4+2\ 3^{1+e^4-x}\right )} \log (3) \, dx=e^{\left (2 e^{\left (- x + 1 + e^{4}\right ) \log {\left (3 \right )}} + 4\right ) e^{- \left (- x + 1 + e^{4}\right ) \log {\left (3 \right )}}} \]

[In]

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

[Out]

exp((2*exp((-x + 1 + exp(4))*log(3)) + 4)*exp(-(-x + 1 + exp(4))*log(3)))

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.50 \[ \int 4\ 3^{-1-e^4+x} e^{3^{-1-e^4+x} \left (4+2\ 3^{1+e^4-x}\right )} \log (3) \, dx=3^{\frac {4 \cdot 3^{x - e^{4} - 1}}{\log \left (3\right )} + \frac {2}{\log \left (3\right )}} \]

[In]

integrate(4*log(3)*exp((2*exp((exp(4)-x+1)*log(3))+4)/exp((exp(4)-x+1)*log(3)))/exp((exp(4)-x+1)*log(3)),x, al
gorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int 4\ 3^{-1-e^4+x} e^{3^{-1-e^4+x} \left (4+2\ 3^{1+e^4-x}\right )} \log (3) \, dx=e^{\left (\frac {4 \cdot 3^{x}}{3 \cdot 3^{e^{4}}} + 2\right )} \]

[In]

integrate(4*log(3)*exp((2*exp((exp(4)-x+1)*log(3))+4)/exp((exp(4)-x+1)*log(3)))/exp((exp(4)-x+1)*log(3)),x, al
gorithm="giac")

[Out]

e^(4/3*3^x/3^e^4 + 2)

Mupad [B] (verification not implemented)

Time = 9.46 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int 4\ 3^{-1-e^4+x} e^{3^{-1-e^4+x} \left (4+2\ 3^{1+e^4-x}\right )} \log (3) \, dx={\mathrm {e}}^2\,{\mathrm {e}}^{4\,3^{x-{\mathrm {e}}^4-1}} \]

[In]

int(4*exp(exp(-log(3)*(exp(4) - x + 1))*(2*exp(log(3)*(exp(4) - x + 1)) + 4))*exp(-log(3)*(exp(4) - x + 1))*lo
g(3),x)

[Out]

exp(2)*exp(4*3^(x - exp(4) - 1))