\(\int -12 e^{4+12 e^x+x} \log (\frac {4}{3}) \, dx\) [1061]

Optimal result

Integrand size = 16, antiderivative size = 15 \[ \int -12 e^{4+12 e^x+x} \log \left (\frac {4}{3}\right ) \, dx=-e^{4+12 e^x} \log \left (\frac {4}{3}\right ) \]

[Out]

exp(-ln(x)+12*exp(x)+4)*ln(3/4)*x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {12, 2320, 2225} \[ \int -12 e^{4+12 e^x+x} \log \left (\frac {4}{3}\right ) \, dx=-e^{12 e^x+4} \log \left (\frac {4}{3}\right ) \]

[In]

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

[Out]

-(E^(4 + 12*E^x)*Log[4/3])

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}& = -\left (\left (12 \log \left (\frac {4}{3}\right )\right ) \int e^{4+12 e^x+x} \, dx\right ) \\ & = -\left (\left (12 \log \left (\frac {4}{3}\right )\right ) \text {Subst}\left (\int e^{4+12 x} \, dx,x,e^x\right )\right ) \\ & = -e^{4+12 e^x} \log \left (\frac {4}{3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int -12 e^{4+12 e^x+x} \log \left (\frac {4}{3}\right ) \, dx=-e^{4+12 e^x} \log \left (\frac {4}{3}\right ) \]

[In]

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

[Out]

-(E^(4 + 12*E^x)*Log[4/3])

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73

method	result	size
default	\(\ln \left (\frac {3}{4}\right ) {\mathrm e}^{4} {\mathrm e}^{12 \,{\mathrm e}^{x}}\)	\(11\)
parallelrisch	\({\mathrm e}^{-\ln \left (x \right )+12 \,{\mathrm e}^{x}+4} \ln \left (\frac {3}{4}\right ) x\)	\(16\)
norman	\(\left (\ln \left (3\right )-2 \ln \left (2\right )\right ) x \,{\mathrm e}^{-\ln \left (x \right )+12 \,{\mathrm e}^{x}+4}\)	\(21\)
risch	\(-2 \,{\mathrm e}^{4+12 \,{\mathrm e}^{x}} \ln \left (2\right )+{\mathrm e}^{4+12 \,{\mathrm e}^{x}} \ln \left (3\right )\)	\(23\)

[In]

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

[Out]

ln(3/4)*exp(4)*exp(exp(x))^12

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int -12 e^{4+12 e^x+x} \log \left (\frac {4}{3}\right ) \, dx=x e^{\left (12 \, e^{x} - \log \left (x\right ) + 4\right )} \log \left (\frac {3}{4}\right ) \]

[In]

integrate(12*x*log(3/4)*exp(x)*exp(-log(x)+12*exp(x)+4),x, algorithm="fricas")

[Out]

x*e^(12*e^x - log(x) + 4)*log(3/4)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int -12 e^{4+12 e^x+x} \log \left (\frac {4}{3}\right ) \, dx=\left (- 2 \log {\left (2 \right )} + \log {\left (3 \right )}\right ) e^{12 e^{x} + 4} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int -12 e^{4+12 e^x+x} \log \left (\frac {4}{3}\right ) \, dx=e^{\left (12 \, e^{x} + 4\right )} \log \left (\frac {3}{4}\right ) \]

[In]

integrate(12*x*log(3/4)*exp(x)*exp(-log(x)+12*exp(x)+4),x, algorithm="maxima")

[Out]

e^(12*e^x + 4)*log(3/4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.67 \[ \int -12 e^{4+12 e^x+x} \log \left (\frac {4}{3}\right ) \, dx=e^{\left (12 \, e^{x} + 4\right )} \log \left (\frac {3}{4}\right ) \]

[In]

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

[Out]

e^(12*e^x + 4)*log(3/4)

Mupad [B] (verification not implemented)

Time = 8.33 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int -12 e^{4+12 e^x+x} \log \left (\frac {4}{3}\right ) \, dx=-{\mathrm {e}}^4\,{\mathrm {e}}^{12\,{\mathrm {e}}^x}\,\left (2\,\ln \left (2\right )-\ln \left (3\right )\right ) \]

[In]

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

[Out]

-exp(4)*exp(12*exp(x))*(2*log(2) - log(3))