\(\int \frac {1}{5} e^{-3-e^{e^{-4+x}}} (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4))) \, dx\) [7801]

Optimal result

Integrand size = 64, antiderivative size = 31 \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=2 \left (2+e^{-3-e^{e^{-4+x}}} \left (-\frac {3}{5}+e^{e^{1+x}}\right ) \log (4)\right ) \]

[Out]

4*(exp(exp(1+x))-3/5)/exp(exp(exp(x-4))+3)*ln(2)+4

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {12, 2320, 2326} \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=-\frac {2}{5} e^{-e^{e^{x-4}}-e^{x-4}-3} \left (3 e^{e^{x-4}}-5 e^{\left (\frac {1}{e^4}+e\right ) e^x}\right ) \log (4) \]

[In]

Int[(E^(-3 - E^E^(-4 + x))*(6*E^(-4 + E^(-4 + x) + x)*Log[4] + E^E^(1 + x)*(10*E^(1 + x)*Log[4] - 10*E^(-4 + E
^(-4 + x) + x)*Log[4])))/5,x]

[Out]

(-2*E^(-3 - E^E^(-4 + x) - E^(-4 + x))*(3*E^E^(-4 + x) - 5*E^(E^x*(E^(-4) + E)))*Log[4])/5

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx \\ & = \frac {1}{5} \text {Subst}\left (\int 2 e^{-7-e^{\frac {x}{e^4}}} \left (3 e^{\frac {x}{e^4}}+5 e^{5+e x}-5 e^{\frac {x}{e^4}+e x}\right ) \log (4) \, dx,x,e^x\right ) \\ & = \frac {1}{5} (2 \log (4)) \text {Subst}\left (\int e^{-7-e^{\frac {x}{e^4}}} \left (3 e^{\frac {x}{e^4}}+5 e^{5+e x}-5 e^{\frac {x}{e^4}+e x}\right ) \, dx,x,e^x\right ) \\ & = -\frac {2}{5} e^{-3-e^{e^{-4+x}}-e^{-4+x}} \left (3 e^{e^{-4+x}}-5 e^{e^x \left (\frac {1}{e^4}+e\right )}\right ) \log (4) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=\frac {2}{5} e^{-3-e^{e^{-4+x}}} \left (-3+5 e^{e^{1+x}}\right ) \log (4) \]

[In]

Integrate[(E^(-3 - E^E^(-4 + x))*(6*E^(-4 + E^(-4 + x) + x)*Log[4] + E^E^(1 + x)*(10*E^(1 + x)*Log[4] - 10*E^(
-4 + E^(-4 + x) + x)*Log[4])))/5,x]

[Out]

(2*E^(-3 - E^E^(-4 + x))*(-3 + 5*E^E^(1 + x))*Log[4])/5

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77

method	result	size
risch	\(\frac {4 \ln \left (2\right ) \left (5 \,{\mathrm e}^{{\mathrm e}^{1+x}}-3\right ) {\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{x -4}}-3}}{5}\)	\(24\)
parallelrisch	\(\frac {\left (20 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{1+x}}-12 \ln \left (2\right )\right ) {\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{x -4}}-3}}{5}\)	\(27\)
norman	\(\left (-\frac {12 \,{\mathrm e}^{-5} {\mathrm e}^{5} \ln \left (2\right )}{5}+4 \,{\mathrm e}^{-5} {\mathrm e}^{5} \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{1+x}}\right ) {\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{x -4}}-3}\)	\(45\)

[In]

int(1/5*((-20*ln(2)*exp(x-4)*exp(exp(x-4))+20*ln(2)*exp(1+x))*exp(exp(1+x))+12*ln(2)*exp(x-4)*exp(exp(x-4)))/e
xp(exp(exp(x-4))+3),x,method=_RETURNVERBOSE)

[Out]

4/5*ln(2)*(5*exp(exp(1+x))-3)*exp(-exp(exp(x-4))-3)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=\frac {4}{5} \, {\left (5 \, e^{\left (e^{\left (x + 1\right )}\right )} \log \left (2\right ) - 3 \, \log \left (2\right )\right )} e^{\left (-{\left (e^{\left ({\left ({\left (x - 4\right )} e^{5} + e^{\left (x + 1\right )}\right )} e^{\left (-5\right )} + 5\right )} + 3 \, e^{\left (x + 1\right )}\right )} e^{\left (-x - 1\right )}\right )} \]

[In]

integrate(1/5*((-20*log(2)*exp(x-4)*exp(exp(x-4))+20*log(2)*exp(1+x))*exp(exp(1+x))+12*log(2)*exp(x-4)*exp(exp
(x-4)))/exp(exp(exp(x-4))+3),x, algorithm="fricas")

[Out]

4/5*(5*e^(e^(x + 1))*log(2) - 3*log(2))*e^(-(e^(((x - 4)*e^5 + e^(x + 1))*e^(-5) + 5) + 3*e^(x + 1))*e^(-x - 1
))

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=\text {Timed out} \]

[In]

integrate(1/5*((-20*ln(2)*exp(x-4)*exp(exp(x-4))+20*ln(2)*exp(1+x))*exp(exp(1+x))+12*ln(2)*exp(x-4)*exp(exp(x-
4)))/exp(exp(exp(x-4))+3),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=4 \, e^{\left (e^{\left (x + 1\right )} - e^{\left (e^{\left (x - 4\right )}\right )} - 3\right )} \log \left (2\right ) - \frac {12}{5} \, e^{\left (-e^{\left (e^{\left (x - 4\right )}\right )} - 3\right )} \log \left (2\right ) \]

[In]

integrate(1/5*((-20*log(2)*exp(x-4)*exp(exp(x-4))+20*log(2)*exp(1+x))*exp(exp(1+x))+12*log(2)*exp(x-4)*exp(exp
(x-4)))/exp(exp(exp(x-4))+3),x, algorithm="maxima")

[Out]

4*e^(e^(x + 1) - e^(e^(x - 4)) - 3)*log(2) - 12/5*e^(-e^(e^(x - 4)) - 3)*log(2)

Giac [F]

\[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=\int { -\frac {4}{5} \, {\left (5 \, {\left (e^{\left (x + e^{\left (x - 4\right )} - 4\right )} \log \left (2\right ) - e^{\left (x + 1\right )} \log \left (2\right )\right )} e^{\left (e^{\left (x + 1\right )}\right )} - 3 \, e^{\left (x + e^{\left (x - 4\right )} - 4\right )} \log \left (2\right )\right )} e^{\left (-e^{\left (e^{\left (x - 4\right )}\right )} - 3\right )} \,d x } \]

[In]

integrate(1/5*((-20*log(2)*exp(x-4)*exp(exp(x-4))+20*log(2)*exp(1+x))*exp(exp(1+x))+12*log(2)*exp(x-4)*exp(exp
(x-4)))/exp(exp(exp(x-4))+3),x, algorithm="giac")

[Out]

integrate(-4/5*(5*(e^(x + e^(x - 4) - 4)*log(2) - e^(x + 1)*log(2))*e^(e^(x + 1)) - 3*e^(x + e^(x - 4) - 4)*lo
g(2))*e^(-e^(e^(x - 4)) - 3), x)

Mupad [B] (verification not implemented)

Time = 13.83 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \frac {1}{5} e^{-3-e^{e^{-4+x}}} \left (6 e^{-4+e^{-4+x}+x} \log (4)+e^{e^{1+x}} \left (10 e^{1+x} \log (4)-10 e^{-4+e^{-4+x}+x} \log (4)\right )\right ) \, dx=\frac {4\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{-{\mathrm {e}}^{{\mathrm {e}}^{-4}\,{\mathrm {e}}^x}}\,\ln \left (2\right )\,\left (5\,{\mathrm {e}}^{\mathrm {e}\,{\mathrm {e}}^x}-3\right )}{5} \]

[In]

int(exp(- exp(exp(x - 4)) - 3)*((exp(exp(x + 1))*(20*exp(x + 1)*log(2) - 20*exp(x - 4)*exp(exp(x - 4))*log(2))
)/5 + (12*exp(x - 4)*exp(exp(x - 4))*log(2))/5),x)

[Out]

(4*exp(-3)*exp(-exp(exp(-4)*exp(x)))*log(2)*(5*exp(exp(1)*exp(x)) - 3))/5