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

Optimal. Leaf size=31 \[ 2 \left (2+e^{-3-e^{e^{-4+x}}} \left (-\frac {3}{5}+e^{e^{1+x}}\right ) \log (4)\right ) \]

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Rubi [A]  time = 0.15, antiderivative size = 49, normalized size of antiderivative = 1.58, number of steps used = 4, number of rules used = 3, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {12, 2282, 2288} \begin {gather*} -\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) \end {gather*}

Antiderivative was successfully verified.

[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 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 2288

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 {gather*} \begin {aligned} \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} \operatorname {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)) \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 30, normalized size = 0.97 \begin {gather*} \frac {2}{5} e^{-3-e^{e^{-4+x}}} \left (-3+5 e^{e^{1+x}}\right ) \log (4) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.83, size = 49, normalized size = 1.58 \begin {gather*} \frac {4}{5} \, {\left (5 \, e^{\left (e^{\left (x + 1\right )}\right )} \log \relax (2) - 3 \, \log \relax (2)\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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-20*log(2)*exp(x-4)*exp(exp(x-4))+20*log(2)*exp(x+1))*exp(exp(x+1))+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
))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {4}{5} \, {\left (5 \, {\left (e^{\left (x + e^{\left (x - 4\right )} - 4\right )} \log \relax (2) - e^{\left (x + 1\right )} \log \relax (2)\right )} e^{\left (e^{\left (x + 1\right )}\right )} - 3 \, e^{\left (x + e^{\left (x - 4\right )} - 4\right )} \log \relax (2)\right )} e^{\left (-e^{\left (e^{\left (x - 4\right )}\right )} - 3\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-20*log(2)*exp(x-4)*exp(exp(x-4))+20*log(2)*exp(x+1))*exp(exp(x+1))+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)

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maple [A]  time = 0.21, size = 27, normalized size = 0.87

method	result	size
risch	\(\frac {\left (20 \ln \relax (2) {\mathrm e}^{{\mathrm e}^{x +1}}-12 \ln \relax (2)\right ) {\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{x -4}}-3}}{5}\)	\(27\)
norman	\(\left (4 \ln \relax (2) {\mathrm e}^{{\mathrm e}^{5} {\mathrm e}^{x -4}}-\frac {12 \ln \relax (2)}{5}\right ) {\mathrm e}^{-{\mathrm e}^{{\mathrm e}^{x -4}}-3}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.58, size = 33, normalized size = 1.06 \begin {gather*} 4 \, e^{\left (e^{\left (x + 1\right )} - e^{\left (e^{\left (x - 4\right )}\right )} - 3\right )} \log \relax (2) - \frac {12}{5} \, e^{\left (-e^{\left (e^{\left (x - 4\right )}\right )} - 3\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-20*log(2)*exp(x-4)*exp(exp(x-4))+20*log(2)*exp(x+1))*exp(exp(x+1))+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)

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mupad [B]  time = 6.26, size = 25, normalized size = 0.81 \begin {gather*} \frac {4\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{-{\mathrm {e}}^{{\mathrm {e}}^{-4}\,{\mathrm {e}}^x}}\,\ln \relax (2)\,\left (5\,{\mathrm {e}}^{\mathrm {e}\,{\mathrm {e}}^x}-3\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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