3.104.31 \(\int -\frac {12 e^{e^{\frac {1}{5} (\log ^2(10)+5 \log (e^{-x} (4+e^x)))}+\frac {1}{5} (\log ^2(10)+5 \log (e^{-x} (4+e^x)))}}{4+e^x} \, dx\)

Optimal. Leaf size=24 \[ 3 e^{e^{\frac {\log ^2(10)}{5}} \left (1+4 e^{-x}\right )} \]

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Rubi [A]  time = 0.19, antiderivative size = 31, normalized size of antiderivative = 1.29, number of steps used = 3, number of rules used = 3, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {12, 2282, 2209} \begin {gather*} 3 e^{4 e^{\frac {\log ^2(10)}{5}-x}+e^{\frac {\log ^2(10)}{5}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-12*E^(E^((Log[10]^2 + 5*Log[(4 + E^x)/E^x])/5) + (Log[10]^2 + 5*Log[(4 + E^x)/E^x])/5))/(4 + E^x),x]

[Out]

3*E^(E^(Log[10]^2/5) + 4*E^(-x + Log[10]^2/5))

Rule 12

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

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

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]]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left (12 \int \frac {\exp \left (e^{\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )}+\frac {1}{5} \left (\log ^2(10)+5 \log \left (e^{-x} \left (4+e^x\right )\right )\right )\right )}{4+e^x} \, dx\right )\\ &=-\left (12 \operatorname {Subst}\left (\int \frac {e^{e^{\frac {\log ^2(10)}{5}}+\frac {4 e^{\frac {\log ^2(10)}{5}}}{x}+\frac {\log ^2(10)}{5}}}{x^2} \, dx,x,e^x\right )\right )\\ &=3 e^{e^{\frac {\log ^2(10)}{5}}+4 e^{-x+\frac {\log ^2(10)}{5}}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 24, normalized size = 1.00 \begin {gather*} 3 e^{e^{-x+\frac {\log ^2(10)}{5}} \left (4+e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12*E^(E^((Log[10]^2 + 5*Log[(4 + E^x)/E^x])/5) + (Log[10]^2 + 5*Log[(4 + E^x)/E^x])/5))/(4 + E^x),
x]

[Out]

3*E^(E^(-x + Log[10]^2/5)*(4 + E^x))

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fricas [B]  time = 0.74, size = 58, normalized size = 2.42 \begin {gather*} \frac {3 \, e^{\left (\frac {1}{5} \, {\left (e^{x} \log \left (10\right )^{2} + 5 \, {\left (e^{x} + 4\right )} e^{\left (\frac {1}{5} \, \log \left (10\right )^{2}\right )} + 5 \, e^{x} \log \left ({\left (e^{x} + 4\right )} e^{\left (-x\right )}\right )\right )} e^{\left (-x\right )} - \frac {1}{5} \, \log \left (10\right )^{2} + x\right )}}{e^{x} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-12*exp(log((exp(x)+4)/exp(x))+1/5*log(10)^2)*exp(exp(log((exp(x)+4)/exp(x))+1/5*log(10)^2))/(exp(x)
+4),x, algorithm="fricas")

[Out]

3*e^(1/5*(e^x*log(10)^2 + 5*(e^x + 4)*e^(1/5*log(10)^2) + 5*e^x*log((e^x + 4)*e^(-x)))*e^(-x) - 1/5*log(10)^2
+ x)/(e^x + 4)

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giac [A]  time = 0.18, size = 24, normalized size = 1.00 \begin {gather*} 3 \, e^{\left (e^{\left (\frac {1}{5} \, \log \left (10\right )^{2}\right )} + 4 \, e^{\left (\frac {1}{5} \, \log \left (10\right )^{2} - x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-12*exp(log((exp(x)+4)/exp(x))+1/5*log(10)^2)*exp(exp(log((exp(x)+4)/exp(x))+1/5*log(10)^2))/(exp(x)
+4),x, algorithm="giac")

[Out]

3*e^(e^(1/5*log(10)^2) + 4*e^(1/5*log(10)^2 - x))

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maple [A]  time = 0.20, size = 20, normalized size = 0.83

method	result	size
norman	\(3 \,{\mathrm e}^{{\mathrm e}^{\frac {\ln \left (10\right )^{2}}{5}} \left ({\mathrm e}^{x}+4\right ) {\mathrm e}^{-x}}\)	\(20\)
derivativedivides	\(3 \,{\mathrm e}^{-\frac {\left (\ln \relax (2)+\ln \relax (5)\right )^{2}}{5}} {\mathrm e}^{\frac {\ln \relax (2)^{2}}{5}+\frac {2 \ln \relax (2) \ln \relax (5)}{5}+\frac {\ln \relax (5)^{2}}{5}+{\mathrm e}^{\frac {\left (\ln \relax (2)+\ln \relax (5)\right )^{2}}{5}}+4 \,{\mathrm e}^{\frac {\left (\ln \relax (2)+\ln \relax (5)\right )^{2}}{5}} {\mathrm e}^{-x}}\)	\(61\)
default	\(3 \,{\mathrm e}^{-\frac {\left (\ln \relax (2)+\ln \relax (5)\right )^{2}}{5}} {\mathrm e}^{\frac {\ln \relax (2)^{2}}{5}+\frac {2 \ln \relax (2) \ln \relax (5)}{5}+\frac {\ln \relax (5)^{2}}{5}+{\mathrm e}^{\frac {\left (\ln \relax (2)+\ln \relax (5)\right )^{2}}{5}}+4 \,{\mathrm e}^{\frac {\left (\ln \relax (2)+\ln \relax (5)\right )^{2}}{5}} {\mathrm e}^{-x}}\)	\(61\)
risch	\(3 \,{\mathrm e}^{2^{\frac {2 \ln \relax (5)}{5}} \left ({\mathrm e}^{x}+4\right ) {\mathrm e}^{-x -\frac {i \pi \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right )^{3}}{2}+\frac {i \pi \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )}{2}+\frac {i \pi \mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right )^{2} \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+4\right )\right )}{2}-\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x} \left ({\mathrm e}^{x}+4\right )\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}+4\right )\right )}{2}+\frac {\ln \relax (2)^{2}}{5}+\frac {\ln \relax (5)^{2}}{5}}}\)	\(134\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-12*exp(ln((exp(x)+4)/exp(x))+1/5*ln(10)^2)*exp(exp(ln((exp(x)+4)/exp(x))+1/5*ln(10)^2))/(exp(x)+4),x,meth
od=_RETURNVERBOSE)

[Out]

3*exp(exp(1/5*ln(10)^2)*(exp(x)+4)/exp(x))

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maxima [B]  time = 0.44, size = 51, normalized size = 2.12 \begin {gather*} 3 \, e^{\left (2^{\frac {2}{5} \, \log \relax (5) + 2} e^{\left (\frac {1}{5} \, \log \relax (5)^{2} + \frac {1}{5} \, \log \relax (2)^{2} - x\right )} + 2^{\frac {2}{5} \, \log \relax (5)} e^{\left (\frac {1}{5} \, \log \relax (5)^{2} + \frac {1}{5} \, \log \relax (2)^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-12*exp(log((exp(x)+4)/exp(x))+1/5*log(10)^2)*exp(exp(log((exp(x)+4)/exp(x))+1/5*log(10)^2))/(exp(x)
+4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 8.46, size = 24, normalized size = 1.00 \begin {gather*} 3\,{\mathrm {e}}^{4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{\frac {{\ln \left (10\right )}^2}{5}}}\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {{\ln \left (10\right )}^2}{5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(12*exp(log(exp(-x)*(exp(x) + 4)) + log(10)^2/5)*exp(exp(log(exp(-x)*(exp(x) + 4)) + log(10)^2/5)))/(exp(
x) + 4),x)

[Out]

3*exp(4*exp(-x)*exp(log(10)^2/5))*exp(exp(log(10)^2/5))

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sympy [A]  time = 0.17, size = 19, normalized size = 0.79 \begin {gather*} 3 e^{\left (e^{x} + 4\right ) e^{- x} e^{\frac {\log {\left (10 \right )}^{2}}{5}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-12*exp(ln((exp(x)+4)/exp(x))+1/5*ln(10)**2)*exp(exp(ln((exp(x)+4)/exp(x))+1/5*ln(10)**2))/(exp(x)+4
),x)

[Out]

3*exp((exp(x) + 4)*exp(-x)*exp(log(10)**2/5))

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