3.93.21 \(\int \frac {e^{e^{-\frac {e^{x/9}}{-18+\log ^2(5)}}+x} (162+e^{\frac {x}{9}-\frac {e^{x/9}}{-18+\log ^2(5)}}-9 \log ^2(5))}{-162+9 \log ^2(5)} \, dx\)

Optimal. Leaf size=30 \[ 2-e^{e^{\frac {e^{x/9}}{18-\log ^2(5)}}+x}+\log (2) \]

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Rubi [A]  time = 0.34, antiderivative size = 26, normalized size of antiderivative = 0.87, number of steps used = 3, number of rules used = 3, integrand size = 67, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {12, 2282, 2288} \begin {gather*} -e^{x+e^{\frac {e^{x/9}}{18-\log ^2(5)}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^(-(E^(x/9)/(-18 + Log[5]^2))) + x)*(162 + E^(x/9 - E^(x/9)/(-18 + Log[5]^2)) - 9*Log[5]^2))/(-162 +
9*Log[5]^2),x]

[Out]

-E^(E^(E^(x/9)/(18 - Log[5]^2)) + 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 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 {\int e^{e^{-\frac {e^{x/9}}{-18+\log ^2(5)}}+x} \left (162+e^{\frac {x}{9}-\frac {e^{x/9}}{-18+\log ^2(5)}}-9 \log ^2(5)\right ) \, dx}{9 \left (18-\log ^2(5)\right )}\\ &=-\frac {\operatorname {Subst}\left (\int e^{e^{-\frac {x}{-18+\log ^2(5)}}} x^8 \left (e^{-\frac {x}{-18+\log ^2(5)}} x+162 \left (1-\frac {\log ^2(5)}{18}\right )\right ) \, dx,x,e^{x/9}\right )}{18-\log ^2(5)}\\ &=-e^{e^{\frac {e^{x/9}}{18-\log ^2(5)}}+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 25, normalized size = 0.83 \begin {gather*} -e^{e^{-\frac {e^{x/9}}{-18+\log ^2(5)}}+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^(-(E^(x/9)/(-18 + Log[5]^2))) + x)*(162 + E^(x/9 - E^(x/9)/(-18 + Log[5]^2)) - 9*Log[5]^2))/(-
162 + 9*Log[5]^2),x]

[Out]

-E^(E^(-(E^(x/9)/(-18 + Log[5]^2))) + x)

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fricas [A]  time = 0.52, size = 42, normalized size = 1.40 \begin {gather*} -e^{\left ({\left (x e^{\left (\frac {1}{9} \, x\right )} + e^{\left (\frac {x \log \relax (5)^{2} - 18 \, x - 9 \, e^{\left (\frac {1}{9} \, x\right )}}{9 \, {\left (\log \relax (5)^{2} - 18\right )}}\right )}\right )} e^{\left (-\frac {1}{9} \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1/9*x)*exp(-exp(1/9*x)/(log(5)^2-18))-9*log(5)^2+162)*exp(exp(-exp(1/9*x)/(log(5)^2-18))+x)/(9*
log(5)^2-162),x, algorithm="fricas")

[Out]

-e^((x*e^(1/9*x) + e^(1/9*(x*log(5)^2 - 18*x - 9*e^(1/9*x))/(log(5)^2 - 18)))*e^(-1/9*x))

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giac [A]  time = 0.28, size = 20, normalized size = 0.67 \begin {gather*} -e^{\left (x + e^{\left (-\frac {e^{\left (\frac {1}{9} \, x\right )}}{\log \relax (5)^{2} - 18}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1/9*x)*exp(-exp(1/9*x)/(log(5)^2-18))-9*log(5)^2+162)*exp(exp(-exp(1/9*x)/(log(5)^2-18))+x)/(9*
log(5)^2-162),x, algorithm="giac")

[Out]

-e^(x + e^(-e^(1/9*x)/(log(5)^2 - 18)))

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maple [A]  time = 0.11, size = 21, normalized size = 0.70

method	result	size
norman	\(-{\mathrm e}^{{\mathrm e}^{-\frac {{\mathrm e}^{\frac {x}{9}}}{\ln \relax (5)^{2}-18}}+x}\)	\(21\)
risch	\(-\frac {9 \,{\mathrm e}^{{\mathrm e}^{-\frac {{\mathrm e}^{\frac {x}{9}}}{\ln \relax (5)^{2}-18}}+x} \ln \relax (5)^{2}}{9 \ln \relax (5)^{2}-162}+\frac {162 \,{\mathrm e}^{{\mathrm e}^{-\frac {{\mathrm e}^{\frac {x}{9}}}{\ln \relax (5)^{2}-18}}+x}}{9 \ln \relax (5)^{2}-162}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1/9*x)*exp(-exp(1/9*x)/(ln(5)^2-18))-9*ln(5)^2+162)*exp(exp(-exp(1/9*x)/(ln(5)^2-18))+x)/(9*ln(5)^2-1
62),x,method=_RETURNVERBOSE)

[Out]

-exp(exp(-exp(1/9*x)/(ln(5)^2-18))+x)

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maxima [A]  time = 0.56, size = 20, normalized size = 0.67 \begin {gather*} -e^{\left (x + e^{\left (-\frac {e^{\left (\frac {1}{9} \, x\right )}}{\log \relax (5)^{2} - 18}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1/9*x)*exp(-exp(1/9*x)/(log(5)^2-18))-9*log(5)^2+162)*exp(exp(-exp(1/9*x)/(log(5)^2-18))+x)/(9*
log(5)^2-162),x, algorithm="maxima")

[Out]

-e^(x + e^(-e^(1/9*x)/(log(5)^2 - 18)))

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mupad [B]  time = 5.98, size = 20, normalized size = 0.67 \begin {gather*} -{\mathrm {e}}^x\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {{\mathrm {e}}^{x/9}}{{\ln \relax (5)}^2-18}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + exp(-exp(x/9)/(log(5)^2 - 18)))*(exp(x/9)*exp(-exp(x/9)/(log(5)^2 - 18)) - 9*log(5)^2 + 162))/(9*
log(5)^2 - 162),x)

[Out]

-exp(x)*exp(exp(-exp(x/9)/(log(5)^2 - 18)))

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sympy [A]  time = 0.40, size = 19, normalized size = 0.63 \begin {gather*} - e^{x + e^{- \frac {e^{\frac {x}{9}}}{-18 + \log {\relax (5 )}^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1/9*x)*exp(-exp(1/9*x)/(ln(5)**2-18))-9*ln(5)**2+162)*exp(exp(-exp(1/9*x)/(ln(5)**2-18))+x)/(9*
ln(5)**2-162),x)

[Out]

-exp(x + exp(-exp(x/9)/(-18 + log(5)**2)))

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