3.32.56 \(\int \frac {(-5-e) e^x+e^{2 x}+(-e^x x \log (x)+(-5 x-e x+e^x x) \log (5+e-e^x) \log (x)) \log (\log (x))}{(e^{2 x} x+e^x (-5 x-e x)) \log (x) \log (\log (x))} \, dx\)

Optimal. Leaf size=21 \[ -e^{-x} \log \left (5+e-e^x\right )+\log (\log (\log (x))) \]

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Rubi [A]  time = 0.78, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6688, 2282, 36, 31, 29, 2194, 2554, 2302} \begin {gather*} \log (\log (\log (x)))-e^{-x} \log \left (-e^x+5+e\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2194

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 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 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 21, normalized size = 1.00 \begin {gather*} -e^{-x} \log \left (5+e-e^x\right )+\log (\log (\log (x))) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 24, normalized size = 1.14 \begin {gather*} {\left (e^{x} \log \left (\log \left (\log \relax (x)\right )\right ) - \log \left (e - e^{x} + 5\right )\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^x*log(log(log(x))) - log(e - e^x + 5))*e^(-x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-((e + 5)*e^x + (x*e^x*log(x) + (x*e - x*e^x + 5*x)*log(x)*log(e - e^x + 5))*log(log(x)) - e^(2*x))/
((x*e^(2*x) - (x*e + 5*x)*e^x)*log(x)*log(log(x))), x)

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maple [A]  time = 0.04, size = 21, normalized size = 1.00

method	result	size
risch	\(\ln \left (\ln \left (\ln \relax (x )\right )\right )-\ln \left (-{\mathrm e}^{x}+{\mathrm e}+5\right ) {\mathrm e}^{-x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(ln(ln(x)))-ln(-exp(x)+exp(1)+5)*exp(-x)

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maxima [A]  time = 0.49, size = 20, normalized size = 0.95 \begin {gather*} -e^{\left (-x\right )} \log \left (e - e^{x} + 5\right ) + \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e^(-x)*log(e - e^x + 5) + log(log(log(x)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {{\mathrm {e}}^x\,\left (\mathrm {e}+5\right )-{\mathrm {e}}^{2\,x}+\ln \left (\ln \relax (x)\right )\,\left (x\,{\mathrm {e}}^x\,\ln \relax (x)+\ln \left (\mathrm {e}-{\mathrm {e}}^x+5\right )\,\ln \relax (x)\,\left (5\,x+x\,\mathrm {e}-x\,{\mathrm {e}}^x\right )\right )}{\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (x\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^x\,\left (5\,x+x\,\mathrm {e}\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(exp(1) + 5) - exp(2*x) + log(log(x))*(x*exp(x)*log(x) + log(exp(1) - exp(x) + 5)*log(x)*(5*x + x
*exp(1) - x*exp(x))))/(log(log(x))*log(x)*(x*exp(2*x) - exp(x)*(5*x + x*exp(1)))),x)

[Out]

int(-(exp(x)*(exp(1) + 5) - exp(2*x) + log(log(x))*(x*exp(x)*log(x) + log(exp(1) - exp(x) + 5)*log(x)*(5*x + x
*exp(1) - x*exp(x))))/(log(log(x))*log(x)*(x*exp(2*x) - exp(x)*(5*x + x*exp(1)))), x)

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sympy [A]  time = 2.41, size = 19, normalized size = 0.90 \begin {gather*} \log {\left (\log {\left (\log {\relax (x )} \right )} \right )} - e^{- x} \log {\left (- e^{x} + e + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(log(x))) - exp(-x)*log(-exp(x) + E + 5)

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