3.80.53 \(\int \frac {(-4-8 e) e^{-1+e^{\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}+\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}}}{(3+x+2 e x) \log (3+x+2 e x) \log ^2(\log (3+x+2 e x))} \, dx\)

Optimal. Leaf size=21 \[ e^{-1+e^{3+\frac {4}{\log (\log (3+x+2 e x))}}} \]

________________________________________________________________________________________

Rubi [A]  time = 1.48, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 92, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.076, Rules used = {6, 12, 2444, 6688, 6715, 2282, 2194} \begin {gather*} e^{e^{\frac {4}{\log (\log (2 e x+x+3))}+3}-1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-4 - 8*E)*E^(-1 + E^((4 + 3*Log[Log[3 + x + 2*E*x]])/Log[Log[3 + x + 2*E*x]]) + (4 + 3*Log[Log[3 + x + 2
*E*x]])/Log[Log[3 + x + 2*E*x]]))/((3 + x + 2*E*x)*Log[3 + x + 2*E*x]*Log[Log[3 + x + 2*E*x]]^2),x]

[Out]

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

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 2444

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rule 6688

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

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(-4-8 e) \exp \left (-1+\exp \left (\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}\right )+\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}\right )}{(3+(1+2 e) x) \log (3+x+2 e x) \log ^2(\log (3+x+2 e x))} \, dx\\ &=-\left ((4 (1+2 e)) \int \frac {\exp \left (-1+\exp \left (\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}\right )+\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}\right )}{(3+(1+2 e) x) \log (3+x+2 e x) \log ^2(\log (3+x+2 e x))} \, dx\right )\\ &=-\left ((4 (1+2 e)) \int \frac {\exp \left (-1+\exp \left (\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}\right )+\frac {4+3 \log (\log (3+x+2 e x))}{\log (\log (3+x+2 e x))}\right )}{(3+(1+2 e) x) \log (3+(1+2 e) x) \log ^2(\log (3+x+2 e x))} \, dx\right )\\ &=-\left ((4 (1+2 e)) \int \frac {\exp \left (2+e^{3+\frac {4}{\log (\log (3+x+2 e x))}}+\frac {4}{\log (\log (3+x+2 e x))}\right )}{(3+(1+2 e) x) \log (3+(1+2 e) x) \log ^2(\log (3+(1+2 e) x))} \, dx\right )\\ &=-\left (4 \operatorname {Subst}\left (\int \frac {e^{2+e^{3+\frac {4}{\log (\log (x))}}+\frac {4}{\log (\log (x))}}}{x \log (x) \log ^2(\log (x))} \, dx,x,3+(1+2 e) x\right )\right )\\ &=-\left (4 \operatorname {Subst}\left (\int \frac {e^{2+e^{3+\frac {4}{\log (x)}}+\frac {4}{\log (x)}}}{x \log ^2(x)} \, dx,x,\log (3+x+2 e x)\right )\right )\\ &=-\left (4 \operatorname {Subst}\left (\int \frac {e^{2+e^{3+\frac {4}{x}}+\frac {4}{x}}}{x^2} \, dx,x,\log (\log (3+x+2 e x))\right )\right )\\ &=4 \operatorname {Subst}\left (\int e^{2+e^{3+4 x}+4 x} \, dx,x,\frac {1}{\log (\log (3+x+2 e x))}\right )\\ &=\operatorname {Subst}\left (\int e^{2+e^3 x} \, dx,x,e^{\frac {4}{\log (\log (3+x+2 e x))}}\right )\\ &=e^{-1+e^{3+\frac {4}{\log (\log (3+x+2 e x))}}}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.43, size = 21, normalized size = 1.00 \begin {gather*} e^{-1+e^{3+\frac {4}{\log (\log (3+x+2 e x))}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-4 - 8*E)*E^(-1 + E^((4 + 3*Log[Log[3 + x + 2*E*x]])/Log[Log[3 + x + 2*E*x]]) + (4 + 3*Log[Log[3 +
x + 2*E*x]])/Log[Log[3 + x + 2*E*x]]))/((3 + x + 2*E*x)*Log[3 + x + 2*E*x]*Log[Log[3 + x + 2*E*x]]^2),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.90, size = 96, normalized size = 4.57 \begin {gather*} e^{\left (\frac {e^{\left (\frac {3 \, \log \left (\log \left (2 \, x e + x + 3\right )\right ) + 4}{\log \left (\log \left (2 \, x e + x + 3\right )\right )}\right )} \log \left (\log \left (2 \, x e + x + 3\right )\right ) + 2 \, \log \left (\log \left (2 \, x e + x + 3\right )\right ) + 4}{\log \left (\log \left (2 \, x e + x + 3\right )\right )} - \frac {3 \, \log \left (\log \left (2 \, x e + x + 3\right )\right ) + 4}{\log \left (\log \left (2 \, x e + x + 3\right )\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)-4)*exp((3*log(log(2*x*exp(1)+3+x))+4)/log(log(2*x*exp(1)+3+x)))*exp(exp((3*log(log(2*x*ex
p(1)+3+x))+4)/log(log(2*x*exp(1)+3+x)))-1)/(2*x*exp(1)+3+x)/log(2*x*exp(1)+3+x)/log(log(2*x*exp(1)+3+x))^2,x,
algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)-4)*exp((3*log(log(2*x*exp(1)+3+x))+4)/log(log(2*x*exp(1)+3+x)))*exp(exp((3*log(log(2*x*ex
p(1)+3+x))+4)/log(log(2*x*exp(1)+3+x)))-1)/(2*x*exp(1)+3+x)/log(2*x*exp(1)+3+x)/log(log(2*x*exp(1)+3+x))^2,x,
algorithm="giac")

[Out]

undef

________________________________________________________________________________________

maple [A]  time = 0.27, size = 38, normalized size = 1.81

method	result	size
derivativedivides	\(-\frac {\left (-2 \,{\mathrm e}-1\right ) {\mathrm e}^{{\mathrm e}^{\frac {4}{\ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )}} {\mathrm e}^{3}} {\mathrm e}^{-1}}{2 \,{\mathrm e}+1}\)	\(38\)
default	\(-\frac {\left (-8 \,{\mathrm e}-4\right ) {\mathrm e}^{{\mathrm e}^{\frac {4}{\ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )}} {\mathrm e}^{3}} {\mathrm e}^{-1}}{4 \left (2 \,{\mathrm e}+1\right )}\)	\(38\)
risch	\(\frac {2 \,{\mathrm e}^{{\mathrm e}^{\frac {3 \ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )+4}{\ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )}}-1} {\mathrm e}}{2 \,{\mathrm e}+1}+\frac {{\mathrm e}^{{\mathrm e}^{\frac {3 \ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )+4}{\ln \left (\ln \left (2 x \,{\mathrm e}+3+x \right )\right )}}-1}}{2 \,{\mathrm e}+1}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*exp(1)-4)*exp((3*ln(ln(2*x*exp(1)+3+x))+4)/ln(ln(2*x*exp(1)+3+x)))*exp(exp((3*ln(ln(2*x*exp(1)+3+x))+4
)/ln(ln(2*x*exp(1)+3+x)))-1)/(2*x*exp(1)+3+x)/ln(2*x*exp(1)+3+x)/ln(ln(2*x*exp(1)+3+x))^2,x,method=_RETURNVERB
OSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 20, normalized size = 0.95 \begin {gather*} e^{\left (e^{\left (\frac {4}{\log \left (\log \left (2 \, x e + x + 3\right )\right )} + 3\right )} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)-4)*exp((3*log(log(2*x*exp(1)+3+x))+4)/log(log(2*x*exp(1)+3+x)))*exp(exp((3*log(log(2*x*ex
p(1)+3+x))+4)/log(log(2*x*exp(1)+3+x)))-1)/(2*x*exp(1)+3+x)/log(2*x*exp(1)+3+x)/log(log(2*x*exp(1)+3+x))^2,x,
algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 6.29, size = 22, normalized size = 1.05 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{\frac {4}{\ln \left (\ln \left (x+2\,x\,\mathrm {e}+3\right )\right )}}\,{\mathrm {e}}^3}\,{\mathrm {e}}^{-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp((3*log(log(x + 2*x*exp(1) + 3)) + 4)/log(log(x + 2*x*exp(1) + 3))) - 1)*exp((3*log(log(x + 2*x*e
xp(1) + 3)) + 4)/log(log(x + 2*x*exp(1) + 3)))*(8*exp(1) + 4))/(log(x + 2*x*exp(1) + 3)*log(log(x + 2*x*exp(1)
+ 3))^2*(x + 2*x*exp(1) + 3)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.54, size = 34, normalized size = 1.62 \begin {gather*} e^{e^{\frac {3 \log {\left (\log {\left (x + 2 e x + 3 \right )} \right )} + 4}{\log {\left (\log {\left (x + 2 e x + 3 \right )} \right )}}} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*exp(1)-4)*exp((3*ln(ln(2*x*exp(1)+3+x))+4)/ln(ln(2*x*exp(1)+3+x)))*exp(exp((3*ln(ln(2*x*exp(1)+3
+x))+4)/ln(ln(2*x*exp(1)+3+x)))-1)/(2*x*exp(1)+3+x)/ln(2*x*exp(1)+3+x)/ln(ln(2*x*exp(1)+3+x))**2,x)

[Out]

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

________________________________________________________________________________________