\(\int -\frac {20 e^{1+e^{e^{1+x}}+e^{1+x}+x}}{49+14 e^2+e^4+e^{2 e^{e^{1+x}}}+e^{e^{e^{1+x}}} (14+2 e^2)} \, dx\) [2209]

Optimal result

Integrand size = 59, antiderivative size = 18 \[ \int -\frac {20 e^{1+e^{e^{1+x}}+e^{1+x}+x}}{49+14 e^2+e^4+e^{2 e^{e^{1+x}}}+e^{e^{e^{1+x}}} \left (14+2 e^2\right )} \, dx=\frac {20}{7+e^2+e^{e^{e^{1+x}}}} \]

[Out]

20/(7+exp(exp(exp(1+x)))+exp(2))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {12, 2320, 2279, 2278, 32} \[ \int -\frac {20 e^{1+e^{e^{1+x}}+e^{1+x}+x}}{49+14 e^2+e^4+e^{2 e^{e^{1+x}}}+e^{e^{e^{1+x}}} \left (14+2 e^2\right )} \, dx=\frac {20}{e^{e^{e^{x+1}}}+7+e^2} \]

[In]

Int[(-20*E^(1 + E^E^(1 + x) + E^(1 + x) + x))/(49 + 14*E^2 + E^4 + E^(2*E^E^(1 + x)) + E^E^E^(1 + x)*(14 + 2*E
^2)),x]

[Out]

20/(7 + E^2 + E^E^E^(1 + 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2278

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

Rule 2279

Int[((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.)*((G_)^((h_.)*((f_.) + (g_.)*(x_))))^(m_.),
x_Symbol] :> Dist[(G^(h*(f + g*x)))^m/(F^(e*(c + d*x)))^n, Int[(F^(e*(c + d*x)))^n*(a + b*(F^(e*(c + d*x)))^n)
^p, x], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, m, n, p}, x] && EqQ[d*e*n*Log[F], g*h*m*Log[G]]

Rule 2320

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{align*} \text {integral}& = -\left (20 \int \frac {e^{1+e^{e^{1+x}}+e^{1+x}+x}}{49+14 e^2+e^4+e^{2 e^{e^{1+x}}}+e^{e^{e^{1+x}}} \left (14+2 e^2\right )} \, dx\right ) \\ & = -\left (20 \text {Subst}\left (\int \frac {e^{1+e^{e x}+e x}}{\left (e^{e^{e x}}+7 \left (1+\frac {e^2}{7}\right )\right )^2} \, dx,x,e^x\right )\right ) \\ & = -\frac {20 \text {Subst}\left (\int \frac {e^{1+x}}{\left (e^x+7 \left (1+\frac {e^2}{7}\right )\right )^2} \, dx,x,e^{e^{1+x}}\right )}{e} \\ & = -\left (20 \text {Subst}\left (\int \frac {e^x}{\left (e^x+7 \left (1+\frac {e^2}{7}\right )\right )^2} \, dx,x,e^{e^{1+x}}\right )\right ) \\ & = -\left (20 \text {Subst}\left (\int \frac {1}{\left (7 \left (1+\frac {e^2}{7}\right )+x\right )^2} \, dx,x,e^{e^{e^{1+x}}}\right )\right ) \\ & = \frac {20}{7+e^2+e^{e^{e^{1+x}}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int -\frac {20 e^{1+e^{e^{1+x}}+e^{1+x}+x}}{49+14 e^2+e^4+e^{2 e^{e^{1+x}}}+e^{e^{e^{1+x}}} \left (14+2 e^2\right )} \, dx=\frac {20}{7+e^2+e^{e^{e^{1+x}}}} \]

[In]

Integrate[(-20*E^(1 + E^E^(1 + x) + E^(1 + x) + x))/(49 + 14*E^2 + E^4 + E^(2*E^E^(1 + x)) + E^E^E^(1 + x)*(14
+ 2*E^2)),x]

[Out]

20/(7 + E^2 + E^E^E^(1 + x))

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

method	result	size
norman	\(\frac {20}{7+{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{1+x}}}+{\mathrm e}^{2}}\)	\(15\)
risch	\(\frac {20}{7+{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{1+x}}}+{\mathrm e}^{2}}\)	\(15\)
parallelrisch	\(\frac {20}{7+{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{1+x}}}+{\mathrm e}^{2}}\)	\(15\)

[In]

int(-20*exp(1+x)*exp(exp(1+x))*exp(exp(exp(1+x)))/(exp(exp(exp(1+x)))^2+(2*exp(2)+14)*exp(exp(exp(1+x)))+exp(2
)^2+14*exp(2)+49),x,method=_RETURNVERBOSE)

[Out]

20/(7+exp(exp(exp(1+x)))+exp(2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (14) = 28\).

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 2.17 \[ \int -\frac {20 e^{1+e^{e^{1+x}}+e^{1+x}+x}}{49+14 e^2+e^4+e^{2 e^{e^{1+x}}}+e^{e^{e^{1+x}}} \left (14+2 e^2\right )} \, dx=\frac {20 \, e^{\left (x + e^{\left (x + 1\right )} + 1\right )}}{{\left (e^{2} + 7\right )} e^{\left (x + e^{\left (x + 1\right )} + 1\right )} + e^{\left (x + e^{\left (x + 1\right )} + e^{\left (e^{\left (x + 1\right )}\right )} + 1\right )}} \]

[In]

integrate(-20*exp(1+x)*exp(exp(1+x))*exp(exp(exp(1+x)))/(exp(exp(exp(1+x)))^2+(2*exp(2)+14)*exp(exp(exp(1+x)))
+exp(2)^2+14*exp(2)+49),x, algorithm="fricas")

[Out]

20*e^(x + e^(x + 1) + 1)/((e^2 + 7)*e^(x + e^(x + 1) + 1) + e^(x + e^(x + 1) + e^(e^(x + 1)) + 1))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int -\frac {20 e^{1+e^{e^{1+x}}+e^{1+x}+x}}{49+14 e^2+e^4+e^{2 e^{e^{1+x}}}+e^{e^{e^{1+x}}} \left (14+2 e^2\right )} \, dx=\frac {20}{e^{e^{e^{x + 1}}} + 7 + e^{2}} \]

[In]

integrate(-20*exp(1+x)*exp(exp(1+x))*exp(exp(exp(1+x)))/(exp(exp(exp(1+x)))**2+(2*exp(2)+14)*exp(exp(exp(1+x))
)+exp(2)**2+14*exp(2)+49),x)

[Out]

20/(exp(exp(exp(x + 1))) + 7 + exp(2))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int -\frac {20 e^{1+e^{e^{1+x}}+e^{1+x}+x}}{49+14 e^2+e^4+e^{2 e^{e^{1+x}}}+e^{e^{e^{1+x}}} \left (14+2 e^2\right )} \, dx=\frac {20}{e^{2} + e^{\left (e^{\left (e^{\left (x + 1\right )}\right )}\right )} + 7} \]

[In]

integrate(-20*exp(1+x)*exp(exp(1+x))*exp(exp(exp(1+x)))/(exp(exp(exp(1+x)))^2+(2*exp(2)+14)*exp(exp(exp(1+x)))
+exp(2)^2+14*exp(2)+49),x, algorithm="maxima")

[Out]

20/(e^2 + e^(e^(e^(x + 1))) + 7)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int -\frac {20 e^{1+e^{e^{1+x}}+e^{1+x}+x}}{49+14 e^2+e^4+e^{2 e^{e^{1+x}}}+e^{e^{e^{1+x}}} \left (14+2 e^2\right )} \, dx=\frac {20}{e^{2} + e^{\left (e^{\left (e^{\left (x + 1\right )}\right )}\right )} + 7} \]

[In]

integrate(-20*exp(1+x)*exp(exp(1+x))*exp(exp(exp(1+x)))/(exp(exp(exp(1+x)))^2+(2*exp(2)+14)*exp(exp(exp(1+x)))
+exp(2)^2+14*exp(2)+49),x, algorithm="giac")

[Out]

20/(e^2 + e^(e^(e^(x + 1))) + 7)

Mupad [B] (verification not implemented)

Time = 9.31 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int -\frac {20 e^{1+e^{e^{1+x}}+e^{1+x}+x}}{49+14 e^2+e^4+e^{2 e^{e^{1+x}}}+e^{e^{e^{1+x}}} \left (14+2 e^2\right )} \, dx=\frac {20}{{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{x+1}}}+{\mathrm {e}}^2+7} \]

[In]

int(-(20*exp(exp(exp(x + 1)))*exp(x + 1)*exp(exp(x + 1)))/(14*exp(2) + exp(4) + exp(2*exp(exp(x + 1))) + exp(e
xp(exp(x + 1)))*(2*exp(2) + 14) + 49),x)

[Out]

20/(exp(exp(exp(x + 1))) + exp(2) + 7)