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

Optimal. Leaf size=18 \[ \frac {20}{7+e^2+e^{e^{e^{1+x}}}} \]

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Rubi [A]  time = 0.22, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {12, 2282, 2247, 2246, 32} \begin {gather*} \frac {20}{e^{e^{e^{x+1}}}+7+e^2} \end {gather*}

Antiderivative was successfully verified.

[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 2246

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 2247

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 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 (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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 18, normalized size = 1.00 \begin {gather*} \frac {20}{7+e^2+e^{e^{e^{1+x}}}} \end {gather*}

Antiderivative was successfully verified.

[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))

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fricas [B]  time = 0.83, size = 39, normalized size = 2.17 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -10/sqrt(exp(2)^2-exp(4))*ln(sqrt((2*exp
(exp(exp(1)*exp(sageVARx)))+2*exp(2)+14)^2+(-2*sqrt(-exp(2)^2+exp(4)))^2)/sqrt((2*exp(exp(exp(1)*exp(sageVARx)
))+2*exp(2)+14)^2+(2*

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maple [A]  time = 0.81, size = 15, normalized size = 0.83

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.47, size = 14, normalized size = 0.78 \begin {gather*} \frac {20}{e^{2} + e^{\left (e^{\left (e^{\left (x + 1\right )}\right )}\right )} + 7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

20/(e^2 + e^(e^(e^(x + 1))) + 7)

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mupad [B]  time = 1.38, size = 14, normalized size = 0.78 \begin {gather*} \frac {20}{{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{x+1}}}+{\mathrm {e}}^2+7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.15, size = 14, normalized size = 0.78 \begin {gather*} \frac {20}{e^{e^{e^{x + 1}}} + 7 + e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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