3.81.91 \(\int -\frac {5 e^{3+e^x+x}}{25+10 e^{3+e^x}+e^{6+2 e^x}} \, dx\)

Optimal. Leaf size=13 \[ \frac {5}{5+e^{3+e^x}} \]

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Rubi [A]  time = 0.08, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2282, 2246, 32} \begin {gather*} \frac {5}{e^{e^x+3}+5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5*E^(3 + E^x + x))/(25 + 10*E^(3 + E^x) + E^(6 + 2*E^x)),x]

[Out]

5/(5 + E^(3 + E^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 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 (5 \int \frac {e^{3+e^x+x}}{25+10 e^{3+e^x}+e^{6+2 e^x}} \, dx\right )\\ &=-\left (5 \operatorname {Subst}\left (\int \frac {e^{3+x}}{\left (5+e^{3+x}\right )^2} \, dx,x,e^x\right )\right )\\ &=-\left (5 \operatorname {Subst}\left (\int \frac {1}{(5+x)^2} \, dx,x,e^{3+e^x}\right )\right )\\ &=\frac {5}{5+e^{3+e^x}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 13, normalized size = 1.00 \begin {gather*} \frac {5}{5+e^{3+e^x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*E^(3 + E^x + x))/(25 + 10*E^(3 + E^x) + E^(6 + 2*E^x)),x]

[Out]

5/(5 + E^(3 + E^x))

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fricas [A]  time = 0.79, size = 17, normalized size = 1.31 \begin {gather*} \frac {5 \, e^{x}}{e^{\left (x + e^{x} + 3\right )} + 5 \, e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*exp(3)*exp(x)*exp(exp(x))/(exp(3)^2*exp(exp(x))^2+10*exp(3)*exp(exp(x))+25),x, algorithm="fricas"
)

[Out]

5*e^x/(e^(x + e^x + 3) + 5*e^x)

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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(-5*exp(3)*exp(x)*exp(exp(x))/(exp(3)^2*exp(exp(x))^2+10*exp(3)*exp(exp(x))+25),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -5*exp(3)/5/sqrt(-exp(3)^2+exp(6))*atan(
(exp(exp(sageVARx))*exp(6)+5*exp(3))/5/sqrt(-exp(3)^2+exp(6)))

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maple [A]  time = 0.13, size = 12, normalized size = 0.92

method	result	size
risch	\(\frac {5}{5+{\mathrm e}^{3+{\mathrm e}^{x}}}\)	\(12\)
norman	\(-\frac {{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}^{x}}}{5+{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}^{x}}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5*exp(3)*exp(x)*exp(exp(x))/(exp(3)^2*exp(exp(x))^2+10*exp(3)*exp(exp(x))+25),x,method=_RETURNVERBOSE)

[Out]

5/(5+exp(3+exp(x)))

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maxima [A]  time = 0.37, size = 11, normalized size = 0.85 \begin {gather*} \frac {5}{e^{\left (e^{x} + 3\right )} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*exp(3)*exp(x)*exp(exp(x))/(exp(3)^2*exp(exp(x))^2+10*exp(3)*exp(exp(x))+25),x, algorithm="maxima"
)

[Out]

5/(e^(e^x + 3) + 5)

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mupad [B]  time = 0.16, size = 11, normalized size = 0.85 \begin {gather*} \frac {5}{{\mathrm {e}}^{{\mathrm {e}}^x+3}+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*exp(exp(x))*exp(3)*exp(x))/(10*exp(exp(x))*exp(3) + exp(6)*exp(2*exp(x)) + 25),x)

[Out]

5/(exp(exp(x) + 3) + 5)

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sympy [A]  time = 0.10, size = 10, normalized size = 0.77 \begin {gather*} \frac {5}{e^{3} e^{e^{x}} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5*exp(3)*exp(x)*exp(exp(x))/(exp(3)**2*exp(exp(x))**2+10*exp(3)*exp(exp(x))+25),x)

[Out]

5/(exp(3)*exp(exp(x)) + 5)

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