3.13.80 \(\int -\frac {5000 e^{5+625 e^{4 x}+4 x}}{e^{10+1250 e^{4 x}}+2 e^{5+625 e^{4 x}} \log (2)+\log ^2(2)} \, dx\) [1280]

Optimal. Leaf size=20 \[ -5+\frac {2}{e^{5+625 e^{4 x}}+\log (2)} \]

[Out]

2/(exp(625*exp(x)^4)/exp(-5)+ln(2))-5

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Rubi [A]
time = 0.08, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.082, Rules used = {12, 2320, 2278, 32} \begin {gather*} \frac {2}{e^{625 e^{4 x}+5}+\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-5000*E^(5 + 625*E^(4*x) + 4*x))/(E^(10 + 1250*E^(4*x)) + 2*E^(5 + 625*E^(4*x))*Log[2] + Log[2]^2),x]

[Out]

2/(E^(5 + 625*E^(4*x)) + Log[2])

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 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 {gather*} \begin {aligned} \text {integral} &=-\left (5000 \int \frac {e^{5+625 e^{4 x}+4 x}}{e^{10+1250 e^{4 x}}+2 e^{5+625 e^{4 x}} \log (2)+\log ^2(2)} \, dx\right )\\ &=-\left (1250 \text {Subst}\left (\int \frac {e^{5+625 x}}{\left (e^{5+625 x}+\log (2)\right )^2} \, dx,x,e^{4 x}\right )\right )\\ &=-\left (2 \text {Subst}\left (\int \frac {1}{(x+\log (2))^2} \, dx,x,e^{5+625 e^{4 x}}\right )\right )\\ &=\frac {2}{e^{5+625 e^{4 x}}+\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.03, size = 18, normalized size = 0.90 \begin {gather*} \frac {2}{e^{5+625 e^{4 x}}+\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5000*E^(5 + 625*E^(4*x) + 4*x))/(E^(10 + 1250*E^(4*x)) + 2*E^(5 + 625*E^(4*x))*Log[2] + Log[2]^2),
x]

[Out]

2/(E^(5 + 625*E^(4*x)) + Log[2])

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Maple [A]
time = 0.78, size = 17, normalized size = 0.85

method	result	size
risch	\(\frac {2}{{\mathrm e}^{5+625 \,{\mathrm e}^{4 x}}+\ln \left (2\right )}\)	\(17\)
norman	\(\frac {2}{{\mathrm e}^{5} {\mathrm e}^{625 \,{\mathrm e}^{4 x}}+\ln \left (2\right )}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-5000*exp(5)*exp(x)^4*exp(625*exp(x)^4)/(exp(5)^2*exp(625*exp(x)^4)^2+2*exp(5)*ln(2)*exp(625*exp(x)^4)+ln(
2)^2),x,method=_RETURNVERBOSE)

[Out]

2/(exp(5+625*exp(4*x))+ln(2))

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Maxima [A]
time = 0.46, size = 16, normalized size = 0.80 \begin {gather*} \frac {2}{e^{\left (625 \, e^{\left (4 \, x\right )} + 5\right )} + \log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5000*exp(5)*exp(x)^4*exp(625*exp(x)^4)/(exp(5)^2*exp(625*exp(x)^4)^2+2*exp(5)*log(2)*exp(625*exp(x)
^4)+log(2)^2),x, algorithm="maxima")

[Out]

2/(e^(625*e^(4*x) + 5) + log(2))

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Fricas [A]
time = 0.33, size = 28, normalized size = 1.40 \begin {gather*} \frac {2 \, e^{\left (4 \, x\right )}}{e^{\left (4 \, x\right )} \log \left (2\right ) + e^{\left (4 \, x + 625 \, e^{\left (4 \, x\right )} + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5000*exp(5)*exp(x)^4*exp(625*exp(x)^4)/(exp(5)^2*exp(625*exp(x)^4)^2+2*exp(5)*log(2)*exp(625*exp(x)
^4)+log(2)^2),x, algorithm="fricas")

[Out]

2*e^(4*x)/(e^(4*x)*log(2) + e^(4*x + 625*e^(4*x) + 5))

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Sympy [A]
time = 0.05, size = 15, normalized size = 0.75 \begin {gather*} \frac {2}{e^{5} e^{625 e^{4 x}} + \log {\left (2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5000*exp(5)*exp(x)**4*exp(625*exp(x)**4)/(exp(5)**2*exp(625*exp(x)**4)**2+2*exp(5)*ln(2)*exp(625*ex
p(x)**4)+ln(2)**2),x)

[Out]

2/(exp(5)*exp(625*exp(4*x)) + log(2))

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Giac [A]
time = 0.39, size = 16, normalized size = 0.80 \begin {gather*} \frac {2}{e^{\left (625 \, e^{\left (4 \, x\right )} + 5\right )} + \log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-5000*exp(5)*exp(x)^4*exp(625*exp(x)^4)/(exp(5)^2*exp(625*exp(x)^4)^2+2*exp(5)*log(2)*exp(625*exp(x)
^4)+log(2)^2),x, algorithm="giac")

[Out]

2/(e^(625*e^(4*x) + 5) + log(2))

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Mupad [B]
time = 1.00, size = 19, normalized size = 0.95 \begin {gather*} \frac {2\,{\mathrm {e}}^{-5}}{{\mathrm {e}}^{625\,{\mathrm {e}}^{4\,x}}+{\mathrm {e}}^{-5}\,\ln \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5000*exp(625*exp(4*x))*exp(4*x)*exp(5))/(exp(1250*exp(4*x))*exp(10) + log(2)^2 + 2*exp(625*exp(4*x))*exp
(5)*log(2)),x)

[Out]

(2*exp(-5))/(exp(625*exp(4*x)) + exp(-5)*log(2))

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