3.38.81 \(\int \frac {16 e^{4+2 e^{4+2 x}+2 x}}{-4+e^{2 e^{4+2 x}}} \, dx\)

Optimal. Leaf size=16 \[ \log \left (\left (-4+e^{2 e^{4+2 x}}\right )^4\right ) \]

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Rubi [A]  time = 0.11, antiderivative size = 18, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 2282, 2247, 2246, 31} \begin {gather*} 4 \log \left (4-e^{2 e^{2 x+4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(16*E^(4 + 2*E^(4 + 2*x) + 2*x))/(-4 + E^(2*E^(4 + 2*x))),x]

[Out]

4*Log[4 - E^(2*E^(4 + 2*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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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} &=16 \int \frac {e^{4+2 e^{4+2 x}+2 x}}{-4+e^{2 e^{4+2 x}}} \, dx\\ &=8 \operatorname {Subst}\left (\int \frac {e^{4+2 e^4 x}}{-4+e^{2 e^4 x}} \, dx,x,e^{2 x}\right )\\ &=\left (8 e^4\right ) \operatorname {Subst}\left (\int \frac {e^{2 e^4 x}}{-4+e^{2 e^4 x}} \, dx,x,e^{2 x}\right )\\ &=4 \operatorname {Subst}\left (\int \frac {1}{-4+x} \, dx,x,e^{2 e^{4+2 x}}\right )\\ &=4 \log \left (4-e^{2 e^{4+2 x}}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 16, normalized size = 1.00 \begin {gather*} 4 \log \left (-4+e^{2 e^{4+2 x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*E^(4 + 2*E^(4 + 2*x) + 2*x))/(-4 + E^(2*E^(4 + 2*x))),x]

[Out]

4*Log[-4 + E^(2*E^(4 + 2*x))]

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fricas [B]  time = 0.72, size = 34, normalized size = 2.12 \begin {gather*} -8 \, x + 4 \, \log \left (e^{\left (2 \, x + e^{\left (2 \, x + \log \relax (2) + 4\right )} + \log \relax (2) + 4\right )} - 4 \, e^{\left (2 \, x + \log \relax (2) + 4\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8*x + 4*log(e^(2*x + e^(2*x + log(2) + 4) + log(2) + 4) - 4*e^(2*x + log(2) + 4))

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giac [A]  time = 0.22, size = 15, normalized size = 0.94 \begin {gather*} 4 \, \log \left ({\left | e^{\left (2 \, e^{\left (2 \, x + 4\right )}\right )} - 4 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*log(abs(e^(2*e^(2*x + 4)) - 4))

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maple [A]  time = 0.04, size = 15, normalized size = 0.94

method	result	size
risch	\(4 \ln \left ({\mathrm e}^{2 \,{\mathrm e}^{2 x +4}}-4\right )\)	\(15\)
derivativedivides	\(4 \ln \left ({\mathrm e}^{{\mathrm e}^{6} {\mathrm e}^{\ln \relax (2)+2 x -2}}-4\right )\)	\(20\)
default	\(4 \ln \left ({\mathrm e}^{{\mathrm e}^{6} {\mathrm e}^{\ln \relax (2)+2 x -2}}-4\right )\)	\(20\)
norman	\(4 \ln \left ({\mathrm e}^{{\mathrm e}^{6} {\mathrm e}^{\ln \relax (2)+2 x -2}}-4\right )\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*exp(3)^2*exp(ln(2)+2*x-2)*exp(exp(3)^2*exp(ln(2)+2*x-2))/(exp(exp(3)^2*exp(ln(2)+2*x-2))-4),x,method=_RE
TURNVERBOSE)

[Out]

4*ln(exp(2*exp(2*x+4))-4)

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maxima [A]  time = 0.35, size = 14, normalized size = 0.88 \begin {gather*} 4 \, \log \left (e^{\left (2 \, e^{\left (2 \, x + 4\right )}\right )} - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*log(e^(2*e^(2*x + 4)) - 4)

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mupad [B]  time = 0.22, size = 14, normalized size = 0.88 \begin {gather*} 4\,\ln \left ({\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^4}-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*exp(2*x + log(2) - 2)*exp(exp(2*x + log(2) - 2)*exp(6))*exp(6))/(exp(exp(2*x + log(2) - 2)*exp(6)) - 4)
,x)

[Out]

4*log(exp(2*exp(2*x)*exp(4)) - 4)

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sympy [A]  time = 0.12, size = 17, normalized size = 1.06 \begin {gather*} 4 \log {\left (e^{2 e^{6} e^{2 x - 2}} - 4 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*log(exp(2*exp(6)*exp(2*x - 2)) - 4)

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