3.21.18 \(\int \frac {e^{e^{\frac {4+\log (4)}{x}}+\frac {4+\log (4)}{x}} (4+\log (4))}{4 x^2} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{4} \left (2+e^2-e^{e^{\frac {4+\log (4)}{x}}}\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 19, normalized size of antiderivative = 0.83, 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, 6715, 2282, 2194} \begin {gather*} -\frac {1}{4} e^{4^{\frac {1}{x}} e^{4/x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*E^(4^x^(-1)*E^(4/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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, 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]]]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 16, normalized size = 0.70 \begin {gather*} -\frac {1}{4} e^{e^{\frac {4+\log (4)}{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*E^E^((4 + Log[4])/x)

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fricas [A]  time = 0.75, size = 35, normalized size = 1.52 \begin {gather*} -\frac {1}{4} \, e^{\left (\frac {x e^{\left (\frac {2 \, {\left (\log \relax (2) + 2\right )}}{x}\right )} + 2 \, \log \relax (2) + 4}{x} - \frac {2 \, {\left (\log \relax (2) + 2\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4+2*log(2))*exp((4+2*log(2))/x)*exp(exp((4+2*log(2))/x))/x^2,x, algorithm="fricas")

[Out]

-1/4*e^((x*e^(2*(log(2) + 2)/x) + 2*log(2) + 4)/x - 2*(log(2) + 2)/x)

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giac [A]  time = 0.45, size = 13, normalized size = 0.57 \begin {gather*} -\frac {1}{4} \, e^{\left (e^{\left (\frac {2 \, {\left (\log \relax (2) + 2\right )}}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4+2*log(2))*exp((4+2*log(2))/x)*exp(exp((4+2*log(2))/x))/x^2,x, algorithm="giac")

[Out]

-1/4*e^(e^(2*(log(2) + 2)/x))

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maple [A]  time = 0.05, size = 15, normalized size = 0.65

method	result	size
derivativedivides	\(-\frac {{\mathrm e}^{{\mathrm e}^{\frac {4+2 \ln \relax (2)}{x}}}}{4}\)	\(15\)
norman	\(-\frac {{\mathrm e}^{{\mathrm e}^{\frac {4+2 \ln \relax (2)}{x}}}}{4}\)	\(15\)
default	\(-\frac {\left (1+\frac {\ln \relax (2)}{2}\right ) {\mathrm e}^{{\mathrm e}^{\frac {4+2 \ln \relax (2)}{x}}}}{4+2 \ln \relax (2)}\)	\(29\)
risch	\(-\frac {{\mathrm e}^{4^{\frac {1}{x}} {\mathrm e}^{\frac {4}{x}}}}{2 \left (\ln \relax (2)+2\right )}-\frac {{\mathrm e}^{4^{\frac {1}{x}} {\mathrm e}^{\frac {4}{x}}} \ln \relax (2)}{4 \left (\ln \relax (2)+2\right )}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(4+2*ln(2))*exp((4+2*ln(2))/x)*exp(exp((4+2*ln(2))/x))/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/4*exp(exp((4+2*ln(2))/x))

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maxima [A]  time = 0.47, size = 17, normalized size = 0.74 \begin {gather*} -\frac {1}{4} \, e^{\left (e^{\left (\frac {2 \, \log \relax (2)}{x} + \frac {4}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4+2*log(2))*exp((4+2*log(2))/x)*exp(exp((4+2*log(2))/x))/x^2,x, algorithm="maxima")

[Out]

-1/4*e^(e^(2*log(2)/x + 4/x))

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mupad [B]  time = 1.31, size = 27, normalized size = 1.17 \begin {gather*} -\frac {{\mathrm {e}}^{2^{2/x}\,{\mathrm {e}}^{4/x}}\,\left (\ln \relax (4)+4\right )}{\ln \left (256\right )+16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp((2*log(2) + 4)/x))*exp((2*log(2) + 4)/x)*(2*log(2) + 4))/(4*x^2),x)

[Out]

-(exp(2^(2/x)*exp(4/x))*(log(4) + 4))/(log(256) + 16)

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sympy [A]  time = 0.33, size = 14, normalized size = 0.61 \begin {gather*} - \frac {e^{e^{\frac {2 \log {\relax (2 )} + 4}{x}}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(4+2*ln(2))*exp((4+2*ln(2))/x)*exp(exp((4+2*ln(2))/x))/x**2,x)

[Out]

-exp(exp((2*log(2) + 4)/x))/4

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