3.77.60 \(\int \frac {5+e^{e^{16 e^2}}}{x} \, dx\)

Optimal. Leaf size=20 \[ \left (5+e^{e^{16 e^2}}\right ) \log \left (\left (3+e^{25}\right ) x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 29} \begin {gather*} \left (5+e^{e^{16 e^2}}\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 + E^E^(16*E^2))/x,x]

[Out]

(5 + E^E^(16*E^2))*Log[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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\left (5+e^{e^{16 e^2}}\right ) \int \frac {1}{x} \, dx\\ &=\left (5+e^{e^{16 e^2}}\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.70 \begin {gather*} \left (5+e^{e^{16 e^2}}\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + E^E^(16*E^2))/x,x]

[Out]

(5 + E^E^(16*E^2))*Log[x]

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fricas [A]  time = 0.92, size = 14, normalized size = 0.70 \begin {gather*} e^{\left (e^{\left (16 \, e^{2}\right )}\right )} \log \relax (x) + 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(exp(16*exp(1)^2))+5)/x,x, algorithm="fricas")

[Out]

e^(e^(16*e^2))*log(x) + 5*log(x)

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giac [A]  time = 0.12, size = 12, normalized size = 0.60 \begin {gather*} {\left (e^{\left (e^{\left (16 \, e^{2}\right )}\right )} + 5\right )} \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(exp(16*exp(1)^2))+5)/x,x, algorithm="giac")

[Out]

(e^(e^(16*e^2)) + 5)*log(abs(x))

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maple [A]  time = 0.02, size = 14, normalized size = 0.70

method	result	size
default	\(\left ({\mathrm e}^{{\mathrm e}^{16 \,{\mathrm e}^{2}}}+5\right ) \ln \relax (x )\)	\(14\)
norman	\(\left ({\mathrm e}^{{\mathrm e}^{16 \,{\mathrm e}^{2}}}+5\right ) \ln \relax (x )\)	\(14\)
risch	\(\ln \relax (x ) {\mathrm e}^{{\mathrm e}^{16 \,{\mathrm e}^{2}}}+5 \ln \relax (x )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(16*exp(1)^2))+5)/x,x,method=_RETURNVERBOSE)

[Out]

(exp(exp(16*exp(1)^2))+5)*ln(x)

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maxima [A]  time = 0.36, size = 11, normalized size = 0.55 \begin {gather*} {\left (e^{\left (e^{\left (16 \, e^{2}\right )}\right )} + 5\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(exp(16*exp(1)^2))+5)/x,x, algorithm="maxima")

[Out]

(e^(e^(16*e^2)) + 5)*log(x)

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mupad [B]  time = 0.04, size = 11, normalized size = 0.55 \begin {gather*} \ln \relax (x)\,\left ({\mathrm {e}}^{{\mathrm {e}}^{16\,{\mathrm {e}}^2}}+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(16*exp(2))) + 5)/x,x)

[Out]

log(x)*(exp(exp(16*exp(2))) + 5)

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sympy [A]  time = 0.07, size = 12, normalized size = 0.60 \begin {gather*} \left (5 + e^{e^{16 e^{2}}}\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(exp(16*exp(1)**2))+5)/x,x)

[Out]

(5 + exp(exp(16*exp(2))))*log(x)

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