\(\int \frac {5+e^{e^{16 e^2}}}{x} \, dx\) [7660]

Optimal result

Integrand size = 15, antiderivative size = 20 \[ \int \frac {5+e^{e^{16 e^2}}}{x} \, dx=\left (5+e^{e^{16 e^2}}\right ) \log \left (\left (3+e^{25}\right ) x\right ) \]

[Out]

(exp(exp(16*exp(1)^2))+5)*ln((exp(25)+3)*x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 29} \[ \int \frac {5+e^{e^{16 e^2}}}{x} \, dx=\left (5+e^{e^{16 e^2}}\right ) \log (x) \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {5+e^{e^{16 e^2}}}{x} \, dx=\left (5+e^{e^{16 e^2}}\right ) \log (x) \]

[In]

Integrate[(5 + E^E^(16*E^2))/x,x]

[Out]

(5 + E^E^(16*E^2))*Log[x]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {5+e^{e^{16 e^2}}}{x} \, dx=e^{\left (e^{\left (16 \, e^{2}\right )}\right )} \log \left (x\right ) + 5 \, \log \left (x\right ) \]

[In]

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

[Out]

e^(e^(16*e^2))*log(x) + 5*log(x)

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {5+e^{e^{16 e^2}}}{x} \, dx=\left (5 + e^{e^{16 e^{2}}}\right ) \log {\left (x \right )} \]

[In]

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

[Out]

(5 + exp(exp(16*exp(2))))*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int \frac {5+e^{e^{16 e^2}}}{x} \, dx={\left (e^{\left (e^{\left (16 \, e^{2}\right )}\right )} + 5\right )} \log \left (x\right ) \]

[In]

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

[Out]

(e^(e^(16*e^2)) + 5)*log(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {5+e^{e^{16 e^2}}}{x} \, dx={\left (e^{\left (e^{\left (16 \, e^{2}\right )}\right )} + 5\right )} \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

(e^(e^(16*e^2)) + 5)*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.55 \[ \int \frac {5+e^{e^{16 e^2}}}{x} \, dx=\ln \left (x\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^{16\,{\mathrm {e}}^2}}+5\right ) \]

[In]

int((exp(exp(16*exp(2))) + 5)/x,x)

[Out]

log(x)*(exp(exp(16*exp(2))) + 5)