3.59.26 \(\int \frac {2}{60 e^{\frac {1}{e^3 \log (5)}}+x} \, dx\)

Optimal. Leaf size=19 \[ \log \left (\left (e^{\frac {1}{e^3 \log (5)}}+\frac {x}{60}\right )^2\right ) \]

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Rubi [A]  time = 0.00, antiderivative size = 17, normalized size of antiderivative = 0.89, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 31} \begin {gather*} 2 \log \left (x+60 e^{\frac {1}{e^3 \log (5)}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2/(60*E^(1/(E^3*Log[5])) + x),x]

[Out]

2*Log[60*E^(1/(E^3*Log[5])) + 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=2 \int \frac {1}{60 e^{\frac {1}{e^3 \log (5)}}+x} \, dx\\ &=2 \log \left (60 e^{\frac {1}{e^3 \log (5)}}+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 0.89 \begin {gather*} 2 \log \left (60 e^{\frac {1}{e^3 \log (5)}}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2/(60*E^(1/(E^3*Log[5])) + x),x]

[Out]

2*Log[60*E^(1/(E^3*Log[5])) + x]

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fricas [A]  time = 0.95, size = 15, normalized size = 0.79 \begin {gather*} 2 \, \log \left (x + 60 \, e^{\left (\frac {e^{\left (-3\right )}}{\log \relax (5)}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(60*exp(1/exp(3)/log(5))+x),x, algorithm="fricas")

[Out]

2*log(x + 60*e^(e^(-3)/log(5)))

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giac [A]  time = 0.17, size = 16, normalized size = 0.84 \begin {gather*} 2 \, \log \left ({\left | x + 60 \, e^{\left (\frac {e^{\left (-3\right )}}{\log \relax (5)}\right )} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(60*exp(1/exp(3)/log(5))+x),x, algorithm="giac")

[Out]

2*log(abs(x + 60*e^(e^(-3)/log(5))))

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maple [A]  time = 0.28, size = 16, normalized size = 0.84

method	result	size
risch	\(2 \ln \left (60 \,{\mathrm e}^{\frac {{\mathrm e}^{-3}}{\ln \relax (5)}}+x \right )\)	\(16\)
default	\(2 \ln \left (60 \,{\mathrm e}^{\frac {{\mathrm e}^{-3}}{\ln \relax (5)}}+x \right )\)	\(18\)
norman	\(2 \ln \left (60 \,{\mathrm e}^{\frac {{\mathrm e}^{-3}}{\ln \relax (5)}}+x \right )\)	\(18\)
meijerg	\(2 \ln \left (1+\frac {x \,{\mathrm e}^{-\frac {{\mathrm e}^{-3}}{\ln \relax (5)}}}{60}\right )\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/(60*exp(1/exp(3)/ln(5))+x),x,method=_RETURNVERBOSE)

[Out]

2*ln(60*exp(exp(-3)/ln(5))+x)

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maxima [A]  time = 0.35, size = 15, normalized size = 0.79 \begin {gather*} 2 \, \log \left (x + 60 \, e^{\left (\frac {e^{\left (-3\right )}}{\log \relax (5)}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(60*exp(1/exp(3)/log(5))+x),x, algorithm="maxima")

[Out]

2*log(x + 60*e^(e^(-3)/log(5)))

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mupad [B]  time = 0.07, size = 15, normalized size = 0.79 \begin {gather*} 2\,\ln \left (x+60\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-3}}{\ln \relax (5)}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2/(x + 60*exp(exp(-3)/log(5))),x)

[Out]

2*log(x + 60*exp(exp(-3)/log(5)))

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sympy [A]  time = 0.06, size = 15, normalized size = 0.79 \begin {gather*} 2 \log {\left (x + 60 e^{\frac {1}{e^{3} \log {\relax (5 )}}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2/(60*exp(1/exp(3)/ln(5))+x),x)

[Out]

2*log(x + 60*exp(exp(-3)/log(5)))

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