3.32.50 \(\int \frac {1}{e^4 (2+x)} \, dx\)

Optimal. Leaf size=10 \[ \frac {\log (3 (2+x))}{e^4} \]

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Rubi [A]  time = 0.00, antiderivative size = 8, normalized size of antiderivative = 0.80, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {12, 31} \begin {gather*} \frac {\log (x+2)}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^4*(2 + x)),x]

[Out]

Log[2 + x]/E^4

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} &=\frac {\int \frac {1}{2+x} \, dx}{e^4}\\ &=\frac {\log (2+x)}{e^4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 8, normalized size = 0.80 \begin {gather*} \frac {\log (2+x)}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^4*(2 + x)),x]

[Out]

Log[2 + x]/E^4

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fricas [A]  time = 0.81, size = 7, normalized size = 0.70 \begin {gather*} e^{\left (-4\right )} \log \left (x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-4)*log(x + 2)

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giac [A]  time = 0.21, size = 8, normalized size = 0.80 \begin {gather*} e^{\left (-4\right )} \log \left ({\left | x + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-4)*log(abs(x + 2))

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maple [A]  time = 0.51, size = 8, normalized size = 0.80

method	result	size
risch	\({\mathrm e}^{-4} \ln \left (2+x \right )\)	\(8\)
default	\({\mathrm e}^{-4} \ln \left (2+x \right )\)	\(10\)
norman	\({\mathrm e}^{-4} \ln \left (2+x \right )\)	\(10\)
meijerg	\({\mathrm e}^{-4} \ln \left (1+\frac {x}{2}\right )\)	\(10\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 7, normalized size = 0.70 \begin {gather*} e^{\left (-4\right )} \log \left (x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-4)*log(x + 2)

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mupad [B]  time = 1.88, size = 7, normalized size = 0.70 \begin {gather*} \ln \left (x+2\right )\,{\mathrm {e}}^{-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-4)/(x + 2),x)

[Out]

log(x + 2)*exp(-4)

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sympy [A]  time = 0.06, size = 14, normalized size = 1.40 \begin {gather*} \frac {\log {\left (x e^{4} + 2 e^{4} \right )}}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(-4)*log(x*exp(4) + 2*exp(4))

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