3.16.32 \(\int \frac {1+4 e^9}{13 e^9 x} \, dx\)

Optimal. Leaf size=14 \[ \frac {1}{13} \left (4+\left (4+\frac {1}{e^9}\right ) \log (x)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 11, normalized size of antiderivative = 0.79, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 29} \begin {gather*} \frac {1}{13} \left (4+\frac {1}{e^9}\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 4*E^9)/(13*E^9*x),x]

[Out]

((4 + E^(-9))*Log[x])/13

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} &=\frac {1}{13} \left (4+\frac {1}{e^9}\right ) \int \frac {1}{x} \, dx\\ &=\frac {1}{13} \left (4+\frac {1}{e^9}\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.14 \begin {gather*} \frac {\left (1+4 e^9\right ) \log (x)}{13 e^9} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 4*E^9)/(13*E^9*x),x]

[Out]

((1 + 4*E^9)*Log[x])/(13*E^9)

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fricas [A]  time = 0.50, size = 12, normalized size = 0.86 \begin {gather*} \frac {1}{13} \, {\left (4 \, e^{9} + 1\right )} e^{\left (-9\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/13*(4*exp(9)+1)/x/exp(9),x, algorithm="fricas")

[Out]

1/13*(4*e^9 + 1)*e^(-9)*log(x)

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giac [A]  time = 0.12, size = 13, normalized size = 0.93 \begin {gather*} \frac {1}{13} \, {\left (4 \, e^{9} + 1\right )} e^{\left (-9\right )} \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/13*(4*exp(9)+1)/x/exp(9),x, algorithm="giac")

[Out]

1/13*(4*e^9 + 1)*e^(-9)*log(abs(x))

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maple [A]  time = 0.02, size = 15, normalized size = 1.07

method	result	size
default	\(\frac {\left (4 \,{\mathrm e}^{9}+1\right ) {\mathrm e}^{-9} \ln \relax (x )}{13}\)	\(15\)
norman	\(\frac {\left (4 \,{\mathrm e}^{9}+1\right ) {\mathrm e}^{-9} \ln \relax (x )}{13}\)	\(15\)
risch	\(\frac {4 \,{\mathrm e}^{-9} \ln \relax (x ) {\mathrm e}^{9}}{13}+\frac {{\mathrm e}^{-9} \ln \relax (x )}{13}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/13*(4*exp(9)+1)/x/exp(9),x,method=_RETURNVERBOSE)

[Out]

1/13*(4*exp(9)+1)/exp(9)*ln(x)

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maxima [A]  time = 0.57, size = 12, normalized size = 0.86 \begin {gather*} \frac {1}{13} \, {\left (4 \, e^{9} + 1\right )} e^{\left (-9\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/13*(4*exp(9)+1)/x/exp(9),x, algorithm="maxima")

[Out]

1/13*(4*e^9 + 1)*e^(-9)*log(x)

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mupad [B]  time = 1.03, size = 12, normalized size = 0.86 \begin {gather*} \frac {{\mathrm {e}}^{-9}\,\ln \relax (x)\,\left (4\,{\mathrm {e}}^9+1\right )}{13} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-9)*((4*exp(9))/13 + 1/13))/x,x)

[Out]

(exp(-9)*log(x)*(4*exp(9) + 1))/13

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sympy [A]  time = 0.06, size = 14, normalized size = 1.00 \begin {gather*} \frac {\left (1 + 4 e^{9}\right ) \log {\relax (x )}}{13 e^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/13*(4*exp(9)+1)/x/exp(9),x)

[Out]

(1 + 4*exp(9))*exp(-9)*log(x)/13

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