3.28.95 \(\int \frac {2-e^{-5+x}}{8+4 e^{-5+x} x} \, dx\)

Optimal. Leaf size=26 \[ \frac {1}{4} \left (x-\log \left (2+e^{-5+x} x\right )\right )+\log \left (\frac {\log (4)}{4}\right ) \]

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Rubi [F]  time = 0.37, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {2-e^{-5+x}}{8+4 e^{-5+x} x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 - E^(-5 + x))/(8 + 4*E^(-5 + x)*x),x]

[Out]

-1/4*Log[x] + (E^5*Defer[Int][(2*E^5 + E^x*x)^(-1), x])/2 + (E^5*Defer[Int][1/(x*(2*E^5 + E^x*x)), x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {1}{4 x}+\frac {e^5 (1+x)}{2 x \left (2 e^5+e^x x\right )}\right ) \, dx\\ &=-\frac {\log (x)}{4}+\frac {1}{2} e^5 \int \frac {1+x}{x \left (2 e^5+e^x x\right )} \, dx\\ &=-\frac {\log (x)}{4}+\frac {1}{2} e^5 \int \left (\frac {1}{2 e^5+e^x x}+\frac {1}{x \left (2 e^5+e^x x\right )}\right ) \, dx\\ &=-\frac {\log (x)}{4}+\frac {1}{2} e^5 \int \frac {1}{2 e^5+e^x x} \, dx+\frac {1}{2} e^5 \int \frac {1}{x \left (2 e^5+e^x x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 22, normalized size = 0.85 \begin {gather*} \frac {x}{4}-\frac {1}{4} \log \left (2 e^5+e^x x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - E^(-5 + x))/(8 + 4*E^(-5 + x)*x),x]

[Out]

x/4 - Log[2*E^5 + E^x*x]/4

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fricas [A]  time = 0.46, size = 23, normalized size = 0.88 \begin {gather*} \frac {1}{4} \, x - \frac {1}{4} \, \log \relax (x) - \frac {1}{4} \, \log \left (\frac {x e^{\left (x - 5\right )} + 2}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(x-5)+2)/(4*x*exp(x-5)+8),x, algorithm="fricas")

[Out]

1/4*x - 1/4*log(x) - 1/4*log((x*e^(x - 5) + 2)/x)

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giac [A]  time = 0.41, size = 16, normalized size = 0.62 \begin {gather*} \frac {1}{4} \, x - \frac {1}{4} \, \log \left (x e^{x} + 2 \, e^{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(x-5)+2)/(4*x*exp(x-5)+8),x, algorithm="giac")

[Out]

1/4*x - 1/4*log(x*e^x + 2*e^5)

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maple [A]  time = 0.08, size = 17, normalized size = 0.65

method	result	size
norman	\(\frac {x}{4}-\frac {\ln \left (4 x \,{\mathrm e}^{x -5}+8\right )}{4}\)	\(17\)
risch	\(-\frac {\ln \relax (x )}{4}+\frac {x}{4}-\frac {5}{4}-\frac {\ln \left ({\mathrm e}^{x -5}+\frac {2}{x}\right )}{4}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-exp(x-5)+2)/(4*x*exp(x-5)+8),x,method=_RETURNVERBOSE)

[Out]

1/4*x-1/4*ln(4*x*exp(x-5)+8)

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maxima [A]  time = 0.57, size = 24, normalized size = 0.92 \begin {gather*} \frac {1}{4} \, x - \frac {1}{4} \, \log \relax (x) - \frac {1}{4} \, \log \left (\frac {x e^{x} + 2 \, e^{5}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(x-5)+2)/(4*x*exp(x-5)+8),x, algorithm="maxima")

[Out]

1/4*x - 1/4*log(x) - 1/4*log((x*e^x + 2*e^5)/x)

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mupad [B]  time = 0.08, size = 15, normalized size = 0.58 \begin {gather*} \frac {x}{4}-\frac {\ln \left (x\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^x+2\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x - 5) - 2)/(4*x*exp(x - 5) + 8),x)

[Out]

x/4 - log(x*exp(-5)*exp(x) + 2)/4

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sympy [A]  time = 0.15, size = 19, normalized size = 0.73 \begin {gather*} \frac {x}{4} - \frac {\log {\relax (x )}}{4} - \frac {\log {\left (e^{x - 5} + \frac {2}{x} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(x-5)+2)/(4*x*exp(x-5)+8),x)

[Out]

x/4 - log(x)/4 - log(exp(x - 5) + 2/x)/4

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