3.40.95 \(\int \frac {e^{-3+e^{-\frac {x}{5-x+\log (2)}}-\frac {x}{5-x+\log (2)}} (-15-3 \log (2))}{50-20 x+2 x^2+(20-4 x) \log (2)+2 \log ^2(2)} \, dx\)

Optimal. Leaf size=21 \[ \frac {3}{2} e^{-3+e^{\frac {x}{-5+x-\log (2)}}} \]

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Rubi [F]  time = 0.53, 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 {e^{-3+e^{-\frac {x}{5-x+\log (2)}}-\frac {x}{5-x+\log (2)}} (-15-3 \log (2))}{50-20 x+2 x^2+(20-4 x) \log (2)+2 \log ^2(2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-3 + E^(-(x/(5 - x + Log[2]))) - x/(5 - x + Log[2]))*(-15 - 3*Log[2]))/(50 - 20*x + 2*x^2 + (20 - 4*x)
*Log[2] + 2*Log[2]^2),x]

[Out]

(-3*(5 + Log[2])*Defer[Int][E^(-3 + E^(x/(-5 + x - Log[2])) + x/(-5 + x - Log[2]))/(5 - x + Log[2])^2, x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=-\left ((15+3 \log (2)) \int \frac {e^{-3+e^{-\frac {x}{5-x+\log (2)}}-\frac {x}{5-x+\log (2)}}}{50-20 x+2 x^2+(20-4 x) \log (2)+2 \log ^2(2)} \, dx\right )\\ &=-\left ((15+3 \log (2)) \int \frac {e^{-3+e^{\frac {x}{-5+x-\log (2)}}+\frac {x}{-5+x-\log (2)}}}{2 (5-x+\log (2))^2} \, dx\right )\\ &=-\left (\frac {1}{2} (3 (5+\log (2))) \int \frac {e^{-3+e^{\frac {x}{-5+x-\log (2)}}+\frac {x}{-5+x-\log (2)}}}{(5-x+\log (2))^2} \, dx\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 21, normalized size = 1.00 \begin {gather*} \frac {3}{2} e^{-3+e^{\frac {x}{-5+x-\log (2)}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-3 + E^(-(x/(5 - x + Log[2]))) - x/(5 - x + Log[2]))*(-15 - 3*Log[2]))/(50 - 20*x + 2*x^2 + (20
- 4*x)*Log[2] + 2*Log[2]^2),x]

[Out]

(3*E^(-3 + E^(x/(-5 + x - Log[2]))))/2

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fricas [B]  time = 0.54, size = 55, normalized size = 2.62 \begin {gather*} \frac {3}{2} \, e^{\left (\frac {{\left (x - \log \relax (2) - 5\right )} e^{\left (\frac {x}{x - \log \relax (2) - 5}\right )} - 2 \, x + 3 \, \log \relax (2) + 15}{x - \log \relax (2) - 5} - \frac {x}{x - \log \relax (2) - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(2)-15)*exp(-x/(log(2)+5-x))*exp(exp(-x/(log(2)+5-x))-3)/(2*log(2)^2+(-4*x+20)*log(2)+2*x^2-2
0*x+50),x, algorithm="fricas")

[Out]

3/2*e^(((x - log(2) - 5)*e^(x/(x - log(2) - 5)) - 2*x + 3*log(2) + 15)/(x - log(2) - 5) - x/(x - log(2) - 5))

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giac [B]  time = 0.25, size = 99, normalized size = 4.71 \begin {gather*} \frac {3 \, {\left (\log \relax (2) + 5\right )} e^{\left (\frac {x e^{\left (\frac {x}{x - \log \relax (2) - 5}\right )} - e^{\left (\frac {x}{x - \log \relax (2) - 5}\right )} \log \relax (2) + x - 5 \, e^{\left (\frac {x}{x - \log \relax (2) - 5}\right )}}{x - \log \relax (2) - 5}\right )}}{2 \, {\left (e^{\left (\frac {x}{x - \log \relax (2) - 5} + 3\right )} \log \relax (2) + 5 \, e^{\left (\frac {x}{x - \log \relax (2) - 5} + 3\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(2)-15)*exp(-x/(log(2)+5-x))*exp(exp(-x/(log(2)+5-x))-3)/(2*log(2)^2+(-4*x+20)*log(2)+2*x^2-2
0*x+50),x, algorithm="giac")

[Out]

3/2*(log(2) + 5)*e^((x*e^(x/(x - log(2) - 5)) - e^(x/(x - log(2) - 5))*log(2) + x - 5*e^(x/(x - log(2) - 5)))/
(x - log(2) - 5))/(e^(x/(x - log(2) - 5) + 3)*log(2) + 5*e^(x/(x - log(2) - 5) + 3))

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maple [B]  time = 0.15, size = 52, normalized size = 2.48

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{{\mathrm e}^{-\frac {x}{\ln \relax (2)+5-x}}-3} \ln \relax (2)}{2 \left (\ln \relax (2)+5\right )}+\frac {15 \,{\mathrm e}^{{\mathrm e}^{-\frac {x}{\ln \relax (2)+5-x}}-3}}{2 \left (\ln \relax (2)+5\right )}\)	\(52\)
norman	\(\frac {\left (\frac {3 \ln \relax (2)}{2}+\frac {15}{2}\right ) {\mathrm e}^{{\mathrm e}^{-\frac {x}{\ln \relax (2)+5-x}}-3}-\frac {3 x \,{\mathrm e}^{{\mathrm e}^{-\frac {x}{\ln \relax (2)+5-x}}-3}}{2}}{\ln \relax (2)+5-x}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*ln(2)-15)*exp(-x/(ln(2)+5-x))*exp(exp(-x/(ln(2)+5-x))-3)/(2*ln(2)^2+(-4*x+20)*ln(2)+2*x^2-20*x+50),x,m
ethod=_RETURNVERBOSE)

[Out]

3/2/(ln(2)+5)*exp(exp(-x/(ln(2)+5-x))-3)*ln(2)+15/2/(ln(2)+5)*exp(exp(-x/(ln(2)+5-x))-3)

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maxima [A]  time = 0.45, size = 31, normalized size = 1.48 \begin {gather*} \frac {3}{2} \, e^{\left (e^{\left (\frac {\log \relax (2)}{x - \log \relax (2) - 5} + \frac {5}{x - \log \relax (2) - 5} + 1\right )} - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*log(2)-15)*exp(-x/(log(2)+5-x))*exp(exp(-x/(log(2)+5-x))-3)/(2*log(2)^2+(-4*x+20)*log(2)+2*x^2-2
0*x+50),x, algorithm="maxima")

[Out]

3/2*e^(e^(log(2)/(x - log(2) - 5) + 5/(x - log(2) - 5) + 1) - 3)

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mupad [B]  time = 3.50, size = 18, normalized size = 0.86 \begin {gather*} \frac {3\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {x}{\ln \relax (2)-x+5}}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(-x/(log(2) - x + 5)) - 3)*exp(-x/(log(2) - x + 5))*(3*log(2) + 15))/(2*log(2)^2 - log(2)*(4*x -
20) - 20*x + 2*x^2 + 50),x)

[Out]

(3*exp(-3)*exp(exp(-x/(log(2) - x + 5))))/2

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sympy [A]  time = 0.65, size = 17, normalized size = 0.81 \begin {gather*} \frac {3 e^{-3 + e^{- \frac {x}{- x + \log {\relax (2 )} + 5}}}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*ln(2)-15)*exp(-x/(ln(2)+5-x))*exp(exp(-x/(ln(2)+5-x))-3)/(2*ln(2)**2+(-4*x+20)*ln(2)+2*x**2-20*x
+50),x)

[Out]

3*exp(-3 + exp(-x/(-x + log(2) + 5)))/2

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