3.15.44 \(\int \frac {e^{-4-x+\frac {e^{-3-x-\log ^2(x)}}{2+\log (\frac {3}{x})}-\log ^2(x)} (e (1-2 x)-e x \log (\frac {3}{x})+(-4 e-2 e \log (\frac {3}{x})) \log (x))}{4 x+4 x \log (\frac {3}{x})+x \log ^2(\frac {3}{x})} \, dx\)

Optimal. Leaf size=28 \[ -4+e^{\frac {e^{-3-x-\log ^2(x)}}{2+\log \left (\frac {3}{x}\right )}} \]

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Rubi [F]  time = 22.91, 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^{-4-x+\frac {e^{-3-x-\log ^2(x)}}{2+\log \left (\frac {3}{x}\right )}-\log ^2(x)} \left (e (1-2 x)-e x \log \left (\frac {3}{x}\right )+\left (-4 e-2 e \log \left (\frac {3}{x}\right )\right ) \log (x)\right )}{4 x+4 x \log \left (\frac {3}{x}\right )+x \log ^2\left (\frac {3}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-4 - x + E^(-3 - x - Log[x]^2)/(2 + Log[3/x]) - Log[x]^2)*(E*(1 - 2*x) - E*x*Log[3/x] + (-4*E - 2*E*Lo
g[3/x])*Log[x]))/(4*x + 4*x*Log[3/x] + x*Log[3/x]^2),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 5.32, size = 26, normalized size = 0.93 \begin {gather*} e^{\frac {e^{-3-x-\log ^2(x)}}{2+\log \left (\frac {3}{x}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-4 - x + E^(-3 - x - Log[x]^2)/(2 + Log[3/x]) - Log[x]^2)*(E*(1 - 2*x) - E*x*Log[3/x] + (-4*E -
2*E*Log[3/x])*Log[x]))/(4*x + 4*x*Log[3/x] + x*Log[3/x]^2),x]

[Out]

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

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fricas [B]  time = 1.53, size = 123, normalized size = 4.39 \begin {gather*} e^{\left (\log \relax (3)^{2} - 2 \, \log \relax (3) \log \left (\frac {3}{x}\right ) + \log \left (\frac {3}{x}\right )^{2} + x + \frac {2 \, {\left (\log \relax (3) - 1\right )} \log \left (\frac {3}{x}\right )^{2} - \log \left (\frac {3}{x}\right )^{3} - 2 \, \log \relax (3)^{2} - {\left (\log \relax (3)^{2} + x - 4 \, \log \relax (3) + 4\right )} \log \left (\frac {3}{x}\right ) - 2 \, x + e^{\left (-\log \relax (3)^{2} + 2 \, \log \relax (3) \log \left (\frac {3}{x}\right ) - \log \left (\frac {3}{x}\right )^{2} - x - 3\right )} - 8}{\log \left (\frac {3}{x}\right ) + 2} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)*log(3/x)-4*exp(1))*log(x)-x*exp(1)*log(3/x)+(1-2*x)*exp(1))*exp(exp(1)/(log(3/x)+2)/exp(
log(x)^2+4+x))/(x*log(3/x)^2+4*x*log(3/x)+4*x)/exp(log(x)^2+4+x),x, algorithm="fricas")

[Out]

e^(log(3)^2 - 2*log(3)*log(3/x) + log(3/x)^2 + x + (2*(log(3) - 1)*log(3/x)^2 - log(3/x)^3 - 2*log(3)^2 - (log
(3)^2 + x - 4*log(3) + 4)*log(3/x) - 2*x + e^(-log(3)^2 + 2*log(3)*log(3/x) - log(3/x)^2 - x - 3) - 8)/(log(3/
x) + 2) + 4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (x e \log \left (\frac {3}{x}\right ) + {\left (2 \, x - 1\right )} e + 2 \, {\left (e \log \left (\frac {3}{x}\right ) + 2 \, e\right )} \log \relax (x)\right )} e^{\left (-\log \relax (x)^{2} - x + \frac {e^{\left (-\log \relax (x)^{2} - x - 3\right )}}{\log \left (\frac {3}{x}\right ) + 2} - 4\right )}}{x \log \left (\frac {3}{x}\right )^{2} + 4 \, x \log \left (\frac {3}{x}\right ) + 4 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)*log(3/x)-4*exp(1))*log(x)-x*exp(1)*log(3/x)+(1-2*x)*exp(1))*exp(exp(1)/(log(3/x)+2)/exp(
log(x)^2+4+x))/(x*log(3/x)^2+4*x*log(3/x)+4*x)/exp(log(x)^2+4+x),x, algorithm="giac")

[Out]

integrate(-(x*e*log(3/x) + (2*x - 1)*e + 2*(e*log(3/x) + 2*e)*log(x))*e^(-log(x)^2 - x + e^(-log(x)^2 - x - 3)
/(log(3/x) + 2) - 4)/(x*log(3/x)^2 + 4*x*log(3/x) + 4*x), x)

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maple [A]  time = 0.82, size = 25, normalized size = 0.89

method	result	size
risch	\({\mathrm e}^{\frac {{\mathrm e}^{-3-\ln \relax (x )^{2}-x}}{\ln \relax (3)-\ln \relax (x )+2}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(1)*ln(3/x)-4*exp(1))*ln(x)-x*exp(1)*ln(3/x)+(1-2*x)*exp(1))*exp(exp(1)/(ln(3/x)+2)/exp(ln(x)^2+4+
x))/(x*ln(3/x)^2+4*x*ln(3/x)+4*x)/exp(ln(x)^2+4+x),x,method=_RETURNVERBOSE)

[Out]

exp(1/(ln(3)-ln(x)+2)*exp(-3-ln(x)^2-x))

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maxima [A]  time = 1.77, size = 34, normalized size = 1.21 \begin {gather*} e^{\left (\frac {1}{{\left (e^{3} \log \relax (3) + 2 \, e^{3}\right )} e^{\left (\log \relax (x)^{2} + x\right )} - e^{\left (\log \relax (x)^{2} + x + 3\right )} \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)*log(3/x)-4*exp(1))*log(x)-x*exp(1)*log(3/x)+(1-2*x)*exp(1))*exp(exp(1)/(log(3/x)+2)/exp(
log(x)^2+4+x))/(x*log(3/x)^2+4*x*log(3/x)+4*x)/exp(log(x)^2+4+x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.48, size = 25, normalized size = 0.89 \begin {gather*} {\mathrm {e}}^{\frac {{\mathrm {e}}^{-{\ln \relax (x)}^2}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-3}}{\ln \left (\frac {1}{x}\right )+\ln \relax (3)+2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- x - log(x)^2 - 4)*exp((exp(- x - log(x)^2 - 4)*exp(1))/(log(3/x) + 2))*(log(x)*(4*exp(1) + 2*exp(1
)*log(3/x)) + exp(1)*(2*x - 1) + x*exp(1)*log(3/x)))/(4*x + 4*x*log(3/x) + x*log(3/x)^2),x)

[Out]

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

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sympy [A]  time = 3.19, size = 24, normalized size = 0.86 \begin {gather*} e^{\frac {e e^{- x - \log {\relax (x )}^{2} - 4}}{- \log {\relax (x )} + \log {\relax (3 )} + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(1)*ln(3/x)-4*exp(1))*ln(x)-x*exp(1)*ln(3/x)+(1-2*x)*exp(1))*exp(exp(1)/(ln(3/x)+2)/exp(ln(x
)**2+4+x))/(x*ln(3/x)**2+4*x*ln(3/x)+4*x)/exp(ln(x)**2+4+x),x)

[Out]

exp(E*exp(-x - log(x)**2 - 4)/(-log(x) + log(3) + 2))

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