\(\int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} (4 x^2+8 x^3-8 x^2 \log (3)+(4 x^2-4 x \log (3)) \log (-x+\log (3))))}{-x+\log (3)} \, dx\) [8187]

Optimal result

Integrand size = 106, antiderivative size = 24 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=e^{-e^{4 x (x+\log (-x+\log (3)))}-x} x \]

[Out]

x/exp(x+exp(4*x*(ln(ln(3)-x)+x)))

Rubi [F]

\[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=\int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx \]

[In]

Int[(E^(-E^(4*x^2 + 4*x*Log[-x + Log[3]]) - x)*(-x + x^2 + (1 - x)*Log[3] + E^(4*x^2 + 4*x*Log[-x + Log[3]])*(
4*x^2 + 8*x^3 - 8*x^2*Log[3] + (4*x^2 - 4*x*Log[3])*Log[-x + Log[3]])))/(-x + Log[3]),x]

[Out]

Defer[Int][E^(-E^(4*x^2 + 4*x*Log[-x + Log[3]]) - x), x] - Defer[Int][E^(-E^(4*x^2 + 4*x*Log[-x + Log[3]]) - x
)*x, x] + 4*(1 - Log[9])*Defer[Int][E^(-E^(4*x^2 + 4*x*Log[-x + Log[3]]) - x + 4*x^2)*x^2*(-x + Log[3])^(-1 +
4*x), x] + 8*Defer[Int][E^(-E^(4*x^2 + 4*x*Log[-x + Log[3]]) - x + 4*x^2)*x^3*(-x + Log[3])^(-1 + 4*x), x] - 4
*Defer[Int][E^(-E^(4*x^2 + 4*x*Log[-x + Log[3]]) - x + 4*x^2)*x*(-x + Log[3])^(4*x)*Log[-x + Log[3]], x]

Rubi steps \begin{align*} \text {integral}& = \int \left (e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x}-e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} x+4 \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x (-x+\log (3))^{-1+4 x} \left (2 x^2+x (1-\log (9))+x \log (-x+\log (3))-\log (3) \log (-x+\log (3))\right )\right ) \, dx \\ & = 4 \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x (-x+\log (3))^{-1+4 x} \left (2 x^2+x (1-\log (9))+x \log (-x+\log (3))-\log (3) \log (-x+\log (3))\right ) \, dx+\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \, dx-\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} x \, dx \\ & = 4 \int \left (\exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x^2 (-x+\log (3))^{-1+4 x} (1+2 x-\log (9))-\exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x (-x+\log (3))^{4 x} \log (-x+\log (3))\right ) \, dx+\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \, dx-\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} x \, dx \\ & = 4 \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x^2 (-x+\log (3))^{-1+4 x} (1+2 x-\log (9)) \, dx-4 \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x (-x+\log (3))^{4 x} \log (-x+\log (3)) \, dx+\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \, dx-\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} x \, dx \\ & = 4 \int \left (2 \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x^3 (-x+\log (3))^{-1+4 x}-\exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x^2 (-x+\log (3))^{-1+4 x} (-1+\log (9))\right ) \, dx-4 \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x (-x+\log (3))^{4 x} \log (-x+\log (3)) \, dx+\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \, dx-\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} x \, dx \\ & = -\left (4 \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x (-x+\log (3))^{4 x} \log (-x+\log (3)) \, dx\right )+8 \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x^3 (-x+\log (3))^{-1+4 x} \, dx+(4 (1-\log (9))) \int \exp \left (-e^{4 x^2+4 x \log (-x+\log (3))}-x+4 x^2\right ) x^2 (-x+\log (3))^{-1+4 x} \, dx+\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \, dx-\int e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} x \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=\int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx \]

[In]

Integrate[(E^(-E^(4*x^2 + 4*x*Log[-x + Log[3]]) - x)*(-x + x^2 + (1 - x)*Log[3] + E^(4*x^2 + 4*x*Log[-x + Log[
3]])*(4*x^2 + 8*x^3 - 8*x^2*Log[3] + (4*x^2 - 4*x*Log[3])*Log[-x + Log[3]])))/(-x + Log[3]),x]

[Out]

Integrate[(E^(-E^(4*x^2 + 4*x*Log[-x + Log[3]]) - x)*(-x + x^2 + (1 - x)*Log[3] + E^(4*x^2 + 4*x*Log[-x + Log[
3]])*(4*x^2 + 8*x^3 - 8*x^2*Log[3] + (4*x^2 - 4*x*Log[3])*Log[-x + Log[3]])))/(-x + Log[3]), x]

Maple [A] (verified)

Time = 1.35 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(x \,{\mathrm e}^{-x -{\mathrm e}^{4 x \left (\ln \left (\ln \left (3\right )-x \right )+x \right )}}\)	\(21\)
risch	\(x \,{\mathrm e}^{-\left (\ln \left (3\right )-x \right )^{4 x} {\mathrm e}^{4 x^{2}}-x}\)	\(26\)

[In]

int((((-4*x*ln(3)+4*x^2)*ln(ln(3)-x)-8*x^2*ln(3)+8*x^3+4*x^2)*exp(4*x*ln(ln(3)-x)+4*x^2)+(1-x)*ln(3)+x^2-x)/(l
n(3)-x)/exp(exp(4*x*ln(ln(3)-x)+4*x^2)+x),x,method=_RETURNVERBOSE)

[Out]

x/exp(x+exp(4*x*(ln(ln(3)-x)+x)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=x e^{\left (-x - e^{\left (4 \, x^{2} + 4 \, x \log \left (-x + \log \left (3\right )\right )\right )}\right )} \]

[In]

integrate((((-4*x*log(3)+4*x^2)*log(log(3)-x)-8*x^2*log(3)+8*x^3+4*x^2)*exp(4*x*log(log(3)-x)+4*x^2)+(1-x)*log
(3)+x^2-x)/(log(3)-x)/exp(exp(4*x*log(log(3)-x)+4*x^2)+x),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=\text {Timed out} \]

[In]

integrate((((-4*x*ln(3)+4*x**2)*ln(ln(3)-x)-8*x**2*ln(3)+8*x**3+4*x**2)*exp(4*x*ln(ln(3)-x)+4*x**2)+(1-x)*ln(3
)+x**2-x)/(ln(3)-x)/exp(exp(4*x*ln(ln(3)-x)+4*x**2)+x),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=x e^{\left (-x - e^{\left (4 \, x^{2} + 4 \, x \log \left (-x + \log \left (3\right )\right )\right )}\right )} \]

[In]

integrate((((-4*x*log(3)+4*x^2)*log(log(3)-x)-8*x^2*log(3)+8*x^3+4*x^2)*exp(4*x*log(log(3)-x)+4*x^2)+(1-x)*log
(3)+x^2-x)/(log(3)-x)/exp(exp(4*x*log(log(3)-x)+4*x^2)+x),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=\int { -\frac {{\left (x^{2} + 4 \, {\left (2 \, x^{3} - 2 \, x^{2} \log \left (3\right ) + x^{2} + {\left (x^{2} - x \log \left (3\right )\right )} \log \left (-x + \log \left (3\right )\right )\right )} e^{\left (4 \, x^{2} + 4 \, x \log \left (-x + \log \left (3\right )\right )\right )} - {\left (x - 1\right )} \log \left (3\right ) - x\right )} e^{\left (-x - e^{\left (4 \, x^{2} + 4 \, x \log \left (-x + \log \left (3\right )\right )\right )}\right )}}{x - \log \left (3\right )} \,d x } \]

[In]

integrate((((-4*x*log(3)+4*x^2)*log(log(3)-x)-8*x^2*log(3)+8*x^3+4*x^2)*exp(4*x*log(log(3)-x)+4*x^2)+(1-x)*log
(3)+x^2-x)/(log(3)-x)/exp(exp(4*x*log(log(3)-x)+4*x^2)+x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.79 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{-e^{4 x^2+4 x \log (-x+\log (3))}-x} \left (-x+x^2+(1-x) \log (3)+e^{4 x^2+4 x \log (-x+\log (3))} \left (4 x^2+8 x^3-8 x^2 \log (3)+\left (4 x^2-4 x \log (3)\right ) \log (-x+\log (3))\right )\right )}{-x+\log (3)} \, dx=x\,{\mathrm {e}}^{-{\mathrm {e}}^{4\,x^2}\,{\left (\ln \left (3\right )-x\right )}^{4\,x}}\,{\mathrm {e}}^{-x} \]

[In]

int((exp(- x - exp(4*x^2 + 4*x*log(log(3) - x)))*(x + log(3)*(x - 1) - x^2 + exp(4*x^2 + 4*x*log(log(3) - x))*
(log(log(3) - x)*(4*x*log(3) - 4*x^2) + 8*x^2*log(3) - 4*x^2 - 8*x^3)))/(x - log(3)),x)

[Out]

x*exp(-exp(4*x^2)*(log(3) - x)^(4*x))*exp(-x)