\(\int \frac {e^{-x^2} (8 e^{x^2}+3 x-3 x^3)}{2 \log (e^{-1+\frac {1}{4} e^{-x^2} (16 e^{x^2} x+3 x^2)})} \, dx\) [6437]

Optimal result

Integrand size = 59, antiderivative size = 22 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=\log \left (\log \left (e^{-1+x \left (4+\frac {3}{4} e^{-x^2} x\right )}\right )\right ) \]

[Out]

ln(ln(exp(x*(3/4*x/exp(x^2)+4))/exp(1)))

Rubi [F]

\[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=\int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx \]

[In]

Int[(8*E^x^2 + 3*x - 3*x^3)/(2*E^x^2*Log[E^(-1 + (16*E^x^2*x + 3*x^2)/(4*E^x^2))]),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{\log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx \\ & = \frac {1}{2} \int \left (\frac {8}{\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )}-\frac {3 e^{-x^2} x \left (-1+x^2\right )}{\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )}\right ) \, dx \\ & = -\left (\frac {3}{2} \int \frac {e^{-x^2} x \left (-1+x^2\right )}{\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )} \, dx\right )+4 \int \frac {1}{\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )} \, dx \\ & = -\left (\frac {3}{2} \int \left (-\frac {e^{-x^2} x}{\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )}+\frac {e^{-x^2} x^3}{\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )}\right ) \, dx\right )+4 \int \frac {1}{\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )} \, dx \\ & = \frac {3}{2} \int \frac {e^{-x^2} x}{\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )} \, dx-\frac {3}{2} \int \frac {e^{-x^2} x^3}{\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )} \, dx+4 \int \frac {1}{\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=\log \left (\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )\right ) \]

[In]

Integrate[(8*E^x^2 + 3*x - 3*x^3)/(2*E^x^2*Log[E^(-1 + (16*E^x^2*x + 3*x^2)/(4*E^x^2))]),x]

[Out]

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

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14

method	result	size
risch	\(\ln \left (\ln \left ({\mathrm e}^{\frac {x \left (16 \,{\mathrm e}^{x^{2}}+3 x \right ) {\mathrm e}^{-x^{2}}}{4}}\right )-1\right )\)	\(25\)
parallelrisch	\(\ln \left (\ln \left ({\mathrm e}^{\frac {x \left (16 \,{\mathrm e}^{x^{2}}+3 x \right ) {\mathrm e}^{-x^{2}}}{4}} {\mathrm e}^{-1}\right )\right )\)	\(28\)
derivativedivides	\(\ln \left (\ln \left ({\mathrm e}^{\frac {\left (16 \,{\mathrm e}^{x^{2}} x +3 x^{2}\right ) {\mathrm e}^{-x^{2}}}{4}} {\mathrm e}^{-1}\right )\right )\)	\(30\)
default	\(\ln \left (\ln \left ({\mathrm e}^{\frac {\left (16 \,{\mathrm e}^{x^{2}} x +3 x^{2}\right ) {\mathrm e}^{-x^{2}}}{4}} {\mathrm e}^{-1}\right )\right )\)	\(30\)

[In]

int(1/2*(8*exp(x^2)-3*x^3+3*x)/exp(x^2)/ln(exp(1/4*(16*exp(x^2)*x+3*x^2)/exp(x^2))/exp(1)),x,method=_RETURNVER
BOSE)

[Out]

ln(ln(exp(1/4*x*(16*exp(x^2)+3*x)*exp(-x^2)))-1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=-x^{2} + \log \left (4 \, x - 1\right ) + \log \left (\frac {3 \, x^{2} + 4 \, {\left (4 \, x - 1\right )} e^{\left (x^{2}\right )}}{4 \, x - 1}\right ) \]

[In]

integrate(1/2*(8*exp(x^2)-3*x^3+3*x)/exp(x^2)/log(exp(1/4*(16*exp(x^2)*x+3*x^2)/exp(x^2))/exp(1)),x, algorithm
="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=2 \log {\left (x \right )} + \log {\left (e^{- x^{2}} + \frac {16 x - 4}{3 x^{2}} \right )} \]

[In]

integrate(1/2*(8*exp(x**2)-3*x**3+3*x)/exp(x**2)/ln(exp(1/4*(16*exp(x**2)*x+3*x**2)/exp(x**2))/exp(1)),x)

[Out]

2*log(x) + log(exp(-x**2) + (16*x - 4)/(3*x**2))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).

Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=-x^{2} + \log \left (4 \, x - 1\right ) + \log \left (\frac {3 \, x^{2} + 4 \, {\left (4 \, x - 1\right )} e^{\left (x^{2}\right )}}{4 \, {\left (4 \, x - 1\right )}}\right ) \]

[In]

integrate(1/2*(8*exp(x^2)-3*x^3+3*x)/exp(x^2)/log(exp(1/4*(16*exp(x^2)*x+3*x^2)/exp(x^2))/exp(1)),x, algorithm
="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=\log \left (3 \, x^{2} e^{\left (-x^{2}\right )} + 16 \, x - 4\right ) \]

[In]

integrate(1/2*(8*exp(x^2)-3*x^3+3*x)/exp(x^2)/log(exp(1/4*(16*exp(x^2)*x+3*x^2)/exp(x^2))/exp(1)),x, algorithm
="giac")

[Out]

log(3*x^2*e^(-x^2) + 16*x - 4)

Mupad [B] (verification not implemented)

Time = 12.59 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=\ln \left (16\,x+3\,x^2\,{\mathrm {e}}^{-x^2}-4\right ) \]

[In]

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

[Out]

log(16*x + 3*x^2*exp(-x^2) - 4)