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

Optimal. Leaf size=22 \[ \log \left (\log \left (e^{-1+x \left (4+\frac {3}{4} e^{-x^2} x\right )}\right )\right ) \]

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Rubi [F]  time = 0.89, 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^{-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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 23, normalized size = 1.05 \begin {gather*} \log \left (\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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))]]

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fricas [A]  time = 0.52, size = 38, normalized size = 1.73 \begin {gather*} -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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.19, size = 26, normalized size = 1.18 \begin {gather*} -x^{2} + \log \left (3 \, x^{2} + 16 \, x e^{\left (x^{2}\right )} - 4 \, e^{\left (x^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

-x^2 + log(3*x^2 + 16*x*e^(x^2) - 4*e^(x^2))

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maple [A]  time = 0.57, size = 25, normalized size = 1.14

method	result	size
risch	\(\ln \left (\ln \left ({\mathrm e}^{\frac {x \left (16 \,{\mathrm e}^{x^{2}}+3 x \right ) {\mathrm e}^{-x^{2}}}{4}}\right )-1\right )\)	\(25\)
default	\(-x^{2}+\ln \left (16 \,{\mathrm e}^{x^{2}} x +4 \left (\ln \left ({\mathrm e}^{\frac {\left (16 \,{\mathrm e}^{x^{2}} x +3 x^{2}\right ) {\mathrm e}^{-x^{2}}}{4}} {\mathrm e}^{-1}\right )-\frac {\left (16 \,{\mathrm e}^{x^{2}} x +3 x^{2}\right ) {\mathrm e}^{-x^{2}}}{4}\right ) {\mathrm e}^{x^{2}}+3 x^{2}\right )\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [B]  time = 0.40, size = 39, normalized size = 1.77 \begin {gather*} -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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 4.71, size = 17, normalized size = 0.77 \begin {gather*} \ln \left (16\,x+3\,x^2\,{\mathrm {e}}^{-x^2}-4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.43, size = 22, normalized size = 1.00 \begin {gather*} 2 \log {\relax (x )} + \log {\left (e^{- x^{2}} + \frac {16 x - 4}{3 x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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