\(\int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} (-36-24 x^2+(-12-8 x^2) \log (5))}{x^4} \, dx\) [2432]

Optimal result

Integrand size = 47, antiderivative size = 18 \[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=e^{\frac {4 e^{-x^2} (3+\log (5))}{x^3}} \]

[Out]

exp(2/x^3*(ln(5)+3)/exp(x^2))^2

Rubi [F]

\[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=\int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx \]

[In]

Int[(E^(-x^2 + (2*(6 + 2*Log[5]))/(E^x^2*x^3))*(-36 - 24*x^2 + (-12 - 8*x^2)*Log[5]))/x^4,x]

[Out]

-12*(3 + Log[5])*Defer[Int][E^(-x^2 + (2*(6 + 2*Log[5]))/(E^x^2*x^3))/x^4, x] - 8*(3 + Log[5])*Defer[Int][E^(-
x^2 + (2*(6 + 2*Log[5]))/(E^x^2*x^3))/x^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-12 (3+\log (5))-8 x^2 (3+\log (5))\right )}{x^4} \, dx \\ & = \int \left (-\frac {12 e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} (3+\log (5))}{x^4}-\frac {8 e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} (3+\log (5))}{x^2}\right ) \, dx \\ & = -\left ((8 (3+\log (5))) \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}}}{x^2} \, dx\right )-(12 (3+\log (5))) \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}}}{x^4} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61 \[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=5^{\frac {4 e^{-x^2}}{x^3}} e^{\frac {12 e^{-x^2}}{x^3}} \]

[In]

Integrate[(E^(-x^2 + (2*(6 + 2*Log[5]))/(E^x^2*x^3))*(-36 - 24*x^2 + (-12 - 8*x^2)*Log[5]))/x^4,x]

[Out]

5^(4/(E^x^2*x^3))*E^(12/(E^x^2*x^3))

Maple [A] (verified)

Time = 4.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
risch	\({\mathrm e}^{\frac {4 \left (\ln \left (5\right )+3\right ) {\mathrm e}^{-x^{2}}}{x^{3}}}\)	\(17\)
norman	\({\mathrm e}^{\frac {2 \left (2 \ln \left (5\right )+6\right ) {\mathrm e}^{-x^{2}}}{x^{3}}}\)	\(20\)

[In]

int(((-8*x^2-12)*ln(5)-24*x^2-36)*exp((2*ln(5)+6)/x^3/exp(x^2))^2/x^4/exp(x^2),x,method=_RETURNVERBOSE)

[Out]

exp(4*(ln(5)+3)*exp(-x^2)/x^3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=e^{\left (x^{2} - \frac {x^{5} - 4 \, {\left (\log \left (5\right ) + 3\right )} e^{\left (-x^{2}\right )}}{x^{3}}\right )} \]

[In]

integrate(((-8*x^2-12)*log(5)-24*x^2-36)*exp((2*log(5)+6)/x^3/exp(x^2))^2/x^4/exp(x^2),x, algorithm="fricas")

[Out]

e^(x^2 - (x^5 - 4*(log(5) + 3)*e^(-x^2))/x^3)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=e^{\frac {2 \cdot \left (2 \log {\left (5 \right )} + 6\right ) e^{- x^{2}}}{x^{3}}} \]

[In]

integrate(((-8*x**2-12)*ln(5)-24*x**2-36)*exp((2*ln(5)+6)/x**3/exp(x**2))**2/x**4/exp(x**2),x)

[Out]

exp(2*(2*log(5) + 6)*exp(-x**2)/x**3)

Maxima [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=e^{\left (\frac {4 \, e^{\left (-x^{2}\right )} \log \left (5\right )}{x^{3}} + \frac {12 \, e^{\left (-x^{2}\right )}}{x^{3}}\right )} \]

[In]

integrate(((-8*x^2-12)*log(5)-24*x^2-36)*exp((2*log(5)+6)/x^3/exp(x^2))^2/x^4/exp(x^2),x, algorithm="maxima")

[Out]

e^(4*e^(-x^2)*log(5)/x^3 + 12*e^(-x^2)/x^3)

Giac [F]

\[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=\int { -\frac {4 \, {\left (6 \, x^{2} + {\left (2 \, x^{2} + 3\right )} \log \left (5\right ) + 9\right )} e^{\left (-x^{2} + \frac {4 \, {\left (\log \left (5\right ) + 3\right )} e^{\left (-x^{2}\right )}}{x^{3}}\right )}}{x^{4}} \,d x } \]

[In]

integrate(((-8*x^2-12)*log(5)-24*x^2-36)*exp((2*log(5)+6)/x^3/exp(x^2))^2/x^4/exp(x^2),x, algorithm="giac")

[Out]

integrate(-4*(6*x^2 + (2*x^2 + 3)*log(5) + 9)*e^(-x^2 + 4*(log(5) + 3)*e^(-x^2)/x^3)/x^4, x)

Mupad [B] (verification not implemented)

Time = 9.55 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.44 \[ \int \frac {e^{-x^2+\frac {2 e^{-x^2} (6+2 \log (5))}{x^3}} \left (-36-24 x^2+\left (-12-8 x^2\right ) \log (5)\right )}{x^4} \, dx=5^{\frac {4\,{\mathrm {e}}^{-x^2}}{x^3}}\,{\mathrm {e}}^{\frac {12\,{\mathrm {e}}^{-x^2}}{x^3}} \]

[In]

int(-(exp((2*exp(-x^2)*(2*log(5) + 6))/x^3)*exp(-x^2)*(log(5)*(8*x^2 + 12) + 24*x^2 + 36))/x^4,x)

[Out]

5^((4*exp(-x^2))/x^3)*exp((12*exp(-x^2))/x^3)