\(\int \frac {e^{-\frac {-e^{5 x^2} \log (4) \log (5)+(1+4 \log (4)) \log (5)}{\log (4)}} (-2+20 e^{5 x^2} x^2 \log (5))}{x^2} \, dx\) [3326]

Optimal result

Integrand size = 51, antiderivative size = 24 \[ \int \frac {e^{-\frac {-e^{5 x^2} \log (4) \log (5)+(1+4 \log (4)) \log (5)}{\log (4)}} \left (-2+20 e^{5 x^2} x^2 \log (5)\right )}{x^2} \, dx=9+\frac {2\ 5^{-4+e^{5 x^2}-\frac {1}{\log (4)}}}{x} \]

[Out]

9+2/exp((1/2/ln(2)+4-exp(5*x^2))*ln(5))/x

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {2306, 2326} \[ \int \frac {e^{-\frac {-e^{5 x^2} \log (4) \log (5)+(1+4 \log (4)) \log (5)}{\log (4)}} \left (-2+20 e^{5 x^2} x^2 \log (5)\right )}{x^2} \, dx=\frac {2\ 5^{e^{5 x^2}-4-\frac {1}{\log (4)}}}{x} \]

[In]

Int[(-2 + 20*E^(5*x^2)*x^2*Log[5])/(E^((-(E^(5*x^2)*Log[4]*Log[5]) + (1 + 4*Log[4])*Log[5])/Log[4])*x^2),x]

[Out]

(2*5^(-4 + E^(5*x^2) - Log[4]^(-1)))/x

Rule 2306

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {5^{e^{5 x^2}-\frac {1+4 \log (4)}{\log (4)}} \left (-2+20 e^{5 x^2} x^2 \log (5)\right )}{x^2} \, dx \\ & = \frac {2\ 5^{-4+e^{5 x^2}-\frac {1}{\log (4)}}}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-\frac {-e^{5 x^2} \log (4) \log (5)+(1+4 \log (4)) \log (5)}{\log (4)}} \left (-2+20 e^{5 x^2} x^2 \log (5)\right )}{x^2} \, dx=\frac {2\ 5^{-4+e^{5 x^2}-\frac {1}{\log (4)}}}{x} \]

[In]

Integrate[(-2 + 20*E^(5*x^2)*x^2*Log[5])/(E^((-(E^(5*x^2)*Log[4]*Log[5]) + (1 + 4*Log[4])*Log[5])/Log[4])*x^2)
,x]

[Out]

(2*5^(-4 + E^(5*x^2) - Log[4]^(-1)))/x

Maple [A] (verified)

Time = 1.40 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33

method	result	size
risch	\(\frac {2 \,5^{\frac {2 \ln \left (2\right ) {\mathrm e}^{5 x^{2}}+\ln \left (\frac {1}{256}\right )-1}{2 \ln \left (2\right )}}}{x}\)	\(32\)
parallelrisch	\(\frac {2 \,{\mathrm e}^{\frac {\ln \left (5\right ) \left (2 \ln \left (2\right ) {\mathrm e}^{5 x^{2}}+\ln \left (\frac {1}{256}\right )-1\right )}{2 \ln \left (2\right )}}}{x}\)	\(33\)
norman	\(\frac {2 \,{\mathrm e}^{-\frac {-2 \ln \left (2\right ) \ln \left (5\right ) {\mathrm e}^{5 x^{2}}+\left (8 \ln \left (2\right )+1\right ) \ln \left (5\right )}{2 \ln \left (2\right )}}}{x}\)	\(37\)

[In]

int((20*x^2*ln(5)*exp(5*x^2)-2)/x^2/exp(1/2*(-2*ln(2)*ln(5)*exp(5*x^2)+(8*ln(2)+1)*ln(5))/ln(2)),x,method=_RET
URNVERBOSE)

[Out]

2/x/(5^(-1/2*(2*ln(2)*exp(5*x^2)-8*ln(2)-1)/ln(2)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^{-\frac {-e^{5 x^2} \log (4) \log (5)+(1+4 \log (4)) \log (5)}{\log (4)}} \left (-2+20 e^{5 x^2} x^2 \log (5)\right )}{x^2} \, dx=\frac {2 \, e^{\left (\frac {2 \, e^{\left (5 \, x^{2}\right )} \log \left (5\right ) \log \left (2\right ) - {\left (8 \, \log \left (2\right ) + 1\right )} \log \left (5\right )}{2 \, \log \left (2\right )}\right )}}{x} \]

[In]

integrate((20*x^2*log(5)*exp(5*x^2)-2)/x^2/exp(1/2*(-2*log(2)*log(5)*exp(5*x^2)+(8*log(2)+1)*log(5))/log(2)),x
, algorithm="fricas")

[Out]

2*e^(1/2*(2*e^(5*x^2)*log(5)*log(2) - (8*log(2) + 1)*log(5))/log(2))/x

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-\frac {-e^{5 x^2} \log (4) \log (5)+(1+4 \log (4)) \log (5)}{\log (4)}} \left (-2+20 e^{5 x^2} x^2 \log (5)\right )}{x^2} \, dx=\frac {2 e^{- \frac {- e^{5 x^{2}} \log {\left (2 \right )} \log {\left (5 \right )} + \frac {\left (1 + 8 \log {\left (2 \right )}\right ) \log {\left (5 \right )}}{2}}{\log {\left (2 \right )}}}}{x} \]

[In]

integrate((20*x**2*ln(5)*exp(5*x**2)-2)/x**2/exp(1/2*(-2*ln(2)*ln(5)*exp(5*x**2)+(8*ln(2)+1)*ln(5))/ln(2)),x)

[Out]

2*exp(-(-exp(5*x**2)*log(2)*log(5) + (1 + 8*log(2))*log(5)/2)/log(2))/x

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-\frac {-e^{5 x^2} \log (4) \log (5)+(1+4 \log (4)) \log (5)}{\log (4)}} \left (-2+20 e^{5 x^2} x^2 \log (5)\right )}{x^2} \, dx=\frac {2 \cdot 5^{-\frac {1}{2 \, \log \left (2\right )} - 4} 5^{e^{\left (5 \, x^{2}\right )}}}{x} \]

[In]

integrate((20*x^2*log(5)*exp(5*x^2)-2)/x^2/exp(1/2*(-2*log(2)*log(5)*exp(5*x^2)+(8*log(2)+1)*log(5))/log(2)),x
, algorithm="maxima")

[Out]

2*5^(-1/2/log(2) - 4)*5^e^(5*x^2)/x

Giac [F]

\[ \int \frac {e^{-\frac {-e^{5 x^2} \log (4) \log (5)+(1+4 \log (4)) \log (5)}{\log (4)}} \left (-2+20 e^{5 x^2} x^2 \log (5)\right )}{x^2} \, dx=\int { \frac {2 \, {\left (10 \, x^{2} e^{\left (5 \, x^{2}\right )} \log \left (5\right ) - 1\right )} e^{\left (\frac {2 \, e^{\left (5 \, x^{2}\right )} \log \left (5\right ) \log \left (2\right ) - {\left (8 \, \log \left (2\right ) + 1\right )} \log \left (5\right )}{2 \, \log \left (2\right )}\right )}}{x^{2}} \,d x } \]

[In]

integrate((20*x^2*log(5)*exp(5*x^2)-2)/x^2/exp(1/2*(-2*log(2)*log(5)*exp(5*x^2)+(8*log(2)+1)*log(5))/log(2)),x
, algorithm="giac")

[Out]

integrate(2*(10*x^2*e^(5*x^2)*log(5) - 1)*e^(1/2*(2*e^(5*x^2)*log(5)*log(2) - (8*log(2) + 1)*log(5))/log(2))/x
^2, x)

Mupad [B] (verification not implemented)

Time = 8.55 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-\frac {-e^{5 x^2} \log (4) \log (5)+(1+4 \log (4)) \log (5)}{\log (4)}} \left (-2+20 e^{5 x^2} x^2 \log (5)\right )}{x^2} \, dx=\frac {2\,5^{{\mathrm {e}}^{5\,x^2}}}{625\,5^{\frac {1}{2\,\ln \left (2\right )}}\,x} \]

[In]

int((exp(-((log(5)*(8*log(2) + 1))/2 - exp(5*x^2)*log(2)*log(5))/log(2))*(20*x^2*exp(5*x^2)*log(5) - 2))/x^2,x
)

[Out]

(2*5^exp(5*x^2))/(625*5^(1/(2*log(2)))*x)