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

Optimal. Leaf size=24 \[ 9+\frac {2\ 5^{-4+e^{5 x^2}-\frac {1}{\log (4)}}}{x} \]

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Rubi [A]  time = 0.24, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {2274, 2288} \begin {gather*} \frac {2\ 5^{e^{5 x^2}-4-\frac {1}{\log (4)}}}{x} \end {gather*}

Antiderivative was successfully verified.

[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 2274

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 2288

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

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Mathematica [A]  time = 0.05, size = 22, normalized size = 0.92 \begin {gather*} \frac {2\ 5^{-4+e^{5 x^2}-\frac {1}{\log (4)}}}{x} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.55, size = 35, normalized size = 1.46 \begin {gather*} \frac {2 \, e^{\left (\frac {2 \, e^{\left (5 \, x^{2}\right )} \log \relax (5) \log \relax (2) - {\left (8 \, \log \relax (2) + 1\right )} \log \relax (5)}{2 \, \log \relax (2)}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (10 \, x^{2} e^{\left (5 \, x^{2}\right )} \log \relax (5) - 1\right )} e^{\left (\frac {2 \, e^{\left (5 \, x^{2}\right )} \log \relax (5) \log \relax (2) - {\left (8 \, \log \relax (2) + 1\right )} \log \relax (5)}{2 \, \log \relax (2)}\right )}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.11, size = 32, normalized size = 1.33

method	result	size
risch	\(\frac {2 \,5^{\frac {{\mathrm e}^{5 x^{2}} \ln \relax (4)-8 \ln \relax (2)-1}{2 \ln \relax (2)}}}{x}\)	\(32\)
norman	\(\frac {2 \,{\mathrm e}^{-\frac {{\mathrm e}^{5 x^{2}} \ln \relax (5) \ln \left (\frac {1}{4}\right )+\left (8 \ln \relax (2)+1\right ) \ln \relax (5)}{2 \ln \relax (2)}}}{x}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maxima [A]  time = 0.85, size = 23, normalized size = 0.96 \begin {gather*} \frac {2 \cdot 5^{-\frac {1}{2 \, \log \relax (2)} - 4} 5^{e^{\left (5 \, x^{2}\right )}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 2.00, size = 23, normalized size = 0.96 \begin {gather*} \frac {2\,5^{{\mathrm {e}}^{5\,x^2}}}{625\,5^{\frac {1}{2\,\ln \relax (2)}}\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.28, size = 32, normalized size = 1.33 \begin {gather*} \frac {2 e^{- \frac {- e^{5 x^{2}} \log {\relax (2 )} \log {\relax (5 )} + \frac {\left (1 + 8 \log {\relax (2 )}\right ) \log {\relax (5 )}}{2}}{\log {\relax (2 )}}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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