3.43.58 \(\int \frac {e^{-2 x^2 \log ^2(5) \log ^4(x)} (e^{2 x^2 \log ^2(5) \log ^4(x)}+e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} (-16 x^2 \log ^2(5) \log ^3(x)-8 x^2 \log ^2(5) \log ^4(x)))}{2 x} \, dx\)

Optimal. Leaf size=26 \[ e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}}+\frac {\log \left (x^2\right )}{4} \]

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Rubi [F]  time = 1.17, 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^{-2 x^2 \log ^2(5) \log ^4(x)} \left (e^{2 x^2 \log ^2(5) \log ^4(x)}+e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} \left (-16 x^2 \log ^2(5) \log ^3(x)-8 x^2 \log ^2(5) \log ^4(x)\right )\right )}{2 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*x^2*Log[5]^2*Log[x]^4) + E^E^(-2*x^2*Log[5]^2*Log[x]^4)*(-16*x^2*Log[5]^2*Log[x]^3 - 8*x^2*Log[5]^2*
Log[x]^4))/(2*E^(2*x^2*Log[5]^2*Log[x]^4)*x),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {e^{-2 x^2 \log ^2(5) \log ^4(x)} \left (e^{2 x^2 \log ^2(5) \log ^4(x)}+e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} \left (-16 x^2 \log ^2(5) \log ^3(x)-8 x^2 \log ^2(5) \log ^4(x)\right )\right )}{x} \, dx\\ &=\frac {1}{2} \int \left (\frac {1}{x}-8 \exp \left (e^{-2 x^2 \log ^2(5) \log ^4(x)}-2 x^2 \log ^2(5) \log ^4(x)\right ) x \log ^2(5) \log ^3(x) (2+\log (x))\right ) \, dx\\ &=\frac {\log (x)}{2}-\left (4 \log ^2(5)\right ) \int \exp \left (e^{-2 x^2 \log ^2(5) \log ^4(x)}-2 x^2 \log ^2(5) \log ^4(x)\right ) x \log ^3(x) (2+\log (x)) \, dx\\ &=\frac {\log (x)}{2}-\left (4 \log ^2(5)\right ) \int \left (2 \exp \left (e^{-2 x^2 \log ^2(5) \log ^4(x)}-2 x^2 \log ^2(5) \log ^4(x)\right ) x \log ^3(x)+\exp \left (e^{-2 x^2 \log ^2(5) \log ^4(x)}-2 x^2 \log ^2(5) \log ^4(x)\right ) x \log ^4(x)\right ) \, dx\\ &=\frac {\log (x)}{2}-\left (4 \log ^2(5)\right ) \int \exp \left (e^{-2 x^2 \log ^2(5) \log ^4(x)}-2 x^2 \log ^2(5) \log ^4(x)\right ) x \log ^4(x) \, dx-\left (8 \log ^2(5)\right ) \int \exp \left (e^{-2 x^2 \log ^2(5) \log ^4(x)}-2 x^2 \log ^2(5) \log ^4(x)\right ) x \log ^3(x) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 24, normalized size = 0.92 \begin {gather*} e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}}+\frac {\log (x)}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x^2*Log[5]^2*Log[x]^4) + E^E^(-2*x^2*Log[5]^2*Log[x]^4)*(-16*x^2*Log[5]^2*Log[x]^3 - 8*x^2*Log
[5]^2*Log[x]^4))/(2*E^(2*x^2*Log[5]^2*Log[x]^4)*x),x]

[Out]

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

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fricas [A]  time = 0.73, size = 20, normalized size = 0.77 \begin {gather*} e^{\left (e^{\left (-2 \, x^{2} \log \relax (5)^{2} \log \relax (x)^{4}\right )}\right )} + \frac {1}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-8*x^2*log(5)^2*log(x)^4-16*x^2*log(5)^2*log(x)^3)*exp(1/exp(x^2*log(5)^2*log(x)^4)^2)+exp(x^2
*log(5)^2*log(x)^4)^2)/x/exp(x^2*log(5)^2*log(x)^4)^2,x, algorithm="fricas")

[Out]

e^(e^(-2*x^2*log(5)^2*log(x)^4)) + 1/2*log(x)

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giac [A]  time = 0.24, size = 20, normalized size = 0.77 \begin {gather*} e^{\left (e^{\left (-2 \, x^{2} \log \relax (5)^{2} \log \relax (x)^{4}\right )}\right )} + \frac {1}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-8*x^2*log(5)^2*log(x)^4-16*x^2*log(5)^2*log(x)^3)*exp(1/exp(x^2*log(5)^2*log(x)^4)^2)+exp(x^2
*log(5)^2*log(x)^4)^2)/x/exp(x^2*log(5)^2*log(x)^4)^2,x, algorithm="giac")

[Out]

e^(e^(-2*x^2*log(5)^2*log(x)^4)) + 1/2*log(x)

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maple [A]  time = 0.05, size = 21, normalized size = 0.81

method	result	size
risch	\(\frac {\ln \relax (x )}{2}+{\mathrm e}^{{\mathrm e}^{-2 x^{2} \ln \relax (5)^{2} \ln \relax (x )^{4}}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-8*x^2*ln(5)^2*ln(x)^4-16*x^2*ln(5)^2*ln(x)^3)*exp(1/exp(x^2*ln(5)^2*ln(x)^4)^2)+exp(x^2*ln(5)^2*ln(
x)^4)^2)/x/exp(x^2*ln(5)^2*ln(x)^4)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.78, size = 20, normalized size = 0.77 \begin {gather*} e^{\left (e^{\left (-2 \, x^{2} \log \relax (5)^{2} \log \relax (x)^{4}\right )}\right )} + \frac {1}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-8*x^2*log(5)^2*log(x)^4-16*x^2*log(5)^2*log(x)^3)*exp(1/exp(x^2*log(5)^2*log(x)^4)^2)+exp(x^2
*log(5)^2*log(x)^4)^2)/x/exp(x^2*log(5)^2*log(x)^4)^2,x, algorithm="maxima")

[Out]

e^(e^(-2*x^2*log(5)^2*log(x)^4)) + 1/2*log(x)

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mupad [B]  time = 3.06, size = 20, normalized size = 0.77 \begin {gather*} {\mathrm {e}}^{{\mathrm {e}}^{-2\,x^2\,{\ln \relax (5)}^2\,{\ln \relax (x)}^4}}+\frac {\ln \relax (x)}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-2*x^2*log(5)^2*log(x)^4)*(exp(2*x^2*log(5)^2*log(x)^4)/2 - (exp(exp(-2*x^2*log(5)^2*log(x)^4))*(16*x
^2*log(5)^2*log(x)^3 + 8*x^2*log(5)^2*log(x)^4))/2))/x,x)

[Out]

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

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sympy [A]  time = 0.74, size = 24, normalized size = 0.92 \begin {gather*} e^{e^{- 2 x^{2} \log {\relax (5 )}^{2} \log {\relax (x )}^{4}}} + \frac {\log {\relax (x )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-8*x**2*ln(5)**2*ln(x)**4-16*x**2*ln(5)**2*ln(x)**3)*exp(1/exp(x**2*ln(5)**2*ln(x)**4)**2)+exp
(x**2*ln(5)**2*ln(x)**4)**2)/x/exp(x**2*ln(5)**2*ln(x)**4)**2,x)

[Out]

exp(exp(-2*x**2*log(5)**2*log(x)**4)) + log(x)/2

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