Integrand size = 83, antiderivative size = 26 \[ \int \frac {e^{-2 x^2 \log ^2(5) \log ^4(x)} \left (e^{2 x^2 \log ^2(5) \log ^4(x)}+e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} \left (-16 x^2 \log ^2(5) \log ^3(x)-8 x^2 \log ^2(5) \log ^4(x)\right )\right )}{2 x} \, dx=e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}}+\frac {\log \left (x^2\right )}{4} \] Output:
exp(1/exp(x^2*ln(5)^2*ln(x)^4)^2)+1/4*ln(x^2)
Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-2 x^2 \log ^2(5) \log ^4(x)} \left (e^{2 x^2 \log ^2(5) \log ^4(x)}+e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} \left (-16 x^2 \log ^2(5) \log ^3(x)-8 x^2 \log ^2(5) \log ^4(x)\right )\right )}{2 x} \, dx=e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}}+\frac {\log (x)}{2} \] Input:
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]
Output:
E^E^(-2*x^2*Log[5]^2*Log[x]^4) + Log[x]/2
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{-2 x^2 \log ^2(5) \log ^4(x)} \left (e^{2 x^2 \log ^2(5) \log ^4(x)}+e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} \left (-8 x^2 \log ^2(5) \log ^4(x)-16 x^2 \log ^2(5) \log ^3(x)\right )\right )}{2 x} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {e^{-2 x^2 \log ^2(5) \log ^4(x)} \left (e^{2 x^2 \log ^2(5) \log ^4(x)}-8 e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} \left (x^2 \log ^2(5) \log ^4(x)+2 x^2 \log ^2(5) \log ^3(x)\right )\right )}{x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{2} \int \left (\frac {1}{x}-8 \exp \left (e^{-2 x^2 \log ^2(5) \log ^4(x)}-2 x^2 \log ^2(5) \log ^4(x)\right ) x \log ^2(5) \log ^3(x) (\log (x)+2)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-8 \log ^2(5) \int \exp \left (e^{-2 x^2 \log ^2(5) \log ^4(x)}-2 x^2 \log ^2(5) \log ^4(x)\right ) x \log ^4(x)dx-16 \log ^2(5) \int \exp \left (e^{-2 x^2 \log ^2(5) \log ^4(x)}-2 x^2 \log ^2(5) \log ^4(x)\right ) x \log ^3(x)dx+\log (x)\right )\) |
Input:
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]
Output:
$Aborted
Time = 34.31 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {\ln \left (x \right )}{2}+{\mathrm e}^{{\mathrm e}^{-2 x^{2} \ln \left (5\right )^{2} \ln \left (x \right )^{4}}}\) | \(21\) |
parallelrisch | \(\frac {\ln \left (x \right )}{2}+{\mathrm e}^{{\mathrm e}^{-2 x^{2} \ln \left (5\right )^{2} \ln \left (x \right )^{4}}}\) | \(22\) |
Input:
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)
Output:
1/2*ln(x)+exp(exp(-2*x^2*ln(5)^2*ln(x)^4))
Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-2 x^2 \log ^2(5) \log ^4(x)} \left (e^{2 x^2 \log ^2(5) \log ^4(x)}+e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} \left (-16 x^2 \log ^2(5) \log ^3(x)-8 x^2 \log ^2(5) \log ^4(x)\right )\right )}{2 x} \, dx=e^{\left (e^{\left (-2 \, x^{2} \log \left (5\right )^{2} \log \left (x\right )^{4}\right )}\right )} + \frac {1}{2} \, \log \left (x\right ) \] Input:
integrate(1/2*((-8*x^2*log(5)^2*log(x)^4-16*x^2*log(5)^2*log(x)^3)*exp(1/e xp(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")
Output:
e^(e^(-2*x^2*log(5)^2*log(x)^4)) + 1/2*log(x)
Time = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{-2 x^2 \log ^2(5) \log ^4(x)} \left (e^{2 x^2 \log ^2(5) \log ^4(x)}+e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} \left (-16 x^2 \log ^2(5) \log ^3(x)-8 x^2 \log ^2(5) \log ^4(x)\right )\right )}{2 x} \, dx=e^{e^{- 2 x^{2} \log {\left (5 \right )}^{2} \log {\left (x \right )}^{4}}} + \frac {\log {\left (x \right )}}{2} \] Input:
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)
Output:
exp(exp(-2*x**2*log(5)**2*log(x)**4)) + log(x)/2
Time = 0.48 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-2 x^2 \log ^2(5) \log ^4(x)} \left (e^{2 x^2 \log ^2(5) \log ^4(x)}+e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} \left (-16 x^2 \log ^2(5) \log ^3(x)-8 x^2 \log ^2(5) \log ^4(x)\right )\right )}{2 x} \, dx=e^{\left (e^{\left (-2 \, x^{2} \log \left (5\right )^{2} \log \left (x\right )^{4}\right )}\right )} + \frac {1}{2} \, \log \left (x\right ) \] Input:
integrate(1/2*((-8*x^2*log(5)^2*log(x)^4-16*x^2*log(5)^2*log(x)^3)*exp(1/e xp(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")
Output:
e^(e^(-2*x^2*log(5)^2*log(x)^4)) + 1/2*log(x)
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-2 x^2 \log ^2(5) \log ^4(x)} \left (e^{2 x^2 \log ^2(5) \log ^4(x)}+e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} \left (-16 x^2 \log ^2(5) \log ^3(x)-8 x^2 \log ^2(5) \log ^4(x)\right )\right )}{2 x} \, dx=e^{\left (e^{\left (-2 \, x^{2} \log \left (5\right )^{2} \log \left (x\right )^{4}\right )}\right )} + \frac {1}{2} \, \log \left (x\right ) \] Input:
integrate(1/2*((-8*x^2*log(5)^2*log(x)^4-16*x^2*log(5)^2*log(x)^3)*exp(1/e xp(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")
Output:
e^(e^(-2*x^2*log(5)^2*log(x)^4)) + 1/2*log(x)
Time = 3.52 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-2 x^2 \log ^2(5) \log ^4(x)} \left (e^{2 x^2 \log ^2(5) \log ^4(x)}+e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} \left (-16 x^2 \log ^2(5) \log ^3(x)-8 x^2 \log ^2(5) \log ^4(x)\right )\right )}{2 x} \, dx={\mathrm {e}}^{{\mathrm {e}}^{-2\,x^2\,{\ln \left (5\right )}^2\,{\ln \left (x\right )}^4}}+\frac {\ln \left (x\right )}{2} \] Input:
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)
Output:
exp(exp(-2*x^2*log(5)^2*log(x)^4)) + log(x)/2
\[ \int \frac {e^{-2 x^2 \log ^2(5) \log ^4(x)} \left (e^{2 x^2 \log ^2(5) \log ^4(x)}+e^{e^{-2 x^2 \log ^2(5) \log ^4(x)}} \left (-16 x^2 \log ^2(5) \log ^3(x)-8 x^2 \log ^2(5) \log ^4(x)\right )\right )}{2 x} \, dx=-4 \left (\int \frac {e^{\frac {1}{e^{2 \mathrm {log}\left (x \right )^{4} \mathrm {log}\left (5\right )^{2} x^{2}}}} \mathrm {log}\left (x \right )^{4} x}{e^{2 \mathrm {log}\left (x \right )^{4} \mathrm {log}\left (5\right )^{2} x^{2}}}d x \right ) \mathrm {log}\left (5\right )^{2}-8 \left (\int \frac {e^{\frac {1}{e^{2 \mathrm {log}\left (x \right )^{4} \mathrm {log}\left (5\right )^{2} x^{2}}}} \mathrm {log}\left (x \right )^{3} x}{e^{2 \mathrm {log}\left (x \right )^{4} \mathrm {log}\left (5\right )^{2} x^{2}}}d x \right ) \mathrm {log}\left (5\right )^{2}+\frac {\mathrm {log}\left (x \right )}{2} \] Input:
int(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*lo g(x)^4)^2,x)
Output:
( - 8*int((e**(1/e**(2*log(x)**4*log(5)**2*x**2))*log(x)**4*x)/e**(2*log(x )**4*log(5)**2*x**2),x)*log(5)**2 - 16*int((e**(1/e**(2*log(x)**4*log(5)** 2*x**2))*log(x)**3*x)/e**(2*log(x)**4*log(5)**2*x**2),x)*log(5)**2 + log(x ))/2