\(\int \frac {5^{\frac {5}{(x+x^2 \log (5)) \log (x)}} e^{-4+5^{\frac {5}{(x+x^2 \log (5)) \log (x)}}} (-5 \log (5)-5 x \log ^2(5)+(-5 \log (5)-10 x \log ^2(5)) \log (x))}{(x^2+2 x^3 \log (5)+x^4 \log ^2(5)) \log ^2(x)} \, dx\) [2630]

Optimal result

Integrand size = 93, antiderivative size = 24 \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=e^{-4+e^{\frac {5}{\left (x^2+\frac {x}{\log (5)}\right ) \log (x)}}} \] Output:

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

Mathematica [F]

\[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=\int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx \] Input:

Integrate[(5^(5/((x + x^2*Log[5])*Log[x]))*E^(-4 + 5^(5/((x + x^2*Log[5])* 
Log[x])))*(-5*Log[5] - 5*x*Log[5]^2 + (-5*Log[5] - 10*x*Log[5]^2)*Log[x])) 
/((x^2 + 2*x^3*Log[5] + x^4*Log[5]^2)*Log[x]^2),x]
 

Output:

Integrate[(5^(5/((x + x^2*Log[5])*Log[x]))*E^(-4 + 5^(5/((x + x^2*Log[5])* 
Log[x])))*(-5*Log[5] - 5*x*Log[5]^2 + (-5*Log[5] - 10*x*Log[5]^2)*Log[x])) 
/((x^2 + 2*x^3*Log[5] + x^4*Log[5]^2)*Log[x]^2), x]
 

Rubi [F]

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.

\(\displaystyle \int \frac {5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}} e^{5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}}-4} \left (-5 x \log ^2(5)+\left (-10 x \log ^2(5)-5 \log (5)\right ) \log (x)-5 \log (5)\right )}{\left (x^4 \log ^2(5)+2 x^3 \log (5)+x^2\right ) \log ^2(x)} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}} e^{5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}}-4} \left (-5 x \log ^2(5)+\left (-10 x \log ^2(5)-5 \log (5)\right ) \log (x)-5 \log (5)\right )}{x^2 \left (x^2 \log ^2(5)+2 x \log (5)+1\right ) \log ^2(x)}dx\)

\(\Big \downarrow \) 2007

\(\displaystyle \int \frac {5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}} e^{5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}}-4} \left (-5 x \log ^2(5)+\left (-10 x \log ^2(5)-5 \log (5)\right ) \log (x)-5 \log (5)\right )}{x^2 (x \log (5)+1)^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {\log (5) 5^{\frac {5}{x (x \log (5)+1) \log (x)}+1} e^{5^{\frac {5}{\left (x^2 \log (5)+x\right ) \log (x)}}-4} (x (-\log (25)) \log (x)-x \log (5)-\log (x)-1)}{x^2 (x \log (5)+1)^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \log (5) \int -\frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}} (\log (25) \log (x) x+\log (5) x+\log (x)+1)}{x^2 (\log (5) x+1)^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\log (5) \int \frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}} (\log (25) \log (x) x+\log (5) x+\log (x)+1)}{x^2 (\log (5) x+1)^2 \log ^2(x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\log (5) \int \left (\frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}} (\log (25) x+1)}{x^2 (\log (5) x+1)^2 \log (x)}+\frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}}}{x^2 (\log (5) x+1) \log ^2(x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\log (5) \left (\int \frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}}}{x^2 \log ^2(x)}dx-\log (5) \int \frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}}}{x \log ^2(x)}dx+\log ^2(5) \int \frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}}}{(\log (5) x+1) \log ^2(x)}dx-\log ^2(5) \int \frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}}}{(\log (5) x+1)^2 \log (x)}dx+\int \frac {5^{1+\frac {5}{x (\log (5) x+1) \log (x)}} e^{-4+5^{\frac {5}{\left (\log (5) x^2+x\right ) \log (x)}}}}{x^2 \log (x)}dx\right )\)

Input:

Int[(5^(5/((x + x^2*Log[5])*Log[x]))*E^(-4 + 5^(5/((x + x^2*Log[5])*Log[x] 
)))*(-5*Log[5] - 5*x*Log[5]^2 + (-5*Log[5] - 10*x*Log[5]^2)*Log[x]))/((x^2 
 + 2*x^3*Log[5] + x^4*Log[5]^2)*Log[x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 153.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92

method	result	size
risch	\({\mathrm e}^{3125^{\frac {1}{x \left (x \ln \left (5\right )+1\right ) \ln \left (x \right )}}-4}\)	\(22\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{\frac {5 \ln \left (5\right )}{x \left (x \ln \left (5\right )+1\right ) \ln \left (x \right )}}-4}\)	\(24\)

Input:

int(((-10*x*ln(5)^2-5*ln(5))*ln(x)-5*x*ln(5)^2-5*ln(5))*exp(5*ln(5)/(x^2*l 
n(5)+x)/ln(x))*exp(exp(5*ln(5)/(x^2*ln(5)+x)/ln(x))-4)/(x^4*ln(5)^2+2*x^3* 
ln(5)+x^2)/ln(x)^2,x,method=_RETURNVERBOSE)
 

Output:

exp(3125^(1/x/(x*ln(5)+1)/ln(x))-4)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=e^{\left (5^{\frac {5}{{\left (x^{2} \log \left (5\right ) + x\right )} \log \left (x\right )}} - 4\right )} \] Input:

integrate(((-10*x*log(5)^2-5*log(5))*log(x)-5*x*log(5)^2-5*log(5))*exp(5*l 
og(5)/(x^2*log(5)+x)/log(x))*exp(exp(5*log(5)/(x^2*log(5)+x)/log(x))-4)/(x 
^4*log(5)^2+2*x^3*log(5)+x^2)/log(x)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(((-10*x*ln(5)**2-5*ln(5))*ln(x)-5*x*ln(5)**2-5*ln(5))*exp(5*ln(5 
)/(x**2*ln(5)+x)/ln(x))*exp(exp(5*ln(5)/(x**2*ln(5)+x)/ln(x))-4)/(x**4*ln( 
5)**2+2*x**3*ln(5)+x**2)/ln(x)**2,x)
 

Output:

Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 

Maxima [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=e^{\left (e^{\left (-\frac {5 \, \log \left (5\right )^{2}}{{\left (x \log \left (5\right ) + 1\right )} \log \left (x\right )} + \frac {5 \, \log \left (5\right )}{x \log \left (x\right )}\right )} - 4\right )} \] Input:

integrate(((-10*x*log(5)^2-5*log(5))*log(x)-5*x*log(5)^2-5*log(5))*exp(5*l 
og(5)/(x^2*log(5)+x)/log(x))*exp(exp(5*log(5)/(x^2*log(5)+x)/log(x))-4)/(x 
^4*log(5)^2+2*x^3*log(5)+x^2)/log(x)^2,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=e^{\left (5^{\frac {5}{x^{2} \log \left (5\right ) \log \left (x\right ) + x \log \left (x\right )}} - 4\right )} \] Input:

integrate(((-10*x*log(5)^2-5*log(5))*log(x)-5*x*log(5)^2-5*log(5))*exp(5*l 
og(5)/(x^2*log(5)+x)/log(x))*exp(exp(5*log(5)/(x^2*log(5)+x)/log(x))-4)/(x 
^4*log(5)^2+2*x^3*log(5)+x^2)/log(x)^2,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 3.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx={\mathrm {e}}^{5^{\frac {5}{x\,\ln \left (x\right )+x^2\,\ln \left (5\right )\,\ln \left (x\right )}}}\,{\mathrm {e}}^{-4} \] Input:

int(-(exp(exp((5*log(5))/(log(x)*(x + x^2*log(5)))) - 4)*exp((5*log(5))/(l 
og(x)*(x + x^2*log(5))))*(5*log(5) + log(x)*(5*log(5) + 10*x*log(5)^2) + 5 
*x*log(5)^2))/(log(x)^2*(x^4*log(5)^2 + 2*x^3*log(5) + x^2)),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}} e^{-4+5^{\frac {5}{\left (x+x^2 \log (5)\right ) \log (x)}}} \left (-5 \log (5)-5 x \log ^2(5)+\left (-5 \log (5)-10 x \log ^2(5)\right ) \log (x)\right )}{\left (x^2+2 x^3 \log (5)+x^4 \log ^2(5)\right ) \log ^2(x)} \, dx=\frac {e^{e^{\frac {5 \,\mathrm {log}\left (5\right )}{\mathrm {log}\left (x \right ) \mathrm {log}\left (5\right ) x^{2}+\mathrm {log}\left (x \right ) x}}}}{e^{4}} \] Input:

int(((-10*x*log(5)^2-5*log(5))*log(x)-5*x*log(5)^2-5*log(5))*exp(5*log(5)/ 
(x^2*log(5)+x)/log(x))*exp(exp(5*log(5)/(x^2*log(5)+x)/log(x))-4)/(x^4*log 
(5)^2+2*x^3*log(5)+x^2)/log(x)^2,x)
 

Output:

e**(e**((5*log(5))/(log(x)*log(5)*x**2 + log(x)*x)))/e**4