\(\int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} (-25 \log (x)-\log (x) \log ^2(3 x)))+(-e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x \log ^2(3 x)+5 x^3 \log ^2(3 x)) \log (\frac {1}{5} (e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}}-5 x^2))}{e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x^2 \log ^2(x) \log ^2(3 x)-5 x^4 \log ^2(x) \log ^2(3 x)} \, dx\) [1422]

Optimal result

Integrand size = 200, antiderivative size = 34 \[ \int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} \left (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} \left (-25 \log (x)-\log (x) \log ^2(3 x)\right )\right )+\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x \log ^2(3 x)+5 x^3 \log ^2(3 x)\right ) \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}}-5 x^2\right )\right )}{e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x^2 \log ^2(x) \log ^2(3 x)-5 x^4 \log ^2(x) \log ^2(3 x)} \, dx=\frac {\log \left (\frac {1}{5} e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-x^2\right )}{\log (x)} \] Output:

ln(1/5*exp(exp(25/ln(3*x))/x+x)-x^2)/ln(x)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} \left (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} \left (-25 \log (x)-\log (x) \log ^2(3 x)\right )\right )+\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x \log ^2(3 x)+5 x^3 \log ^2(3 x)\right ) \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}}-5 x^2\right )\right )}{e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x^2 \log ^2(x) \log ^2(3 x)-5 x^4 \log ^2(x) \log ^2(3 x)} \, dx=\frac {\log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}}{x}+x}-5 x^2\right )\right )}{\log (x)} \] Input:

Integrate[(-10*x^3*Log[x]*Log[3*x]^2 + E^((E^(25/Log[3*x]) + x^2)/x)*(x^2* 
Log[x]*Log[3*x]^2 + E^(25/Log[3*x])*(-25*Log[x] - Log[x]*Log[3*x]^2)) + (- 
(E^((E^(25/Log[3*x]) + x^2)/x)*x*Log[3*x]^2) + 5*x^3*Log[3*x]^2)*Log[(E^(( 
E^(25/Log[3*x]) + x^2)/x) - 5*x^2)/5])/(E^((E^(25/Log[3*x]) + x^2)/x)*x^2* 
Log[x]^2*Log[3*x]^2 - 5*x^4*Log[x]^2*Log[3*x]^2),x]
 

Output:

Log[(E^(E^(25/Log[3*x])/x + x) - 5*x^2)/5]/Log[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.

\(\displaystyle \int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {x^2+e^{\frac {25}{\log (3 x)}}}{x}} \left (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} \left (-\log (x) \log ^2(3 x)-25 \log (x)\right )\right )+\left (5 x^3 \log ^2(3 x)-x e^{\frac {x^2+e^{\frac {25}{\log (3 x)}}}{x}} \log ^2(3 x)\right ) \log \left (\frac {1}{5} \left (e^{\frac {x^2+e^{\frac {25}{\log (3 x)}}}{x}}-5 x^2\right )\right )}{x^2 e^{\frac {x^2+e^{\frac {25}{\log (3 x)}}}{x}} \log ^2(x) \log ^2(3 x)-5 x^4 \log ^2(x) \log ^2(3 x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {x^2+e^{\frac {25}{\log (3 x)}}}{x}} \left (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} \left (-\log (x) \log ^2(3 x)-25 \log (x)\right )\right )+\left (5 x^3 \log ^2(3 x)-x e^{\frac {x^2+e^{\frac {25}{\log (3 x)}}}{x}} \log ^2(3 x)\right ) \log \left (\frac {1}{5} \left (e^{\frac {x^2+e^{\frac {25}{\log (3 x)}}}{x}}-5 x^2\right )\right )}{x^2 \left (e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}}-5 x^2\right ) \log ^2(x) \log ^2(3 x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {10 x^2 \log (x)-5 x^2 \log \left (\frac {1}{5} \left (e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}}-5 x^2\right )\right )+e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}} \log \left (\frac {1}{5} \left (e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}}-5 x^2\right )\right )-x e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}} \log (x)}{x \left (5 x^2-e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}}\right ) \log ^2(x)}-\frac {e^{x+\frac {25}{\log (3 x)}+\frac {e^{\frac {25}{\log (3 x)}}}{x}} \left (\log ^2(3 x)+25\right )}{x^2 \left (e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}}-5 x^2\right ) \log (x) \log ^2(3 x)}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (\frac {10 x^2 \log (x)-5 x^2 \log \left (\frac {1}{5} \left (e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}}-5 x^2\right )\right )+e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}} \log \left (\frac {1}{5} \left (e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}}-5 x^2\right )\right )-x e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}} \log (x)}{x \left (5 x^2-e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}}\right ) \log ^2(x)}-\frac {e^{x+\frac {25}{\log (3 x)}+\frac {e^{\frac {25}{\log (3 x)}}}{x}} \left (\log ^2(3 x)+25\right )}{x^2 \left (e^{x+\frac {e^{\frac {25}{\log (3 x)}}}{x}}-5 x^2\right ) \log (x) \log ^2(3 x)}\right )dx\)

Input:

Int[(-10*x^3*Log[x]*Log[3*x]^2 + E^((E^(25/Log[3*x]) + x^2)/x)*(x^2*Log[x] 
*Log[3*x]^2 + E^(25/Log[3*x])*(-25*Log[x] - Log[x]*Log[3*x]^2)) + (-(E^((E 
^(25/Log[3*x]) + x^2)/x)*x*Log[3*x]^2) + 5*x^3*Log[3*x]^2)*Log[(E^((E^(25/ 
Log[3*x]) + x^2)/x) - 5*x^2)/5])/(E^((E^(25/Log[3*x]) + x^2)/x)*x^2*Log[x] 
^2*Log[3*x]^2 - 5*x^4*Log[x]^2*Log[3*x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00

\[\frac {\ln \left (\frac {{\mathrm e}^{\frac {{\mathrm e}^{\frac {25}{\ln \left (3\right )+\ln \left (x \right )}}+x^{2}}{x}}}{5}-x^{2}\right )}{\ln \left (x \right )}\]

Input:

int(((-x*ln(3*x)^2*exp((exp(25/ln(3*x))+x^2)/x)+5*x^3*ln(3*x)^2)*ln(1/5*ex 
p((exp(25/ln(3*x))+x^2)/x)-x^2)+((-ln(x)*ln(3*x)^2-25*ln(x))*exp(25/ln(3*x 
))+x^2*ln(x)*ln(3*x)^2)*exp((exp(25/ln(3*x))+x^2)/x)-10*x^3*ln(x)*ln(3*x)^ 
2)/(x^2*ln(x)^2*ln(3*x)^2*exp((exp(25/ln(3*x))+x^2)/x)-5*x^4*ln(x)^2*ln(3* 
x)^2),x)
 

Output:

1/ln(x)*ln(1/5*exp((exp(25/(ln(3)+ln(x)))+x^2)/x)-x^2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} \left (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} \left (-25 \log (x)-\log (x) \log ^2(3 x)\right )\right )+\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x \log ^2(3 x)+5 x^3 \log ^2(3 x)\right ) \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}}-5 x^2\right )\right )}{e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x^2 \log ^2(x) \log ^2(3 x)-5 x^4 \log ^2(x) \log ^2(3 x)} \, dx=\frac {\log \left (-x^{2} + \frac {1}{5} \, e^{\left (\frac {x^{2} + e^{\left (\frac {25}{\log \left (3\right ) + \log \left (x\right )}\right )}}{x}\right )}\right )}{\log \left (x\right )} \] Input:

integrate(((-x*log(3*x)^2*exp((exp(25/log(3*x))+x^2)/x)+5*x^3*log(3*x)^2)* 
log(1/5*exp((exp(25/log(3*x))+x^2)/x)-x^2)+((-log(x)*log(3*x)^2-25*log(x)) 
*exp(25/log(3*x))+x^2*log(x)*log(3*x)^2)*exp((exp(25/log(3*x))+x^2)/x)-10* 
x^3*log(x)*log(3*x)^2)/(x^2*log(x)^2*log(3*x)^2*exp((exp(25/log(3*x))+x^2) 
/x)-5*x^4*log(x)^2*log(3*x)^2),x, algorithm="fricas")
 

Output:

log(-x^2 + 1/5*e^((x^2 + e^(25/(log(3) + log(x))))/x))/log(x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} \left (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} \left (-25 \log (x)-\log (x) \log ^2(3 x)\right )\right )+\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x \log ^2(3 x)+5 x^3 \log ^2(3 x)\right ) \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}}-5 x^2\right )\right )}{e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x^2 \log ^2(x) \log ^2(3 x)-5 x^4 \log ^2(x) \log ^2(3 x)} \, dx=\text {Timed out} \] Input:

integrate(((-x*ln(3*x)**2*exp((exp(25/ln(3*x))+x**2)/x)+5*x**3*ln(3*x)**2) 
*ln(1/5*exp((exp(25/ln(3*x))+x**2)/x)-x**2)+((-ln(x)*ln(3*x)**2-25*ln(x))* 
exp(25/ln(3*x))+x**2*ln(x)*ln(3*x)**2)*exp((exp(25/ln(3*x))+x**2)/x)-10*x* 
*3*ln(x)*ln(3*x)**2)/(x**2*ln(x)**2*ln(3*x)**2*exp((exp(25/ln(3*x))+x**2)/ 
x)-5*x**4*ln(x)**2*ln(3*x)**2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} \left (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} \left (-25 \log (x)-\log (x) \log ^2(3 x)\right )\right )+\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x \log ^2(3 x)+5 x^3 \log ^2(3 x)\right ) \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}}-5 x^2\right )\right )}{e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x^2 \log ^2(x) \log ^2(3 x)-5 x^4 \log ^2(x) \log ^2(3 x)} \, dx=-\frac {\log \left (5\right ) - \log \left (-5 \, x^{2} + e^{\left (x + \frac {e^{\left (\frac {25}{\log \left (3\right ) + \log \left (x\right )}\right )}}{x}\right )}\right )}{\log \left (x\right )} \] Input:

integrate(((-x*log(3*x)^2*exp((exp(25/log(3*x))+x^2)/x)+5*x^3*log(3*x)^2)* 
log(1/5*exp((exp(25/log(3*x))+x^2)/x)-x^2)+((-log(x)*log(3*x)^2-25*log(x)) 
*exp(25/log(3*x))+x^2*log(x)*log(3*x)^2)*exp((exp(25/log(3*x))+x^2)/x)-10* 
x^3*log(x)*log(3*x)^2)/(x^2*log(x)^2*log(3*x)^2*exp((exp(25/log(3*x))+x^2) 
/x)-5*x^4*log(x)^2*log(3*x)^2),x, algorithm="maxima")
 

Output:

-(log(5) - log(-5*x^2 + e^(x + e^(25/(log(3) + log(x)))/x)))/log(x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} \left (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} \left (-25 \log (x)-\log (x) \log ^2(3 x)\right )\right )+\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x \log ^2(3 x)+5 x^3 \log ^2(3 x)\right ) \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}}-5 x^2\right )\right )}{e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x^2 \log ^2(x) \log ^2(3 x)-5 x^4 \log ^2(x) \log ^2(3 x)} \, dx=\text {Timed out} \] Input:

integrate(((-x*log(3*x)^2*exp((exp(25/log(3*x))+x^2)/x)+5*x^3*log(3*x)^2)* 
log(1/5*exp((exp(25/log(3*x))+x^2)/x)-x^2)+((-log(x)*log(3*x)^2-25*log(x)) 
*exp(25/log(3*x))+x^2*log(x)*log(3*x)^2)*exp((exp(25/log(3*x))+x^2)/x)-10* 
x^3*log(x)*log(3*x)^2)/(x^2*log(x)^2*log(3*x)^2*exp((exp(25/log(3*x))+x^2) 
/x)-5*x^4*log(x)^2*log(3*x)^2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} \left (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} \left (-25 \log (x)-\log (x) \log ^2(3 x)\right )\right )+\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x \log ^2(3 x)+5 x^3 \log ^2(3 x)\right ) \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}}-5 x^2\right )\right )}{e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x^2 \log ^2(x) \log ^2(3 x)-5 x^4 \log ^2(x) \log ^2(3 x)} \, dx=\int \frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {25}{\ln \left (3\,x\right )}}+x^2}{x}}\,\left ({\mathrm {e}}^{\frac {25}{\ln \left (3\,x\right )}}\,\left (\ln \left (x\right )\,{\ln \left (3\,x\right )}^2+25\,\ln \left (x\right )\right )-x^2\,{\ln \left (3\,x\right )}^2\,\ln \left (x\right )\right )-\ln \left (\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {25}{\ln \left (3\,x\right )}}+x^2}{x}}}{5}-x^2\right )\,\left (5\,x^3\,{\ln \left (3\,x\right )}^2-x\,{\ln \left (3\,x\right )}^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {25}{\ln \left (3\,x\right )}}+x^2}{x}}\right )+10\,x^3\,{\ln \left (3\,x\right )}^2\,\ln \left (x\right )}{5\,x^4\,{\ln \left (3\,x\right )}^2\,{\ln \left (x\right )}^2-x^2\,{\ln \left (3\,x\right )}^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{\frac {25}{\ln \left (3\,x\right )}}+x^2}{x}}\,{\ln \left (x\right )}^2} \,d x \] Input:

int((exp((exp(25/log(3*x)) + x^2)/x)*(exp(25/log(3*x))*(25*log(x) + log(3* 
x)^2*log(x)) - x^2*log(3*x)^2*log(x)) - log(exp((exp(25/log(3*x)) + x^2)/x 
)/5 - x^2)*(5*x^3*log(3*x)^2 - x*log(3*x)^2*exp((exp(25/log(3*x)) + x^2)/x 
)) + 10*x^3*log(3*x)^2*log(x))/(5*x^4*log(3*x)^2*log(x)^2 - x^2*log(3*x)^2 
*exp((exp(25/log(3*x)) + x^2)/x)*log(x)^2),x)
 

Output:

int((exp((exp(25/log(3*x)) + x^2)/x)*(exp(25/log(3*x))*(25*log(x) + log(3* 
x)^2*log(x)) - x^2*log(3*x)^2*log(x)) - log(exp((exp(25/log(3*x)) + x^2)/x 
)/5 - x^2)*(5*x^3*log(3*x)^2 - x*log(3*x)^2*exp((exp(25/log(3*x)) + x^2)/x 
)) + 10*x^3*log(3*x)^2*log(x))/(5*x^4*log(3*x)^2*log(x)^2 - x^2*log(3*x)^2 
*exp((exp(25/log(3*x)) + x^2)/x)*log(x)^2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {-10 x^3 \log (x) \log ^2(3 x)+e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} \left (x^2 \log (x) \log ^2(3 x)+e^{\frac {25}{\log (3 x)}} \left (-25 \log (x)-\log (x) \log ^2(3 x)\right )\right )+\left (-e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x \log ^2(3 x)+5 x^3 \log ^2(3 x)\right ) \log \left (\frac {1}{5} \left (e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}}-5 x^2\right )\right )}{e^{\frac {e^{\frac {25}{\log (3 x)}}+x^2}{x}} x^2 \log ^2(x) \log ^2(3 x)-5 x^4 \log ^2(x) \log ^2(3 x)} \, dx=\frac {\mathrm {log}\left (\frac {e^{\frac {e^{\frac {25}{\mathrm {log}\left (3 x \right )}}+x^{2}}{x}}}{5}-x^{2}\right )}{\mathrm {log}\left (x \right )} \] Input:

int(((-x*log(3*x)^2*exp((exp(25/log(3*x))+x^2)/x)+5*x^3*log(3*x)^2)*log(1/ 
5*exp((exp(25/log(3*x))+x^2)/x)-x^2)+((-log(x)*log(3*x)^2-25*log(x))*exp(2 
5/log(3*x))+x^2*log(x)*log(3*x)^2)*exp((exp(25/log(3*x))+x^2)/x)-10*x^3*lo 
g(x)*log(3*x)^2)/(x^2*log(x)^2*log(3*x)^2*exp((exp(25/log(3*x))+x^2)/x)-5* 
x^4*log(x)^2*log(3*x)^2),x)
 

Output:

log((e**((e**(25/log(3*x)) + x**2)/x) - 5*x**2)/5)/log(x)