\(\int \frac {e^{e^x} (250-200 x^2) \log (\log (\frac {x}{20+16 x^2}))+e^{e^x+x} (125 x+100 x^3) \log (\frac {x}{20+16 x^2}) \log ^2(\log (\frac {x}{20+16 x^2}))}{(5 x+4 x^3) \log (\frac {x}{20+16 x^2})} \, dx\) [1550]

Optimal result

Integrand size = 97, antiderivative size = 25 \[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=25 e^{e^x} \log ^2\left (\log \left (\frac {x}{4 \left (5+4 x^2\right )}\right )\right ) \] Output:

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

Mathematica [F]

\[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=\int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx \] Input:

Integrate[(E^E^x*(250 - 200*x^2)*Log[Log[x/(20 + 16*x^2)]] + E^(E^x + x)*( 
125*x + 100*x^3)*Log[x/(20 + 16*x^2)]*Log[Log[x/(20 + 16*x^2)]]^2)/((5*x + 
 4*x^3)*Log[x/(20 + 16*x^2)]),x]
 

Output:

Integrate[(E^E^x*(250 - 200*x^2)*Log[Log[x/(20 + 16*x^2)]] + E^(E^x + x)*( 
125*x + 100*x^3)*Log[x/(20 + 16*x^2)]*Log[Log[x/(20 + 16*x^2)]]^2)/((5*x + 
 4*x^3)*Log[x/(20 + 16*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.

\(\displaystyle \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{16 x^2+20}\right )\right )+e^{x+e^x} \left (100 x^3+125 x\right ) \log \left (\frac {x}{16 x^2+20}\right ) \log ^2\left (\log \left (\frac {x}{16 x^2+20}\right )\right )}{\left (4 x^3+5 x\right ) \log \left (\frac {x}{16 x^2+20}\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{16 x^2+20}\right )\right )+e^{x+e^x} \left (100 x^3+125 x\right ) \log \left (\frac {x}{16 x^2+20}\right ) \log ^2\left (\log \left (\frac {x}{16 x^2+20}\right )\right )}{x \left (4 x^2+5\right ) \log \left (\frac {x}{16 x^2+20}\right )}dx\)

\(\Big \downarrow \) 7276

\(\displaystyle \int \left (25 e^{x+e^x} \log ^2\left (\log \left (\frac {x}{16 x^2+20}\right )\right )-\frac {50 e^{e^x} \left (4 x^2-5\right ) \log \left (\log \left (\frac {x}{16 x^2+20}\right )\right )}{x \left (4 x^2+5\right ) \log \left (\frac {x}{16 x^2+20}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 25 \int e^{x+e^x} \log ^2\left (\log \left (\frac {x}{16 x^2+20}\right )\right )dx+100 \int \frac {e^{e^x} \log \left (\log \left (\frac {x}{16 x^2+20}\right )\right )}{\left (i \sqrt {5}-2 x\right ) \log \left (\frac {x}{16 x^2+20}\right )}dx+50 \int \frac {e^{e^x} \log \left (\log \left (\frac {x}{16 x^2+20}\right )\right )}{x \log \left (\frac {x}{16 x^2+20}\right )}dx-100 \int \frac {e^{e^x} \log \left (\log \left (\frac {x}{16 x^2+20}\right )\right )}{\left (2 x+i \sqrt {5}\right ) \log \left (\frac {x}{16 x^2+20}\right )}dx\)

Input:

Int[(E^E^x*(250 - 200*x^2)*Log[Log[x/(20 + 16*x^2)]] + E^(E^x + x)*(125*x 
+ 100*x^3)*Log[x/(20 + 16*x^2)]*Log[Log[x/(20 + 16*x^2)]]^2)/((5*x + 4*x^3 
)*Log[x/(20 + 16*x^2)]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 59.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(25 {\ln \left (\ln \left (\frac {x}{16 x^{2}+20}\right )\right )}^{2} {\mathrm e}^{{\mathrm e}^{x}}\)	\(22\)
risch	\(25 \,{\mathrm e}^{{\mathrm e}^{x}} {\ln \left (-4 \ln \left (2\right )+\ln \left (x \right )-\ln \left (x^{2}+\frac {5}{4}\right )-\frac {i \pi \,\operatorname {csgn}\left (\frac {i x}{x^{2}+\frac {5}{4}}\right ) \left (-\operatorname {csgn}\left (\frac {i x}{x^{2}+\frac {5}{4}}\right )+\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (\frac {i x}{x^{2}+\frac {5}{4}}\right )+\operatorname {csgn}\left (\frac {i}{x^{2}+\frac {5}{4}}\right )\right )}{2}\right )}^{2}\)	\(86\)

Input:

int(((100*x^3+125*x)*exp(x)*ln(x/(16*x^2+20))*exp(exp(x))*ln(ln(x/(16*x^2+ 
20)))^2+(-200*x^2+250)*exp(exp(x))*ln(ln(x/(16*x^2+20))))/(4*x^3+5*x)/ln(x 
/(16*x^2+20)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=25 \, e^{\left (e^{x}\right )} \log \left (\log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right )\right )^{2} \] Input:

integrate(((100*x^3+125*x)*exp(x)*log(x/(16*x^2+20))*exp(exp(x))*log(log(x 
/(16*x^2+20)))^2+(-200*x^2+250)*exp(exp(x))*log(log(x/(16*x^2+20))))/(4*x^ 
3+5*x)/log(x/(16*x^2+20)),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=\text {Timed out} \] Input:

integrate(((100*x**3+125*x)*exp(x)*ln(x/(16*x**2+20))*exp(exp(x))*ln(ln(x/ 
(16*x**2+20)))**2+(-200*x**2+250)*exp(exp(x))*ln(ln(x/(16*x**2+20))))/(4*x 
**3+5*x)/ln(x/(16*x**2+20)),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=25 \, e^{\left (e^{x}\right )} \log \left (-2 \, \log \left (2\right ) - \log \left (4 \, x^{2} + 5\right ) + \log \left (x\right )\right )^{2} \] Input:

integrate(((100*x^3+125*x)*exp(x)*log(x/(16*x^2+20))*exp(exp(x))*log(log(x 
/(16*x^2+20)))^2+(-200*x^2+250)*exp(exp(x))*log(log(x/(16*x^2+20))))/(4*x^ 
3+5*x)/log(x/(16*x^2+20)),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=\int { \frac {25 \, {\left ({\left (4 \, x^{3} + 5 \, x\right )} e^{\left (x + e^{x}\right )} \log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right ) \log \left (\log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right )\right )^{2} - 2 \, {\left (4 \, x^{2} - 5\right )} e^{\left (e^{x}\right )} \log \left (\log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right )\right )\right )}}{{\left (4 \, x^{3} + 5 \, x\right )} \log \left (\frac {x}{4 \, {\left (4 \, x^{2} + 5\right )}}\right )} \,d x } \] Input:

integrate(((100*x^3+125*x)*exp(x)*log(x/(16*x^2+20))*exp(exp(x))*log(log(x 
/(16*x^2+20)))^2+(-200*x^2+250)*exp(exp(x))*log(log(x/(16*x^2+20))))/(4*x^ 
3+5*x)/log(x/(16*x^2+20)),x, algorithm="giac")
 

Output:

integrate(25*((4*x^3 + 5*x)*e^(x + e^x)*log(1/4*x/(4*x^2 + 5))*log(log(1/4 
*x/(4*x^2 + 5)))^2 - 2*(4*x^2 - 5)*e^(e^x)*log(log(1/4*x/(4*x^2 + 5))))/(( 
4*x^3 + 5*x)*log(1/4*x/(4*x^2 + 5))), x)
 

Mupad [B] (verification not implemented)

Time = 3.80 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=25\,{\ln \left (\ln \left (x\right )-\ln \left (16\,x^2+20\right )\right )}^2\,{\mathrm {e}}^{{\mathrm {e}}^x} \] Input:

int(-(exp(exp(x))*log(log(x/(16*x^2 + 20)))*(200*x^2 - 250) - log(x/(16*x^ 
2 + 20))*exp(exp(x))*exp(x)*log(log(x/(16*x^2 + 20)))^2*(125*x + 100*x^3)) 
/(log(x/(16*x^2 + 20))*(5*x + 4*x^3)),x)
 

Output:

25*log(log(x) - log(16*x^2 + 20))^2*exp(exp(x))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^x} \left (250-200 x^2\right ) \log \left (\log \left (\frac {x}{20+16 x^2}\right )\right )+e^{e^x+x} \left (125 x+100 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right ) \log ^2\left (\log \left (\frac {x}{20+16 x^2}\right )\right )}{\left (5 x+4 x^3\right ) \log \left (\frac {x}{20+16 x^2}\right )} \, dx=25 e^{e^{x}} {\mathrm {log}\left (\mathrm {log}\left (\frac {x}{16 x^{2}+20}\right )\right )}^{2} \] Input:

int(((100*x^3+125*x)*exp(x)*log(x/(16*x^2+20))*exp(exp(x))*log(log(x/(16*x 
^2+20)))^2+(-200*x^2+250)*exp(exp(x))*log(log(x/(16*x^2+20))))/(4*x^3+5*x) 
/log(x/(16*x^2+20)),x)
 

Output:

25*e**(e**x)*log(log(x/(16*x**2 + 20)))**2