\(\int \frac {(-2-2 e^{4 x+2 x^2}+4 x+4 x^2) \log ^2(\frac {25 e^{-x}}{4})+(-2 x-2 e^{4 x+2 x^2} x) \log (\frac {25 e^{-x}}{4}) \log (e^{-4 x-2 x^2} (x^2+e^{4 x+2 x^2} x^2))}{(x+e^{4 x+2 x^2} x) \log ^2(e^{-4 x-2 x^2} (x^2+e^{4 x+2 x^2} x^2))} \, dx\) [172]

Optimal result

Integrand size = 150, antiderivative size = 32 \[ \int \frac {\left (-2-2 e^{4 x+2 x^2}+4 x+4 x^2\right ) \log ^2\left (\frac {25 e^{-x}}{4}\right )+\left (-2 x-2 e^{4 x+2 x^2} x\right ) \log \left (\frac {25 e^{-x}}{4}\right ) \log \left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )}{\left (x+e^{4 x+2 x^2} x\right ) \log ^2\left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )} \, dx=\frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{\log \left (x \left (x+e^{-x (4+2 x)} x\right )\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {\left (-2-2 e^{4 x+2 x^2}+4 x+4 x^2\right ) \log ^2\left (\frac {25 e^{-x}}{4}\right )+\left (-2 x-2 e^{4 x+2 x^2} x\right ) \log \left (\frac {25 e^{-x}}{4}\right ) \log \left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )}{\left (x+e^{4 x+2 x^2} x\right ) \log ^2\left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )} \, dx=\frac {\log ^2\left (\frac {25 e^{-x}}{4}\right )}{\log \left (\left (1+e^{-4 x-2 x^2}\right ) x^2\right )} \] Input:

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

Output:

Log[25/(4*E^x)]^2/Log[(1 + E^(-4*x - 2*x^2))*x^2]
 

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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.97 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31

method	result	size
parallelrisch	\(\frac {\ln \left (\frac {25 \,{\mathrm e}^{-x}}{4}\right )^{2}}{\ln \left (x^{2} \left ({\mathrm e}^{2 x^{2}+4 x}+1\right ) {\mathrm e}^{-2 x^{2}-4 x}\right )}\)	\(42\)
risch	\(\frac {i \left (-32 \ln \left (2\right ) \ln \left (5\right )+16 \ln \left (5\right )^{2}+16 \ln \left (2\right )^{2}-16 \ln \left (5\right ) \ln \left ({\mathrm e}^{x}\right )+16 \ln \left (2\right ) \ln \left ({\mathrm e}^{x}\right )+4 \ln \left ({\mathrm e}^{x}\right )^{2}\right )}{2 \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-4 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+2 \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right )-2 \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right ) \operatorname {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right )^{2}-2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )}\right ) \operatorname {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right )^{2}+2 \pi \operatorname {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right )^{3}-2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right ) \operatorname {csgn}\left (i x^{2} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right ) {\mathrm e}^{-2 x \left (2+x \right )}\right )^{2}+2 \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-2 x \left (2+x \right )} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )\right ) \operatorname {csgn}\left (i x^{2} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right ) {\mathrm e}^{-2 x \left (2+x \right )}\right ) \operatorname {csgn}\left (i x^{2}\right )+2 \pi \operatorname {csgn}\left (i x^{2} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right ) {\mathrm e}^{-2 x \left (2+x \right )}\right )^{3}-2 \pi \operatorname {csgn}\left (i x^{2} \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right ) {\mathrm e}^{-2 x \left (2+x \right )}\right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+8 i \ln \left (x \right )+4 i \ln \left ({\mathrm e}^{2 x \left (2+x \right )}+1\right )-4 i \ln \left ({\mathrm e}^{2 x \left (2+x \right )}\right )}\)	\(426\)

Input:

int(((-2*x*exp(2*x^2+4*x)-2*x)*ln(25/4/exp(x))*ln((x^2*exp(2*x^2+4*x)+x^2) 
/exp(2*x^2+4*x))+(-2*exp(2*x^2+4*x)+4*x^2+4*x-2)*ln(25/4/exp(x))^2)/(x*exp 
(2*x^2+4*x)+x)/ln((x^2*exp(2*x^2+4*x)+x^2)/exp(2*x^2+4*x))^2,x,method=_RET 
URNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \frac {\left (-2-2 e^{4 x+2 x^2}+4 x+4 x^2\right ) \log ^2\left (\frac {25 e^{-x}}{4}\right )+\left (-2 x-2 e^{4 x+2 x^2} x\right ) \log \left (\frac {25 e^{-x}}{4}\right ) \log \left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )}{\left (x+e^{4 x+2 x^2} x\right ) \log ^2\left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )} \, dx=\frac {x^{2} - 2 \, x \log \left (\frac {25}{4}\right ) + \log \left (\frac {25}{4}\right )^{2}}{\log \left ({\left (x^{2} e^{\left (2 \, x^{2} + 4 \, x\right )} + x^{2}\right )} e^{\left (-2 \, x^{2} - 4 \, x\right )}\right )} \] Input:

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

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (24) = 48\).

Time = 0.34 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.12 \[ \int \frac {\left (-2-2 e^{4 x+2 x^2}+4 x+4 x^2\right ) \log ^2\left (\frac {25 e^{-x}}{4}\right )+\left (-2 x-2 e^{4 x+2 x^2} x\right ) \log \left (\frac {25 e^{-x}}{4}\right ) \log \left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )}{\left (x+e^{4 x+2 x^2} x\right ) \log ^2\left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )} \, dx=\frac {x^{2} - 4 x \log {\left (5 \right )} + 4 x \log {\left (2 \right )} - 8 \log {\left (2 \right )} \log {\left (5 \right )} + 4 \log {\left (2 \right )}^{2} + 4 \log {\left (5 \right )}^{2}}{\log {\left (\left (x^{2} e^{2 x^{2} + 4 x} + x^{2}\right ) e^{- 2 x^{2} - 4 x} \right )}} \] Input:

integrate(((-2*x*exp(2*x**2+4*x)-2*x)*ln(25/4/exp(x))*ln((x**2*exp(2*x**2+ 
4*x)+x**2)/exp(2*x**2+4*x))+(-2*exp(2*x**2+4*x)+4*x**2+4*x-2)*ln(25/4/exp( 
x))**2)/(x*exp(2*x**2+4*x)+x)/ln((x**2*exp(2*x**2+4*x)+x**2)/exp(2*x**2+4* 
x))**2,x)
 

Output:

(x**2 - 4*x*log(5) + 4*x*log(2) - 8*log(2)*log(5) + 4*log(2)**2 + 4*log(5) 
**2)/log((x**2*exp(2*x**2 + 4*x) + x**2)*exp(-2*x**2 - 4*x))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (26) = 52\).

Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.00 \[ \int \frac {\left (-2-2 e^{4 x+2 x^2}+4 x+4 x^2\right ) \log ^2\left (\frac {25 e^{-x}}{4}\right )+\left (-2 x-2 e^{4 x+2 x^2} x\right ) \log \left (\frac {25 e^{-x}}{4}\right ) \log \left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )}{\left (x+e^{4 x+2 x^2} x\right ) \log ^2\left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )} \, dx=-\frac {x^{2} - 4 \, x {\left (\log \left (5\right ) - \log \left (2\right )\right )} + 4 \, \log \left (5\right )^{2} - 8 \, \log \left (5\right ) \log \left (2\right ) + 4 \, \log \left (2\right )^{2}}{2 \, x^{2} + 4 \, x - 2 \, \log \left (x\right ) - \log \left (e^{\left (2 \, x^{2} + 4 \, x\right )} + 1\right )} \] Input:

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

Output:

-(x^2 - 4*x*(log(5) - log(2)) + 4*log(5)^2 - 8*log(5)*log(2) + 4*log(2)^2) 
/(2*x^2 + 4*x - 2*log(x) - log(e^(2*x^2 + 4*x) + 1))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (26) = 52\).

Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.03 \[ \int \frac {\left (-2-2 e^{4 x+2 x^2}+4 x+4 x^2\right ) \log ^2\left (\frac {25 e^{-x}}{4}\right )+\left (-2 x-2 e^{4 x+2 x^2} x\right ) \log \left (\frac {25 e^{-x}}{4}\right ) \log \left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )}{\left (x+e^{4 x+2 x^2} x\right ) \log ^2\left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )} \, dx=\frac {x^{2} - 4 \, x \log \left (5\right ) + 4 \, \log \left (5\right )^{2} + 4 \, x \log \left (2\right ) - 8 \, \log \left (5\right ) \log \left (2\right ) + 4 \, \log \left (2\right )^{2}}{\log \left ({\left (x^{2} e^{\left (2 \, x^{2} + 4 \, x\right )} + x^{2}\right )} e^{\left (-2 \, x^{2} - 4 \, x\right )}\right )} \] Input:

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

Output:

(x^2 - 4*x*log(5) + 4*log(5)^2 + 4*x*log(2) - 8*log(5)*log(2) + 4*log(2)^2 
)/log((x^2*e^(2*x^2 + 4*x) + x^2)*e^(-2*x^2 - 4*x))
 

Mupad [B] (verification not implemented)

Time = 2.58 (sec) , antiderivative size = 226, normalized size of antiderivative = 7.06 \[ \int \frac {\left (-2-2 e^{4 x+2 x^2}+4 x+4 x^2\right ) \log ^2\left (\frac {25 e^{-x}}{4}\right )+\left (-2 x-2 e^{4 x+2 x^2} x\right ) \log \left (\frac {25 e^{-x}}{4}\right ) \log \left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )}{\left (x+e^{4 x+2 x^2} x\right ) \log ^2\left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )} \, dx=\frac {{\left (x-\ln \left (\frac {25}{4}\right )\right )}^2+\frac {x\,\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{-2\,x^2}\,\left (x^2+x^2\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{2\,x^2}\right )\right )\,\left ({\mathrm {e}}^{2\,x^2+4\,x}+1\right )\,\left (x-\ln \left (\frac {25}{4}\right )\right )}{2\,x-{\mathrm {e}}^{2\,x^2+4\,x}+2\,x^2-1}}{\ln \left ({\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^{-2\,x^2}\,\left (x^2+x^2\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{2\,x^2}\right )\right )}-x\,\ln \left (\frac {25}{4}\right )+x^2-\frac {2\,\left (3\,x^2\,\ln \left (\frac {25}{4}\right )+3\,x^3\,\ln \left (\frac {25}{4}\right )-8\,x^4\,\ln \left (\frac {25}{4}\right )-12\,x^5\,\ln \left (\frac {25}{4}\right )-4\,x^6\,\ln \left (\frac {25}{4}\right )-3\,x^3-3\,x^4+8\,x^5+12\,x^6+4\,x^7\right )}{\left (4\,x^3+8\,x^2-3\right )\,\left (2\,x-{\mathrm {e}}^{2\,x^2+4\,x}+2\,x^2-1\right )} \] Input:

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

Output:

((x - log(25/4))^2 + (x*log(exp(-4*x)*exp(-2*x^2)*(x^2 + x^2*exp(4*x)*exp( 
2*x^2)))*(exp(4*x + 2*x^2) + 1)*(x - log(25/4)))/(2*x - exp(4*x + 2*x^2) + 
 2*x^2 - 1))/log(exp(-4*x)*exp(-2*x^2)*(x^2 + x^2*exp(4*x)*exp(2*x^2))) - 
x*log(25/4) + x^2 - (2*(3*x^2*log(25/4) + 3*x^3*log(25/4) - 8*x^4*log(25/4 
) - 12*x^5*log(25/4) - 4*x^6*log(25/4) - 3*x^3 - 3*x^4 + 8*x^5 + 12*x^6 + 
4*x^7))/((8*x^2 + 4*x^3 - 3)*(2*x - exp(4*x + 2*x^2) + 2*x^2 - 1))
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.47 \[ \int \frac {\left (-2-2 e^{4 x+2 x^2}+4 x+4 x^2\right ) \log ^2\left (\frac {25 e^{-x}}{4}\right )+\left (-2 x-2 e^{4 x+2 x^2} x\right ) \log \left (\frac {25 e^{-x}}{4}\right ) \log \left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )}{\left (x+e^{4 x+2 x^2} x\right ) \log ^2\left (e^{-4 x-2 x^2} \left (x^2+e^{4 x+2 x^2} x^2\right )\right )} \, dx=\frac {\mathrm {log}\left (\frac {25}{4 e^{x}}\right )^{2}}{\mathrm {log}\left (\frac {e^{2 x^{2}+4 x} x^{2}+x^{2}}{e^{2 x^{2}+4 x}}\right )} \] Input:

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

Output:

log(25/(4*e**x))**2/log((e**(2*x**2 + 4*x)*x**2 + x**2)/e**(2*x**2 + 4*x))