\(\int \frac {-25 x+e^{4 x^2} (-50 x^3-200 x^5)-25 x \log (e^{e^{4 x^2} x^2} x)+(2+45 x) \log ^2(e^{e^{4 x^2} x^2} x)}{5 x^3 \log ^2(e^{e^{4 x^2} x^2} x)} \, dx\) [1176]

Optimal result

Integrand size = 91, antiderivative size = 33 \[ \int \frac {-25 x+e^{4 x^2} \left (-50 x^3-200 x^5\right )-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+(2+45 x) \log ^2\left (e^{e^{4 x^2} x^2} x\right )}{5 x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx=\frac {-9-\frac {1}{5 x}+\frac {5}{\log \left (e^{e^{4 x^2} x^2} x\right )}}{x} \] Output:

(5/ln(x*exp(x^2*exp(x^2)^4))-1/5/x-9)/x
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {-25 x+e^{4 x^2} \left (-50 x^3-200 x^5\right )-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+(2+45 x) \log ^2\left (e^{e^{4 x^2} x^2} x\right )}{5 x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx=\frac {1}{5} \left (-\frac {1}{x^2}-\frac {45}{x}+\frac {25}{x \log \left (e^{e^{4 x^2} x^2} x\right )}\right ) \] Input:

Integrate[(-25*x + E^(4*x^2)*(-50*x^3 - 200*x^5) - 25*x*Log[E^(E^(4*x^2)*x 
^2)*x] + (2 + 45*x)*Log[E^(E^(4*x^2)*x^2)*x]^2)/(5*x^3*Log[E^(E^(4*x^2)*x^ 
2)*x]^2),x]
 

Output:

(-x^(-2) - 45/x + 25/(x*Log[E^(E^(4*x^2)*x^2)*x]))/5
 

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.

\(\displaystyle \int \frac {(45 x+2) \log ^2\left (e^{e^{4 x^2} x^2} x\right )-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+e^{4 x^2} \left (-200 x^5-50 x^3\right )-25 x}{5 x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int -\frac {-\left ((45 x+2) \log ^2\left (e^{e^{4 x^2} x^2} x\right )\right )+25 x \log \left (e^{e^{4 x^2} x^2} x\right )+25 x+50 e^{4 x^2} \left (4 x^5+x^3\right )}{x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{5} \int \frac {-\left ((45 x+2) \log ^2\left (e^{e^{4 x^2} x^2} x\right )\right )+25 x \log \left (e^{e^{4 x^2} x^2} x\right )+25 x+50 e^{4 x^2} \left (4 x^5+x^3\right )}{x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{5} \int \left (\frac {50 e^{4 x^2} \left (4 x^2+1\right )}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )}+\frac {-45 x \log ^2\left (e^{e^{4 x^2} x^2} x\right )-2 \log ^2\left (e^{e^{4 x^2} x^2} x\right )+25 x \log \left (e^{e^{4 x^2} x^2} x\right )+25 x}{x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{5} \left (-50 \int \frac {e^{4 x^2}}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )}dx-25 \int \frac {1}{x^2 \log ^2\left (e^{e^{4 x^2} x^2} x\right )}dx-200 \int \frac {e^{4 x^2} x^2}{\log ^2\left (e^{e^{4 x^2} x^2} x\right )}dx-25 \int \frac {1}{x^2 \log \left (e^{e^{4 x^2} x^2} x\right )}dx-\frac {(45 x+2)^2}{4 x^2}\right )\)

Input:

Int[(-25*x + E^(4*x^2)*(-50*x^3 - 200*x^5) - 25*x*Log[E^(E^(4*x^2)*x^2)*x] 
 + (2 + 45*x)*Log[E^(E^(4*x^2)*x^2)*x]^2)/(5*x^3*Log[E^(E^(4*x^2)*x^2)*x]^ 
2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.79

method	result	size
parallelrisch	\(\frac {-45 x \ln \left (x \,{\mathrm e}^{{\mathrm e}^{4 x^{2}} x^{2}}\right )+25 x -\ln \left (x \,{\mathrm e}^{{\mathrm e}^{4 x^{2}} x^{2}}\right )}{5 x^{2} \ln \left (x \,{\mathrm e}^{{\mathrm e}^{4 x^{2}} x^{2}}\right )}\)	\(59\)
risch	\(-\frac {45 x +1}{5 x^{2}}-\frac {10 i}{x \left (\pi \,\operatorname {csgn}\left (i {\mathrm e}^{{\mathrm e}^{4 x^{2}} x^{2}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{4 x^{2}} x^{2}}\right )^{2}-\pi \,\operatorname {csgn}\left (i {\mathrm e}^{{\mathrm e}^{4 x^{2}} x^{2}}\right ) \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{4 x^{2}} x^{2}}\right ) \operatorname {csgn}\left (i x \right )-\pi \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{4 x^{2}} x^{2}}\right )^{3}+\pi \operatorname {csgn}\left (i x \,{\mathrm e}^{{\mathrm e}^{4 x^{2}} x^{2}}\right )^{2} \operatorname {csgn}\left (i x \right )-2 i \ln \left (x \right )-2 i \ln \left ({\mathrm e}^{{\mathrm e}^{4 x^{2}} x^{2}}\right )\right )}\)	\(161\)

Input:

int(1/5*((45*x+2)*ln(x*exp(x^2*exp(x^2)^4))^2-25*x*ln(x*exp(x^2*exp(x^2)^4 
))+(-200*x^5-50*x^3)*exp(x^2)^4-25*x)/x^3/ln(x*exp(x^2*exp(x^2)^4))^2,x,me 
thod=_RETURNVERBOSE)
 

Output:

1/5/x^2*(-45*x*ln(x*exp(x^2*exp(x^2)^4))+25*x-ln(x*exp(x^2*exp(x^2)^4)))/l 
n(x*exp(x^2*exp(x^2)^4))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36 \[ \int \frac {-25 x+e^{4 x^2} \left (-50 x^3-200 x^5\right )-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+(2+45 x) \log ^2\left (e^{e^{4 x^2} x^2} x\right )}{5 x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx=-\frac {{\left (45 \, x + 1\right )} \log \left (x e^{\left (x^{2} e^{\left (4 \, x^{2}\right )}\right )}\right ) - 25 \, x}{5 \, x^{2} \log \left (x e^{\left (x^{2} e^{\left (4 \, x^{2}\right )}\right )}\right )} \] Input:

integrate(1/5*((45*x+2)*log(x*exp(x^2*exp(x^2)^4))^2-25*x*log(x*exp(x^2*ex 
p(x^2)^4))+(-200*x^5-50*x^3)*exp(x^2)^4-25*x)/x^3/log(x*exp(x^2*exp(x^2)^4 
))^2,x, algorithm="fricas")
 

Output:

-1/5*((45*x + 1)*log(x*e^(x^2*e^(4*x^2))) - 25*x)/(x^2*log(x*e^(x^2*e^(4*x 
^2))))
 

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {-25 x+e^{4 x^2} \left (-50 x^3-200 x^5\right )-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+(2+45 x) \log ^2\left (e^{e^{4 x^2} x^2} x\right )}{5 x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx=\frac {5}{x \log {\left (x e^{x^{2} e^{4 x^{2}}} \right )}} + \frac {- 45 x - 1}{5 x^{2}} \] Input:

integrate(1/5*((45*x+2)*ln(x*exp(x**2*exp(x**2)**4))**2-25*x*ln(x*exp(x**2 
*exp(x**2)**4))+(-200*x**5-50*x**3)*exp(x**2)**4-25*x)/x**3/ln(x*exp(x**2* 
exp(x**2)**4))**2,x)
 

Output:

5/(x*log(x*exp(x**2*exp(4*x**2)))) + (-45*x - 1)/(5*x**2)
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {-25 x+e^{4 x^2} \left (-50 x^3-200 x^5\right )-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+(2+45 x) \log ^2\left (e^{e^{4 x^2} x^2} x\right )}{5 x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx=\frac {5}{x^{3} e^{\left (4 \, x^{2}\right )} + x \log \left (x\right )} - \frac {9}{x} - \frac {1}{5 \, x^{2}} \] Input:

integrate(1/5*((45*x+2)*log(x*exp(x^2*exp(x^2)^4))^2-25*x*log(x*exp(x^2*ex 
p(x^2)^4))+(-200*x^5-50*x^3)*exp(x^2)^4-25*x)/x^3/log(x*exp(x^2*exp(x^2)^4 
))^2,x, algorithm="maxima")
 

Output:

5/(x^3*e^(4*x^2) + x*log(x)) - 9/x - 1/5/x^2
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.61 \[ \int \frac {-25 x+e^{4 x^2} \left (-50 x^3-200 x^5\right )-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+(2+45 x) \log ^2\left (e^{e^{4 x^2} x^2} x\right )}{5 x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx=-\frac {45 \, x^{3} e^{\left (4 \, x^{2}\right )} + x^{2} e^{\left (4 \, x^{2}\right )} + 45 \, x \log \left (x\right ) - 25 \, x + \log \left (x\right )}{5 \, {\left (x^{4} e^{\left (4 \, x^{2}\right )} + x^{2} \log \left (x\right )\right )}} \] Input:

integrate(1/5*((45*x+2)*log(x*exp(x^2*exp(x^2)^4))^2-25*x*log(x*exp(x^2*ex 
p(x^2)^4))+(-200*x^5-50*x^3)*exp(x^2)^4-25*x)/x^3/log(x*exp(x^2*exp(x^2)^4 
))^2,x, algorithm="giac")
 

Output:

-1/5*(45*x^3*e^(4*x^2) + x^2*e^(4*x^2) + 45*x*log(x) - 25*x + log(x))/(x^4 
*e^(4*x^2) + x^2*log(x))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-25 x+e^{4 x^2} \left (-50 x^3-200 x^5\right )-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+(2+45 x) \log ^2\left (e^{e^{4 x^2} x^2} x\right )}{5 x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx=-\int \frac {5\,x+5\,x\,\ln \left (x\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{4\,x^2}}\right )-\frac {{\ln \left (x\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{4\,x^2}}\right )}^2\,\left (45\,x+2\right )}{5}+\frac {{\mathrm {e}}^{4\,x^2}\,\left (200\,x^5+50\,x^3\right )}{5}}{x^3\,{\ln \left (x\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^{4\,x^2}}\right )}^2} \,d x \] Input:

int(-(5*x + 5*x*log(x*exp(x^2*exp(4*x^2))) - (log(x*exp(x^2*exp(4*x^2)))^2 
*(45*x + 2))/5 + (exp(4*x^2)*(50*x^3 + 200*x^5))/5)/(x^3*log(x*exp(x^2*exp 
(4*x^2)))^2),x)
 

Output:

-int((5*x + 5*x*log(x*exp(x^2*exp(4*x^2))) - (log(x*exp(x^2*exp(4*x^2)))^2 
*(45*x + 2))/5 + (exp(4*x^2)*(50*x^3 + 200*x^5))/5)/(x^3*log(x*exp(x^2*exp 
(4*x^2)))^2), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.94 \[ \int \frac {-25 x+e^{4 x^2} \left (-50 x^3-200 x^5\right )-25 x \log \left (e^{e^{4 x^2} x^2} x\right )+(2+45 x) \log ^2\left (e^{e^{4 x^2} x^2} x\right )}{5 x^3 \log ^2\left (e^{e^{4 x^2} x^2} x\right )} \, dx=\frac {-45 \,\mathrm {log}\left (e^{e^{4 x^{2}} x^{2}} x \right ) x -\mathrm {log}\left (e^{e^{4 x^{2}} x^{2}} x \right )+25 x}{5 \,\mathrm {log}\left (e^{e^{4 x^{2}} x^{2}} x \right ) x^{2}} \] Input:

int(1/5*((45*x+2)*log(x*exp(x^2*exp(x^2)^4))^2-25*x*log(x*exp(x^2*exp(x^2) 
^4))+(-200*x^5-50*x^3)*exp(x^2)^4-25*x)/x^3/log(x*exp(x^2*exp(x^2)^4))^2,x 
)
 

Output:

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