\(\int \frac {-5 x+x^2+e^{x^2} (10 x^3-2 x^4+e^3 (-10 x^2+2 x^3))+(-20 e^3+e^{x^2} (5 e^3-5 x)+20 x+(-5 e^3+5 x) \log (e^3-x)) \log (\frac {5}{4-e^{x^2}+\log (e^3-x)})}{(4 e^3 x^2-4 x^3+e^{x^2} (-e^3 x^2+x^3)+(e^3 x^2-x^3) \log (e^3-x)) \log ^2(\frac {5}{4-e^{x^2}+\log (e^3-x)})} \, dx\) [2426]

Optimal result

Integrand size = 185, antiderivative size = 35 \[ \int \frac {-5 x+x^2+e^{x^2} \left (10 x^3-2 x^4+e^3 \left (-10 x^2+2 x^3\right )\right )+\left (-20 e^3+e^{x^2} \left (5 e^3-5 x\right )+20 x+\left (-5 e^3+5 x\right ) \log \left (e^3-x\right )\right ) \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}{\left (4 e^3 x^2-4 x^3+e^{x^2} \left (-e^3 x^2+x^3\right )+\left (e^3 x^2-x^3\right ) \log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx=3+\frac {5-x}{x \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \] Output:

3+(5-x)/ln(5/(ln(-x+exp(3))+4-exp(x^2)))/x
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {-5 x+x^2+e^{x^2} \left (10 x^3-2 x^4+e^3 \left (-10 x^2+2 x^3\right )\right )+\left (-20 e^3+e^{x^2} \left (5 e^3-5 x\right )+20 x+\left (-5 e^3+5 x\right ) \log \left (e^3-x\right )\right ) \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}{\left (4 e^3 x^2-4 x^3+e^{x^2} \left (-e^3 x^2+x^3\right )+\left (e^3 x^2-x^3\right ) \log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx=\frac {5-x}{x \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \] Input:

Integrate[(-5*x + x^2 + E^x^2*(10*x^3 - 2*x^4 + E^3*(-10*x^2 + 2*x^3)) + ( 
-20*E^3 + E^x^2*(5*E^3 - 5*x) + 20*x + (-5*E^3 + 5*x)*Log[E^3 - x])*Log[5/ 
(4 - E^x^2 + Log[E^3 - x])])/((4*E^3*x^2 - 4*x^3 + E^x^2*(-(E^3*x^2) + x^3 
) + (E^3*x^2 - x^3)*Log[E^3 - x])*Log[5/(4 - E^x^2 + Log[E^3 - x])]^2),x]
 

Output:

(5 - x)/(x*Log[5/(4 - E^x^2 + Log[E^3 - 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.

\(\displaystyle \int \frac {x^2+\left (e^{x^2} \left (5 e^3-5 x\right )+20 x+\left (5 x-5 e^3\right ) \log \left (e^3-x\right )-20 e^3\right ) \log \left (\frac {5}{-e^{x^2}+\log \left (e^3-x\right )+4}\right )+e^{x^2} \left (-2 x^4+10 x^3+e^3 \left (2 x^3-10 x^2\right )\right )-5 x}{\left (-4 x^3+4 e^3 x^2+e^{x^2} \left (x^3-e^3 x^2\right )+\left (e^3 x^2-x^3\right ) \log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{-e^{x^2}+\log \left (e^3-x\right )+4}\right )} \, dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {-2 x^3+10 x^2-5 \log \left (\frac {5}{-e^{x^2}+\log \left (e^3-x\right )+4}\right )}{x^2 \log ^2\left (\frac {5}{-e^{x^2}+\log \left (e^3-x\right )+4}\right )}-\frac {(x-5) \left (8 x^2+2 x^2 \log \left (e^3-x\right )-8 e^3 x-2 e^3 x \log \left (e^3-x\right )-1\right )}{x \left (x-e^3\right ) \left (e^{x^2}-\log \left (e^3-x\right )-4\right ) \log ^2\left (\frac {5}{-e^{x^2}+\log \left (e^3-x\right )+4}\right )}\right )dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \left (\frac {-2 x^3+10 x^2-5 \log \left (\frac {5}{-e^{x^2}+\log \left (e^3-x\right )+4}\right )}{x^2 \log ^2\left (\frac {5}{-e^{x^2}+\log \left (e^3-x\right )+4}\right )}-\frac {(x-5) \left (8 x^2+2 x^2 \log \left (e^3-x\right )-8 e^3 x-2 e^3 x \log \left (e^3-x\right )-1\right )}{x \left (x-e^3\right ) \left (e^{x^2}-\log \left (e^3-x\right )-4\right ) \log ^2\left (\frac {5}{-e^{x^2}+\log \left (e^3-x\right )+4}\right )}\right )dx\)

Input:

Int[(-5*x + x^2 + E^x^2*(10*x^3 - 2*x^4 + E^3*(-10*x^2 + 2*x^3)) + (-20*E^ 
3 + E^x^2*(5*E^3 - 5*x) + 20*x + (-5*E^3 + 5*x)*Log[E^3 - x])*Log[5/(4 - E 
^x^2 + Log[E^3 - x])])/((4*E^3*x^2 - 4*x^3 + E^x^2*(-(E^3*x^2) + x^3) + (E 
^3*x^2 - x^3)*Log[E^3 - x])*Log[5/(4 - E^x^2 + Log[E^3 - x])]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 31.72 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91

method	result	size
parallelrisch	\(\frac {5-x}{\ln \left (\frac {5}{\ln \left (-x +{\mathrm e}^{3}\right )+4-{\mathrm e}^{x^{2}}}\right ) x}\)	\(32\)
risch	\(\frac {2 i \left (-5+x \right )}{x \left (-2 \pi {\operatorname {csgn}\left (\frac {i}{-\ln \left (-x +{\mathrm e}^{3}\right )-4+{\mathrm e}^{x^{2}}}\right )}^{2}+2 \pi {\operatorname {csgn}\left (\frac {i}{-\ln \left (-x +{\mathrm e}^{3}\right )-4+{\mathrm e}^{x^{2}}}\right )}^{3}+2 \pi -2 i \ln \left (5\right )+2 i \ln \left (-\ln \left (-x +{\mathrm e}^{3}\right )-4+{\mathrm e}^{x^{2}}\right )\right )}\)	\(92\)

Input:

int((((-5*exp(3)+5*x)*ln(-x+exp(3))+(5*exp(3)-5*x)*exp(x^2)-20*exp(3)+20*x 
)*ln(5/(ln(-x+exp(3))+4-exp(x^2)))+((2*x^3-10*x^2)*exp(3)-2*x^4+10*x^3)*ex 
p(x^2)+x^2-5*x)/((x^2*exp(3)-x^3)*ln(-x+exp(3))+(-x^2*exp(3)+x^3)*exp(x^2) 
+4*x^2*exp(3)-4*x^3)/ln(5/(ln(-x+exp(3))+4-exp(x^2)))^2,x,method=_RETURNVE 
RBOSE)
 

Output:

(5-x)/ln(5/(ln(-x+exp(3))+4-exp(x^2)))/x
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int \frac {-5 x+x^2+e^{x^2} \left (10 x^3-2 x^4+e^3 \left (-10 x^2+2 x^3\right )\right )+\left (-20 e^3+e^{x^2} \left (5 e^3-5 x\right )+20 x+\left (-5 e^3+5 x\right ) \log \left (e^3-x\right )\right ) \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}{\left (4 e^3 x^2-4 x^3+e^{x^2} \left (-e^3 x^2+x^3\right )+\left (e^3 x^2-x^3\right ) \log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx=-\frac {x - 5}{x \log \left (-\frac {5}{e^{\left (x^{2}\right )} - \log \left (-x + e^{3}\right ) - 4}\right )} \] Input:

integrate((((-5*exp(3)+5*x)*log(-x+exp(3))+(5*exp(3)-5*x)*exp(x^2)-20*exp( 
3)+20*x)*log(5/(log(-x+exp(3))+4-exp(x^2)))+((2*x^3-10*x^2)*exp(3)-2*x^4+1 
0*x^3)*exp(x^2)+x^2-5*x)/((x^2*exp(3)-x^3)*log(-x+exp(3))+(-x^2*exp(3)+x^3 
)*exp(x^2)+4*x^2*exp(3)-4*x^3)/log(5/(log(-x+exp(3))+4-exp(x^2)))^2,x, alg 
orithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 1.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.57 \[ \int \frac {-5 x+x^2+e^{x^2} \left (10 x^3-2 x^4+e^3 \left (-10 x^2+2 x^3\right )\right )+\left (-20 e^3+e^{x^2} \left (5 e^3-5 x\right )+20 x+\left (-5 e^3+5 x\right ) \log \left (e^3-x\right )\right ) \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}{\left (4 e^3 x^2-4 x^3+e^{x^2} \left (-e^3 x^2+x^3\right )+\left (e^3 x^2-x^3\right ) \log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx=\frac {5 - x}{x \log {\left (\frac {5}{- e^{x^{2}} + \log {\left (- x + e^{3} \right )} + 4} \right )}} \] Input:

integrate((((-5*exp(3)+5*x)*ln(-x+exp(3))+(5*exp(3)-5*x)*exp(x**2)-20*exp( 
3)+20*x)*ln(5/(ln(-x+exp(3))+4-exp(x**2)))+((2*x**3-10*x**2)*exp(3)-2*x**4 
+10*x**3)*exp(x**2)+x**2-5*x)/((x**2*exp(3)-x**3)*ln(-x+exp(3))+(-x**2*exp 
(3)+x**3)*exp(x**2)+4*x**2*exp(3)-4*x**3)/ln(5/(ln(-x+exp(3))+4-exp(x**2)) 
)**2,x)
 

Output:

(5 - x)/(x*log(5/(-exp(x**2) + log(-x + exp(3)) + 4)))
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {-5 x+x^2+e^{x^2} \left (10 x^3-2 x^4+e^3 \left (-10 x^2+2 x^3\right )\right )+\left (-20 e^3+e^{x^2} \left (5 e^3-5 x\right )+20 x+\left (-5 e^3+5 x\right ) \log \left (e^3-x\right )\right ) \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}{\left (4 e^3 x^2-4 x^3+e^{x^2} \left (-e^3 x^2+x^3\right )+\left (e^3 x^2-x^3\right ) \log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx=-\frac {x - 5}{x \log \left (5\right ) - x \log \left (-e^{\left (x^{2}\right )} + \log \left (-x + e^{3}\right ) + 4\right )} \] Input:

integrate((((-5*exp(3)+5*x)*log(-x+exp(3))+(5*exp(3)-5*x)*exp(x^2)-20*exp( 
3)+20*x)*log(5/(log(-x+exp(3))+4-exp(x^2)))+((2*x^3-10*x^2)*exp(3)-2*x^4+1 
0*x^3)*exp(x^2)+x^2-5*x)/((x^2*exp(3)-x^3)*log(-x+exp(3))+(-x^2*exp(3)+x^3 
)*exp(x^2)+4*x^2*exp(3)-4*x^3)/log(5/(log(-x+exp(3))+4-exp(x^2)))^2,x, alg 
orithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 569 vs. \(2 (32) = 64\).

Time = 0.53 (sec) , antiderivative size = 569, normalized size of antiderivative = 16.26 \[ \int \frac {-5 x+x^2+e^{x^2} \left (10 x^3-2 x^4+e^3 \left (-10 x^2+2 x^3\right )\right )+\left (-20 e^3+e^{x^2} \left (5 e^3-5 x\right )+20 x+\left (-5 e^3+5 x\right ) \log \left (e^3-x\right )\right ) \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}{\left (4 e^3 x^2-4 x^3+e^{x^2} \left (-e^3 x^2+x^3\right )+\left (e^3 x^2-x^3\right ) \log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx =\text {Too large to display} \] Input:

integrate((((-5*exp(3)+5*x)*log(-x+exp(3))+(5*exp(3)-5*x)*exp(x^2)-20*exp( 
3)+20*x)*log(5/(log(-x+exp(3))+4-exp(x^2)))+((2*x^3-10*x^2)*exp(3)-2*x^4+1 
0*x^3)*exp(x^2)+x^2-5*x)/((x^2*exp(3)-x^3)*log(-x+exp(3))+(-x^2*exp(3)+x^3 
)*exp(x^2)+4*x^2*exp(3)-4*x^3)/log(5/(log(-x+exp(3))+4-exp(x^2)))^2,x, alg 
orithm="giac")
 

Output:

2*(2*x*log(5) - x*log(1/2*pi^2*sgn(x - e^3) + 1/2*pi^2 - 2*e^(x^2)*log(abs 
(x - e^3)) + log(abs(x - e^3))^2 + e^(2*x^2) - 8*e^(x^2) + 8*log(abs(x - e 
^3)) + 16) - 10*log(5) + 5*log(1/2*pi^2*sgn(x - e^3) + 1/2*pi^2 - 2*e^(x^2 
)*log(abs(x - e^3)) + log(abs(x - e^3))^2 + e^(2*x^2) - 8*e^(x^2) + 8*log( 
abs(x - e^3)) + 16))/(4*pi^2*x*sgn(pi + pi*sgn(x - e^3))*sgn(e^(x^2) - log 
(abs(x - e^3)) - 4) + 4*pi*x*arctan(1/2*(pi + pi*sgn(x - e^3))/(e^(x^2) - 
log(abs(x - e^3)) - 4))*sgn(pi + pi*sgn(x - e^3))*sgn(e^(x^2) - log(abs(x 
- e^3)) - 4) - 4*pi^2*x*sgn(pi + pi*sgn(x - e^3)) - 4*pi*x*arctan(1/2*(pi 
+ pi*sgn(x - e^3))/(e^(x^2) - log(abs(x - e^3)) - 4))*sgn(pi + pi*sgn(x - 
e^3)) + 2*pi^2*x*sgn(e^(x^2) - log(abs(x - e^3)) - 4) - 6*pi^2*x - 8*pi*x* 
arctan(1/2*(pi + pi*sgn(x - e^3))/(e^(x^2) - log(abs(x - e^3)) - 4)) - 4*x 
*arctan(1/2*(pi + pi*sgn(x - e^3))/(e^(x^2) - log(abs(x - e^3)) - 4))^2 - 
4*x*log(5)^2 + 4*x*log(5)*log(1/2*pi^2*sgn(x - e^3) + 1/2*pi^2 - 2*e^(x^2) 
*log(abs(x - e^3)) + log(abs(x - e^3))^2 + e^(2*x^2) - 8*e^(x^2) + 8*log(a 
bs(x - e^3)) + 16) - x*log(1/2*pi^2*sgn(x - e^3) + 1/2*pi^2 - 2*e^(x^2)*lo 
g(abs(x - e^3)) + log(abs(x - e^3))^2 + e^(2*x^2) - 8*e^(x^2) + 8*log(abs( 
x - e^3)) + 16)^2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-5 x+x^2+e^{x^2} \left (10 x^3-2 x^4+e^3 \left (-10 x^2+2 x^3\right )\right )+\left (-20 e^3+e^{x^2} \left (5 e^3-5 x\right )+20 x+\left (-5 e^3+5 x\right ) \log \left (e^3-x\right )\right ) \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}{\left (4 e^3 x^2-4 x^3+e^{x^2} \left (-e^3 x^2+x^3\right )+\left (e^3 x^2-x^3\right ) \log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx=\int -\frac {5\,x-\ln \left (\frac {5}{\ln \left ({\mathrm {e}}^3-x\right )-{\mathrm {e}}^{x^2}+4}\right )\,\left (20\,x-20\,{\mathrm {e}}^3+\ln \left ({\mathrm {e}}^3-x\right )\,\left (5\,x-5\,{\mathrm {e}}^3\right )-{\mathrm {e}}^{x^2}\,\left (5\,x-5\,{\mathrm {e}}^3\right )\right )-x^2+{\mathrm {e}}^{x^2}\,\left ({\mathrm {e}}^3\,\left (10\,x^2-2\,x^3\right )-10\,x^3+2\,x^4\right )}{{\ln \left (\frac {5}{\ln \left ({\mathrm {e}}^3-x\right )-{\mathrm {e}}^{x^2}+4}\right )}^2\,\left (\ln \left ({\mathrm {e}}^3-x\right )\,\left (x^2\,{\mathrm {e}}^3-x^3\right )+4\,x^2\,{\mathrm {e}}^3-{\mathrm {e}}^{x^2}\,\left (x^2\,{\mathrm {e}}^3-x^3\right )-4\,x^3\right )} \,d x \] Input:

int(-(5*x - log(5/(log(exp(3) - x) - exp(x^2) + 4))*(20*x - 20*exp(3) + lo 
g(exp(3) - x)*(5*x - 5*exp(3)) - exp(x^2)*(5*x - 5*exp(3))) - x^2 + exp(x^ 
2)*(exp(3)*(10*x^2 - 2*x^3) - 10*x^3 + 2*x^4))/(log(5/(log(exp(3) - x) - e 
xp(x^2) + 4))^2*(log(exp(3) - x)*(x^2*exp(3) - x^3) + 4*x^2*exp(3) - exp(x 
^2)*(x^2*exp(3) - x^3) - 4*x^3)),x)
 

Output:

int(-(5*x - log(5/(log(exp(3) - x) - exp(x^2) + 4))*(20*x - 20*exp(3) + lo 
g(exp(3) - x)*(5*x - 5*exp(3)) - exp(x^2)*(5*x - 5*exp(3))) - x^2 + exp(x^ 
2)*(exp(3)*(10*x^2 - 2*x^3) - 10*x^3 + 2*x^4))/(log(5/(log(exp(3) - x) - e 
xp(x^2) + 4))^2*(log(exp(3) - x)*(x^2*exp(3) - x^3) + 4*x^2*exp(3) - exp(x 
^2)*(x^2*exp(3) - x^3) - 4*x^3)), x)
 

Reduce [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {-5 x+x^2+e^{x^2} \left (10 x^3-2 x^4+e^3 \left (-10 x^2+2 x^3\right )\right )+\left (-20 e^3+e^{x^2} \left (5 e^3-5 x\right )+20 x+\left (-5 e^3+5 x\right ) \log \left (e^3-x\right )\right ) \log \left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )}{\left (4 e^3 x^2-4 x^3+e^{x^2} \left (-e^3 x^2+x^3\right )+\left (e^3 x^2-x^3\right ) \log \left (e^3-x\right )\right ) \log ^2\left (\frac {5}{4-e^{x^2}+\log \left (e^3-x\right )}\right )} \, dx=\frac {-x +5}{\mathrm {log}\left (-\frac {5}{e^{x^{2}}-\mathrm {log}\left (e^{3}-x \right )-4}\right ) x} \] Input:

int((((-5*exp(3)+5*x)*log(-x+exp(3))+(5*exp(3)-5*x)*exp(x^2)-20*exp(3)+20* 
x)*log(5/(log(-x+exp(3))+4-exp(x^2)))+((2*x^3-10*x^2)*exp(3)-2*x^4+10*x^3) 
*exp(x^2)+x^2-5*x)/((x^2*exp(3)-x^3)*log(-x+exp(3))+(-x^2*exp(3)+x^3)*exp( 
x^2)+4*x^2*exp(3)-4*x^3)/log(5/(log(-x+exp(3))+4-exp(x^2)))^2,x)
 

Output:

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