\(\int \frac {e^{-2-\frac {-4 x+e^2 x \log (x^2)}{e^2}} (-e^2 \log (\frac {25}{2})+(-20+e^2 (10-2 x)+4 x) \log (\frac {25}{2}) \log (5-x)+e^2 (5-x) \log (\frac {25}{2}) \log (5-x) \log (x^2))}{(-5+x) \log ^2(5-x)} \, dx\) [484]

Optimal result

Integrand size = 94, antiderivative size = 28 \[ \int \frac {e^{-2-\frac {-4 x+e^2 x \log \left (x^2\right )}{e^2}} \left (-e^2 \log \left (\frac {25}{2}\right )+\left (-20+e^2 (10-2 x)+4 x\right ) \log \left (\frac {25}{2}\right ) \log (5-x)+e^2 (5-x) \log \left (\frac {25}{2}\right ) \log (5-x) \log \left (x^2\right )\right )}{(-5+x) \log ^2(5-x)} \, dx=\frac {e^{-x \left (-\frac {4}{e^2}+\log \left (x^2\right )\right )} \log \left (\frac {25}{2}\right )}{\log (5-x)} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2-\frac {-4 x+e^2 x \log \left (x^2\right )}{e^2}} \left (-e^2 \log \left (\frac {25}{2}\right )+\left (-20+e^2 (10-2 x)+4 x\right ) \log \left (\frac {25}{2}\right ) \log (5-x)+e^2 (5-x) \log \left (\frac {25}{2}\right ) \log (5-x) \log \left (x^2\right )\right )}{(-5+x) \log ^2(5-x)} \, dx=\frac {e^{\frac {4 x}{e^2}} \left (x^2\right )^{-x} \log \left (\frac {25}{2}\right )}{\log (5-x)} \] Input:

Integrate[(E^(-2 - (-4*x + E^2*x*Log[x^2])/E^2)*(-(E^2*Log[25/2]) + (-20 + 
 E^2*(10 - 2*x) + 4*x)*Log[25/2]*Log[5 - x] + E^2*(5 - x)*Log[25/2]*Log[5 
- x]*Log[x^2]))/((-5 + x)*Log[5 - x]^2),x]
 

Output:

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(104\) vs. \(2(28)=56\).

Time = 0.55 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.71, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {2726}

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.

\(\displaystyle \int \frac {e^{-\frac {e^2 x \log \left (x^2\right )-4 x}{e^2}-2} \left (e^2 (5-x) \log \left (\frac {25}{2}\right ) \log \left (x^2\right ) \log (5-x)+\left (e^2 (10-2 x)+4 x-20\right ) \log \left (\frac {25}{2}\right ) \log (5-x)-e^2 \log \left (\frac {25}{2}\right )\right )}{(x-5) \log ^2(5-x)} \, dx\)

\(\Big \downarrow \) 2726

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

Input:

Int[(E^(-2 - (-4*x + E^2*x*Log[x^2])/E^2)*(-(E^2*Log[25/2]) + (-20 + E^2*( 
10 - 2*x) + 4*x)*Log[25/2]*Log[5 - x] + E^2*(5 - x)*Log[25/2]*Log[5 - x]*L 
og[x^2]))/((-5 + x)*Log[5 - x]^2),x]
 

Output:

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

Defintions of rubi rules used

rule 2726

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, 
 x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
 

Maple [A] (verified)

Time = 2.60 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07

method	result	size
parallelrisch	\(\frac {\ln \left (\frac {25}{2}\right ) {\mathrm e}^{-x \left ({\mathrm e}^{2} \ln \left (x^{2}\right )-4\right ) {\mathrm e}^{-2}}}{\ln \left (5-x \right )}\)	\(30\)
risch	\(-\frac {\left (-2 \ln \left (5\right )+\ln \left (2\right )\right ) {\mathrm e}^{-\frac {x \left (-i {\mathrm e}^{2} \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i {\mathrm e}^{2} \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i {\mathrm e}^{2} \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+4 \,{\mathrm e}^{2} \ln \left (x \right )-8\right ) {\mathrm e}^{-2}}{2}}}{\ln \left (5-x \right )}\)	\(87\)

Input:

int(((5-x)*exp(2)*ln(25/2)*ln(5-x)*ln(x^2)+((-2*x+10)*exp(2)+4*x-20)*ln(25 
/2)*ln(5-x)-exp(2)*ln(25/2))/(-5+x)/exp(2)/ln(5-x)^2/exp((x*exp(2)*ln(x^2) 
-4*x)/exp(2)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-2-\frac {-4 x+e^2 x \log \left (x^2\right )}{e^2}} \left (-e^2 \log \left (\frac {25}{2}\right )+\left (-20+e^2 (10-2 x)+4 x\right ) \log \left (\frac {25}{2}\right ) \log (5-x)+e^2 (5-x) \log \left (\frac {25}{2}\right ) \log (5-x) \log \left (x^2\right )\right )}{(-5+x) \log ^2(5-x)} \, dx=\frac {e^{\left (-{\left (x e^{2} \log \left (x^{2}\right ) - 4 \, x + 2 \, e^{2}\right )} e^{\left (-2\right )} + 2\right )} \log \left (\frac {25}{2}\right )}{\log \left (-x + 5\right )} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-2-\frac {-4 x+e^2 x \log \left (x^2\right )}{e^2}} \left (-e^2 \log \left (\frac {25}{2}\right )+\left (-20+e^2 (10-2 x)+4 x\right ) \log \left (\frac {25}{2}\right ) \log (5-x)+e^2 (5-x) \log \left (\frac {25}{2}\right ) \log (5-x) \log \left (x^2\right )\right )}{(-5+x) \log ^2(5-x)} \, dx=\frac {\left (- \log {\left (2 \right )} + 2 \log {\left (5 \right )}\right ) e^{- \frac {x e^{2} \log {\left (x^{2} \right )} - 4 x}{e^{2}}}}{\log {\left (5 - x \right )}} \] Input:

integrate(((5-x)*exp(2)*ln(25/2)*ln(5-x)*ln(x**2)+((-2*x+10)*exp(2)+4*x-20 
)*ln(25/2)*ln(5-x)-exp(2)*ln(25/2))/(-5+x)/exp(2)/ln(5-x)**2/exp((x*exp(2) 
*ln(x**2)-4*x)/exp(2)),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-2-\frac {-4 x+e^2 x \log \left (x^2\right )}{e^2}} \left (-e^2 \log \left (\frac {25}{2}\right )+\left (-20+e^2 (10-2 x)+4 x\right ) \log \left (\frac {25}{2}\right ) \log (5-x)+e^2 (5-x) \log \left (\frac {25}{2}\right ) \log (5-x) \log \left (x^2\right )\right )}{(-5+x) \log ^2(5-x)} \, dx=\frac {{\left (2 \, \log \left (5\right ) - \log \left (2\right )\right )} e^{\left (4 \, x e^{\left (-2\right )} - 2 \, x \log \left (x\right )\right )}}{\log \left (-x + 5\right )} \] Input:

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-2-\frac {-4 x+e^2 x \log \left (x^2\right )}{e^2}} \left (-e^2 \log \left (\frac {25}{2}\right )+\left (-20+e^2 (10-2 x)+4 x\right ) \log \left (\frac {25}{2}\right ) \log (5-x)+e^2 (5-x) \log \left (\frac {25}{2}\right ) \log (5-x) \log \left (x^2\right )\right )}{(-5+x) \log ^2(5-x)} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{-4,[0,0,5,4,1,0]%%%}+%%%{4,[0,0,5,4,0,1]%%%}+%%%{1,[0,0,4, 
5,1,0]%%%
 

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-2-\frac {-4 x+e^2 x \log \left (x^2\right )}{e^2}} \left (-e^2 \log \left (\frac {25}{2}\right )+\left (-20+e^2 (10-2 x)+4 x\right ) \log \left (\frac {25}{2}\right ) \log (5-x)+e^2 (5-x) \log \left (\frac {25}{2}\right ) \log (5-x) \log \left (x^2\right )\right )}{(-5+x) \log ^2(5-x)} \, dx=\int -\frac {{\mathrm {e}}^{-2}\,{\mathrm {e}}^{{\mathrm {e}}^{-2}\,\left (4\,x-x\,\ln \left (x^2\right )\,{\mathrm {e}}^2\right )}\,\left ({\mathrm {e}}^2\,\ln \left (\frac {25}{2}\right )+\ln \left (\frac {25}{2}\right )\,\ln \left (5-x\right )\,\left ({\mathrm {e}}^2\,\left (2\,x-10\right )-4\,x+20\right )+\ln \left (x^2\right )\,{\mathrm {e}}^2\,\ln \left (\frac {25}{2}\right )\,\ln \left (5-x\right )\,\left (x-5\right )\right )}{{\ln \left (5-x\right )}^2\,\left (x-5\right )} \,d x \] Input:

int(-(exp(-2)*exp(exp(-2)*(4*x - x*log(x^2)*exp(2)))*(exp(2)*log(25/2) + l 
og(25/2)*log(5 - x)*(exp(2)*(2*x - 10) - 4*x + 20) + log(x^2)*exp(2)*log(2 
5/2)*log(5 - x)*(x - 5)))/(log(5 - x)^2*(x - 5)),x)
 

Output:

int(-(exp(-2)*exp(exp(-2)*(4*x - x*log(x^2)*exp(2)))*(exp(2)*log(25/2) + l 
og(25/2)*log(5 - x)*(exp(2)*(2*x - 10) - 4*x + 20) + log(x^2)*exp(2)*log(2 
5/2)*log(5 - x)*(x - 5)))/(log(5 - x)^2*(x - 5)), x)
 

Reduce [F]

\[ \int \frac {e^{-2-\frac {-4 x+e^2 x \log \left (x^2\right )}{e^2}} \left (-e^2 \log \left (\frac {25}{2}\right )+\left (-20+e^2 (10-2 x)+4 x\right ) \log \left (\frac {25}{2}\right ) \log (5-x)+e^2 (5-x) \log \left (\frac {25}{2}\right ) \log (5-x) \log \left (x^2\right )\right )}{(-5+x) \log ^2(5-x)} \, dx=\frac {\mathrm {log}\left (\frac {25}{2}\right ) \left (-\left (\int \frac {e^{\frac {4 x}{e^{2}}}}{x^{2 x} \mathrm {log}\left (-x +5\right )^{2} x -5 x^{2 x} \mathrm {log}\left (-x +5\right )^{2}}d x \right ) e^{2}+10 \left (\int \frac {e^{\frac {4 x}{e^{2}}}}{x^{2 x} \mathrm {log}\left (-x +5\right ) x -5 x^{2 x} \mathrm {log}\left (-x +5\right )}d x \right ) e^{2}-20 \left (\int \frac {e^{\frac {4 x}{e^{2}}}}{x^{2 x} \mathrm {log}\left (-x +5\right ) x -5 x^{2 x} \mathrm {log}\left (-x +5\right )}d x \right )-\left (\int \frac {e^{\frac {4 x}{e^{2}}} \mathrm {log}\left (x^{2}\right ) x}{x^{2 x} \mathrm {log}\left (-x +5\right ) x -5 x^{2 x} \mathrm {log}\left (-x +5\right )}d x \right ) e^{2}+5 \left (\int \frac {e^{\frac {4 x}{e^{2}}} \mathrm {log}\left (x^{2}\right )}{x^{2 x} \mathrm {log}\left (-x +5\right ) x -5 x^{2 x} \mathrm {log}\left (-x +5\right )}d x \right ) e^{2}-2 \left (\int \frac {e^{\frac {4 x}{e^{2}}} x}{x^{2 x} \mathrm {log}\left (-x +5\right ) x -5 x^{2 x} \mathrm {log}\left (-x +5\right )}d x \right ) e^{2}+4 \left (\int \frac {e^{\frac {4 x}{e^{2}}} x}{x^{2 x} \mathrm {log}\left (-x +5\right ) x -5 x^{2 x} \mathrm {log}\left (-x +5\right )}d x \right )\right )}{e^{2}} \] Input:

int(((5-x)*exp(2)*log(25/2)*log(5-x)*log(x^2)+((-2*x+10)*exp(2)+4*x-20)*lo 
g(25/2)*log(5-x)-exp(2)*log(25/2))/(-5+x)/exp(2)/log(5-x)^2/exp((x*exp(2)* 
log(x^2)-4*x)/exp(2)),x)
 

Output:

(log(25/2)*( - int(e**((4*x)/e**2)/(x**(2*x)*log( - x + 5)**2*x - 5*x**(2* 
x)*log( - x + 5)**2),x)*e**2 + 10*int(e**((4*x)/e**2)/(x**(2*x)*log( - x + 
 5)*x - 5*x**(2*x)*log( - x + 5)),x)*e**2 - 20*int(e**((4*x)/e**2)/(x**(2* 
x)*log( - x + 5)*x - 5*x**(2*x)*log( - x + 5)),x) - int((e**((4*x)/e**2)*l 
og(x**2)*x)/(x**(2*x)*log( - x + 5)*x - 5*x**(2*x)*log( - x + 5)),x)*e**2 
+ 5*int((e**((4*x)/e**2)*log(x**2))/(x**(2*x)*log( - x + 5)*x - 5*x**(2*x) 
*log( - x + 5)),x)*e**2 - 2*int((e**((4*x)/e**2)*x)/(x**(2*x)*log( - x + 5 
)*x - 5*x**(2*x)*log( - x + 5)),x)*e**2 + 4*int((e**((4*x)/e**2)*x)/(x**(2 
*x)*log( - x + 5)*x - 5*x**(2*x)*log( - x + 5)),x)))/e**2