3.13.12 \(\int \frac {e^{-\frac {2 e^2 (4-4 x+x^2)}{100-120 x+36 x^2+(20-12 x) \log (x)+\log ^2(x)}} (-2000 x+3600 x^2-2160 x^3+432 x^4+e^2 (-16 x+32 x^2-12 x^3)+(-600 x+720 x^2-216 x^3+e^2 (-8 x^2+4 x^3)) \log (x)+(-60 x+36 x^2) \log ^2(x)-2 x \log ^3(x))}{1000-1800 x+1080 x^2-216 x^3+(300-360 x+108 x^2) \log (x)+(30-18 x) \log ^2(x)+\log ^3(x)} \, dx\) [1212]

3.13.12.1 Optimal result
3.13.12.2 Mathematica [A] (verified)
3.13.12.3 Rubi [F]
3.13.12.4 Maple [A] (verified)
3.13.12.5 Fricas [A] (verification not implemented)
3.13.12.6 Sympy [A] (verification not implemented)
3.13.12.7 Maxima [F]
3.13.12.8 Giac [B] (verification not implemented)
3.13.12.9 Mupad [B] (verification not implemented)

3.13.12.1 Optimal result

Integrand size = 174, antiderivative size = 30 \[ \int \frac {e^{-\frac {2 e^2 \left (4-4 x+x^2\right )}{100-120 x+36 x^2+(20-12 x) \log (x)+\log ^2(x)}} \left (-2000 x+3600 x^2-2160 x^3+432 x^4+e^2 \left (-16 x+32 x^2-12 x^3\right )+\left (-600 x+720 x^2-216 x^3+e^2 \left (-8 x^2+4 x^3\right )\right ) \log (x)+\left (-60 x+36 x^2\right ) \log ^2(x)-2 x \log ^3(x)\right )}{1000-1800 x+1080 x^2-216 x^3+\left (300-360 x+108 x^2\right ) \log (x)+(30-18 x) \log ^2(x)+\log ^3(x)} \, dx=5-e^{-\frac {2 e^2}{\left (5+\frac {x-\log (x)}{-2+x}\right )^2}} x^2 \]

output
5-x^2/exp(exp(1)^2/(5+(x-ln(x))/(-2+x))^2)^2
 
3.13.12.2 Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-\frac {2 e^2 \left (4-4 x+x^2\right )}{100-120 x+36 x^2+(20-12 x) \log (x)+\log ^2(x)}} \left (-2000 x+3600 x^2-2160 x^3+432 x^4+e^2 \left (-16 x+32 x^2-12 x^3\right )+\left (-600 x+720 x^2-216 x^3+e^2 \left (-8 x^2+4 x^3\right )\right ) \log (x)+\left (-60 x+36 x^2\right ) \log ^2(x)-2 x \log ^3(x)\right )}{1000-1800 x+1080 x^2-216 x^3+\left (300-360 x+108 x^2\right ) \log (x)+(30-18 x) \log ^2(x)+\log ^3(x)} \, dx=-e^{-\frac {2 e^2 (-2+x)^2}{(10-6 x+\log (x))^2}} x^2 \]

input
Integrate[(-2000*x + 3600*x^2 - 2160*x^3 + 432*x^4 + E^2*(-16*x + 32*x^2 - 
 12*x^3) + (-600*x + 720*x^2 - 216*x^3 + E^2*(-8*x^2 + 4*x^3))*Log[x] + (- 
60*x + 36*x^2)*Log[x]^2 - 2*x*Log[x]^3)/(E^((2*E^2*(4 - 4*x + x^2))/(100 - 
 120*x + 36*x^2 + (20 - 12*x)*Log[x] + Log[x]^2))*(1000 - 1800*x + 1080*x^ 
2 - 216*x^3 + (300 - 360*x + 108*x^2)*Log[x] + (30 - 18*x)*Log[x]^2 + Log[ 
x]^3)),x]
 
output
-(x^2/E^((2*E^2*(-2 + x)^2)/(10 - 6*x + Log[x])^2))
 
3.13.12.3 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.

\(\displaystyle \int \frac {\left (432 x^4-2160 x^3+3600 x^2+\left (36 x^2-60 x\right ) \log ^2(x)+e^2 \left (-12 x^3+32 x^2-16 x\right )+\left (-216 x^3+720 x^2+e^2 \left (4 x^3-8 x^2\right )-600 x\right ) \log (x)-2000 x-2 x \log ^3(x)\right ) \exp \left (-\frac {2 e^2 \left (x^2-4 x+4\right )}{36 x^2-120 x+\log ^2(x)+(20-12 x) \log (x)+100}\right )}{-216 x^3+1080 x^2+\left (108 x^2-360 x+300\right ) \log (x)-1800 x+\log ^3(x)+(30-18 x) \log ^2(x)+1000} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {2 x e^{-\frac {2 e^2 (x-2)^2}{(-6 x+\log (x)+10)^2}} \left (-2 e^2 \left (3 x^2-8 x+4\right )+2 \left (\left (e^2-54\right ) x^2-2 \left (e^2-90\right ) x-150\right ) \log (x)+8 (3 x-5)^3-\log ^3(x)+6 (3 x-5) \log ^2(x)\right )}{(-6 x+\log (x)+10)^3}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \int -\frac {e^{-\frac {2 e^2 (2-x)^2}{(-6 x+\log (x)+10)^2}} x \left (8 (5-3 x)^3+6 \log ^2(x) (5-3 x)+\log ^3(x)+2 e^2 \left (3 x^2-8 x+4\right )+2 \left (\left (54-e^2\right ) x^2-2 \left (90-e^2\right ) x+150\right ) \log (x)\right )}{(-6 x+\log (x)+10)^3}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -2 \int \frac {e^{-\frac {2 e^2 (2-x)^2}{(-6 x+\log (x)+10)^2}} x \left (8 (5-3 x)^3+6 \log ^2(x) (5-3 x)+\log ^3(x)+2 e^2 \left (3 x^2-8 x+4\right )+2 \left (\left (54-e^2\right ) x^2-2 \left (90-e^2\right ) x+150\right ) \log (x)\right )}{(-6 x+\log (x)+10)^3}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -2 \int \left (\frac {2 e^{2-\frac {2 e^2 (2-x)^2}{(-6 x+\log (x)+10)^2}} x (6 x-1) (x-2)^2}{(6 x-\log (x)-10)^3}-\frac {2 e^{2-\frac {2 e^2 (2-x)^2}{(-6 x+\log (x)+10)^2}} x^2 (x-2)}{(6 x-\log (x)-10)^2}+e^{-\frac {2 e^2 (2-x)^2}{(-6 x+\log (x)+10)^2}} x\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -2 \left (12 \int \frac {e^{2-\frac {2 e^2 (2-x)^2}{(-6 x+\log (x)+10)^2}} x^4}{(6 x-\log (x)-10)^3}dx-50 \int \frac {e^{2-\frac {2 e^2 (2-x)^2}{(-6 x+\log (x)+10)^2}} x^3}{(6 x-\log (x)-10)^3}dx-2 \int \frac {e^{2-\frac {2 e^2 (2-x)^2}{(-6 x+\log (x)+10)^2}} x^3}{(6 x-\log (x)-10)^2}dx+56 \int \frac {e^{2-\frac {2 e^2 (2-x)^2}{(-6 x+\log (x)+10)^2}} x^2}{(6 x-\log (x)-10)^3}dx+4 \int \frac {e^{2-\frac {2 e^2 (2-x)^2}{(-6 x+\log (x)+10)^2}} x^2}{(6 x-\log (x)-10)^2}dx+\int e^{-\frac {2 e^2 (2-x)^2}{(-6 x+\log (x)+10)^2}} xdx-8 \int \frac {e^{2-\frac {2 e^2 (2-x)^2}{(-6 x+\log (x)+10)^2}} x}{(6 x-\log (x)-10)^3}dx\right )\)

input
Int[(-2000*x + 3600*x^2 - 2160*x^3 + 432*x^4 + E^2*(-16*x + 32*x^2 - 12*x^ 
3) + (-600*x + 720*x^2 - 216*x^3 + E^2*(-8*x^2 + 4*x^3))*Log[x] + (-60*x + 
 36*x^2)*Log[x]^2 - 2*x*Log[x]^3)/(E^((2*E^2*(4 - 4*x + x^2))/(100 - 120*x 
 + 36*x^2 + (20 - 12*x)*Log[x] + Log[x]^2))*(1000 - 1800*x + 1080*x^2 - 21 
6*x^3 + (300 - 360*x + 108*x^2)*Log[x] + (30 - 18*x)*Log[x]^2 + Log[x]^3)) 
,x]
 
output
$Aborted
 

3.13.12.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7239
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7293
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 
3.13.12.4 Maple [A] (verified)

Time = 7.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method result size
risch \(-x^{2} {\mathrm e}^{-\frac {2 \left (-2+x \right )^{2} {\mathrm e}^{2}}{\left (10+\ln \left (x \right )-6 x \right )^{2}}}\) \(25\)
parallelrisch \(-\frac {\left (x^{2} \ln \left (x \right )^{2}+36 x^{4}-120 x^{3}+100 x^{2}-12 x^{3} \ln \left (x \right )+20 x^{2} \ln \left (x \right )\right ) {\mathrm e}^{-\frac {2 \left (x^{2}-4 x +4\right ) {\mathrm e}^{2}}{\ln \left (x \right )^{2}-12 x \ln \left (x \right )+36 x^{2}+20 \ln \left (x \right )-120 x +100}}}{\left (6 x -\ln \left (x \right )-10\right )^{2}}\) \(93\)

input
int((-2*x*ln(x)^3+(36*x^2-60*x)*ln(x)^2+((4*x^3-8*x^2)*exp(1)^2-216*x^3+72 
0*x^2-600*x)*ln(x)+(-12*x^3+32*x^2-16*x)*exp(1)^2+432*x^4-2160*x^3+3600*x^ 
2-2000*x)/(ln(x)^3+(-18*x+30)*ln(x)^2+(108*x^2-360*x+300)*ln(x)-216*x^3+10 
80*x^2-1800*x+1000)/exp((x^2-4*x+4)*exp(1)^2/(ln(x)^2+(-12*x+20)*ln(x)+36* 
x^2-120*x+100))^2,x,method=_RETURNVERBOSE)
 
output
-x^2*exp(-2*(-2+x)^2*exp(2)/(10+ln(x)-6*x)^2)
 
3.13.12.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.43 \[ \int \frac {e^{-\frac {2 e^2 \left (4-4 x+x^2\right )}{100-120 x+36 x^2+(20-12 x) \log (x)+\log ^2(x)}} \left (-2000 x+3600 x^2-2160 x^3+432 x^4+e^2 \left (-16 x+32 x^2-12 x^3\right )+\left (-600 x+720 x^2-216 x^3+e^2 \left (-8 x^2+4 x^3\right )\right ) \log (x)+\left (-60 x+36 x^2\right ) \log ^2(x)-2 x \log ^3(x)\right )}{1000-1800 x+1080 x^2-216 x^3+\left (300-360 x+108 x^2\right ) \log (x)+(30-18 x) \log ^2(x)+\log ^3(x)} \, dx=-x^{2} e^{\left (-\frac {2 \, {\left (x^{2} - 4 \, x + 4\right )} e^{2}}{36 \, x^{2} - 4 \, {\left (3 \, x - 5\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 120 \, x + 100}\right )} \]

input
integrate((-2*x*log(x)^3+(36*x^2-60*x)*log(x)^2+((4*x^3-8*x^2)*exp(1)^2-21 
6*x^3+720*x^2-600*x)*log(x)+(-12*x^3+32*x^2-16*x)*exp(1)^2+432*x^4-2160*x^ 
3+3600*x^2-2000*x)/(log(x)^3+(-18*x+30)*log(x)^2+(108*x^2-360*x+300)*log(x 
)-216*x^3+1080*x^2-1800*x+1000)/exp((x^2-4*x+4)*exp(1)^2/(log(x)^2+(-12*x+ 
20)*log(x)+36*x^2-120*x+100))^2,x, algorithm=\
 
output
-x^2*e^(-2*(x^2 - 4*x + 4)*e^2/(36*x^2 - 4*(3*x - 5)*log(x) + log(x)^2 - 1 
20*x + 100))
 
3.13.12.6 Sympy [A] (verification not implemented)

Time = 32.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {e^{-\frac {2 e^2 \left (4-4 x+x^2\right )}{100-120 x+36 x^2+(20-12 x) \log (x)+\log ^2(x)}} \left (-2000 x+3600 x^2-2160 x^3+432 x^4+e^2 \left (-16 x+32 x^2-12 x^3\right )+\left (-600 x+720 x^2-216 x^3+e^2 \left (-8 x^2+4 x^3\right )\right ) \log (x)+\left (-60 x+36 x^2\right ) \log ^2(x)-2 x \log ^3(x)\right )}{1000-1800 x+1080 x^2-216 x^3+\left (300-360 x+108 x^2\right ) \log (x)+(30-18 x) \log ^2(x)+\log ^3(x)} \, dx=- x^{2} e^{- \frac {2 \left (x^{2} - 4 x + 4\right ) e^{2}}{36 x^{2} - 120 x + \left (20 - 12 x\right ) \log {\left (x \right )} + \log {\left (x \right )}^{2} + 100}} \]

input
integrate((-2*x*ln(x)**3+(36*x**2-60*x)*ln(x)**2+((4*x**3-8*x**2)*exp(1)** 
2-216*x**3+720*x**2-600*x)*ln(x)+(-12*x**3+32*x**2-16*x)*exp(1)**2+432*x** 
4-2160*x**3+3600*x**2-2000*x)/(ln(x)**3+(-18*x+30)*ln(x)**2+(108*x**2-360* 
x+300)*ln(x)-216*x**3+1080*x**2-1800*x+1000)/exp((x**2-4*x+4)*exp(1)**2/(l 
n(x)**2+(-12*x+20)*ln(x)+36*x**2-120*x+100))**2,x)
 
output
-x**2*exp(-2*(x**2 - 4*x + 4)*exp(2)/(36*x**2 - 120*x + (20 - 12*x)*log(x) 
 + log(x)**2 + 100))
 
3.13.12.7 Maxima [F]

\[ \int \frac {e^{-\frac {2 e^2 \left (4-4 x+x^2\right )}{100-120 x+36 x^2+(20-12 x) \log (x)+\log ^2(x)}} \left (-2000 x+3600 x^2-2160 x^3+432 x^4+e^2 \left (-16 x+32 x^2-12 x^3\right )+\left (-600 x+720 x^2-216 x^3+e^2 \left (-8 x^2+4 x^3\right )\right ) \log (x)+\left (-60 x+36 x^2\right ) \log ^2(x)-2 x \log ^3(x)\right )}{1000-1800 x+1080 x^2-216 x^3+\left (300-360 x+108 x^2\right ) \log (x)+(30-18 x) \log ^2(x)+\log ^3(x)} \, dx=\int { -\frac {2 \, {\left (216 \, x^{4} - x \log \left (x\right )^{3} - 1080 \, x^{3} + 6 \, {\left (3 \, x^{2} - 5 \, x\right )} \log \left (x\right )^{2} + 1800 \, x^{2} - 2 \, {\left (3 \, x^{3} - 8 \, x^{2} + 4 \, x\right )} e^{2} - 2 \, {\left (54 \, x^{3} - 180 \, x^{2} - {\left (x^{3} - 2 \, x^{2}\right )} e^{2} + 150 \, x\right )} \log \left (x\right ) - 1000 \, x\right )} e^{\left (-\frac {2 \, {\left (x^{2} - 4 \, x + 4\right )} e^{2}}{36 \, x^{2} - 4 \, {\left (3 \, x - 5\right )} \log \left (x\right ) + \log \left (x\right )^{2} - 120 \, x + 100}\right )}}{216 \, x^{3} + 6 \, {\left (3 \, x - 5\right )} \log \left (x\right )^{2} - \log \left (x\right )^{3} - 1080 \, x^{2} - 12 \, {\left (9 \, x^{2} - 30 \, x + 25\right )} \log \left (x\right ) + 1800 \, x - 1000} \,d x } \]

input
integrate((-2*x*log(x)^3+(36*x^2-60*x)*log(x)^2+((4*x^3-8*x^2)*exp(1)^2-21 
6*x^3+720*x^2-600*x)*log(x)+(-12*x^3+32*x^2-16*x)*exp(1)^2+432*x^4-2160*x^ 
3+3600*x^2-2000*x)/(log(x)^3+(-18*x+30)*log(x)^2+(108*x^2-360*x+300)*log(x 
)-216*x^3+1080*x^2-1800*x+1000)/exp((x^2-4*x+4)*exp(1)^2/(log(x)^2+(-12*x+ 
20)*log(x)+36*x^2-120*x+100))^2,x, algorithm=\
 
output
-2*integrate((216*x^4 - x*log(x)^3 - 1080*x^3 + 6*(3*x^2 - 5*x)*log(x)^2 + 
 1800*x^2 - 2*(3*x^3 - 8*x^2 + 4*x)*e^2 - 2*(54*x^3 - 180*x^2 - (x^3 - 2*x 
^2)*e^2 + 150*x)*log(x) - 1000*x)*e^(-2*(x^2 - 4*x + 4)*e^2/(36*x^2 - 4*(3 
*x - 5)*log(x) + log(x)^2 - 120*x + 100))/(216*x^3 + 6*(3*x - 5)*log(x)^2 
- log(x)^3 - 1080*x^2 - 12*(9*x^2 - 30*x + 25)*log(x) + 1800*x - 1000), x)
 
3.13.12.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (28) = 56\).

Time = 1.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.37 \[ \int \frac {e^{-\frac {2 e^2 \left (4-4 x+x^2\right )}{100-120 x+36 x^2+(20-12 x) \log (x)+\log ^2(x)}} \left (-2000 x+3600 x^2-2160 x^3+432 x^4+e^2 \left (-16 x+32 x^2-12 x^3\right )+\left (-600 x+720 x^2-216 x^3+e^2 \left (-8 x^2+4 x^3\right )\right ) \log (x)+\left (-60 x+36 x^2\right ) \log ^2(x)-2 x \log ^3(x)\right )}{1000-1800 x+1080 x^2-216 x^3+\left (300-360 x+108 x^2\right ) \log (x)+(30-18 x) \log ^2(x)+\log ^3(x)} \, dx=-x^{2} e^{\left (\frac {2 \, {\left (11 \, x^{2} e^{2} - 12 \, x e^{2} \log \left (x\right ) + e^{2} \log \left (x\right )^{2} - 20 \, x e^{2} + 20 \, e^{2} \log \left (x\right )\right )}}{25 \, {\left (36 \, x^{2} - 12 \, x \log \left (x\right ) + \log \left (x\right )^{2} - 120 \, x + 20 \, \log \left (x\right ) + 100\right )}} - \frac {2}{25} \, e^{2}\right )} \]

input
integrate((-2*x*log(x)^3+(36*x^2-60*x)*log(x)^2+((4*x^3-8*x^2)*exp(1)^2-21 
6*x^3+720*x^2-600*x)*log(x)+(-12*x^3+32*x^2-16*x)*exp(1)^2+432*x^4-2160*x^ 
3+3600*x^2-2000*x)/(log(x)^3+(-18*x+30)*log(x)^2+(108*x^2-360*x+300)*log(x 
)-216*x^3+1080*x^2-1800*x+1000)/exp((x^2-4*x+4)*exp(1)^2/(log(x)^2+(-12*x+ 
20)*log(x)+36*x^2-120*x+100))^2,x, algorithm=\
 
output
-x^2*e^(2/25*(11*x^2*e^2 - 12*x*e^2*log(x) + e^2*log(x)^2 - 20*x*e^2 + 20* 
e^2*log(x))/(36*x^2 - 12*x*log(x) + log(x)^2 - 120*x + 20*log(x) + 100) - 
2/25*e^2)
 
3.13.12.9 Mupad [B] (verification not implemented)

Time = 11.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.67 \[ \int \frac {e^{-\frac {2 e^2 \left (4-4 x+x^2\right )}{100-120 x+36 x^2+(20-12 x) \log (x)+\log ^2(x)}} \left (-2000 x+3600 x^2-2160 x^3+432 x^4+e^2 \left (-16 x+32 x^2-12 x^3\right )+\left (-600 x+720 x^2-216 x^3+e^2 \left (-8 x^2+4 x^3\right )\right ) \log (x)+\left (-60 x+36 x^2\right ) \log ^2(x)-2 x \log ^3(x)\right )}{1000-1800 x+1080 x^2-216 x^3+\left (300-360 x+108 x^2\right ) \log (x)+(30-18 x) \log ^2(x)+\log ^3(x)} \, dx=-x^2\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^2\,x^2-8\,{\mathrm {e}}^2\,x+8\,{\mathrm {e}}^2}{36\,x^2-12\,x\,\ln \left (x\right )-120\,x+{\ln \left (x\right )}^2+20\,\ln \left (x\right )+100}} \]

input
int(-(exp(-(2*exp(2)*(x^2 - 4*x + 4))/(log(x)^2 - 120*x - log(x)*(12*x - 2 
0) + 36*x^2 + 100))*(2000*x + log(x)^2*(60*x - 36*x^2) + 2*x*log(x)^3 + ex 
p(2)*(16*x - 32*x^2 + 12*x^3) + log(x)*(600*x + exp(2)*(8*x^2 - 4*x^3) - 7 
20*x^2 + 216*x^3) - 3600*x^2 + 2160*x^3 - 432*x^4))/(log(x)^3 - 1800*x + l 
og(x)*(108*x^2 - 360*x + 300) + 1080*x^2 - 216*x^3 - log(x)^2*(18*x - 30) 
+ 1000),x)
 
output
-x^2*exp(-(8*exp(2) - 8*x*exp(2) + 2*x^2*exp(2))/(20*log(x) - 120*x + log( 
x)^2 - 12*x*log(x) + 36*x^2 + 100))