3.21.73 \(\int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} (-6 e^{2 x^2} \log (5)+(-6 x-24 x^2) \log (5)+e^x (-2 x+2 x^2) \log (5)+e^{x^2} (e^x (-x+2 x^2) \log (5)+(24 x+6 x^2) \log (5)))}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx\) [2073]

3.21.73.1 Optimal result
3.21.73.2 Mathematica [A] (verified)
3.21.73.3 Rubi [F]
3.21.73.4 Maple [A] (verified)
3.21.73.5 Fricas [A] (verification not implemented)
3.21.73.6 Sympy [A] (verification not implemented)
3.21.73.7 Maxima [F]
3.21.73.8 Giac [A] (verification not implemented)
3.21.73.9 Mupad [B] (verification not implemented)

3.21.73.1 Optimal result

Integrand size = 126, antiderivative size = 31 \[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \log (5)}{16 x^2} \]

output
1/16*ln(5)/x^2/exp(1/3*(3+exp(x))/(exp(x^2)-2*x))
 
3.21.73.2 Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\frac {e^{\frac {-3-e^x}{3 \left (e^{x^2}-2 x\right )}} \log (5)}{16 x^2} \]

input
Integrate[(-6*E^(2*x^2)*Log[5] + (-6*x - 24*x^2)*Log[5] + E^x*(-2*x + 2*x^ 
2)*Log[5] + E^x^2*(E^x*(-x + 2*x^2)*Log[5] + (24*x + 6*x^2)*Log[5]))/(E^(( 
3 + E^x)/(3*E^x^2 - 6*x))*(48*E^(2*x^2)*x^3 - 192*E^x^2*x^4 + 192*x^5)),x]
 
output
(E^((-3 - E^x)/(3*(E^x^2 - 2*x)))*Log[5])/(16*x^2)
 
3.21.73.3 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.

\(\displaystyle \int \frac {e^{-\frac {e^x+3}{3 e^{x^2}-6 x}} \left (\left (-24 x^2-6 x\right ) \log (5)+e^{x^2} \left (e^x \left (2 x^2-x\right ) \log (5)+\left (6 x^2+24 x\right ) \log (5)\right )-6 e^{2 x^2} \log (5)+e^x \left (2 x^2-2 x\right ) \log (5)\right )}{192 x^5-192 e^{x^2} x^4+48 e^{2 x^2} x^3} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{-\frac {e^x+3}{3 \left (e^{x^2}-2 x\right )}} \left (6 e^{x^2} x^2+2 e^x x^2+2 e^{x^2+x} x^2-24 x^2+24 e^{x^2} x-e^{x^2+x} x-6 e^{2 x^2}-2 e^x x-6 x\right ) \log (5)}{48 \left (e^{x^2}-2 x\right )^2 x^3}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{48} \log (5) \int -\frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \left (-6 e^{x^2} x^2-2 e^x x^2-2 e^{x^2+x} x^2+24 x^2-24 e^{x^2} x+2 e^x x+e^{x^2+x} x+6 x+6 e^{2 x^2}\right )}{\left (e^{x^2}-2 x\right )^2 x^3}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{48} \log (5) \int \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \left (-6 e^{x^2} x^2-2 e^x x^2-2 e^{x^2+x} x^2+24 x^2-24 e^{x^2} x+2 e^x x+e^{x^2+x} x+6 x+6 e^{2 x^2}\right )}{\left (e^{x^2}-2 x\right )^2 x^3}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{48} \log (5) \int \left (\frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \left (2 e^x x+6 x-e^x\right )}{x^2 \left (2 x-e^{x^2}\right )}-\frac {2 e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \left (3+e^x\right ) \left (2 x^2-1\right )}{\left (e^{x^2}-2 x\right )^2 x^2}+\frac {6 e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{x^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {1}{48} \log (5) \left (-12 \int \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right )^2}dx-4 \int \frac {e^{x-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right )^2}dx+6 \int \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right )^2 x^2}dx+2 \int \frac {e^{x-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right )^2 x^2}dx+\int \frac {e^{x-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right ) x^2}dx-6 \int \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right ) x}dx-2 \int \frac {e^{x-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right ) x}dx+6 \int \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{x^3}dx\right )\)

input
Int[(-6*E^(2*x^2)*Log[5] + (-6*x - 24*x^2)*Log[5] + E^x*(-2*x + 2*x^2)*Log 
[5] + E^x^2*(E^x*(-x + 2*x^2)*Log[5] + (24*x + 6*x^2)*Log[5]))/(E^((3 + E^ 
x)/(3*E^x^2 - 6*x))*(48*E^(2*x^2)*x^3 - 192*E^x^2*x^4 + 192*x^5)),x]
 
output
$Aborted
 

3.21.73.3.1 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 7292
Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

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

Time = 12.89 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87

method result size
risch \(\frac {\ln \left (5\right ) {\mathrm e}^{\frac {3+{\mathrm e}^{x}}{-3 \,{\mathrm e}^{x^{2}}+6 x}}}{16 x^{2}}\) \(27\)
parallelrisch \(\frac {\left (6 x \ln \left (5\right )-3 \ln \left (5\right ) {\mathrm e}^{x^{2}}\right ) {\mathrm e}^{-\frac {3+{\mathrm e}^{x}}{3 \left ({\mathrm e}^{x^{2}}-2 x \right )}}}{48 x^{2} \left (-{\mathrm e}^{x^{2}}+2 x \right )}\) \(51\)

input
int((-6*ln(5)*exp(x^2)^2+((2*x^2-x)*ln(5)*exp(x)+(6*x^2+24*x)*ln(5))*exp(x 
^2)+(2*x^2-2*x)*ln(5)*exp(x)+(-24*x^2-6*x)*ln(5))/(48*x^3*exp(x^2)^2-192*x 
^4*exp(x^2)+192*x^5)/exp((3+exp(x))/(3*exp(x^2)-6*x)),x,method=_RETURNVERB 
OSE)
 
output
1/16*ln(5)/x^2*exp(1/3*(3+exp(x))/(-exp(x^2)+2*x))
 
3.21.73.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\frac {e^{\left (\frac {e^{x} + 3}{3 \, {\left (2 \, x - e^{\left (x^{2}\right )}\right )}}\right )} \log \left (5\right )}{16 \, x^{2}} \]

input
integrate((-6*log(5)*exp(x^2)^2+((2*x^2-x)*log(5)*exp(x)+(6*x^2+24*x)*log( 
5))*exp(x^2)+(2*x^2-2*x)*log(5)*exp(x)+(-24*x^2-6*x)*log(5))/(48*x^3*exp(x 
^2)^2-192*x^4*exp(x^2)+192*x^5)/exp((3+exp(x))/(3*exp(x^2)-6*x)),x, algori 
thm=\
 
output
1/16*e^(1/3*(e^x + 3)/(2*x - e^(x^2)))*log(5)/x^2
 
3.21.73.6 Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\frac {e^{- \frac {e^{x} + 3}{- 6 x + 3 e^{x^{2}}}} \log {\left (5 \right )}}{16 x^{2}} \]

input
integrate((-6*ln(5)*exp(x**2)**2+((2*x**2-x)*ln(5)*exp(x)+(6*x**2+24*x)*ln 
(5))*exp(x**2)+(2*x**2-2*x)*ln(5)*exp(x)+(-24*x**2-6*x)*ln(5))/(48*x**3*ex 
p(x**2)**2-192*x**4*exp(x**2)+192*x**5)/exp((3+exp(x))/(3*exp(x**2)-6*x)), 
x)
 
output
exp(-(exp(x) + 3)/(-6*x + 3*exp(x**2)))*log(5)/(16*x**2)
 
3.21.73.7 Maxima [F]

\[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\int { \frac {{\left (2 \, {\left (x^{2} - x\right )} e^{x} \log \left (5\right ) + {\left ({\left (2 \, x^{2} - x\right )} e^{x} \log \left (5\right ) + 6 \, {\left (x^{2} + 4 \, x\right )} \log \left (5\right )\right )} e^{\left (x^{2}\right )} - 6 \, {\left (4 \, x^{2} + x\right )} \log \left (5\right ) - 6 \, e^{\left (2 \, x^{2}\right )} \log \left (5\right )\right )} e^{\left (\frac {e^{x} + 3}{3 \, {\left (2 \, x - e^{\left (x^{2}\right )}\right )}}\right )}}{48 \, {\left (4 \, x^{5} - 4 \, x^{4} e^{\left (x^{2}\right )} + x^{3} e^{\left (2 \, x^{2}\right )}\right )}} \,d x } \]

input
integrate((-6*log(5)*exp(x^2)^2+((2*x^2-x)*log(5)*exp(x)+(6*x^2+24*x)*log( 
5))*exp(x^2)+(2*x^2-2*x)*log(5)*exp(x)+(-24*x^2-6*x)*log(5))/(48*x^3*exp(x 
^2)^2-192*x^4*exp(x^2)+192*x^5)/exp((3+exp(x))/(3*exp(x^2)-6*x)),x, algori 
thm=\
 
output
1/48*integrate((2*(x^2 - x)*e^x*log(5) + ((2*x^2 - x)*e^x*log(5) + 6*(x^2 
+ 4*x)*log(5))*e^(x^2) - 6*(4*x^2 + x)*log(5) - 6*e^(2*x^2)*log(5))*e^(1/3 
*(e^x + 3)/(2*x - e^(x^2)))/(4*x^5 - 4*x^4*e^(x^2) + x^3*e^(2*x^2)), x)
 
3.21.73.8 Giac [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\frac {e^{\left (\frac {e^{x} + 3}{3 \, {\left (2 \, x - e^{\left (x^{2}\right )}\right )}}\right )} \log \left (5\right )}{16 \, x^{2}} \]

input
integrate((-6*log(5)*exp(x^2)^2+((2*x^2-x)*log(5)*exp(x)+(6*x^2+24*x)*log( 
5))*exp(x^2)+(2*x^2-2*x)*log(5)*exp(x)+(-24*x^2-6*x)*log(5))/(48*x^3*exp(x 
^2)^2-192*x^4*exp(x^2)+192*x^5)/exp((3+exp(x))/(3*exp(x^2)-6*x)),x, algori 
thm=\
 
output
1/16*e^(1/3*(e^x + 3)/(2*x - e^(x^2)))*log(5)/x^2
 
3.21.73.9 Mupad [B] (verification not implemented)

Time = 13.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\frac {{\mathrm {e}}^{\frac {1}{2\,x-{\mathrm {e}}^{x^2}}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{3\,\left (2\,x-{\mathrm {e}}^{x^2}\right )}}\,\ln \left (5\right )}{16\,x^2} \]

input
int(-(exp((exp(x) + 3)/(6*x - 3*exp(x^2)))*(log(5)*(6*x + 24*x^2) + 6*exp( 
2*x^2)*log(5) - exp(x^2)*(log(5)*(24*x + 6*x^2) - exp(x)*log(5)*(x - 2*x^2 
)) + exp(x)*log(5)*(2*x - 2*x^2)))/(48*x^3*exp(2*x^2) - 192*x^4*exp(x^2) + 
 192*x^5),x)
 
output
(exp(1/(2*x - exp(x^2)))*exp(exp(x)/(3*(2*x - exp(x^2))))*log(5))/(16*x^2)