Integrand size = 175, antiderivative size = 32 \[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=2 \left (-e^{\frac {2}{3-e^x \left (\log (3)-\frac {x}{\log (4)}\right )^2}}+\log (2)\right ) \] Output:
2*ln(2)-2*exp(2/(3-exp(x)*(ln(3)-1/2*x/ln(2))^2))
\[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=\int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx \] Input:
Integrate[(E^(x - (2*Log[4]^2)/(-3*Log[4]^2 + E^x*(x^2 - 2*x*Log[3]*Log[4] + Log[3]^2*Log[4]^2)))*((-8*x - 4*x^2)*Log[4]^2 + (8 + 8*x)*Log[3]*Log[4] ^3 - 4*Log[3]^2*Log[4]^4))/(9*Log[4]^4 + E^x*(-6*x^2*Log[4]^2 + 12*x*Log[3 ]*Log[4]^3 - 6*Log[3]^2*Log[4]^4) + E^(2*x)*(x^4 - 4*x^3*Log[3]*Log[4] + 6 *x^2*Log[3]^2*Log[4]^2 - 4*x*Log[3]^3*Log[4]^3 + Log[3]^4*Log[4]^4)),x]
Output:
Integrate[(E^(x - (2*Log[4]^2)/(-3*Log[4]^2 + E^x*(x^2 - 2*x*Log[3]*Log[4] + Log[3]^2*Log[4]^2)))*((-8*x - 4*x^2)*Log[4]^2 + (8 + 8*x)*Log[3]*Log[4] ^3 - 4*Log[3]^2*Log[4]^4))/(9*Log[4]^4 + E^x*(-6*x^2*Log[4]^2 + 12*x*Log[3 ]*Log[4]^3 - 6*Log[3]^2*Log[4]^4) + E^(2*x)*(x^4 - 4*x^3*Log[3]*Log[4] + 6 *x^2*Log[3]^2*Log[4]^2 - 4*x*Log[3]^3*Log[4]^3 + Log[3]^4*Log[4]^4)), x]
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.
| \(\displaystyle \int \frac {\left (\left (-4 x^2-8 x\right ) \log ^2(4)+(8 x+8) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right ) \exp \left (x-\frac {2 \log ^2(4)}{e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )-3 \log ^2(4)}\right )}{e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )+9 \log ^4(4)} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {4 \log ^2(4) \left (-x^2-x (2-\log (4) \log (9))+\log (3) \log (4) (2-\log (3) \log (4))\right ) \exp \left (x-\frac {2 \log ^2(4)}{e^x (x-\log (3) \log (4))^2-3 \log ^2(4)}\right )}{\left (3 \log ^2(4)-e^x (x-\log (3) \log (4))^2\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 4 \log ^2(4) \int -\frac {\exp \left (x+\frac {2 \log ^2(4)}{3 \log ^2(4)-e^x (x-\log (3) \log (4))^2}\right ) \left (x^2+(2-\log (4) \log (9)) x-\log (3) \log (4) (2-\log (3) \log (4))\right )}{\left (3 \log ^2(4)-e^x (x-\log (3) \log (4))^2\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -4 \log ^2(4) \int \frac {\exp \left (x+\frac {2 \log ^2(4)}{3 \log ^2(4)-e^x (x-\log (3) \log (4))^2}\right ) \left (x^2+(2-\log (4) \log (9)) x-\log (3) \log (4) (2-\log (3) \log (4))\right )}{\left (3 \log ^2(4)-e^x (x-\log (3) \log (4))^2\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -4 \log ^2(4) \int \left (\frac {\exp \left (x+\frac {2 \log ^2(4)}{3 \log ^2(4)-e^x (x-\log (3) \log (4))^2}\right ) x^2}{\left (e^x x^2-2 e^x \log (3) \log (4) x+e^x \log ^2(3) \log ^2(4)-3 \log ^2(4)\right )^2}-\frac {\exp \left (x+\frac {2 \log ^2(4)}{3 \log ^2(4)-e^x (x-\log (3) \log (4))^2}\right ) (-2+\log (4) \log (9)) x}{\left (e^x x^2-2 e^x \log (3) \log (4) x+e^x \log ^2(3) \log ^2(4)-3 \log ^2(4)\right )^2}+\frac {\exp \left (x+\frac {2 \log ^2(4)}{3 \log ^2(4)-e^x (x-\log (3) \log (4))^2}\right ) \log (3) \log (4) (-2+\log (3) \log (4))}{\left (e^x x^2-2 e^x \log (3) \log (4) x+e^x \log ^2(3) \log ^2(4)-3 \log ^2(4)\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -4 \log ^2(4) \left (-\log (3) \log (4) (2-\log (3) \log (4)) \int \frac {\exp \left (x+\frac {2 \log ^2(4)}{3 \log ^2(4)-e^x (x-\log (3) \log (4))^2}\right )}{\left (e^x x^2-2 e^x \log (3) \log (4) x+e^x \log ^2(3) \log ^2(4)-3 \log ^2(4)\right )^2}dx+(2-\log (4) \log (9)) \int \frac {\exp \left (x+\frac {2 \log ^2(4)}{3 \log ^2(4)-e^x (x-\log (3) \log (4))^2}\right ) x}{\left (e^x x^2-2 e^x \log (3) \log (4) x+e^x \log ^2(3) \log ^2(4)-3 \log ^2(4)\right )^2}dx+\int \frac {\exp \left (x+\frac {2 \log ^2(4)}{3 \log ^2(4)-e^x (x-\log (3) \log (4))^2}\right ) x^2}{\left (e^x x^2-2 e^x \log (3) \log (4) x+e^x \log ^2(3) \log ^2(4)-3 \log ^2(4)\right )^2}dx\right )\) |
Input:
Int[(E^(x - (2*Log[4]^2)/(-3*Log[4]^2 + E^x*(x^2 - 2*x*Log[3]*Log[4] + Log [3]^2*Log[4]^2)))*((-8*x - 4*x^2)*Log[4]^2 + (8 + 8*x)*Log[3]*Log[4]^3 - 4 *Log[3]^2*Log[4]^4))/(9*Log[4]^4 + E^x*(-6*x^2*Log[4]^2 + 12*x*Log[3]*Log[ 4]^3 - 6*Log[3]^2*Log[4]^4) + E^(2*x)*(x^4 - 4*x^3*Log[3]*Log[4] + 6*x^2*L og[3]^2*Log[4]^2 - 4*x*Log[3]^3*Log[4]^3 + Log[3]^4*Log[4]^4)),x]
Output:
$Aborted
Time = 11.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44
| method | result | size |
| risch | \(-2 \,{\mathrm e}^{-\frac {8 \ln \left (2\right )^{2}}{4 \ln \left (2\right )^{2} \ln \left (3\right )^{2} {\mathrm e}^{x}-4 \ln \left (2\right ) \ln \left (3\right ) {\mathrm e}^{x} x +{\mathrm e}^{x} x^{2}-12 \ln \left (2\right )^{2}}}\) | \(46\) |
| parallelrisch | \(-2 \,{\mathrm e}^{-\frac {8 \ln \left (2\right )^{2}}{4 \ln \left (2\right )^{2} \ln \left (3\right )^{2} {\mathrm e}^{x}-4 \ln \left (2\right ) \ln \left (3\right ) {\mathrm e}^{x} x +{\mathrm e}^{x} x^{2}-12 \ln \left (2\right )^{2}}}\) | \(46\) |
| norman | \(\frac {24 \ln \left (2\right )^{2} {\mathrm e}^{-\frac {8 \ln \left (2\right )^{2}}{\left (4 \ln \left (3\right )^{2} \ln \left (2\right )^{2}-4 x \ln \left (2\right ) \ln \left (3\right )+x^{2}\right ) {\mathrm e}^{x}-12 \ln \left (2\right )^{2}}}-2 \,{\mathrm e}^{x} x^{2} {\mathrm e}^{-\frac {8 \ln \left (2\right )^{2}}{\left (4 \ln \left (3\right )^{2} \ln \left (2\right )^{2}-4 x \ln \left (2\right ) \ln \left (3\right )+x^{2}\right ) {\mathrm e}^{x}-12 \ln \left (2\right )^{2}}}-8 \ln \left (2\right )^{2} \ln \left (3\right )^{2} {\mathrm e}^{x} {\mathrm e}^{-\frac {8 \ln \left (2\right )^{2}}{\left (4 \ln \left (3\right )^{2} \ln \left (2\right )^{2}-4 x \ln \left (2\right ) \ln \left (3\right )+x^{2}\right ) {\mathrm e}^{x}-12 \ln \left (2\right )^{2}}}+8 \ln \left (2\right ) \ln \left (3\right ) {\mathrm e}^{x} x \,{\mathrm e}^{-\frac {8 \ln \left (2\right )^{2}}{\left (4 \ln \left (3\right )^{2} \ln \left (2\right )^{2}-4 x \ln \left (2\right ) \ln \left (3\right )+x^{2}\right ) {\mathrm e}^{x}-12 \ln \left (2\right )^{2}}}}{4 \ln \left (2\right )^{2} \ln \left (3\right )^{2} {\mathrm e}^{x}-4 \ln \left (2\right ) \ln \left (3\right ) {\mathrm e}^{x} x +{\mathrm e}^{x} x^{2}-12 \ln \left (2\right )^{2}}\) | \(233\) |
Input:
int((-64*ln(3)^2*ln(2)^4+8*(8*x+8)*ln(3)*ln(2)^3+4*(-4*x^2-8*x)*ln(2)^2)*e xp(x)*exp(-8*ln(2)^2/((4*ln(3)^2*ln(2)^2-4*x*ln(2)*ln(3)+x^2)*exp(x)-12*ln (2)^2))/((16*ln(3)^4*ln(2)^4-32*x*ln(3)^3*ln(2)^3+24*x^2*ln(3)^2*ln(2)^2-8 *x^3*ln(3)*ln(2)+x^4)*exp(x)^2+(-96*ln(3)^2*ln(2)^4+96*x*ln(3)*ln(2)^3-24* x^2*ln(2)^2)*exp(x)+144*ln(2)^4),x,method=_RETURNVERBOSE)
Output:
-2*exp(-8*ln(2)^2/(4*ln(2)^2*ln(3)^2*exp(x)-4*ln(2)*ln(3)*exp(x)*x+exp(x)* x^2-12*ln(2)^2))
Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (30) = 60\).
Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.56 \[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=-2 \, e^{\left (-x - \frac {4 \, {\left (3 \, x + 2\right )} \log \left (2\right )^{2} - {\left (4 \, x \log \left (3\right )^{2} \log \left (2\right )^{2} - 4 \, x^{2} \log \left (3\right ) \log \left (2\right ) + x^{3}\right )} e^{x}}{{\left (4 \, \log \left (3\right )^{2} \log \left (2\right )^{2} - 4 \, x \log \left (3\right ) \log \left (2\right ) + x^{2}\right )} e^{x} - 12 \, \log \left (2\right )^{2}}\right )} \] Input:
integrate((-64*log(3)^2*log(2)^4+8*(8*x+8)*log(3)*log(2)^3+4*(-4*x^2-8*x)* log(2)^2)*exp(x)*exp(-8*log(2)^2/((4*log(3)^2*log(2)^2-4*x*log(2)*log(3)+x ^2)*exp(x)-12*log(2)^2))/((16*log(3)^4*log(2)^4-32*x*log(3)^3*log(2)^3+24* x^2*log(3)^2*log(2)^2-8*x^3*log(3)*log(2)+x^4)*exp(x)^2+(-96*log(3)^2*log( 2)^4+96*x*log(3)*log(2)^3-24*x^2*log(2)^2)*exp(x)+144*log(2)^4),x, algorit hm="fricas")
Output:
-2*e^(-x - (4*(3*x + 2)*log(2)^2 - (4*x*log(3)^2*log(2)^2 - 4*x^2*log(3)*l og(2) + x^3)*e^x)/((4*log(3)^2*log(2)^2 - 4*x*log(3)*log(2) + x^2)*e^x - 1 2*log(2)^2))
Time = 0.43 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=- 2 e^{- \frac {8 \log {\left (2 \right )}^{2}}{\left (x^{2} - 4 x \log {\left (2 \right )} \log {\left (3 \right )} + 4 \log {\left (2 \right )}^{2} \log {\left (3 \right )}^{2}\right ) e^{x} - 12 \log {\left (2 \right )}^{2}}} \] Input:
integrate((-64*ln(3)**2*ln(2)**4+8*(8*x+8)*ln(3)*ln(2)**3+4*(-4*x**2-8*x)* ln(2)**2)*exp(x)*exp(-8*ln(2)**2/((4*ln(3)**2*ln(2)**2-4*x*ln(2)*ln(3)+x** 2)*exp(x)-12*ln(2)**2))/((16*ln(3)**4*ln(2)**4-32*x*ln(3)**3*ln(2)**3+24*x **2*ln(3)**2*ln(2)**2-8*x**3*ln(3)*ln(2)+x**4)*exp(x)**2+(-96*ln(3)**2*ln( 2)**4+96*x*ln(3)*ln(2)**3-24*x**2*ln(2)**2)*exp(x)+144*ln(2)**4),x)
Output:
-2*exp(-8*log(2)**2/((x**2 - 4*x*log(2)*log(3) + 4*log(2)**2*log(3)**2)*ex p(x) - 12*log(2)**2))
Time = 0.42 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=-2 \, e^{\left (-\frac {8 \, \log \left (2\right )^{2}}{{\left (4 \, \log \left (3\right )^{2} \log \left (2\right )^{2} - 4 \, x \log \left (3\right ) \log \left (2\right ) + x^{2}\right )} e^{x} - 12 \, \log \left (2\right )^{2}}\right )} \] Input:
integrate((-64*log(3)^2*log(2)^4+8*(8*x+8)*log(3)*log(2)^3+4*(-4*x^2-8*x)* log(2)^2)*exp(x)*exp(-8*log(2)^2/((4*log(3)^2*log(2)^2-4*x*log(2)*log(3)+x ^2)*exp(x)-12*log(2)^2))/((16*log(3)^4*log(2)^4-32*x*log(3)^3*log(2)^3+24* x^2*log(3)^2*log(2)^2-8*x^3*log(3)*log(2)+x^4)*exp(x)^2+(-96*log(3)^2*log( 2)^4+96*x*log(3)*log(2)^3-24*x^2*log(2)^2)*exp(x)+144*log(2)^4),x, algorit hm="maxima")
Output:
-2*e^(-8*log(2)^2/((4*log(3)^2*log(2)^2 - 4*x*log(3)*log(2) + x^2)*e^x - 1 2*log(2)^2))
Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=-2 \, e^{\left (-\frac {8 \, \log \left (2\right )^{2}}{4 \, e^{x} \log \left (3\right )^{2} \log \left (2\right )^{2} - 4 \, x e^{x} \log \left (3\right ) \log \left (2\right ) + x^{2} e^{x} - 12 \, \log \left (2\right )^{2}}\right )} \] Input:
integrate((-64*log(3)^2*log(2)^4+8*(8*x+8)*log(3)*log(2)^3+4*(-4*x^2-8*x)* log(2)^2)*exp(x)*exp(-8*log(2)^2/((4*log(3)^2*log(2)^2-4*x*log(2)*log(3)+x ^2)*exp(x)-12*log(2)^2))/((16*log(3)^4*log(2)^4-32*x*log(3)^3*log(2)^3+24* x^2*log(3)^2*log(2)^2-8*x^3*log(3)*log(2)+x^4)*exp(x)^2+(-96*log(3)^2*log( 2)^4+96*x*log(3)*log(2)^3-24*x^2*log(2)^2)*exp(x)+144*log(2)^4),x, algorit hm="giac")
Output:
-2*e^(-8*log(2)^2/(4*e^x*log(3)^2*log(2)^2 - 4*x*e^x*log(3)*log(2) + x^2*e ^x - 12*log(2)^2))
Time = 3.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=-2\,{\mathrm {e}}^{-\frac {8\,{\ln \left (2\right )}^2}{x^2\,{\mathrm {e}}^x-12\,{\ln \left (2\right )}^2+4\,{\mathrm {e}}^x\,{\ln \left (2\right )}^2\,{\ln \left (3\right )}^2-4\,x\,{\mathrm {e}}^x\,\ln \left (2\right )\,\ln \left (3\right )}} \] Input:
int(-(exp(-(8*log(2)^2)/(exp(x)*(x^2 + 4*log(2)^2*log(3)^2 - 4*x*log(2)*lo g(3)) - 12*log(2)^2))*exp(x)*(4*log(2)^2*(8*x + 4*x^2) + 64*log(2)^4*log(3 )^2 - 8*log(2)^3*log(3)*(8*x + 8)))/(exp(2*x)*(x^4 + 16*log(2)^4*log(3)^4 - 8*x^3*log(2)*log(3) - 32*x*log(2)^3*log(3)^3 + 24*x^2*log(2)^2*log(3)^2) - exp(x)*(24*x^2*log(2)^2 + 96*log(2)^4*log(3)^2 - 96*x*log(2)^3*log(3)) + 144*log(2)^4),x)
Output:
-2*exp(-(8*log(2)^2)/(x^2*exp(x) - 12*log(2)^2 + 4*exp(x)*log(2)^2*log(3)^ 2 - 4*x*exp(x)*log(2)*log(3)))
\[ \int \frac {e^{x-\frac {2 \log ^2(4)}{-3 \log ^2(4)+e^x \left (x^2-2 x \log (3) \log (4)+\log ^2(3) \log ^2(4)\right )}} \left (\left (-8 x-4 x^2\right ) \log ^2(4)+(8+8 x) \log (3) \log ^3(4)-4 \log ^2(3) \log ^4(4)\right )}{9 \log ^4(4)+e^x \left (-6 x^2 \log ^2(4)+12 x \log (3) \log ^3(4)-6 \log ^2(3) \log ^4(4)\right )+e^{2 x} \left (x^4-4 x^3 \log (3) \log (4)+6 x^2 \log ^2(3) \log ^2(4)-4 x \log ^3(3) \log ^3(4)+\log ^4(3) \log ^4(4)\right )} \, dx=\text {too large to display} \] Input:
int((-64*log(3)^2*log(2)^4+8*(8*x+8)*log(3)*log(2)^3+4*(-4*x^2-8*x)*log(2) ^2)*exp(x)*exp(-8*log(2)^2/((4*log(3)^2*log(2)^2-4*x*log(2)*log(3)+x^2)*ex p(x)-12*log(2)^2))/((16*log(3)^4*log(2)^4-32*x*log(3)^3*log(2)^3+24*x^2*lo g(3)^2*log(2)^2-8*x^3*log(3)*log(2)+x^4)*exp(x)^2+(-96*log(3)^2*log(2)^4+9 6*x*log(3)*log(2)^3-24*x^2*log(2)^2)*exp(x)+144*log(2)^4),x)
Output:
16*log(2)**2*( - 4*int(e**x/(16*e**((8*e**x*log(3)**2*log(2)**2*x - 8*e**x *log(3)*log(2)*x**2 + 2*e**x*x**3 - 24*log(2)**2*x + 8*log(2)**2)/(4*e**x* log(3)**2*log(2)**2 - 4*e**x*log(3)*log(2)*x + e**x*x**2 - 12*log(2)**2))* log(3)**4*log(2)**4 - 32*e**((8*e**x*log(3)**2*log(2)**2*x - 8*e**x*log(3) *log(2)*x**2 + 2*e**x*x**3 - 24*log(2)**2*x + 8*log(2)**2)/(4*e**x*log(3)* *2*log(2)**2 - 4*e**x*log(3)*log(2)*x + e**x*x**2 - 12*log(2)**2))*log(3)* *3*log(2)**3*x + 24*e**((8*e**x*log(3)**2*log(2)**2*x - 8*e**x*log(3)*log( 2)*x**2 + 2*e**x*x**3 - 24*log(2)**2*x + 8*log(2)**2)/(4*e**x*log(3)**2*lo g(2)**2 - 4*e**x*log(3)*log(2)*x + e**x*x**2 - 12*log(2)**2))*log(3)**2*lo g(2)**2*x**2 - 8*e**((8*e**x*log(3)**2*log(2)**2*x - 8*e**x*log(3)*log(2)* x**2 + 2*e**x*x**3 - 24*log(2)**2*x + 8*log(2)**2)/(4*e**x*log(3)**2*log(2 )**2 - 4*e**x*log(3)*log(2)*x + e**x*x**2 - 12*log(2)**2))*log(3)*log(2)*x **3 + e**((8*e**x*log(3)**2*log(2)**2*x - 8*e**x*log(3)*log(2)*x**2 + 2*e* *x*x**3 - 24*log(2)**2*x + 8*log(2)**2)/(4*e**x*log(3)**2*log(2)**2 - 4*e* *x*log(3)*log(2)*x + e**x*x**2 - 12*log(2)**2))*x**4 - 96*e**((4*e**x*log( 3)**2*log(2)**2*x - 4*e**x*log(3)*log(2)*x**2 + e**x*x**3 - 12*log(2)**2*x + 8*log(2)**2)/(4*e**x*log(3)**2*log(2)**2 - 4*e**x*log(3)*log(2)*x + e** x*x**2 - 12*log(2)**2))*log(3)**2*log(2)**4 + 96*e**((4*e**x*log(3)**2*log (2)**2*x - 4*e**x*log(3)*log(2)*x**2 + e**x*x**3 - 12*log(2)**2*x + 8*log( 2)**2)/(4*e**x*log(3)**2*log(2)**2 - 4*e**x*log(3)*log(2)*x + e**x*x**2...