Integrand size = 146, antiderivative size = 28 \[ \int \frac {e^{\frac {16+16 e^x+3 x^2+16 \log ^2(3)-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}} \left (6 x+e^x \left (6 x-3 x^2\right )+6 x \log ^2(3)+\left (-6 x-6 e^x x-6 x \log ^2(3)\right ) \log (x)+\left (-6 x+e^x \left (-6 x+3 x^2\right )-6 x \log ^2(3)\right ) \log ^2(x)\right )}{1+e^{2 x}+2 \log ^2(3)+\log ^4(3)+e^x \left (2+2 \log ^2(3)\right )} \, dx=e^{16-\frac {x^2 \left (-3+3 \log ^2(x)\right )}{1+e^x+\log ^2(3)}} \] Output:
exp(16-x^2*(3*ln(x)^2-3)/(exp(x)+ln(3)^2+1))
Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {e^{\frac {16+16 e^x+3 x^2+16 \log ^2(3)-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}} \left (6 x+e^x \left (6 x-3 x^2\right )+6 x \log ^2(3)+\left (-6 x-6 e^x x-6 x \log ^2(3)\right ) \log (x)+\left (-6 x+e^x \left (-6 x+3 x^2\right )-6 x \log ^2(3)\right ) \log ^2(x)\right )}{1+e^{2 x}+2 \log ^2(3)+\log ^4(3)+e^x \left (2+2 \log ^2(3)\right )} \, dx=e^{16+\frac {3 x^2}{1+e^x+\log ^2(3)}-\frac {3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}} \] Input:
Integrate[(E^((16 + 16*E^x + 3*x^2 + 16*Log[3]^2 - 3*x^2*Log[x]^2)/(1 + E^ x + Log[3]^2))*(6*x + E^x*(6*x - 3*x^2) + 6*x*Log[3]^2 + (-6*x - 6*E^x*x - 6*x*Log[3]^2)*Log[x] + (-6*x + E^x*(-6*x + 3*x^2) - 6*x*Log[3]^2)*Log[x]^ 2))/(1 + E^(2*x) + 2*Log[3]^2 + Log[3]^4 + E^x*(2 + 2*Log[3]^2)),x]
Output:
E^(16 + (3*x^2)/(1 + E^x + Log[3]^2) - (3*x^2*Log[x]^2)/(1 + E^x + Log[3]^ 2))
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.
| \(\displaystyle \int \frac {\left (e^x \left (6 x-3 x^2\right )+\left (e^x \left (3 x^2-6 x\right )-6 x-6 x \log ^2(3)\right ) \log ^2(x)+6 x+\left (-6 e^x x-6 x-6 x \log ^2(3)\right ) \log (x)+6 x \log ^2(3)\right ) \exp \left (\frac {3 x^2-3 x^2 \log ^2(x)+16 e^x+16+16 \log ^2(3)}{e^x+1+\log ^2(3)}\right )}{e^{2 x}+e^x \left (2+2 \log ^2(3)\right )+1+\log ^4(3)+2 \log ^2(3)} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {\left (e^x \left (6 x-3 x^2\right )+\left (e^x \left (3 x^2-6 x\right )-6 x-6 x \log ^2(3)\right ) \log ^2(x)+\left (-6 e^x x-6 x-6 x \log ^2(3)\right ) \log (x)+x \left (6+6 \log ^2(3)\right )\right ) \exp \left (\frac {3 x^2-3 x^2 \log ^2(x)+16 e^x+16+16 \log ^2(3)}{e^x+1+\log ^2(3)}\right )}{e^{2 x}+e^x \left (2+2 \log ^2(3)\right )+1+\log ^4(3)+2 \log ^2(3)}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {3 x \left (2 e^x-e^x x-2 e^x \log ^2(x)+e^x x \log ^2(x)-2 \left (1+\log ^2(3)\right ) \log ^2(x)-2 \left (1+\log ^2(3)\right ) \log (x)-2 e^x \log (x)+2 \left (1+\log ^2(3)\right )\right ) \exp \left (\frac {3 x^2-3 x^2 \log ^2(x)+16 e^x+16 \left (1+\log ^2(3)\right )}{e^x+1+\log ^2(3)}\right )}{\left (e^x+1+\log ^2(3)\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 3 \int \frac {\exp \left (\frac {-3 \log ^2(x) x^2+3 x^2+16 e^x+16 \left (1+\log ^2(3)\right )}{1+e^x+\log ^2(3)}\right ) x \left (-2 e^x \log ^2(x)+e^x x \log ^2(x)-2 \left (1+\log ^2(3)\right ) \log ^2(x)-2 e^x \log (x)-2 \left (1+\log ^2(3)\right ) \log (x)+2 e^x-e^x x+2 \left (1+\log ^2(3)\right )\right )}{\left (1+e^x+\log ^2(3)\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 3 \int \left (\frac {\exp \left (\frac {-3 \log ^2(x) x^2+3 x^2+16 e^x+16 \left (1+\log ^2(3)\right )}{1+e^x+\log ^2(3)}\right ) x \left (x \log ^2(x)-2 \log ^2(x)-2 \log (x)-x+2\right )}{1+e^x+\log ^2(3)}-\frac {\exp \left (\frac {-3 \log ^2(x) x^2+3 x^2+16 e^x+16 \left (1+\log ^2(3)\right )}{1+e^x+\log ^2(3)}\right ) x^2 \left (1+\log ^2(3)\right ) \left (\log ^2(x)-1\right )}{\left (1+e^x+\log ^2(3)\right )^2}\right )dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 3 \int \left (\frac {\exp \left (\frac {-3 \log ^2(x) x^2+3 x^2+16 e^x+16 \left (1+\log ^2(3)\right )}{1+e^x+\log ^2(3)}\right ) \left (1+\log ^2(3)\right ) \left (1-\log ^2(x)\right ) x^2}{\left (1+e^x+\log ^2(3)\right )^2}+\frac {\exp \left (\frac {-3 \log ^2(x) x^2+3 x^2+16 e^x+16 \left (1+\log ^2(3)\right )}{1+e^x+\log ^2(3)}\right ) \left (x \log ^2(x)-2 \log ^2(x)-2 \log (x)-x+2\right ) x}{1+e^x+\log ^2(3)}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle 3 \int \left (\frac {\exp \left (\frac {-3 \log ^2(x) x^2+3 x^2+16 e^x+16 \left (1+\log ^2(3)\right )}{1+e^x+\log ^2(3)}\right ) \left (1+\log ^2(3)\right ) \left (1-\log ^2(x)\right ) x^2}{\left (1+e^x+\log ^2(3)\right )^2}+\frac {\exp \left (\frac {-3 \log ^2(x) x^2+3 x^2+16 e^x+16 \left (1+\log ^2(3)\right )}{1+e^x+\log ^2(3)}\right ) \left (x \log ^2(x)-2 \log ^2(x)-2 \log (x)-x+2\right ) x}{1+e^x+\log ^2(3)}\right )dx\) |
Input:
Int[(E^((16 + 16*E^x + 3*x^2 + 16*Log[3]^2 - 3*x^2*Log[x]^2)/(1 + E^x + Lo g[3]^2))*(6*x + E^x*(6*x - 3*x^2) + 6*x*Log[3]^2 + (-6*x - 6*E^x*x - 6*x*L og[3]^2)*Log[x] + (-6*x + E^x*(-6*x + 3*x^2) - 6*x*Log[3]^2)*Log[x]^2))/(1 + E^(2*x) + 2*Log[3]^2 + Log[3]^4 + E^x*(2 + 2*Log[3]^2)),x]
Output:
$Aborted
Time = 36.41 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39
| method | result | size |
| risch | \({\mathrm e}^{\frac {-3 x^{2} \ln \left (x \right )^{2}+16 \,{\mathrm e}^{x}+16 \ln \left (3\right )^{2}+3 x^{2}+16}{{\mathrm e}^{x}+\ln \left (3\right )^{2}+1}}\) | \(39\) |
| parallelrisch | \({\mathrm e}^{\frac {-3 x^{2} \ln \left (x \right )^{2}+16 \,{\mathrm e}^{x}+16 \ln \left (3\right )^{2}+3 x^{2}+16}{{\mathrm e}^{x}+\ln \left (3\right )^{2}+1}}\) | \(39\) |
Input:
int((((3*x^2-6*x)*exp(x)-6*x*ln(3)^2-6*x)*ln(x)^2+(-6*exp(x)*x-6*x*ln(3)^2 -6*x)*ln(x)+(-3*x^2+6*x)*exp(x)+6*x*ln(3)^2+6*x)*exp((-3*x^2*ln(x)^2+16*ex p(x)+16*ln(3)^2+3*x^2+16)/(exp(x)+ln(3)^2+1))/(exp(x)^2+(2*ln(3)^2+2)*exp( x)+ln(3)^4+2*ln(3)^2+1),x,method=_RETURNVERBOSE)
Output:
exp((-3*x^2*ln(x)^2+16*exp(x)+16*ln(3)^2+3*x^2+16)/(exp(x)+ln(3)^2+1))
Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {e^{\frac {16+16 e^x+3 x^2+16 \log ^2(3)-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}} \left (6 x+e^x \left (6 x-3 x^2\right )+6 x \log ^2(3)+\left (-6 x-6 e^x x-6 x \log ^2(3)\right ) \log (x)+\left (-6 x+e^x \left (-6 x+3 x^2\right )-6 x \log ^2(3)\right ) \log ^2(x)\right )}{1+e^{2 x}+2 \log ^2(3)+\log ^4(3)+e^x \left (2+2 \log ^2(3)\right )} \, dx=e^{\left (-\frac {3 \, x^{2} \log \left (x\right )^{2} - 3 \, x^{2} - 16 \, \log \left (3\right )^{2} - 16 \, e^{x} - 16}{\log \left (3\right )^{2} + e^{x} + 1}\right )} \] Input:
integrate((((3*x^2-6*x)*exp(x)-6*x*log(3)^2-6*x)*log(x)^2+(-6*exp(x)*x-6*x *log(3)^2-6*x)*log(x)+(-3*x^2+6*x)*exp(x)+6*x*log(3)^2+6*x)*exp((-3*x^2*lo g(x)^2+16*exp(x)+16*log(3)^2+3*x^2+16)/(exp(x)+log(3)^2+1))/(exp(x)^2+(2*l og(3)^2+2)*exp(x)+log(3)^4+2*log(3)^2+1),x, algorithm="fricas")
Output:
e^(-(3*x^2*log(x)^2 - 3*x^2 - 16*log(3)^2 - 16*e^x - 16)/(log(3)^2 + e^x + 1))
Time = 0.65 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {e^{\frac {16+16 e^x+3 x^2+16 \log ^2(3)-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}} \left (6 x+e^x \left (6 x-3 x^2\right )+6 x \log ^2(3)+\left (-6 x-6 e^x x-6 x \log ^2(3)\right ) \log (x)+\left (-6 x+e^x \left (-6 x+3 x^2\right )-6 x \log ^2(3)\right ) \log ^2(x)\right )}{1+e^{2 x}+2 \log ^2(3)+\log ^4(3)+e^x \left (2+2 \log ^2(3)\right )} \, dx=e^{\frac {- 3 x^{2} \log {\left (x \right )}^{2} + 3 x^{2} + 16 e^{x} + 16 + 16 \log {\left (3 \right )}^{2}}{e^{x} + 1 + \log {\left (3 \right )}^{2}}} \] Input:
integrate((((3*x**2-6*x)*exp(x)-6*x*ln(3)**2-6*x)*ln(x)**2+(-6*exp(x)*x-6* x*ln(3)**2-6*x)*ln(x)+(-3*x**2+6*x)*exp(x)+6*x*ln(3)**2+6*x)*exp((-3*x**2* ln(x)**2+16*exp(x)+16*ln(3)**2+3*x**2+16)/(exp(x)+ln(3)**2+1))/(exp(x)**2+ (2*ln(3)**2+2)*exp(x)+ln(3)**4+2*ln(3)**2+1),x)
Output:
exp((-3*x**2*log(x)**2 + 3*x**2 + 16*exp(x) + 16 + 16*log(3)**2)/(exp(x) + 1 + log(3)**2))
Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (24) = 48\).
Time = 0.34 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.79 \[ \int \frac {e^{\frac {16+16 e^x+3 x^2+16 \log ^2(3)-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}} \left (6 x+e^x \left (6 x-3 x^2\right )+6 x \log ^2(3)+\left (-6 x-6 e^x x-6 x \log ^2(3)\right ) \log (x)+\left (-6 x+e^x \left (-6 x+3 x^2\right )-6 x \log ^2(3)\right ) \log ^2(x)\right )}{1+e^{2 x}+2 \log ^2(3)+\log ^4(3)+e^x \left (2+2 \log ^2(3)\right )} \, dx=e^{\left (-\frac {3 \, x^{2} \log \left (x\right )^{2}}{\log \left (3\right )^{2} + e^{x} + 1} + \frac {3 \, x^{2}}{\log \left (3\right )^{2} + e^{x} + 1} + \frac {16 \, \log \left (3\right )^{2}}{\log \left (3\right )^{2} + e^{x} + 1} + \frac {16 \, e^{x}}{\log \left (3\right )^{2} + e^{x} + 1} + \frac {16}{\log \left (3\right )^{2} + e^{x} + 1}\right )} \] Input:
integrate((((3*x^2-6*x)*exp(x)-6*x*log(3)^2-6*x)*log(x)^2+(-6*exp(x)*x-6*x *log(3)^2-6*x)*log(x)+(-3*x^2+6*x)*exp(x)+6*x*log(3)^2+6*x)*exp((-3*x^2*lo g(x)^2+16*exp(x)+16*log(3)^2+3*x^2+16)/(exp(x)+log(3)^2+1))/(exp(x)^2+(2*l og(3)^2+2)*exp(x)+log(3)^4+2*log(3)^2+1),x, algorithm="maxima")
Output:
e^(-3*x^2*log(x)^2/(log(3)^2 + e^x + 1) + 3*x^2/(log(3)^2 + e^x + 1) + 16* log(3)^2/(log(3)^2 + e^x + 1) + 16*e^x/(log(3)^2 + e^x + 1) + 16/(log(3)^2 + e^x + 1))
\[ \int \frac {e^{\frac {16+16 e^x+3 x^2+16 \log ^2(3)-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}} \left (6 x+e^x \left (6 x-3 x^2\right )+6 x \log ^2(3)+\left (-6 x-6 e^x x-6 x \log ^2(3)\right ) \log (x)+\left (-6 x+e^x \left (-6 x+3 x^2\right )-6 x \log ^2(3)\right ) \log ^2(x)\right )}{1+e^{2 x}+2 \log ^2(3)+\log ^4(3)+e^x \left (2+2 \log ^2(3)\right )} \, dx=\int { \frac {3 \, {\left (2 \, x \log \left (3\right )^{2} - {\left (2 \, x \log \left (3\right )^{2} - {\left (x^{2} - 2 \, x\right )} e^{x} + 2 \, x\right )} \log \left (x\right )^{2} - {\left (x^{2} - 2 \, x\right )} e^{x} - 2 \, {\left (x \log \left (3\right )^{2} + x e^{x} + x\right )} \log \left (x\right ) + 2 \, x\right )} e^{\left (-\frac {3 \, x^{2} \log \left (x\right )^{2} - 3 \, x^{2} - 16 \, \log \left (3\right )^{2} - 16 \, e^{x} - 16}{\log \left (3\right )^{2} + e^{x} + 1}\right )}}{\log \left (3\right )^{4} + 2 \, {\left (\log \left (3\right )^{2} + 1\right )} e^{x} + 2 \, \log \left (3\right )^{2} + e^{\left (2 \, x\right )} + 1} \,d x } \] Input:
integrate((((3*x^2-6*x)*exp(x)-6*x*log(3)^2-6*x)*log(x)^2+(-6*exp(x)*x-6*x *log(3)^2-6*x)*log(x)+(-3*x^2+6*x)*exp(x)+6*x*log(3)^2+6*x)*exp((-3*x^2*lo g(x)^2+16*exp(x)+16*log(3)^2+3*x^2+16)/(exp(x)+log(3)^2+1))/(exp(x)^2+(2*l og(3)^2+2)*exp(x)+log(3)^4+2*log(3)^2+1),x, algorithm="giac")
Output:
integrate(3*(2*x*log(3)^2 - (2*x*log(3)^2 - (x^2 - 2*x)*e^x + 2*x)*log(x)^ 2 - (x^2 - 2*x)*e^x - 2*(x*log(3)^2 + x*e^x + x)*log(x) + 2*x)*e^(-(3*x^2* log(x)^2 - 3*x^2 - 16*log(3)^2 - 16*e^x - 16)/(log(3)^2 + e^x + 1))/(log(3 )^4 + 2*(log(3)^2 + 1)*e^x + 2*log(3)^2 + e^(2*x) + 1), x)
Time = 4.25 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.93 \[ \int \frac {e^{\frac {16+16 e^x+3 x^2+16 \log ^2(3)-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}} \left (6 x+e^x \left (6 x-3 x^2\right )+6 x \log ^2(3)+\left (-6 x-6 e^x x-6 x \log ^2(3)\right ) \log (x)+\left (-6 x+e^x \left (-6 x+3 x^2\right )-6 x \log ^2(3)\right ) \log ^2(x)\right )}{1+e^{2 x}+2 \log ^2(3)+\log ^4(3)+e^x \left (2+2 \log ^2(3)\right )} \, dx={\mathrm {e}}^{\frac {3\,x^2}{{\mathrm {e}}^x+{\ln \left (3\right )}^2+1}}\,{\mathrm {e}}^{\frac {16\,{\mathrm {e}}^x}{{\mathrm {e}}^x+{\ln \left (3\right )}^2+1}}\,{\mathrm {e}}^{\frac {16}{{\mathrm {e}}^x+{\ln \left (3\right )}^2+1}}\,{\mathrm {e}}^{-\frac {3\,x^2\,{\ln \left (x\right )}^2}{{\mathrm {e}}^x+{\ln \left (3\right )}^2+1}}\,{\mathrm {e}}^{\frac {16\,{\ln \left (3\right )}^2}{{\mathrm {e}}^x+{\ln \left (3\right )}^2+1}} \] Input:
int((exp((16*exp(x) - 3*x^2*log(x)^2 + 16*log(3)^2 + 3*x^2 + 16)/(exp(x) + log(3)^2 + 1))*(6*x - log(x)^2*(6*x + 6*x*log(3)^2 + exp(x)*(6*x - 3*x^2) ) + 6*x*log(3)^2 + exp(x)*(6*x - 3*x^2) - log(x)*(6*x + 6*x*log(3)^2 + 6*x *exp(x))))/(exp(2*x) + exp(x)*(2*log(3)^2 + 2) + 2*log(3)^2 + log(3)^4 + 1 ),x)
Output:
exp((3*x^2)/(exp(x) + log(3)^2 + 1))*exp((16*exp(x))/(exp(x) + log(3)^2 + 1))*exp(16/(exp(x) + log(3)^2 + 1))*exp(-(3*x^2*log(x)^2)/(exp(x) + log(3) ^2 + 1))*exp((16*log(3)^2)/(exp(x) + log(3)^2 + 1))
Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {e^{\frac {16+16 e^x+3 x^2+16 \log ^2(3)-3 x^2 \log ^2(x)}{1+e^x+\log ^2(3)}} \left (6 x+e^x \left (6 x-3 x^2\right )+6 x \log ^2(3)+\left (-6 x-6 e^x x-6 x \log ^2(3)\right ) \log (x)+\left (-6 x+e^x \left (-6 x+3 x^2\right )-6 x \log ^2(3)\right ) \log ^2(x)\right )}{1+e^{2 x}+2 \log ^2(3)+\log ^4(3)+e^x \left (2+2 \log ^2(3)\right )} \, dx=\frac {e^{\frac {3 x^{2}}{e^{x}+\mathrm {log}\left (3\right )^{2}+1}} e^{16}}{e^{\frac {3 \mathrm {log}\left (x \right )^{2} x^{2}}{e^{x}+\mathrm {log}\left (3\right )^{2}+1}}} \] Input:
int((((3*x^2-6*x)*exp(x)-6*x*log(3)^2-6*x)*log(x)^2+(-6*exp(x)*x-6*x*log(3 )^2-6*x)*log(x)+(-3*x^2+6*x)*exp(x)+6*x*log(3)^2+6*x)*exp((-3*x^2*log(x)^2 +16*exp(x)+16*log(3)^2+3*x^2+16)/(exp(x)+log(3)^2+1))/(exp(x)^2+(2*log(3)^ 2+2)*exp(x)+log(3)^4+2*log(3)^2+1),x)
Output:
(e**((3*x**2)/(e**x + log(3)**2 + 1))*e**16)/e**((3*log(x)**2*x**2)/(e**x + log(3)**2 + 1))