Integrand size = 68, antiderivative size = 33 \[ \int \frac {1}{24} e^{-1-\frac {x^2+x^3}{2 e}} \left (12 e+48 e^{1+\frac {x^2+x^3}{2 e}}-12 x^2-18 x^3+\left (2 x+3 x^2\right ) \log (3)\right ) \, dx=2 x-\frac {1}{4} e^{-\frac {x \left (x+x^2\right )}{2 e}} \left (-2 x+\frac {\log (3)}{3}\right ) \]
Time = 2.47 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.45 \[ \int \frac {1}{24} e^{-1-\frac {x^2+x^3}{2 e}} \left (12 e+48 e^{1+\frac {x^2+x^3}{2 e}}-12 x^2-18 x^3+\left (2 x+3 x^2\right ) \log (3)\right ) \, dx=\frac {1}{24} \left (48 x+\frac {e^{-\frac {x^2 (1+x)}{2 e}} \left (36 x^2-2 \log (9)-2 x (-12+\log (27))\right )}{2+3 x}\right ) \]
Integrate[(E^(-1 - (x^2 + x^3)/(2*E))*(12*E + 48*E^(1 + (x^2 + x^3)/(2*E)) - 12*x^2 - 18*x^3 + (2*x + 3*x^2)*Log[3]))/24,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.
\(\displaystyle \int \frac {1}{24} e^{-\frac {x^3+x^2}{2 e}-1} \left (-18 x^3-12 x^2+\left (3 x^2+2 x\right ) \log (3)+48 e^{\frac {x^3+x^2}{2 e}+1}+12 e\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{24} \int e^{-\frac {x^3+x^2}{2 e}-1} \left (-18 x^3-12 x^2+48 e^{\frac {x^3+x^2}{2 e}+1}+\left (3 x^2+2 x\right ) \log (3)+12 e\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{24} \int e^{-\frac {x^3}{2 e}-\frac {x^2}{2 e}-1} \left (-18 x^3-12 x^2+48 e^{\frac {x^3+x^2}{2 e}+1}+\left (3 x^2+2 x\right ) \log (3)+12 e\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{24} \int \left (-18 e^{-\frac {x^3}{2 e}-\frac {x^2}{2 e}-1} x^3-12 e^{-\frac {x^3}{2 e}-\frac {x^2}{2 e}-1} x^2+e^{-\frac {x^3}{2 e}-\frac {x^2}{2 e}-1} (3 x+2) \log (3) x+12 e^{-\frac {x^3}{2 e}-\frac {x^2}{2 e}}+48\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{24} \left (12 \int e^{-\frac {x^3}{2 e}-\frac {x^2}{2 e}}dx-12 \int e^{-\frac {x^3}{2 e}-\frac {x^2}{2 e}-1} x^2dx-18 \int e^{-\frac {x^3}{2 e}-\frac {x^2}{2 e}-1} x^3dx-2 e^{-\frac {x^3}{2 e}-\frac {x^2}{2 e}} \log (3)+48 x\right )\) |
Int[(E^(-1 - (x^2 + x^3)/(2*E))*(12*E + 48*E^(1 + (x^2 + x^3)/(2*E)) - 12* x^2 - 18*x^3 + (2*x + 3*x^2)*Log[3]))/24,x]
3.17.42.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91
method | result | size |
parts | \(2 x +\left (\frac {x}{2}-\frac {\ln \left (3\right )}{12}\right ) {\mathrm e}^{-\frac {\left (x^{3}+x^{2}\right ) {\mathrm e}^{-1}}{2}}\) | \(30\) |
risch | \(2 x +\frac {\left (-2 \ln \left (3\right ) {\mathrm e}+12 x \,{\mathrm e}\right ) {\mathrm e}^{-1-\frac {{\mathrm e}^{-1} x^{3}}{2}-\frac {x^{2} {\mathrm e}^{-1}}{2}}}{24}\) | \(36\) |
norman | \(\left (\frac {x}{2}+2 x \,{\mathrm e}^{\frac {\left (x^{3}+x^{2}\right ) {\mathrm e}^{-1}}{2}}-\frac {\ln \left (3\right )}{12}\right ) {\mathrm e}^{-\frac {\left (x^{3}+x^{2}\right ) {\mathrm e}^{-1}}{2}}\) | \(43\) |
parallelrisch | \(-\frac {{\mathrm e}^{-1} \left (-48 \,{\mathrm e}^{\frac {x^{2} \left (1+x \right ) {\mathrm e}^{-1}}{2}} {\mathrm e} x +2 \ln \left (3\right ) {\mathrm e}-12 x \,{\mathrm e}\right ) {\mathrm e}^{-\frac {x^{2} \left (1+x \right ) {\mathrm e}^{-1}}{2}}}{24}\) | \(52\) |
int(1/24*(48*exp(1)*exp(1/2*(x^3+x^2)/exp(1))+(3*x^2+2*x)*ln(3)+12*exp(1)- 18*x^3-12*x^2)/exp(1)/exp(1/2*(x^3+x^2)/exp(1)),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.48 \[ \int \frac {1}{24} e^{-1-\frac {x^2+x^3}{2 e}} \left (12 e+48 e^{1+\frac {x^2+x^3}{2 e}}-12 x^2-18 x^3+\left (2 x+3 x^2\right ) \log (3)\right ) \, dx=\frac {1}{12} \, {\left (6 \, x e + 24 \, x e^{\left (\frac {1}{2} \, {\left (x^{3} + x^{2} + 2 \, e\right )} e^{\left (-1\right )}\right )} - e \log \left (3\right )\right )} e^{\left (-\frac {1}{2} \, {\left (x^{3} + x^{2} + 2 \, e\right )} e^{\left (-1\right )}\right )} \]
integrate(1/24*(48*exp(1)*exp(1/2*(x^3+x^2)/exp(1))+(3*x^2+2*x)*log(3)+12* exp(1)-18*x^3-12*x^2)/exp(1)/exp(1/2*(x^3+x^2)/exp(1)),x, algorithm=\
1/12*(6*x*e + 24*x*e^(1/2*(x^3 + x^2 + 2*e)*e^(-1)) - e*log(3))*e^(-1/2*(x ^3 + x^2 + 2*e)*e^(-1))
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {1}{24} e^{-1-\frac {x^2+x^3}{2 e}} \left (12 e+48 e^{1+\frac {x^2+x^3}{2 e}}-12 x^2-18 x^3+\left (2 x+3 x^2\right ) \log (3)\right ) \, dx=2 x + \frac {\left (6 x - \log {\left (3 \right )}\right ) e^{- \frac {\frac {x^{3}}{2} + \frac {x^{2}}{2}}{e}}}{12} \]
integrate(1/24*(48*exp(1)*exp(1/2*(x**3+x**2)/exp(1))+(3*x**2+2*x)*ln(3)+1 2*exp(1)-18*x**3-12*x**2)/exp(1)/exp(1/2*(x**3+x**2)/exp(1)),x)
Time = 0.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.91 \[ \int \frac {1}{24} e^{-1-\frac {x^2+x^3}{2 e}} \left (12 e+48 e^{1+\frac {x^2+x^3}{2 e}}-12 x^2-18 x^3+\left (2 x+3 x^2\right ) \log (3)\right ) \, dx=\frac {1}{12} \, {\left (6 \, x - \log \left (3\right )\right )} e^{\left (-\frac {1}{2} \, x^{3} e^{\left (-1\right )} - \frac {1}{2} \, x^{2} e^{\left (-1\right )}\right )} + 2 \, x \]
integrate(1/24*(48*exp(1)*exp(1/2*(x^3+x^2)/exp(1))+(3*x^2+2*x)*log(3)+12* exp(1)-18*x^3-12*x^2)/exp(1)/exp(1/2*(x^3+x^2)/exp(1)),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {1}{24} e^{-1-\frac {x^2+x^3}{2 e}} \left (12 e+48 e^{1+\frac {x^2+x^3}{2 e}}-12 x^2-18 x^3+\left (2 x+3 x^2\right ) \log (3)\right ) \, dx=\frac {1}{2} \, x e^{\left (-\frac {1}{2} \, {\left (x^{3} + x^{2}\right )} e^{\left (-1\right )}\right )} - \frac {1}{12} \, e^{\left (-\frac {1}{2} \, {\left (x^{3} + x^{2}\right )} e^{\left (-1\right )}\right )} \log \left (3\right ) + 2 \, x \]
integrate(1/24*(48*exp(1)*exp(1/2*(x^3+x^2)/exp(1))+(3*x^2+2*x)*log(3)+12* exp(1)-18*x^3-12*x^2)/exp(1)/exp(1/2*(x^3+x^2)/exp(1)),x, algorithm=\
Time = 0.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {1}{24} e^{-1-\frac {x^2+x^3}{2 e}} \left (12 e+48 e^{1+\frac {x^2+x^3}{2 e}}-12 x^2-18 x^3+\left (2 x+3 x^2\right ) \log (3)\right ) \, dx=2\,x-\frac {{\mathrm {e}}^{-\frac {{\mathrm {e}}^{-1}\,x^3}{2}-\frac {{\mathrm {e}}^{-1}\,x^2}{2}}\,\ln \left (3\right )}{12}+\frac {x\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{-1}\,x^3}{2}-\frac {{\mathrm {e}}^{-1}\,x^2}{2}}}{2} \]
int(exp(-1)*exp(-exp(-1)*(x^2/2 + x^3/2))*(exp(1)/2 + (log(3)*(2*x + 3*x^2 ))/24 + 2*exp(1)*exp(exp(-1)*(x^2/2 + x^3/2)) - x^2/2 - (3*x^3)/4),x)