Integrand size = 63, antiderivative size = 26 \[ \int e^{-4+\frac {2-4 e^{4+x}-\log (x)}{e^4}} \left (-x+2 e^4 x+\left (1-e^4\right ) \log (20)+e^x \left (-4 e^4 x^2+4 e^4 x \log (20)\right )\right ) \, dx=e^{-4 e^x+\frac {2-\log (x)}{e^4}} x (x-\log (20)) \]
Time = 3.52 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int e^{-4+\frac {2-4 e^{4+x}-\log (x)}{e^4}} \left (-x+2 e^4 x+\left (1-e^4\right ) \log (20)+e^x \left (-4 e^4 x^2+4 e^4 x \log (20)\right )\right ) \, dx=e^{\frac {2}{e^4}-4 e^x} x^{1-\frac {1}{e^4}} (x-\log (20)) \]
Integrate[E^(-4 + (2 - 4*E^(4 + x) - Log[x])/E^4)*(-x + 2*E^4*x + (1 - E^4 )*Log[20] + E^x*(-4*E^4*x^2 + 4*E^4*x*Log[20])),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.
\(\displaystyle \int e^{\frac {-4 e^{x+4}-\log (x)+2}{e^4}-4} \left (e^x \left (4 e^4 x \log (20)-4 e^4 x^2\right )+2 e^4 x-x+\left (1-e^4\right ) \log (20)\right ) \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int e^{\frac {-4 e^{x+4}-\log (x)+2}{e^4}-4} \left (e^x \left (4 e^4 x \log (20)-4 e^4 x^2\right )+\left (2 e^4-1\right ) x+\left (1-e^4\right ) \log (20)\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int e^{-\frac {4 e^{x+4}+\log (x)-2 \left (1-2 e^4\right )}{e^4}} \left (e^x \left (4 e^4 x \log (20)-4 e^4 x^2\right )+\left (2 e^4-1\right ) x+\left (1-e^4\right ) \log (20)\right )dx\) |
\(\Big \downarrow \) 2704 |
\(\displaystyle \int e^{-\frac {4 e^{x+4}-2 \left (1-2 e^4\right )}{e^4}} x^{-\frac {1}{e^4}} \left (e^x \left (4 e^4 x \log (20)-4 e^4 x^2\right )+\left (2 e^4-1\right ) x+\left (1-e^4\right ) \log (20)\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int e^{-\frac {2 \left (2 e^{x+4}-1+2 e^4\right )}{e^4}} x^{-\frac {1}{e^4}} \left (e^x \left (4 e^4 x \log (20)-4 e^4 x^2\right )+\left (2 e^4-1\right ) x+\left (1-e^4\right ) \log (20)\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\left (2 e^4-1\right ) e^{-\frac {2 \left (2 e^{x+4}-1+2 e^4\right )}{e^4}} x^{1-\frac {1}{e^4}}+\left (e^4-1\right ) \left (-e^{-\frac {2 \left (2 e^{x+4}-1+2 e^4\right )}{e^4}}\right ) x^{-\frac {1}{e^4}} \log (20)-4 e^{-\frac {2 \left (2 e^{x+4}-1+2 e^4\right )}{e^4}+x+4} x^{1-\frac {1}{e^4}} (x-\log (20))\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\left (\left (1-2 e^4\right ) \int e^{-\frac {2 \left (-1+2 e^4+2 e^{x+4}\right )}{e^4}} x^{1-\frac {1}{e^4}}dx\right )-4 \int e^{x-4 e^x+\frac {2}{e^4}} x^{2-\frac {1}{e^4}}dx+4 \log (20) \int e^{x-4 e^x+\frac {2}{e^4}} x^{1-\frac {1}{e^4}}dx+\left (1-e^4\right ) \log (20) \int e^{-\frac {2 \left (-1+2 e^4+2 e^{x+4}\right )}{e^4}} x^{-\frac {1}{e^4}}dx\) |
Int[E^(-4 + (2 - 4*E^(4 + x) - Log[x])/E^4)*(-x + 2*E^4*x + (1 - E^4)*Log[ 20] + E^x*(-4*E^4*x^2 + 4*E^4*x*Log[20])),x]
3.18.84.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)* z^(a*b*Log[F]), x] /; FreeQ[{F, a, b}, x]
Time = 0.72 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.69
method | result | size |
risch | \(\left (-\ln \left (5\right ) {\mathrm e}^{4} x -2 x \,{\mathrm e}^{4} \ln \left (2\right )+x^{2} {\mathrm e}^{4}\right ) {\mathrm e}^{-{\mathrm e}^{-4} \ln \left (x \right )-4 \,{\mathrm e}^{-4} {\mathrm e}^{4+x}+2 \,{\mathrm e}^{-4}-4}\) | \(44\) |
norman | \(x^{2} {\mathrm e}^{\left (-\ln \left (x \right )-4 \,{\mathrm e}^{4} {\mathrm e}^{x}+2\right ) {\mathrm e}^{-4}}-x \ln \left (20\right ) {\mathrm e}^{\left (-\ln \left (x \right )-4 \,{\mathrm e}^{4} {\mathrm e}^{x}+2\right ) {\mathrm e}^{-4}}\) | \(47\) |
parallelrisch | \({\mathrm e}^{-4} \left (-{\mathrm e}^{4} \ln \left (20\right ) x \,{\mathrm e}^{-\left (4 \,{\mathrm e}^{4} {\mathrm e}^{x}+\ln \left (x \right )-2\right ) {\mathrm e}^{-4}}+{\mathrm e}^{-\left (4 \,{\mathrm e}^{4} {\mathrm e}^{x}+\ln \left (x \right )-2\right ) {\mathrm e}^{-4}} x^{2} {\mathrm e}^{4}\right )\) | \(54\) |
int(((4*x*exp(4)*ln(20)-4*x^2*exp(4))*exp(x)+(1-exp(4))*ln(20)+2*x*exp(4)- x)*exp((-ln(x)-4*exp(4)*exp(x)+2)/exp(4))/exp(4),x,method=_RETURNVERBOSE)
(-ln(5)*exp(4)*x-2*x*exp(4)*ln(2)+x^2*exp(4))*exp(-exp(-4)*ln(x)-4*exp(-4) *exp(4+x)+2*exp(-4)-4)
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int e^{-4+\frac {2-4 e^{4+x}-\log (x)}{e^4}} \left (-x+2 e^4 x+\left (1-e^4\right ) \log (20)+e^x \left (-4 e^4 x^2+4 e^4 x \log (20)\right )\right ) \, dx={\left (x^{2} e^{4} - x e^{4} \log \left (20\right )\right )} e^{\left (-{\left (4 \, e^{4} + 4 \, e^{\left (x + 4\right )} + \log \left (x\right ) - 2\right )} e^{\left (-4\right )}\right )} \]
integrate(((4*x*exp(4)*log(20)-4*x^2*exp(4))*exp(x)+(1-exp(4))*log(20)+2*x *exp(4)-x)*exp((-log(x)-4*exp(4)*exp(x)+2)/exp(4))/exp(4),x, algorithm=\
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int e^{-4+\frac {2-4 e^{4+x}-\log (x)}{e^4}} \left (-x+2 e^4 x+\left (1-e^4\right ) \log (20)+e^x \left (-4 e^4 x^2+4 e^4 x \log (20)\right )\right ) \, dx=\left (x^{2} - x \log {\left (20 \right )}\right ) e^{\frac {- 4 e^{4} e^{x} - \log {\left (x \right )} + 2}{e^{4}}} \]
integrate(((4*x*exp(4)*ln(20)-4*x**2*exp(4))*exp(x)+(1-exp(4))*ln(20)+2*x* exp(4)-x)*exp((-ln(x)-4*exp(4)*exp(x)+2)/exp(4))/exp(4),x)
Time = 0.34 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int e^{-4+\frac {2-4 e^{4+x}-\log (x)}{e^4}} \left (-x+2 e^4 x+\left (1-e^4\right ) \log (20)+e^x \left (-4 e^4 x^2+4 e^4 x \log (20)\right )\right ) \, dx={\left (x^{2} e^{\left (2 \, e^{\left (-4\right )}\right )} - x {\left (\log \left (5\right ) + 2 \, \log \left (2\right )\right )} e^{\left (2 \, e^{\left (-4\right )}\right )}\right )} e^{\left (-e^{\left (-4\right )} \log \left (x\right ) - 4 \, e^{x}\right )} \]
integrate(((4*x*exp(4)*log(20)-4*x^2*exp(4))*exp(x)+(1-exp(4))*log(20)+2*x *exp(4)-x)*exp((-log(x)-4*exp(4)*exp(x)+2)/exp(4))/exp(4),x, algorithm=\
\[ \int e^{-4+\frac {2-4 e^{4+x}-\log (x)}{e^4}} \left (-x+2 e^4 x+\left (1-e^4\right ) \log (20)+e^x \left (-4 e^4 x^2+4 e^4 x \log (20)\right )\right ) \, dx=\int { {\left (2 \, x e^{4} - 4 \, {\left (x^{2} e^{4} - x e^{4} \log \left (20\right )\right )} e^{x} - {\left (e^{4} - 1\right )} \log \left (20\right ) - x\right )} e^{\left (-{\left (4 \, e^{\left (x + 4\right )} + \log \left (x\right ) - 2\right )} e^{\left (-4\right )} - 4\right )} \,d x } \]
integrate(((4*x*exp(4)*log(20)-4*x^2*exp(4))*exp(x)+(1-exp(4))*log(20)+2*x *exp(4)-x)*exp((-log(x)-4*exp(4)*exp(x)+2)/exp(4))/exp(4),x, algorithm=\
integrate((2*x*e^4 - 4*(x^2*e^4 - x*e^4*log(20))*e^x - (e^4 - 1)*log(20) - x)*e^(-(4*e^(x + 4) + log(x) - 2)*e^(-4) - 4), x)
Time = 9.49 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int e^{-4+\frac {2-4 e^{4+x}-\log (x)}{e^4}} \left (-x+2 e^4 x+\left (1-e^4\right ) \log (20)+e^x \left (-4 e^4 x^2+4 e^4 x \log (20)\right )\right ) \, dx=\frac {x\,{\mathrm {e}}^{2\,{\mathrm {e}}^{-4}-4\,{\mathrm {e}}^x}\,\left (x-\ln \left (20\right )\right )}{x^{{\mathrm {e}}^{-4}}} \]