Integrand size = 71, antiderivative size = 26 \begin {dmath*} \int \frac {1}{5} e^{-x} \left (e^{\frac {2 e^{-x} x}{5}} \left (5 e^x+2 x-2 x^2\right )+e^{\left (-4+e^5\right )^x} \left (-5 e^x-5 \left (e \left (-4+e^5\right )\right )^x x \log \left (-4+e^5\right )\right )\right ) \, dx=\left (-e^{\left (-4+e^5\right )^x}+e^{\frac {2 e^{-x} x}{5}}\right ) x \end {dmath*}
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 3.94 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.27 \begin {dmath*} \int \frac {1}{5} e^{-x} \left (e^{\frac {2 e^{-x} x}{5}} \left (5 e^x+2 x-2 x^2\right )+e^{\left (-4+e^5\right )^x} \left (-5 e^x-5 \left (e \left (-4+e^5\right )\right )^x x \log \left (-4+e^5\right )\right )\right ) \, dx=\frac {1}{5} \left (5 e^{\frac {2 e^{-x} x}{5}} x-\frac {5 \operatorname {ExpIntegralEi}\left (\left (-4+e^5\right )^x\right )}{\log \left (-4+e^5\right )}\right )-\left (-\frac {\operatorname {ExpIntegralEi}\left (\left (-4+e^5\right )^x\right )}{\log ^2\left (-4+e^5\right )}+\frac {e^{\left (-4+e^5\right )^x} x}{\log \left (-4+e^5\right )}\right ) \log \left (-4+e^5\right ) \end {dmath*}
Integrate[(E^((2*x)/(5*E^x))*(5*E^x + 2*x - 2*x^2) + E^(-4 + E^5)^x*(-5*E^ x - 5*(E*(-4 + E^5))^x*x*Log[-4 + E^5]))/(5*E^x),x]
(5*E^((2*x)/(5*E^x))*x - (5*ExpIntegralEi[(-4 + E^5)^x])/Log[-4 + E^5])/5 - (-(ExpIntegralEi[(-4 + E^5)^x]/Log[-4 + E^5]^2) + (E^(-4 + E^5)^x*x)/Log [-4 + E^5])*Log[-4 + E^5]
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.
\(\displaystyle \int \frac {1}{5} e^{-x} \left (e^{\frac {2 e^{-x} x}{5}} \left (-2 x^2+2 x+5 e^x\right )+e^{\left (e^5-4\right )^x} \left (-5 e^x-5 \left (e \left (e^5-4\right )\right )^x x \log \left (e^5-4\right )\right )\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int e^{-x} \left (e^{\frac {2 e^{-x} x}{5}} \left (-2 x^2+2 x+5 e^x\right )-5 e^{\left (-4+e^5\right )^x} \left (\left (e \left (-4+e^5\right )\right )^x \log \left (-4+e^5\right ) x+e^x\right )\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (e^{\frac {2 e^{-x} x}{5}-x} \left (-2 x^2+2 x+5 e^x\right )-5 e^{\left (-4+e^5\right )^x-x} \left (\left (e \left (-4+e^5\right )\right )^x \log \left (-4+e^5\right ) x+e^x\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-2 \int e^{\frac {2 e^{-x} x}{5}-x} x^2dx+5 \int e^{\frac {2 e^{-x} x}{5}}dx+2 \int e^{\frac {2 e^{-x} x}{5}-x} xdx-5 \log \left (e^5-4\right ) \int e^{\left (-4+e^5\right )^x} \left (-4+e^5\right )^x xdx-\frac {5 \operatorname {ExpIntegralEi}\left (\left (-4+e^5\right )^x\right )}{\log \left (e^5-4\right )}\right )\) |
Int[(E^((2*x)/(5*E^x))*(5*E^x + 2*x - 2*x^2) + E^(-4 + E^5)^x*(-5*E^x - 5* (E*(-4 + E^5))^x*x*Log[-4 + E^5]))/(5*E^x),x]
3.6.95.3.1 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 = 3.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85
method | result | size |
risch | \({\mathrm e}^{\frac {2 x \,{\mathrm e}^{-x}}{5}} x -x \,{\mathrm e}^{\left ({\mathrm e}^{5}-4\right )^{x}}\) | \(22\) |
parallelrisch | \({\mathrm e}^{\frac {2 x \,{\mathrm e}^{-x}}{5}} x -x \,{\mathrm e}^{{\mathrm e}^{x \ln \left ({\mathrm e}^{5}-4\right )}}\) | \(26\) |
int(1/5*((-5*x*exp(x)*ln(exp(5)-4)*exp(x*ln(exp(5)-4))-5*exp(x))*exp(exp(x *ln(exp(5)-4)))+(5*exp(x)-2*x^2+2*x)*exp(1/5*x/exp(x))^2)/exp(x),x,method= _RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \begin {dmath*} \int \frac {1}{5} e^{-x} \left (e^{\frac {2 e^{-x} x}{5}} \left (5 e^x+2 x-2 x^2\right )+e^{\left (-4+e^5\right )^x} \left (-5 e^x-5 \left (e \left (-4+e^5\right )\right )^x x \log \left (-4+e^5\right )\right )\right ) \, dx=x e^{\left (\frac {2}{5} \, x e^{\left (-x\right )}\right )} - x e^{\left ({\left (e^{5} - 4\right )}^{x}\right )} \end {dmath*}
integrate(1/5*((-5*x*exp(x)*log(exp(5)-4)*exp(x*log(exp(5)-4))-5*exp(x))*e xp(exp(x*log(exp(5)-4)))+(5*exp(x)-2*x^2+2*x)*exp(1/5*x/exp(x))^2)/exp(x), x, algorithm=\
Time = 18.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \begin {dmath*} \int \frac {1}{5} e^{-x} \left (e^{\frac {2 e^{-x} x}{5}} \left (5 e^x+2 x-2 x^2\right )+e^{\left (-4+e^5\right )^x} \left (-5 e^x-5 \left (e \left (-4+e^5\right )\right )^x x \log \left (-4+e^5\right )\right )\right ) \, dx=x e^{\frac {2 x e^{- x}}{5}} - x e^{e^{x \log {\left (-4 + e^{5} \right )}}} \end {dmath*}
integrate(1/5*((-5*x*exp(x)*ln(exp(5)-4)*exp(x*ln(exp(5)-4))-5*exp(x))*exp (exp(x*ln(exp(5)-4)))+(5*exp(x)-2*x**2+2*x)*exp(1/5*x/exp(x))**2)/exp(x),x )
\begin {dmath*} \int \frac {1}{5} e^{-x} \left (e^{\frac {2 e^{-x} x}{5}} \left (5 e^x+2 x-2 x^2\right )+e^{\left (-4+e^5\right )^x} \left (-5 e^x-5 \left (e \left (-4+e^5\right )\right )^x x \log \left (-4+e^5\right )\right )\right ) \, dx=\int { -\frac {1}{5} \, {\left ({\left (2 \, x^{2} - 2 \, x - 5 \, e^{x}\right )} e^{\left (\frac {2}{5} \, x e^{\left (-x\right )}\right )} + 5 \, {\left (x {\left (e^{5} - 4\right )}^{x} e^{x} \log \left (e^{5} - 4\right ) + e^{x}\right )} e^{\left ({\left (e^{5} - 4\right )}^{x}\right )}\right )} e^{\left (-x\right )} \,d x } \end {dmath*}
integrate(1/5*((-5*x*exp(x)*log(exp(5)-4)*exp(x*log(exp(5)-4))-5*exp(x))*e xp(exp(x*log(exp(5)-4)))+(5*exp(x)-2*x^2+2*x)*exp(1/5*x/exp(x))^2)/exp(x), x, algorithm=\
-integrate(x*e^(x*log(e^5 - 4) + (e^5 - 4)^x), x)*log(e^5 - 4) - Ei((e^5 - 4)^x)/log(e^5 - 4) + 1/5*integrate(-(2*x^2 - 2*x - 5*e^x)*e^(2/5*x*e^(-x) - x), x)
\begin {dmath*} \int \frac {1}{5} e^{-x} \left (e^{\frac {2 e^{-x} x}{5}} \left (5 e^x+2 x-2 x^2\right )+e^{\left (-4+e^5\right )^x} \left (-5 e^x-5 \left (e \left (-4+e^5\right )\right )^x x \log \left (-4+e^5\right )\right )\right ) \, dx=\int { -\frac {1}{5} \, {\left ({\left (2 \, x^{2} - 2 \, x - 5 \, e^{x}\right )} e^{\left (\frac {2}{5} \, x e^{\left (-x\right )}\right )} + 5 \, {\left (x {\left (e^{5} - 4\right )}^{x} e^{x} \log \left (e^{5} - 4\right ) + e^{x}\right )} e^{\left ({\left (e^{5} - 4\right )}^{x}\right )}\right )} e^{\left (-x\right )} \,d x } \end {dmath*}
integrate(1/5*((-5*x*exp(x)*log(exp(5)-4)*exp(x*log(exp(5)-4))-5*exp(x))*e xp(exp(x*log(exp(5)-4)))+(5*exp(x)-2*x^2+2*x)*exp(1/5*x/exp(x))^2)/exp(x), x, algorithm=\
integrate(-1/5*((2*x^2 - 2*x - 5*e^x)*e^(2/5*x*e^(-x)) + 5*(x*(e^5 - 4)^x* e^x*log(e^5 - 4) + e^x)*e^((e^5 - 4)^x))*e^(-x), x)
Time = 15.76 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \begin {dmath*} \int \frac {1}{5} e^{-x} \left (e^{\frac {2 e^{-x} x}{5}} \left (5 e^x+2 x-2 x^2\right )+e^{\left (-4+e^5\right )^x} \left (-5 e^x-5 \left (e \left (-4+e^5\right )\right )^x x \log \left (-4+e^5\right )\right )\right ) \, dx=-x\,\left ({\mathrm {e}}^{{\left ({\mathrm {e}}^5-4\right )}^x}-{\mathrm {e}}^{\frac {2\,x\,{\mathrm {e}}^{-x}}{5}}\right ) \end {dmath*}