Integrand size = 29, antiderivative size = 22 \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=x+2 e^{-x} x \log \left (e^{12+2 e^2} x\right ) \]
[Out]
Leaf count is larger than twice the leaf count of optimal. \(54\) vs. \(2(22)=44\).
Time = 0.17 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.45, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {6874, 2225, 6820, 2207, 2634} \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=-4 \left (6+e^2\right ) e^{-x} (1-x)+x+4 \left (6+e^2\right ) e^{-x}-2 e^{-x} (1-x) \log (x)+2 e^{-x} \log (x) \]
[In]
[Out]
Rule 2207
Rule 2225
Rule 2634
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \left (1+2 e^{-x}+2 e^{-x} (1-x) \left (12 \left (1+\frac {e^2}{6}\right )+\log (x)\right )\right ) \, dx \\ & = x+2 \int e^{-x} \, dx+2 \int e^{-x} (1-x) \left (12 \left (1+\frac {e^2}{6}\right )+\log (x)\right ) \, dx \\ & = -2 e^{-x}+x+2 \int e^{-x} (1-x) \left (2 \left (6+e^2\right )+\log (x)\right ) \, dx \\ & = -2 e^{-x}+x+2 \int \left (-2 e^{-x} \left (6+e^2\right ) (-1+x)-e^{-x} (-1+x) \log (x)\right ) \, dx \\ & = -2 e^{-x}+x-2 \int e^{-x} (-1+x) \log (x) \, dx-\left (4 \left (6+e^2\right )\right ) \int e^{-x} (-1+x) \, dx \\ & = -2 e^{-x}-4 e^{-x} \left (6+e^2\right ) (1-x)+x+2 e^{-x} \log (x)-2 e^{-x} (1-x) \log (x)-2 \int e^{-x} \, dx-\left (4 \left (6+e^2\right )\right ) \int e^{-x} \, dx \\ & = 4 e^{-x} \left (6+e^2\right )-4 e^{-x} \left (6+e^2\right ) (1-x)+x+2 e^{-x} \log (x)-2 e^{-x} (1-x) \log (x) \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=x+2 e^{-x} \left (2 \left (6+e^2\right ) x+x \log (x)\right ) \]
[In]
[Out]
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91
method | result | size |
default | \(2 x \,{\mathrm e}^{-x} \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{2}+12}\right )+x\) | \(20\) |
risch | \(2 x \,{\mathrm e}^{-x} \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{2}+12}\right )+x\) | \(20\) |
parts | \(2 x \,{\mathrm e}^{-x} \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{2}+12}\right )+x\) | \(20\) |
norman | \(\left ({\mathrm e}^{x} x +2 \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{2}+12}\right ) x \right ) {\mathrm e}^{-x}\) | \(24\) |
parallelrisch | \(\left ({\mathrm e}^{x} x +2 \ln \left (x \,{\mathrm e}^{2 \,{\mathrm e}^{2}+12}\right ) x \right ) {\mathrm e}^{-x}\) | \(24\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx={\left (x e^{x} + 2 \, x \log \left (x e^{\left (2 \, e^{2} + 12\right )}\right )\right )} e^{\left (-x\right )} \]
[In]
[Out]
Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=x + 2 x e^{- x} \log {\left (x e^{12 + 2 e^{2}} \right )} \]
[In]
[Out]
\[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=\int { -{\left (2 \, {\left (x - 1\right )} \log \left (x e^{\left (2 \, e^{2} + 12\right )}\right ) - e^{x} - 2\right )} e^{\left (-x\right )} \,d x } \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=2 \, x e^{\left (-x\right )} \log \left (x\right ) + 24 \, x e^{\left (-x\right )} + 4 \, x e^{\left (-x + 2\right )} + x \]
[In]
[Out]
Time = 11.33 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int e^{-x} \left (2+e^x+(2-2 x) \log \left (e^{12+2 e^2} x\right )\right ) \, dx=x\,{\mathrm {e}}^{-x}\,\left (4\,{\mathrm {e}}^2+{\mathrm {e}}^x+2\,\ln \left (x\right )+24\right ) \]
[In]
[Out]