Integrand size = 40, antiderivative size = 25 \[ \int \frac {e^{-3-e^x} \left (-2+4 \log (2)-4 e^x x \log (2)+\left (-1+e^x x\right ) \log \left (x^2\right )\right )}{\log (2)} \, dx=e^{-3-e^x} \left (4 x-\frac {x \log \left (x^2\right )}{\log (2)}\right ) \]
Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-3-e^x} \left (-2+4 \log (2)-4 e^x x \log (2)+\left (-1+e^x x\right ) \log \left (x^2\right )\right )}{\log (2)} \, dx=-\frac {e^{-3-e^x} x \log \left (\frac {x^2}{16}\right )}{\log (2)} \]
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.
\(\displaystyle \int \frac {e^{-e^x-3} \left (\left (e^x x-1\right ) \log \left (x^2\right )-4 e^x x \log (2)-2+4 \log (2)\right )}{\log (2)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -e^{-3-e^x} \left (4 e^x \log (2) x+\left (1-e^x x\right ) \log \left (x^2\right )-\log (16)+2\right )dx}{\log (2)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int e^{-3-e^x} \left (4 e^x \log (2) x+\left (1-e^x x\right ) \log \left (x^2\right )-\log (16)+2\right )dx}{\log (2)}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {\int e^{-3-e^x} \left (4 e^x \log (2) x+\left (1-e^x x\right ) \log \left (x^2\right )+2 (1-\log (4))\right )dx}{\log (2)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (e^{x-e^x-3} \log (16) x-e^{-3-e^x} \left (e^x x-1\right ) \log \left (x^2\right )-2 e^{-3-e^x} (-1+\log (4))\right )dx}{\log (2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\log (16) \int e^{x-e^x-3} xdx-\frac {2 \operatorname {ExpIntegralEi}\left (-e^x\right )}{e^3}+\frac {2 (1-\log (4)) \operatorname {ExpIntegralEi}\left (-e^x\right )}{e^3}+e^{-e^x-3} x \log \left (x^2\right )}{\log (2)}\) |
3.9.43.3.1 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 = 0.31 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96
method | result | size |
norman | \(\left (4 x -\frac {x \ln \left (x^{2}\right )}{\ln \left (2\right )}\right ) {\mathrm e}^{-{\mathrm e}^{x}-3}\) | \(24\) |
parallelrisch | \(\frac {\left (8 x \ln \left (2\right )-2 x \ln \left (x^{2}\right )\right ) {\mathrm e}^{-{\mathrm e}^{x}-3}}{2 \ln \left (2\right )}\) | \(27\) |
default | \(\frac {\left (\left (4 \ln \left (2\right )-\ln \left (x^{2}\right )+2 \ln \left (x \right )\right ) x -2 x \ln \left (x \right )\right ) {\mathrm e}^{-{\mathrm e}^{x}-3}}{\ln \left (2\right )}\) | \(36\) |
risch | \(\frac {\left (\frac {i x \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}}{2}-i x \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+\frac {i x \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}+4 x \ln \left (2\right )-2 x \ln \left (x \right )\right ) {\mathrm e}^{-{\mathrm e}^{x}-3}}{\ln \left (2\right )}\) | \(76\) |
int(((exp(x)*x-1)*ln(x^2)-4*x*ln(2)*exp(x)+4*ln(2)-2)/ln(2)/exp(3+exp(x)), x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3-e^x} \left (-2+4 \log (2)-4 e^x x \log (2)+\left (-1+e^x x\right ) \log \left (x^2\right )\right )}{\log (2)} \, dx=\frac {{\left (4 \, x \log \left (2\right ) - x \log \left (x^{2}\right )\right )} e^{\left (-e^{x} - 3\right )}}{\log \left (2\right )} \]
integrate(((exp(x)*x-1)*log(x^2)-4*x*log(2)*exp(x)+4*log(2)-2)/log(2)/exp( 3+exp(x)),x, algorithm=\
Time = 1.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-3-e^x} \left (-2+4 \log (2)-4 e^x x \log (2)+\left (-1+e^x x\right ) \log \left (x^2\right )\right )}{\log (2)} \, dx=\frac {\left (- x \log {\left (x^{2} \right )} + 4 x \log {\left (2 \right )}\right ) e^{- e^{x} - 3}}{\log {\left (2 \right )}} \]
\[ \int \frac {e^{-3-e^x} \left (-2+4 \log (2)-4 e^x x \log (2)+\left (-1+e^x x\right ) \log \left (x^2\right )\right )}{\log (2)} \, dx=\int { -\frac {{\left (4 \, x e^{x} \log \left (2\right ) - {\left (x e^{x} - 1\right )} \log \left (x^{2}\right ) - 4 \, \log \left (2\right ) + 2\right )} e^{\left (-e^{x} - 3\right )}}{\log \left (2\right )} \,d x } \]
integrate(((exp(x)*x-1)*log(x^2)-4*x*log(2)*exp(x)+4*log(2)-2)/log(2)/exp( 3+exp(x)),x, algorithm=\
2*(2*Ei(-e^x)*e^(-3)*log(2) - Ei(-e^x)*e^(-3) + (2*x*log(2) - x*log(x))*e^ (-e^x - 3) - (2*log(2) - 1)*integrate(e^(-e^x - 3), x))/log(2)
Time = 0.29 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {e^{-3-e^x} \left (-2+4 \log (2)-4 e^x x \log (2)+\left (-1+e^x x\right ) \log \left (x^2\right )\right )}{\log (2)} \, dx=\frac {{\left (4 \, x e^{\left (x - e^{x}\right )} \log \left (2\right ) - x e^{\left (x - e^{x}\right )} \log \left (x^{2}\right )\right )} e^{\left (-x - 3\right )}}{\log \left (2\right )} \]
integrate(((exp(x)*x-1)*log(x^2)-4*x*log(2)*exp(x)+4*log(2)-2)/log(2)/exp( 3+exp(x)),x, algorithm=\
Timed out. \[ \int \frac {e^{-3-e^x} \left (-2+4 \log (2)-4 e^x x \log (2)+\left (-1+e^x x\right ) \log \left (x^2\right )\right )}{\log (2)} \, dx=\int \frac {{\mathrm {e}}^{-{\mathrm {e}}^x-3}\,\left (4\,\ln \left (2\right )+\ln \left (x^2\right )\,\left (x\,{\mathrm {e}}^x-1\right )-4\,x\,{\mathrm {e}}^x\,\ln \left (2\right )-2\right )}{\ln \left (2\right )} \,d x \]
int((exp(- exp(x) - 3)*(4*log(2) + log(x^2)*(x*exp(x) - 1) - 4*x*exp(x)*lo g(2) - 2))/log(2),x)