Integrand size = 87, antiderivative size = 29 \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx=-\log \left (\frac {5}{2}\right )+e^{e^x} \left (2-\log \left (4+e^{x-x^2}\right )\right ) \] Output:
exp(exp(x))*(2-ln(exp(-x^2+x)+4))+ln(2/5)
Time = 2.95 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx=-e^{e^x} \left (-2+\log \left (4+e^{x-x^2}\right )\right ) \] Input:
Integrate[(E^E^x*(E^x*(8 + 2*E^(x - x^2)) + E^(x - x^2)*(-1 + 2*x)) + E^(E ^x + x)*(-4 - E^(x - x^2))*Log[4 + E^(x - x^2)])/(4 + E^(x - x^2)),x]
Output:
-(E^E^x*(-2 + Log[4 + E^(x - x^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.
\(\displaystyle \int \frac {e^{e^x} \left (e^x \left (2 e^{x-x^2}+8\right )+e^{x-x^2} (2 x-1)\right )+e^{x+e^x} \left (-e^{x-x^2}-4\right ) \log \left (e^{x-x^2}+4\right )}{e^{x-x^2}+4} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-x^2+x+e^x} \left (8 e^{x^2}-4 e^{x^2} \log \left (e^{x-x^2}+4\right )-e^x \log \left (e^{x-x^2}+4\right )+2 x+2 e^x-1\right )}{e^{x-x^2}+4}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {1}{4} e^{-x^2+x+e^x} (2 x-1)-\frac {e^{-x^2+2 x+e^x} (2 x-1)}{4 \left (4 e^{x^2}+e^x\right )}-e^{x+e^x} \left (\log \left (e^{x-x^2}+4\right )-2\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{4} \int e^{-x^2+x+e^x}dx+\int \frac {e^{x+e^x}}{e^x+4 e^{x^2}}dx+\frac {1}{4} \int \frac {e^{-x^2+2 x+e^x}}{e^x+4 e^{x^2}}dx+\frac {1}{2} \int e^{-x^2+x+e^x} xdx-2 \int \frac {e^{x+e^x} x}{e^x+4 e^{x^2}}dx-\frac {1}{2} \int \frac {e^{-x^2+2 x+e^x} x}{e^x+4 e^{x^2}}dx-e^{e^x} \log \left (e^{x-x^2}+4\right )+2 e^{e^x}\) |
Input:
Int[(E^E^x*(E^x*(8 + 2*E^(x - x^2)) + E^(x - x^2)*(-1 + 2*x)) + E^(E^x + x )*(-4 - E^(x - x^2))*Log[4 + E^(x - x^2)])/(4 + E^(x - x^2)),x]
Output:
$Aborted
Time = 0.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76
method | result | size |
risch | \(-{\mathrm e}^{{\mathrm e}^{x}} \ln \left ({\mathrm e}^{-x \left (-1+x \right )}+4\right )+2 \,{\mathrm e}^{{\mathrm e}^{x}}\) | \(22\) |
parallelrisch | \(-{\mathrm e}^{{\mathrm e}^{x}} \ln \left ({\mathrm e}^{-x^{2}+x}+4\right )+2 \,{\mathrm e}^{{\mathrm e}^{x}}\) | \(23\) |
Input:
int(((-exp(-x^2+x)-4)*exp(x)*exp(exp(x))*ln(exp(-x^2+x)+4)+((2*exp(-x^2+x) +8)*exp(x)+(-1+2*x)*exp(-x^2+x))*exp(exp(x)))/(exp(-x^2+x)+4),x,method=_RE TURNVERBOSE)
Output:
-exp(exp(x))*ln(exp(-x*(-1+x))+4)+2*exp(exp(x))
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx=-{\left (e^{\left (x + e^{x}\right )} \log \left (e^{\left (-x^{2} + x\right )} + 4\right ) - 2 \, e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \] Input:
integrate(((-exp(-x^2+x)-4)*exp(x)*exp(exp(x))*log(exp(-x^2+x)+4)+((2*exp( -x^2+x)+8)*exp(x)+(-1+2*x)*exp(-x^2+x))*exp(exp(x)))/(exp(-x^2+x)+4),x, al gorithm="fricas")
Output:
-(e^(x + e^x)*log(e^(-x^2 + x) + 4) - 2*e^(x + e^x))*e^(-x)
Timed out. \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx=\text {Timed out} \] Input:
integrate(((-exp(-x**2+x)-4)*exp(x)*exp(exp(x))*ln(exp(-x**2+x)+4)+((2*exp (-x**2+x)+8)*exp(x)+(-1+2*x)*exp(-x**2+x))*exp(exp(x)))/(exp(-x**2+x)+4),x )
Output:
Timed out
Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx={\left (x^{2} + 2\right )} e^{\left (e^{x}\right )} - e^{\left (e^{x}\right )} \log \left (4 \, e^{\left (x^{2}\right )} + e^{x}\right ) \] Input:
integrate(((-exp(-x^2+x)-4)*exp(x)*exp(exp(x))*log(exp(-x^2+x)+4)+((2*exp( -x^2+x)+8)*exp(x)+(-1+2*x)*exp(-x^2+x))*exp(exp(x)))/(exp(-x^2+x)+4),x, al gorithm="maxima")
Output:
(x^2 + 2)*e^(e^x) - e^(e^x)*log(4*e^(x^2) + e^x)
Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx={\left (x^{2} e^{\left (x + e^{x}\right )} - e^{\left (x + e^{x}\right )} \log \left (4 \, e^{\left (x^{2}\right )} + e^{x}\right ) + 2 \, e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )} \] Input:
integrate(((-exp(-x^2+x)-4)*exp(x)*exp(exp(x))*log(exp(-x^2+x)+4)+((2*exp( -x^2+x)+8)*exp(x)+(-1+2*x)*exp(-x^2+x))*exp(exp(x)))/(exp(-x^2+x)+4),x, al gorithm="giac")
Output:
(x^2*e^(x + e^x) - e^(x + e^x)*log(4*e^(x^2) + e^x) + 2*e^(x + e^x))*e^(-x )
Timed out. \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx=\int \frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left ({\mathrm {e}}^{x-x^2}\,\left (2\,x-1\right )+{\mathrm {e}}^x\,\left (2\,{\mathrm {e}}^{x-x^2}+8\right )\right )-\ln \left ({\mathrm {e}}^{x-x^2}+4\right )\,{\mathrm {e}}^{x+{\mathrm {e}}^x}\,\left ({\mathrm {e}}^{x-x^2}+4\right )}{{\mathrm {e}}^{x-x^2}+4} \,d x \] Input:
int((exp(exp(x))*(exp(x - x^2)*(2*x - 1) + exp(x)*(2*exp(x - x^2) + 8)) - log(exp(x - x^2) + 4)*exp(exp(x))*exp(x)*(exp(x - x^2) + 4))/(exp(x - x^2) + 4),x)
Output:
int((exp(exp(x))*(exp(x - x^2)*(2*x - 1) + exp(x)*(2*exp(x - x^2) + 8)) - log(exp(x - x^2) + 4)*exp(x + exp(x))*(exp(x - x^2) + 4))/(exp(x - x^2) + 4), x)
Time = 0.21 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03 \[ \int \frac {e^{e^x} \left (e^x \left (8+2 e^{x-x^2}\right )+e^{x-x^2} (-1+2 x)\right )+e^{e^x+x} \left (-4-e^{x-x^2}\right ) \log \left (4+e^{x-x^2}\right )}{4+e^{x-x^2}} \, dx=e^{e^{x}} \left (-\mathrm {log}\left (\frac {4 e^{x^{2}}+e^{x}}{e^{x^{2}}}\right )+2\right ) \] Input:
int(((-exp(-x^2+x)-4)*exp(x)*exp(exp(x))*log(exp(-x^2+x)+4)+((2*exp(-x^2+x )+8)*exp(x)+(-1+2*x)*exp(-x^2+x))*exp(exp(x)))/(exp(-x^2+x)+4),x)
Output:
e**(e**x)*( - log((4*e**(x**2) + e**x)/e**(x**2)) + 2)