Integrand size = 76, antiderivative size = 28 \[ \int \frac {x+2 e^{e^{2 x}+2 x} x+\left (5-e^{e^{2 x}}-x\right ) \log \left (5-e^{e^{2 x}}-x\right )}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx=-e^{9/2}-\frac {x}{\log \left (5-e^{e^{2 x}}-x\right )} \] Output:
-x/ln(-exp(exp(x)^2)+5-x)-exp(9/2)
Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {x+2 e^{e^{2 x}+2 x} x+\left (5-e^{e^{2 x}}-x\right ) \log \left (5-e^{e^{2 x}}-x\right )}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx=-\frac {x}{\log \left (5-e^{e^{2 x}}-x\right )} \] Input:
Integrate[(x + 2*E^(E^(2*x) + 2*x)*x + (5 - E^E^(2*x) - x)*Log[5 - E^E^(2* x) - x])/((-5 + E^E^(2*x) + x)*Log[5 - E^E^(2*x) - x]^2),x]
Output:
-(x/Log[5 - E^E^(2*x) - 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.
\(\displaystyle \int \frac {2 e^{2 x+e^{2 x}} x+x+\left (-x-e^{e^{2 x}}+5\right ) \log \left (-x-e^{e^{2 x}}+5\right )}{\left (x+e^{e^{2 x}}-5\right ) \log ^2\left (-x-e^{e^{2 x}}+5\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 e^{2 x+e^{2 x}} x}{\left (x+e^{e^{2 x}}-5\right ) \log ^2\left (-x-e^{e^{2 x}}+5\right )}-\frac {-x+x \log \left (-x-e^{e^{2 x}}+5\right )+e^{e^{2 x}} \log \left (-x-e^{e^{2 x}}+5\right )-5 \log \left (-x-e^{e^{2 x}}+5\right )}{\left (x+e^{e^{2 x}}-5\right ) \log ^2\left (-x-e^{e^{2 x}}+5\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {x}{\left (x+e^{e^{2 x}}-5\right ) \log ^2\left (-x-e^{e^{2 x}}+5\right )}dx+2 \int \frac {e^{2 x+e^{2 x}} x}{\left (x+e^{e^{2 x}}-5\right ) \log ^2\left (-x-e^{e^{2 x}}+5\right )}dx-\int \frac {1}{\log \left (-x-e^{e^{2 x}}+5\right )}dx\) |
Input:
Int[(x + 2*E^(E^(2*x) + 2*x)*x + (5 - E^E^(2*x) - x)*Log[5 - E^E^(2*x) - x ])/((-5 + E^E^(2*x) + x)*Log[5 - E^E^(2*x) - x]^2),x]
Output:
$Aborted
Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68
method | result | size |
risch | \(-\frac {x}{\ln \left (-{\mathrm e}^{{\mathrm e}^{2 x}}+5-x \right )}\) | \(19\) |
parallelrisch | \(-\frac {x}{\ln \left (-{\mathrm e}^{{\mathrm e}^{2 x}}+5-x \right )}\) | \(19\) |
Input:
int(((-exp(exp(x)^2)+5-x)*ln(-exp(exp(x)^2)+5-x)+2*x*exp(x)^2*exp(exp(x)^2 )+x)/(exp(exp(x)^2)+x-5)/ln(-exp(exp(x)^2)+5-x)^2,x,method=_RETURNVERBOSE)
Output:
-x/ln(-exp(exp(2*x))+5-x)
Time = 0.10 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {x+2 e^{e^{2 x}+2 x} x+\left (5-e^{e^{2 x}}-x\right ) \log \left (5-e^{e^{2 x}}-x\right )}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx=-\frac {x}{\log \left (-{\left ({\left (x - 5\right )} e^{\left (2 \, x\right )} + e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )}\right )} \] Input:
integrate(((-exp(exp(x)^2)+5-x)*log(-exp(exp(x)^2)+5-x)+2*x*exp(x)^2*exp(e xp(x)^2)+x)/(exp(exp(x)^2)+x-5)/log(-exp(exp(x)^2)+5-x)^2,x, algorithm="fr icas")
Output:
-x/log(-((x - 5)*e^(2*x) + e^(2*x + e^(2*x)))*e^(-2*x))
Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.50 \[ \int \frac {x+2 e^{e^{2 x}+2 x} x+\left (5-e^{e^{2 x}}-x\right ) \log \left (5-e^{e^{2 x}}-x\right )}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx=- \frac {x}{\log {\left (- x - e^{e^{2 x}} + 5 \right )}} \] Input:
integrate(((-exp(exp(x)**2)+5-x)*ln(-exp(exp(x)**2)+5-x)+2*x*exp(x)**2*exp (exp(x)**2)+x)/(exp(exp(x)**2)+x-5)/ln(-exp(exp(x)**2)+5-x)**2,x)
Output:
-x/log(-x - exp(exp(2*x)) + 5)
Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {x+2 e^{e^{2 x}+2 x} x+\left (5-e^{e^{2 x}}-x\right ) \log \left (5-e^{e^{2 x}}-x\right )}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx=-\frac {x}{\log \left (-x - e^{\left (e^{\left (2 \, x\right )}\right )} + 5\right )} \] Input:
integrate(((-exp(exp(x)^2)+5-x)*log(-exp(exp(x)^2)+5-x)+2*x*exp(x)^2*exp(e xp(x)^2)+x)/(exp(exp(x)^2)+x-5)/log(-exp(exp(x)^2)+5-x)^2,x, algorithm="ma xima")
Output:
-x/log(-x - e^(e^(2*x)) + 5)
Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (23) = 46\).
Time = 0.14 (sec) , antiderivative size = 281, normalized size of antiderivative = 10.04 \[ \int \frac {x+2 e^{e^{2 x}+2 x} x+\left (5-e^{e^{2 x}}-x\right ) \log \left (5-e^{e^{2 x}}-x\right )}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx=-\frac {2 \, x e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )} \log \left (-{\left (x e^{\left (2 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )}\right ) + 4 \, x e^{\left (4 \, x + 2 \, e^{\left (2 \, x\right )}\right )} \log \left (-x - e^{\left (e^{\left (2 \, x\right )}\right )} + 5\right ) + 2 \, x e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )} \log \left (-x - e^{\left (e^{\left (2 \, x\right )}\right )} + 5\right ) + x \log \left (-{\left (x e^{\left (2 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )}\right )}{4 \, e^{\left (4 \, x + 2 \, e^{\left (2 \, x\right )}\right )} \log \left (-{\left (x e^{\left (2 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )}\right ) \log \left (-x - e^{\left (e^{\left (2 \, x\right )}\right )} + 5\right ) + 4 \, e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )} \log \left (-{\left (x e^{\left (2 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )}\right ) \log \left (-x - e^{\left (e^{\left (2 \, x\right )}\right )} + 5\right ) + \log \left (-{\left (x e^{\left (2 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + e^{\left (2 \, x + e^{\left (2 \, x\right )}\right )}\right )} e^{\left (-2 \, x\right )}\right ) \log \left (-x - e^{\left (e^{\left (2 \, x\right )}\right )} + 5\right )} \] Input:
integrate(((-exp(exp(x)^2)+5-x)*log(-exp(exp(x)^2)+5-x)+2*x*exp(x)^2*exp(e xp(x)^2)+x)/(exp(exp(x)^2)+x-5)/log(-exp(exp(x)^2)+5-x)^2,x, algorithm="gi ac")
Output:
-(2*x*e^(2*x + e^(2*x))*log(-(x*e^(2*x) - 5*e^(2*x) + e^(2*x + e^(2*x)))*e ^(-2*x)) + 4*x*e^(4*x + 2*e^(2*x))*log(-x - e^(e^(2*x)) + 5) + 2*x*e^(2*x + e^(2*x))*log(-x - e^(e^(2*x)) + 5) + x*log(-(x*e^(2*x) - 5*e^(2*x) + e^( 2*x + e^(2*x)))*e^(-2*x)))/(4*e^(4*x + 2*e^(2*x))*log(-(x*e^(2*x) - 5*e^(2 *x) + e^(2*x + e^(2*x)))*e^(-2*x))*log(-x - e^(e^(2*x)) + 5) + 4*e^(2*x + e^(2*x))*log(-(x*e^(2*x) - 5*e^(2*x) + e^(2*x + e^(2*x)))*e^(-2*x))*log(-x - e^(e^(2*x)) + 5) + log(-(x*e^(2*x) - 5*e^(2*x) + e^(2*x + e^(2*x)))*e^( -2*x))*log(-x - e^(e^(2*x)) + 5))
Time = 7.34 (sec) , antiderivative size = 106, normalized size of antiderivative = 3.79 \[ \int \frac {x+2 e^{e^{2 x}+2 x} x+\left (5-e^{e^{2 x}}-x\right ) \log \left (5-e^{e^{2 x}}-x\right )}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx=-\frac {{\mathrm {e}}^{-2\,x}\,\left (x\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{2\,x+{\mathrm {e}}^{2\,x}}\,\ln \left (5-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}-x\right )-{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}\,\ln \left (5-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}-x\right )+2\,x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{2\,x+{\mathrm {e}}^{2\,x}}\right )}{\ln \left (5-{\mathrm {e}}^{{\mathrm {e}}^{2\,x}}-x\right )\,\left (2\,{\mathrm {e}}^{2\,x+{\mathrm {e}}^{2\,x}}+1\right )} \] Input:
int((x - log(5 - exp(exp(2*x)) - x)*(x + exp(exp(2*x)) - 5) + 2*x*exp(2*x) *exp(exp(2*x)))/(log(5 - exp(exp(2*x)) - x)^2*(x + exp(exp(2*x)) - 5)),x)
Output:
-(exp(-2*x)*(x*exp(2*x) + exp(2*x + exp(2*x))*log(5 - exp(exp(2*x)) - x) - exp(2*x)*exp(exp(2*x))*log(5 - exp(exp(2*x)) - x) + 2*x*exp(2*x)*exp(2*x + exp(2*x))))/(log(5 - exp(exp(2*x)) - x)*(2*exp(2*x + exp(2*x)) + 1))
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {x+2 e^{e^{2 x}+2 x} x+\left (5-e^{e^{2 x}}-x\right ) \log \left (5-e^{e^{2 x}}-x\right )}{\left (-5+e^{e^{2 x}}+x\right ) \log ^2\left (5-e^{e^{2 x}}-x\right )} \, dx=-\frac {x}{\mathrm {log}\left (-e^{e^{2 x}}-x +5\right )} \] Input:
int(((-exp(exp(x)^2)+5-x)*log(-exp(exp(x)^2)+5-x)+2*x*exp(x)^2*exp(exp(x)^ 2)+x)/(exp(exp(x)^2)+x-5)/log(-exp(exp(x)^2)+5-x)^2,x)
Output:
( - x)/log( - e**(e**(2*x)) - x + 5)