Integrand size = 71, antiderivative size = 23 \[ \int \frac {e^{e^x x} \left (5+e^x+x\right ) \left (1+e^{2 x} (1+x)+e^x \left (6+6 x+x^2\right )\right )+2 e^x x \log ^2(7)+\left (10 x+2 x^2\right ) \log ^2(7)}{5+e^x+x} \, dx=e^{e^x x} \left (5+e^x+x\right )+x^2 \log ^2(7) \] Output:
exp(ln(exp(x)+5+x)+exp(x)*x)+x^2*ln(7)^2
Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26 \[ \int \frac {e^{e^x x} \left (5+e^x+x\right ) \left (1+e^{2 x} (1+x)+e^x \left (6+6 x+x^2\right )\right )+2 e^x x \log ^2(7)+\left (10 x+2 x^2\right ) \log ^2(7)}{5+e^x+x} \, dx=e^{x+e^x x}+e^{e^x x} (5+x)+x^2 \log ^2(7) \] Input:
Integrate[(E^(E^x*x)*(5 + E^x + x)*(1 + E^(2*x)*(1 + x) + E^x*(6 + 6*x + x ^2)) + 2*E^x*x*Log[7]^2 + (10*x + 2*x^2)*Log[7]^2)/(5 + E^x + x),x]
Output:
E^(x + E^x*x) + E^(E^x*x)*(5 + x) + x^2*Log[7]^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.
\(\displaystyle \int \frac {e^{e^x x} \left (x+e^x+5\right ) \left (e^x \left (x^2+6 x+6\right )+e^{2 x} (x+1)+1\right )+\left (2 x^2+10 x\right ) \log ^2(7)+2 e^x x \log ^2(7)}{x+e^x+5} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (e^{e^x x} \left (e^x x^2+6 e^x x+e^{2 x} x+6 e^x+e^{2 x}+1\right )+2 x \log ^2(7)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int e^{e^x x+x} x^2dx+\int e^{e^x x}dx+6 \int e^{e^x x+x}dx+\int e^{e^x x+2 x}dx+6 \int e^{e^x x+x} xdx+\int e^{e^x x+2 x} xdx+x^2 \log ^2(7)\) |
Input:
Int[(E^(E^x*x)*(5 + E^x + x)*(1 + E^(2*x)*(1 + x) + E^x*(6 + 6*x + x^2)) + 2*E^x*x*Log[7]^2 + (10*x + 2*x^2)*Log[7]^2)/(5 + E^x + x),x]
Output:
$Aborted
Time = 0.83 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\left ({\mathrm e}^{x}+5+x \right ) {\mathrm e}^{{\mathrm e}^{x} x}+x^{2} \ln \left (7\right )^{2}\) | \(21\) |
parallelrisch | \({\mathrm e}^{\ln \left ({\mathrm e}^{x}+5+x \right )+{\mathrm e}^{x} x}+x^{2} \ln \left (7\right )^{2}\) | \(22\) |
Input:
int((((1+x)*exp(x)^2+(x^2+6*x+6)*exp(x)+1)*exp(ln(exp(x)+5+x)+exp(x)*x)+2* x*ln(7)^2*exp(x)+(2*x^2+10*x)*ln(7)^2)/(exp(x)+5+x),x,method=_RETURNVERBOS E)
Output:
(exp(x)+5+x)*exp(exp(x)*x)+x^2*ln(7)^2
Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{e^x x} \left (5+e^x+x\right ) \left (1+e^{2 x} (1+x)+e^x \left (6+6 x+x^2\right )\right )+2 e^x x \log ^2(7)+\left (10 x+2 x^2\right ) \log ^2(7)}{5+e^x+x} \, dx=x^{2} \log \left (7\right )^{2} + e^{\left (x e^{x} + \log \left (x + e^{x} + 5\right )\right )} \] Input:
integrate((((1+x)*exp(x)^2+(x^2+6*x+6)*exp(x)+1)*exp(log(exp(x)+5+x)+exp(x )*x)+2*x*log(7)^2*exp(x)+(2*x^2+10*x)*log(7)^2)/(exp(x)+5+x),x, algorithm= "fricas")
Output:
x^2*log(7)^2 + e^(x*e^x + log(x + e^x + 5))
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{e^x x} \left (5+e^x+x\right ) \left (1+e^{2 x} (1+x)+e^x \left (6+6 x+x^2\right )\right )+2 e^x x \log ^2(7)+\left (10 x+2 x^2\right ) \log ^2(7)}{5+e^x+x} \, dx=x^{2} \log {\left (7 \right )}^{2} + \left (x + e^{x} + 5\right ) e^{x e^{x}} \] Input:
integrate((((1+x)*exp(x)**2+(x**2+6*x+6)*exp(x)+1)*exp(ln(exp(x)+5+x)+exp( x)*x)+2*x*ln(7)**2*exp(x)+(2*x**2+10*x)*ln(7)**2)/(exp(x)+5+x),x)
Output:
x**2*log(7)**2 + (x + exp(x) + 5)*exp(x*exp(x))
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {e^{e^x x} \left (5+e^x+x\right ) \left (1+e^{2 x} (1+x)+e^x \left (6+6 x+x^2\right )\right )+2 e^x x \log ^2(7)+\left (10 x+2 x^2\right ) \log ^2(7)}{5+e^x+x} \, dx=x^{2} \log \left (7\right )^{2} + {\left (x + e^{x} + 5\right )} e^{\left (x e^{x}\right )} \] Input:
integrate((((1+x)*exp(x)^2+(x^2+6*x+6)*exp(x)+1)*exp(log(exp(x)+5+x)+exp(x )*x)+2*x*log(7)^2*exp(x)+(2*x^2+10*x)*log(7)^2)/(exp(x)+5+x),x, algorithm= "maxima")
Output:
x^2*log(7)^2 + (x + e^x + 5)*e^(x*e^x)
\[ \int \frac {e^{e^x x} \left (5+e^x+x\right ) \left (1+e^{2 x} (1+x)+e^x \left (6+6 x+x^2\right )\right )+2 e^x x \log ^2(7)+\left (10 x+2 x^2\right ) \log ^2(7)}{5+e^x+x} \, dx=\int { \frac {2 \, x e^{x} \log \left (7\right )^{2} + 2 \, {\left (x^{2} + 5 \, x\right )} \log \left (7\right )^{2} + {\left ({\left (x + 1\right )} e^{\left (2 \, x\right )} + {\left (x^{2} + 6 \, x + 6\right )} e^{x} + 1\right )} e^{\left (x e^{x} + \log \left (x + e^{x} + 5\right )\right )}}{x + e^{x} + 5} \,d x } \] Input:
integrate((((1+x)*exp(x)^2+(x^2+6*x+6)*exp(x)+1)*exp(log(exp(x)+5+x)+exp(x )*x)+2*x*log(7)^2*exp(x)+(2*x^2+10*x)*log(7)^2)/(exp(x)+5+x),x, algorithm= "giac")
Output:
integrate((2*x*e^x*log(7)^2 + 2*(x^2 + 5*x)*log(7)^2 + ((x + 1)*e^(2*x) + (x^2 + 6*x + 6)*e^x + 1)*e^(x*e^x + log(x + e^x + 5)))/(x + e^x + 5), x)
Time = 3.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {e^{e^x x} \left (5+e^x+x\right ) \left (1+e^{2 x} (1+x)+e^x \left (6+6 x+x^2\right )\right )+2 e^x x \log ^2(7)+\left (10 x+2 x^2\right ) \log ^2(7)}{5+e^x+x} \, dx=5\,{\mathrm {e}}^{x\,{\mathrm {e}}^x}+{\mathrm {e}}^{x+x\,{\mathrm {e}}^x}+x^2\,{\ln \left (7\right )}^2+x\,{\mathrm {e}}^{x\,{\mathrm {e}}^x} \] Input:
int((exp(log(x + exp(x) + 5) + x*exp(x))*(exp(x)*(6*x + x^2 + 6) + exp(2*x )*(x + 1) + 1) + log(7)^2*(10*x + 2*x^2) + 2*x*exp(x)*log(7)^2)/(x + exp(x ) + 5),x)
Output:
5*exp(x*exp(x)) + exp(x + x*exp(x)) + x^2*log(7)^2 + x*exp(x*exp(x))
Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.57 \[ \int \frac {e^{e^x x} \left (5+e^x+x\right ) \left (1+e^{2 x} (1+x)+e^x \left (6+6 x+x^2\right )\right )+2 e^x x \log ^2(7)+\left (10 x+2 x^2\right ) \log ^2(7)}{5+e^x+x} \, dx=e^{e^{x} x +x}+e^{e^{x} x} x +5 e^{e^{x} x}+\mathrm {log}\left (7\right )^{2} x^{2} \] Input:
int((((1+x)*exp(x)^2+(x^2+6*x+6)*exp(x)+1)*exp(log(exp(x)+5+x)+exp(x)*x)+2 *x*log(7)^2*exp(x)+(2*x^2+10*x)*log(7)^2)/(exp(x)+5+x),x)
Output:
e**(e**x*x + x) + e**(e**x*x)*x + 5*e**(e**x*x) + log(7)**2*x**2