Integrand size = 81, antiderivative size = 24 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {e^{e^x} \left (4+\frac {5 x^2}{4}\right )}{(2+x) \log (x)} \] Output:
1/ln(x)*(4+5/4*x^2)*exp(exp(x))/(2+x)
Time = 5.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {e^{e^x} \left (16+5 x^2\right )}{4 (2+x) \log (x)} \] Input:
Integrate[(E^E^x*(-32 - 16*x - 10*x^2 - 5*x^3 + (-16*x + 20*x^2 + 5*x^3 + E^x*(32*x + 16*x^2 + 10*x^3 + 5*x^4))*Log[x]))/((16*x + 16*x^2 + 4*x^3)*Lo g[x]^2),x]
Output:
(E^E^x*(16 + 5*x^2))/(4*(2 + x)*Log[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.
\(\displaystyle \int \frac {e^{e^x} \left (-5 x^3-10 x^2+\left (5 x^3+20 x^2+e^x \left (5 x^4+10 x^3+16 x^2+32 x\right )-16 x\right ) \log (x)-16 x-32\right )}{\left (4 x^3+16 x^2+16 x\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{e^x} \left (-5 x^3-10 x^2+\left (5 x^3+20 x^2+e^x \left (5 x^4+10 x^3+16 x^2+32 x\right )-16 x\right ) \log (x)-16 x-32\right )}{x \left (4 x^2+16 x+16\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{e^x} \left (-5 x^3-10 x^2+\left (5 x^3+20 x^2+e^x \left (5 x^4+10 x^3+16 x^2+32 x\right )-16 x\right ) \log (x)-16 x-32\right )}{x (2 x+4)^2 \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {5 e^{e^x} x^2}{4 (x+2)^2 \log ^2(x)}+\frac {5 e^{e^x} x^2}{4 (x+2)^2 \log (x)}+\frac {e^{x+e^x} \left (5 x^2+16\right )}{4 (x+2) \log (x)}-\frac {5 e^{e^x} x}{2 (x+2)^2 \log ^2(x)}-\frac {4 e^{e^x}}{(x+2)^2 \log ^2(x)}-\frac {8 e^{e^x}}{(x+2)^2 x \log ^2(x)}+\frac {5 e^{e^x} x}{(x+2)^2 \log (x)}-\frac {4 e^{e^x}}{(x+2)^2 \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {5}{4} \int \frac {e^{e^x}}{\log ^2(x)}dx-2 \int \frac {e^{e^x}}{x \log ^2(x)}dx+\frac {9}{2} \int \frac {e^{e^x}}{(x+2) \log ^2(x)}dx+\frac {5}{4} \int \frac {e^{e^x}}{\log (x)}dx-\frac {5}{2} \int \frac {e^{x+e^x}}{\log (x)}dx+\frac {5}{4} \int \frac {e^{x+e^x} x}{\log (x)}dx-9 \int \frac {e^{e^x}}{(x+2)^2 \log (x)}dx+9 \int \frac {e^{x+e^x}}{(x+2) \log (x)}dx\) |
Input:
Int[(E^E^x*(-32 - 16*x - 10*x^2 - 5*x^3 + (-16*x + 20*x^2 + 5*x^3 + E^x*(3 2*x + 16*x^2 + 10*x^3 + 5*x^4))*Log[x]))/((16*x + 16*x^2 + 4*x^3)*Log[x]^2 ),x]
Output:
$Aborted
Time = 24.85 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
risch | \(\frac {\left (5 x^{2}+16\right ) {\mathrm e}^{{\mathrm e}^{x}}}{4 \left (2+x \right ) \ln \left (x \right )}\) | \(22\) |
parallelrisch | \(-\frac {-10 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}-32 \,{\mathrm e}^{{\mathrm e}^{x}}}{8 \ln \left (x \right ) \left (2+x \right )}\) | \(26\) |
Input:
int((((5*x^4+10*x^3+16*x^2+32*x)*exp(x)+5*x^3+20*x^2-16*x)*ln(x)-5*x^3-10* x^2-16*x-32)*exp(exp(x))/(4*x^3+16*x^2+16*x)/ln(x)^2,x,method=_RETURNVERBO SE)
Output:
1/4*(5*x^2+16)/(2+x)/ln(x)*exp(exp(x))
Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {{\left (5 \, x^{2} + 16\right )} e^{\left (e^{x}\right )}}{4 \, {\left (x + 2\right )} \log \left (x\right )} \] Input:
integrate((((5*x^4+10*x^3+16*x^2+32*x)*exp(x)+5*x^3+20*x^2-16*x)*log(x)-5* x^3-10*x^2-16*x-32)*exp(exp(x))/(4*x^3+16*x^2+16*x)/log(x)^2,x, algorithm= "fricas")
Output:
1/4*(5*x^2 + 16)*e^(e^x)/((x + 2)*log(x))
Time = 0.16 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {\left (5 x^{2} + 16\right ) e^{e^{x}}}{4 x \log {\left (x \right )} + 8 \log {\left (x \right )}} \] Input:
integrate((((5*x**4+10*x**3+16*x**2+32*x)*exp(x)+5*x**3+20*x**2-16*x)*ln(x )-5*x**3-10*x**2-16*x-32)*exp(exp(x))/(4*x**3+16*x**2+16*x)/ln(x)**2,x)
Output:
(5*x**2 + 16)*exp(exp(x))/(4*x*log(x) + 8*log(x))
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {{\left (5 \, x^{2} + 16\right )} e^{\left (e^{x}\right )}}{4 \, {\left (x + 2\right )} \log \left (x\right )} \] Input:
integrate((((5*x^4+10*x^3+16*x^2+32*x)*exp(x)+5*x^3+20*x^2-16*x)*log(x)-5* x^3-10*x^2-16*x-32)*exp(exp(x))/(4*x^3+16*x^2+16*x)/log(x)^2,x, algorithm= "maxima")
Output:
1/4*(5*x^2 + 16)*e^(e^x)/((x + 2)*log(x))
Time = 0.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {5 \, x^{2} e^{\left (x + e^{x}\right )} + 16 \, e^{\left (x + e^{x}\right )}}{4 \, {\left (x e^{x} \log \left (x\right ) + 2 \, e^{x} \log \left (x\right )\right )}} \] Input:
integrate((((5*x^4+10*x^3+16*x^2+32*x)*exp(x)+5*x^3+20*x^2-16*x)*log(x)-5* x^3-10*x^2-16*x-32)*exp(exp(x))/(4*x^3+16*x^2+16*x)/log(x)^2,x, algorithm= "giac")
Output:
1/4*(5*x^2*e^(x + e^x) + 16*e^(x + e^x))/(x*e^x*log(x) + 2*e^x*log(x))
Time = 3.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (5\,x^2+16\right )}{4\,\ln \left (x\right )\,\left (x+2\right )} \] Input:
int(-(exp(exp(x))*(16*x - log(x)*(exp(x)*(32*x + 16*x^2 + 10*x^3 + 5*x^4) - 16*x + 20*x^2 + 5*x^3) + 10*x^2 + 5*x^3 + 32))/(log(x)^2*(16*x + 16*x^2 + 4*x^3)),x)
Output:
(exp(exp(x))*(5*x^2 + 16))/(4*log(x)*(x + 2))
Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{e^x} \left (-32-16 x-10 x^2-5 x^3+\left (-16 x+20 x^2+5 x^3+e^x \left (32 x+16 x^2+10 x^3+5 x^4\right )\right ) \log (x)\right )}{\left (16 x+16 x^2+4 x^3\right ) \log ^2(x)} \, dx=\frac {e^{e^{x}} \left (5 x^{2}+16\right )}{4 \,\mathrm {log}\left (x \right ) \left (x +2\right )} \] Input:
int((((5*x^4+10*x^3+16*x^2+32*x)*exp(x)+5*x^3+20*x^2-16*x)*log(x)-5*x^3-10 *x^2-16*x-32)*exp(exp(x))/(4*x^3+16*x^2+16*x)/log(x)^2,x)
Output:
(e**(e**x)*(5*x**2 + 16))/(4*log(x)*(x + 2))