Integrand size = 109, antiderivative size = 25 \[ \int \frac {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (25 x^2+10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \log ^2(5)\right ) \log ^2(x)} \, dx=\frac {58 \left (e^{e^x}+x\right )}{x (-5-x+\log (5)) \log (x)} \] Output:
58/x*(x+exp(exp(x)))/ln(x)/(ln(5)-x-5)
Time = 1.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (25 x^2+10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \log ^2(5)\right ) \log ^2(x)} \, dx=-\frac {58 \left (e^{e^x}+x\right )}{x (5+x-\log (5)) \log (x)} \] Input:
Integrate[(290*x + 58*x^2 - 58*x*Log[5] + 58*x^2*Log[x] + E^E^x*(290 + 58* x - 58*Log[5] + (290 + 116*x - 58*Log[5] + E^x*(-290*x - 58*x^2 + 58*x*Log [5]))*Log[x]))/((25*x^2 + 10*x^3 + x^4 + (-10*x^2 - 2*x^3)*Log[5] + x^2*Lo g[5]^2)*Log[x]^2),x]
Output:
(-58*(E^E^x + x))/(x*(5 + x - Log[5])*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.
| \(\displaystyle \int \frac {58 x^2+58 x^2 \log (x)+e^{e^x} \left (\left (e^x \left (-58 x^2-290 x+58 x \log (5)\right )+116 x+290-58 \log (5)\right ) \log (x)+58 x+290-58 \log (5)\right )+290 x-58 x \log (5)}{\left (x^4+10 x^3+25 x^2+x^2 \log ^2(5)+\left (-2 x^3-10 x^2\right ) \log (5)\right ) \log ^2(x)} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {58 x^2+58 x^2 \log (x)+e^{e^x} \left (\left (e^x \left (-58 x^2-290 x+58 x \log (5)\right )+116 x+290-58 \log (5)\right ) \log (x)+58 x+290-58 \log (5)\right )+290 x-58 x \log (5)}{\left (x^4+10 x^3+x^2 \left (25+\log ^2(5)\right )+\left (-2 x^3-10 x^2\right ) \log (5)\right ) \log ^2(x)}dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {58 x^2+58 x^2 \log (x)+e^{e^x} \left (\left (e^x \left (-58 x^2-290 x+58 x \log (5)\right )+116 x+290-58 \log (5)\right ) \log (x)+58 x+290-58 \log (5)\right )+x (290-58 \log (5))}{\left (x^4+10 x^3+x^2 \left (25+\log ^2(5)\right )+\left (-2 x^3-10 x^2\right ) \log (5)\right ) \log ^2(x)}dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {58 x^2+58 x^2 \log (x)+e^{e^x} \left (\left (e^x \left (-58 x^2-290 x+58 x \log (5)\right )+116 x+290-58 \log (5)\right ) \log (x)+58 x+290-58 \log (5)\right )+x (290-58 \log (5))}{x^2 \left (x^2+2 x (5-\log (5))+(\log (5)-5)^2\right ) \log ^2(x)}dx\) |
| \(\Big \downarrow \) 7277 |
| \(\displaystyle 4 \int \frac {29 \left (\log (x) x^2+x^2+(5-\log (5)) x+e^{e^x} \left (x+\left (2 x-e^x \left (x^2-\log (5) x+5 x\right )-\log (5)+5\right ) \log (x)-\log (5)+5\right )\right )}{2 x^2 (x-\log (5)+5)^2 \log ^2(x)}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 58 \int \frac {\log (x) x^2+x^2+(5-\log (5)) x+e^{e^x} \left (x+\left (2 x-e^x \left (x^2-\log (5) x+5 x\right )-\log (5)+5\right ) \log (x)-\log (5)+5\right )}{x^2 (x-\log (5)+5)^2 \log ^2(x)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle 58 \int \left (\frac {1}{(x-\log (5)+5)^2 \log (x)}+\frac {1}{(x-\log (5)+5)^2 \log ^2(x)}-\frac {e^{x+e^x}}{(x-\log (5)+5) \log (x) x}+\frac {2 e^{e^x}}{(x-\log (5)+5)^2 \log (x) x}+\frac {e^{e^x}}{(x-\log (5)+5)^2 \log ^2(x) x}+\frac {5-\log (5)}{(x-\log (5)+5)^2 \log ^2(x) x}+\frac {5 e^{e^x} \left (1-\frac {\log (5)}{5}\right )}{(x-\log (5)+5)^2 \log (x) x^2}+\frac {5 e^{e^x} \left (1-\frac {\log (5)}{5}\right )}{(x-\log (5)+5)^2 \log ^2(x) x^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 58 \left (\frac {\int \frac {e^{e^x}}{x^2 \log ^2(x)}dx}{5-\log (5)}+\frac {\int \frac {e^{e^x}}{x^2 \log (x)}dx}{5-\log (5)}-\frac {\int \frac {e^{e^x}}{x \log ^2(x)}dx}{(5-\log (5))^2}+\int \frac {1}{(x-\log (5)+5)^2 \log ^2(x)}dx+(5-\log (5)) \int \frac {1}{x (x-\log (5)+5)^2 \log ^2(x)}dx+\frac {\int \frac {e^{e^x}}{(x-\log (5)+5) \log ^2(x)}dx}{(5-\log (5))^2}-\frac {\int \frac {e^{x+e^x}}{x \log (x)}dx}{5-\log (5)}+\int \frac {1}{(x-\log (5)+5)^2 \log (x)}dx-\frac {\int \frac {e^{e^x}}{(x-\log (5)+5)^2 \log (x)}dx}{5-\log (5)}+\frac {\int \frac {e^{x+e^x}}{(x-\log (5)+5) \log (x)}dx}{5-\log (5)}\right )\) |
Input:
Int[(290*x + 58*x^2 - 58*x*Log[5] + 58*x^2*Log[x] + E^E^x*(290 + 58*x - 58 *Log[5] + (290 + 116*x - 58*Log[5] + E^x*(-290*x - 58*x^2 + 58*x*Log[5]))* Log[x]))/((25*x^2 + 10*x^3 + x^4 + (-10*x^2 - 2*x^3)*Log[5] + x^2*Log[5]^2 )*Log[x]^2),x]
Output:
$Aborted
Time = 25.99 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08
| method | result | size |
| parallelrisch | \(\frac {58 x +58 \,{\mathrm e}^{{\mathrm e}^{x}}}{x \ln \left (x \right ) \left (\ln \left (5\right )-x -5\right )}\) | \(27\) |
| risch | \(\frac {58}{\ln \left (x \right ) \left (\ln \left (5\right )-x -5\right )}+\frac {58 \,{\mathrm e}^{{\mathrm e}^{x}}}{x \left (\ln \left (5\right )-x -5\right ) \ln \left (x \right )}\) | \(38\) |
Input:
int(((((58*x*ln(5)-58*x^2-290*x)*exp(x)-58*ln(5)+116*x+290)*ln(x)-58*ln(5) +58*x+290)*exp(exp(x))+58*x^2*ln(x)-58*x*ln(5)+58*x^2+290*x)/(x^2*ln(5)^2+ (-2*x^3-10*x^2)*ln(5)+x^4+10*x^3+25*x^2)/ln(x)^2,x,method=_RETURNVERBOSE)
Output:
1/x*(58*x+58*exp(exp(x)))/ln(x)/(ln(5)-x-5)
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (25 x^2+10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \log ^2(5)\right ) \log ^2(x)} \, dx=-\frac {58 \, {\left (x + e^{\left (e^{x}\right )}\right )}}{{\left (x^{2} - x \log \left (5\right ) + 5 \, x\right )} \log \left (x\right )} \] Input:
integrate(((((58*x*log(5)-58*x^2-290*x)*exp(x)-58*log(5)+116*x+290)*log(x) -58*log(5)+58*x+290)*exp(exp(x))+58*x^2*log(x)-58*x*log(5)+58*x^2+290*x)/( x^2*log(5)^2+(-2*x^3-10*x^2)*log(5)+x^4+10*x^3+25*x^2)/log(x)^2,x, algorit hm="fricas")
Output:
-58*(x + e^(e^x))/((x^2 - x*log(5) + 5*x)*log(x))
Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (25 x^2+10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \log ^2(5)\right ) \log ^2(x)} \, dx=- \frac {58 e^{e^{x}}}{x^{2} \log {\left (x \right )} - x \log {\left (5 \right )} \log {\left (x \right )} + 5 x \log {\left (x \right )}} - \frac {58}{\left (x - \log {\left (5 \right )} + 5\right ) \log {\left (x \right )}} \] Input:
integrate(((((58*x*ln(5)-58*x**2-290*x)*exp(x)-58*ln(5)+116*x+290)*ln(x)-5 8*ln(5)+58*x+290)*exp(exp(x))+58*x**2*ln(x)-58*x*ln(5)+58*x**2+290*x)/(x** 2*ln(5)**2+(-2*x**3-10*x**2)*ln(5)+x**4+10*x**3+25*x**2)/ln(x)**2,x)
Output:
-58*exp(exp(x))/(x**2*log(x) - x*log(5)*log(x) + 5*x*log(x)) - 58/((x - lo g(5) + 5)*log(x))
Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (25 x^2+10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \log ^2(5)\right ) \log ^2(x)} \, dx=-\frac {58 \, {\left (x + e^{\left (e^{x}\right )}\right )}}{{\left (x^{2} - x {\left (\log \left (5\right ) - 5\right )}\right )} \log \left (x\right )} \] Input:
integrate(((((58*x*log(5)-58*x^2-290*x)*exp(x)-58*log(5)+116*x+290)*log(x) -58*log(5)+58*x+290)*exp(exp(x))+58*x^2*log(x)-58*x*log(5)+58*x^2+290*x)/( x^2*log(5)^2+(-2*x^3-10*x^2)*log(5)+x^4+10*x^3+25*x^2)/log(x)^2,x, algorit hm="maxima")
Output:
-58*(x + e^(e^x))/((x^2 - x*(log(5) - 5))*log(x))
Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (25 x^2+10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \log ^2(5)\right ) \log ^2(x)} \, dx=-\frac {58 \, {\left (x e^{x} + e^{\left (x + e^{x}\right )}\right )}}{x^{2} e^{x} \log \left (x\right ) - x e^{x} \log \left (5\right ) \log \left (x\right ) + 5 \, x e^{x} \log \left (x\right )} \] Input:
integrate(((((58*x*log(5)-58*x^2-290*x)*exp(x)-58*log(5)+116*x+290)*log(x) -58*log(5)+58*x+290)*exp(exp(x))+58*x^2*log(x)-58*x*log(5)+58*x^2+290*x)/( x^2*log(5)^2+(-2*x^3-10*x^2)*log(5)+x^4+10*x^3+25*x^2)/log(x)^2,x, algorit hm="giac")
Output:
-58*(x*e^x + e^(x + e^x))/(x^2*e^x*log(x) - x*e^x*log(5)*log(x) + 5*x*e^x* log(x))
Timed out. \[ \int \frac {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (25 x^2+10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \log ^2(5)\right ) \log ^2(x)} \, dx=\int \frac {290\,x+58\,x^2\,\ln \left (x\right )-58\,x\,\ln \left (5\right )+58\,x^2+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (58\,x-58\,\ln \left (5\right )+\ln \left (x\right )\,\left (116\,x-58\,\ln \left (5\right )-{\mathrm {e}}^x\,\left (290\,x-58\,x\,\ln \left (5\right )+58\,x^2\right )+290\right )+290\right )}{{\ln \left (x\right )}^2\,\left (x^2\,{\ln \left (5\right )}^2-\ln \left (5\right )\,\left (2\,x^3+10\,x^2\right )+25\,x^2+10\,x^3+x^4\right )} \,d x \] Input:
int((290*x + 58*x^2*log(x) - 58*x*log(5) + 58*x^2 + exp(exp(x))*(58*x - 58 *log(5) + log(x)*(116*x - 58*log(5) - exp(x)*(290*x - 58*x*log(5) + 58*x^2 ) + 290) + 290))/(log(x)^2*(x^2*log(5)^2 - log(5)*(10*x^2 + 2*x^3) + 25*x^ 2 + 10*x^3 + x^4)),x)
Output:
int((290*x + 58*x^2*log(x) - 58*x*log(5) + 58*x^2 + exp(exp(x))*(58*x - 58 *log(5) + log(x)*(116*x - 58*log(5) - exp(x)*(290*x - 58*x*log(5) + 58*x^2 ) + 290) + 290))/(log(x)^2*(x^2*log(5)^2 - log(5)*(10*x^2 + 2*x^3) + 25*x^ 2 + 10*x^3 + x^4)), x)
Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {290 x+58 x^2-58 x \log (5)+58 x^2 \log (x)+e^{e^x} \left (290+58 x-58 \log (5)+\left (290+116 x-58 \log (5)+e^x \left (-290 x-58 x^2+58 x \log (5)\right )\right ) \log (x)\right )}{\left (25 x^2+10 x^3+x^4+\left (-10 x^2-2 x^3\right ) \log (5)+x^2 \log ^2(5)\right ) \log ^2(x)} \, dx=\frac {58 e^{e^{x}}+58 x}{\mathrm {log}\left (x \right ) x \left (\mathrm {log}\left (5\right )-x -5\right )} \] Input:
int(((((58*x*log(5)-58*x^2-290*x)*exp(x)-58*log(5)+116*x+290)*log(x)-58*lo g(5)+58*x+290)*exp(exp(x))+58*x^2*log(x)-58*x*log(5)+58*x^2+290*x)/(x^2*lo g(5)^2+(-2*x^3-10*x^2)*log(5)+x^4+10*x^3+25*x^2)/log(x)^2,x)
Output:
(58*(e**(e**x) + x))/(log(x)*x*(log(5) - x - 5))