Integrand size = 82, antiderivative size = 22 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=-1+\frac {-1+5 \left (e^4+x+\log (x)\right )^x}{2 x} \] Output:
1/2*(5*exp(x*ln(ln(x)+x+exp(4)))-1)/x-1
Time = 0.23 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\frac {-1+5 \left (e^4+x+\log (x)\right )^x}{2 x} \] Input:
Integrate[(E^4 + x + Log[x] + (E^4 + x + Log[x])^x*(-5*E^4 + 5*x^2 - 5*Log [x] + (5*E^4*x + 5*x^2 + 5*x*Log[x])*Log[E^4 + x + Log[x]]))/(2*E^4*x^2 + 2*x^3 + 2*x^2*Log[x]),x]
Output:
(-1 + 5*(E^4 + x + Log[x])^x)/(2*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.
\(\displaystyle \int \frac {\left (5 x^2+\left (5 x^2+5 e^4 x+5 x \log (x)\right ) \log \left (x+\log (x)+e^4\right )-5 \log (x)-5 e^4\right ) \left (x+\log (x)+e^4\right )^x+x+\log (x)+e^4}{2 x^3+2 e^4 x^2+2 x^2 \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (5 x^2+\left (5 x^2+5 e^4 x+5 x \log (x)\right ) \log \left (x+\log (x)+e^4\right )-5 \log (x)-5 e^4\right ) \left (x+\log (x)+e^4\right )^x+x+\log (x)+e^4}{2 x^2 \left (x+\log (x)+e^4\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {-5 \left (-x^2+\log (x)-\left (x^2+\log (x) x+e^4 x\right ) \log \left (x+\log (x)+e^4\right )+e^4\right ) \left (x+\log (x)+e^4\right )^x+x+\log (x)+e^4}{x^2 \left (x+\log (x)+e^4\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{2} \int \left (\frac {5 \left (\log \left (x+\log (x)+e^4\right ) x^2+x^2+\log (x) \log \left (x+\log (x)+e^4\right ) x+e^4 \log \left (x+\log (x)+e^4\right ) x-\log (x)-e^4\right ) \left (x+\log (x)+e^4\right )^{x-1}}{x^2}+\frac {1}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-5 e^4 \int \frac {\left (x+\log (x)+e^4\right )^{x-1}}{x^2}dx-5 \int \frac {\log (x) \left (x+\log (x)+e^4\right )^{x-1}}{x^2}dx+5 \int \left (x+\log (x)+e^4\right )^{x-1}dx+5 \int \frac {\left (x+\log (x)+e^4\right )^x \log \left (x+\log (x)+e^4\right )}{x}dx-\frac {1}{x}\right )\) |
Input:
Int[(E^4 + x + Log[x] + (E^4 + x + Log[x])^x*(-5*E^4 + 5*x^2 - 5*Log[x] + (5*E^4*x + 5*x^2 + 5*x*Log[x])*Log[E^4 + x + Log[x]]))/(2*E^4*x^2 + 2*x^3 + 2*x^2*Log[x]),x]
Output:
$Aborted
Time = 2.63 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91
method | result | size |
risch | \(-\frac {1}{2 x}+\frac {5 \left (\ln \left (x \right )+x +{\mathrm e}^{4}\right )^{x}}{2 x}\) | \(20\) |
parallelrisch | \(\frac {5 \,{\mathrm e}^{x \ln \left (\ln \left (x \right )+x +{\mathrm e}^{4}\right )}-1}{2 x}\) | \(20\) |
Input:
int((((5*x*ln(x)+5*x*exp(4)+5*x^2)*ln(ln(x)+x+exp(4))-5*ln(x)-5*exp(4)+5*x ^2)*exp(x*ln(ln(x)+x+exp(4)))+ln(x)+x+exp(4))/(2*x^2*ln(x)+2*x^2*exp(4)+2* x^3),x,method=_RETURNVERBOSE)
Output:
-1/2/x+5/2*(ln(x)+x+exp(4))^x/x
Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\frac {5 \, {\left (x + e^{4} + \log \left (x\right )\right )}^{x} - 1}{2 \, x} \] Input:
integrate((((5*x*log(x)+5*x*exp(4)+5*x^2)*log(log(x)+x+exp(4))-5*log(x)-5* exp(4)+5*x^2)*exp(x*log(log(x)+x+exp(4)))+log(x)+x+exp(4))/(2*x^2*log(x)+2 *x^2*exp(4)+2*x^3),x, algorithm="fricas")
Output:
1/2*(5*(x + e^4 + log(x))^x - 1)/x
Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\frac {5 e^{x \log {\left (x + \log {\left (x \right )} + e^{4} \right )}}}{2 x} - \frac {1}{2 x} \] Input:
integrate((((5*x*ln(x)+5*x*exp(4)+5*x**2)*ln(ln(x)+x+exp(4))-5*ln(x)-5*exp (4)+5*x**2)*exp(x*ln(ln(x)+x+exp(4)))+ln(x)+x+exp(4))/(2*x**2*ln(x)+2*x**2 *exp(4)+2*x**3),x)
Output:
5*exp(x*log(x + log(x) + exp(4)))/(2*x) - 1/(2*x)
Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\frac {5 \, {\left (x + e^{4} + \log \left (x\right )\right )}^{x} - 1}{2 \, x} \] Input:
integrate((((5*x*log(x)+5*x*exp(4)+5*x^2)*log(log(x)+x+exp(4))-5*log(x)-5* exp(4)+5*x^2)*exp(x*log(log(x)+x+exp(4)))+log(x)+x+exp(4))/(2*x^2*log(x)+2 *x^2*exp(4)+2*x^3),x, algorithm="maxima")
Output:
1/2*(5*(x + e^4 + log(x))^x - 1)/x
\[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\int { \frac {5 \, {\left (x^{2} + {\left (x^{2} + x e^{4} + x \log \left (x\right )\right )} \log \left (x + e^{4} + \log \left (x\right )\right ) - e^{4} - \log \left (x\right )\right )} {\left (x + e^{4} + \log \left (x\right )\right )}^{x} + x + e^{4} + \log \left (x\right )}{2 \, {\left (x^{3} + x^{2} e^{4} + x^{2} \log \left (x\right )\right )}} \,d x } \] Input:
integrate((((5*x*log(x)+5*x*exp(4)+5*x^2)*log(log(x)+x+exp(4))-5*log(x)-5* exp(4)+5*x^2)*exp(x*log(log(x)+x+exp(4)))+log(x)+x+exp(4))/(2*x^2*log(x)+2 *x^2*exp(4)+2*x^3),x, algorithm="giac")
Output:
integrate(1/2*(5*(x^2 + (x^2 + x*e^4 + x*log(x))*log(x + e^4 + log(x)) - e ^4 - log(x))*(x + e^4 + log(x))^x + x + e^4 + log(x))/(x^3 + x^2*e^4 + x^2 *log(x)), x)
Time = 3.62 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\frac {5\,{\left (x+{\mathrm {e}}^4+\ln \left (x\right )\right )}^x-1}{2\,x} \] Input:
int((x + exp(4) + log(x) - exp(x*log(x + exp(4) + log(x)))*(5*exp(4) + 5*l og(x) - log(x + exp(4) + log(x))*(5*x*exp(4) + 5*x*log(x) + 5*x^2) - 5*x^2 ))/(2*x^2*log(x) + 2*x^2*exp(4) + 2*x^3),x)
Output:
(5*(x + exp(4) + log(x))^x - 1)/(2*x)
Time = 0.16 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {e^4+x+\log (x)+\left (e^4+x+\log (x)\right )^x \left (-5 e^4+5 x^2-5 \log (x)+\left (5 e^4 x+5 x^2+5 x \log (x)\right ) \log \left (e^4+x+\log (x)\right )\right )}{2 e^4 x^2+2 x^3+2 x^2 \log (x)} \, dx=\frac {5 \left (\mathrm {log}\left (x \right )+e^{4}+x \right )^{x}-1}{2 x} \] Input:
int((((5*x*log(x)+5*x*exp(4)+5*x^2)*log(log(x)+x+exp(4))-5*log(x)-5*exp(4) +5*x^2)*exp(x*log(log(x)+x+exp(4)))+log(x)+x+exp(4))/(2*x^2*log(x)+2*x^2*e xp(4)+2*x^3),x)
Output:
(5*(log(x) + e**4 + x)**x - 1)/(2*x)