Integrand size = 97, antiderivative size = 25 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=\frac {e^5 \left (2+\frac {x}{x+e^x x}\right )}{2-\log (x)} \] Output:
exp(5)*(x/(exp(x)*x+x)+2)/(-ln(x)+2)
Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=-\frac {e^5 \left (3+2 e^x\right )}{\left (1+e^x\right ) (-2+\log (x))} \] Input:
Integrate[(3*E^5 + 2*E^(5 + 2*x) + E^(5 + x)*(5 - 2*x) + E^(5 + x)*x*Log[x ])/(4*x + 8*E^x*x + 4*E^(2*x)*x + (-4*x - 8*E^x*x - 4*E^(2*x)*x)*Log[x] + (x + 2*E^x*x + E^(2*x)*x)*Log[x]^2),x]
Output:
-((E^5*(3 + 2*E^x))/((1 + E^x)*(-2 + 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.
\(\displaystyle \int \frac {e^{x+5} (5-2 x)+2 e^{2 x+5}+e^{x+5} x \log (x)+3 e^5}{8 e^x x+4 e^{2 x} x+4 x+\left (2 e^x x+e^{2 x} x+x\right ) \log ^2(x)+\left (-8 e^x x-4 e^{2 x} x-4 x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^5 \left (e^x (5-2 x)+2 e^{2 x}+e^x x \log (x)+3\right )}{\left (e^x+1\right )^2 x (2-\log (x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^5 \int \frac {e^x (5-2 x)+2 e^{2 x}+e^x x \log (x)+3}{\left (1+e^x\right )^2 x (2-\log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle e^5 \int \left (\frac {\log (x) x-2 x+1}{\left (1+e^x\right ) x (\log (x)-2)^2}-\frac {1}{\left (1+e^x\right )^2 (\log (x)-2)}+\frac {2}{x (\log (x)-2)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^5 \left (-2 \int \frac {1}{\left (1+e^x\right ) (\log (x)-2)^2}dx+\int \frac {1}{\left (1+e^x\right ) x (\log (x)-2)^2}dx-\int \frac {1}{\left (1+e^x\right )^2 (\log (x)-2)}dx+\int \frac {\log (x)}{\left (1+e^x\right ) (\log (x)-2)^2}dx+\frac {2}{2-\log (x)}\right )\) |
Input:
Int[(3*E^5 + 2*E^(5 + 2*x) + E^(5 + x)*(5 - 2*x) + E^(5 + x)*x*Log[x])/(4* x + 8*E^x*x + 4*E^(2*x)*x + (-4*x - 8*E^x*x - 4*E^(2*x)*x)*Log[x] + (x + 2 *E^x*x + E^(2*x)*x)*Log[x]^2),x]
Output:
$Aborted
Time = 0.58 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
risch | \(-\frac {\left (2 \,{\mathrm e}^{x}+3\right ) {\mathrm e}^{5}}{\left ({\mathrm e}^{x}+1\right ) \left (\ln \left (x \right )-2\right )}\) | \(23\) |
norman | \(\frac {-2 \,{\mathrm e}^{5} {\mathrm e}^{x}-3 \,{\mathrm e}^{5}}{\left (\ln \left (x \right )-2\right ) \left ({\mathrm e}^{x}+1\right )}\) | \(25\) |
parallelrisch | \(-\frac {2 \,{\mathrm e}^{5} {\mathrm e}^{x}+3 \,{\mathrm e}^{5}}{\left ({\mathrm e}^{x}+1\right ) \left (\ln \left (x \right )-2\right )}\) | \(26\) |
Input:
int((x*exp(5)*exp(x)*ln(x)+2*exp(5)*exp(x)^2+(5-2*x)*exp(5)*exp(x)+3*exp(5 ))/((x*exp(x)^2+2*exp(x)*x+x)*ln(x)^2+(-4*x*exp(x)^2-8*exp(x)*x-4*x)*ln(x) +4*x*exp(x)^2+8*exp(x)*x+4*x),x,method=_RETURNVERBOSE)
Output:
-(2*exp(x)+3)*exp(5)/(exp(x)+1)/(ln(x)-2)
Time = 0.07 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=-\frac {3 \, e^{10} + 2 \, e^{\left (x + 10\right )}}{{\left (e^{5} + e^{\left (x + 5\right )}\right )} \log \left (x\right ) - 2 \, e^{5} - 2 \, e^{\left (x + 5\right )}} \] Input:
integrate((x*exp(5)*exp(x)*log(x)+2*exp(5)*exp(x)^2+(5-2*x)*exp(5)*exp(x)+ 3*exp(5))/((x*exp(x)^2+2*exp(x)*x+x)*log(x)^2+(-4*x*exp(x)^2-8*exp(x)*x-4* x)*log(x)+4*x*exp(x)^2+8*exp(x)*x+4*x),x, algorithm="fricas")
Output:
-(3*e^10 + 2*e^(x + 10))/((e^5 + e^(x + 5))*log(x) - 2*e^5 - 2*e^(x + 5))
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=- \frac {e^{5}}{\left (\log {\left (x \right )} - 2\right ) e^{x} + \log {\left (x \right )} - 2} - \frac {2 e^{5}}{\log {\left (x \right )} - 2} \] Input:
integrate((x*exp(5)*exp(x)*ln(x)+2*exp(5)*exp(x)**2+(5-2*x)*exp(5)*exp(x)+ 3*exp(5))/((x*exp(x)**2+2*exp(x)*x+x)*ln(x)**2+(-4*x*exp(x)**2-8*exp(x)*x- 4*x)*ln(x)+4*x*exp(x)**2+8*exp(x)*x+4*x),x)
Output:
-exp(5)/((log(x) - 2)*exp(x) + log(x) - 2) - 2*exp(5)/(log(x) - 2)
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=-\frac {3 \, e^{5} + 2 \, e^{\left (x + 5\right )}}{{\left (\log \left (x\right ) - 2\right )} e^{x} + \log \left (x\right ) - 2} \] Input:
integrate((x*exp(5)*exp(x)*log(x)+2*exp(5)*exp(x)^2+(5-2*x)*exp(5)*exp(x)+ 3*exp(5))/((x*exp(x)^2+2*exp(x)*x+x)*log(x)^2+(-4*x*exp(x)^2-8*exp(x)*x-4* x)*log(x)+4*x*exp(x)^2+8*exp(x)*x+4*x),x, algorithm="maxima")
Output:
-(3*e^5 + 2*e^(x + 5))/((log(x) - 2)*e^x + log(x) - 2)
Time = 0.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=-\frac {3 \, e^{10} + 2 \, e^{\left (x + 10\right )}}{e^{5} \log \left (x\right ) + e^{\left (x + 5\right )} \log \left (x\right ) - 2 \, e^{5} - 2 \, e^{\left (x + 5\right )}} \] Input:
integrate((x*exp(5)*exp(x)*log(x)+2*exp(5)*exp(x)^2+(5-2*x)*exp(5)*exp(x)+ 3*exp(5))/((x*exp(x)^2+2*exp(x)*x+x)*log(x)^2+(-4*x*exp(x)^2-8*exp(x)*x-4* x)*log(x)+4*x*exp(x)^2+8*exp(x)*x+4*x),x, algorithm="giac")
Output:
-(3*e^10 + 2*e^(x + 10))/(e^5*log(x) + e^(x + 5)*log(x) - 2*e^5 - 2*e^(x + 5))
Time = 1.77 (sec) , antiderivative size = 100, normalized size of antiderivative = 4.00 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=\frac {\frac {{\mathrm {e}}^5}{2}+\frac {x\,{\mathrm {e}}^5}{2}}{{\mathrm {e}}^x+1}-\frac {\frac {{\mathrm {e}}^5\,\left (2\,{\mathrm {e}}^{2\,x}+5\,{\mathrm {e}}^x-2\,x\,{\mathrm {e}}^x+3\right )}{{\left ({\mathrm {e}}^x+1\right )}^2}+\frac {x\,{\mathrm {e}}^{x+5}\,\ln \left (x\right )}{{\left ({\mathrm {e}}^x+1\right )}^2}}{\ln \left (x\right )-2}-\frac {\frac {{\mathrm {e}}^5\,\left (x+1\right )}{2}-{\mathrm {e}}^{x+5}\,\left (\frac {x}{2}-\frac {1}{2}\right )}{{\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1} \] Input:
int((3*exp(5) + 2*exp(2*x)*exp(5) - exp(5)*exp(x)*(2*x - 5) + x*exp(5)*exp (x)*log(x))/(4*x + 4*x*exp(2*x) - log(x)*(4*x + 4*x*exp(2*x) + 8*x*exp(x)) + log(x)^2*(x + x*exp(2*x) + 2*x*exp(x)) + 8*x*exp(x)),x)
Output:
(exp(5)/2 + (x*exp(5))/2)/(exp(x) + 1) - ((exp(5)*(2*exp(2*x) + 5*exp(x) - 2*x*exp(x) + 3))/(exp(x) + 1)^2 + (x*exp(x + 5)*log(x))/(exp(x) + 1)^2)/( log(x) - 2) - ((exp(5)*(x + 1))/2 - exp(x + 5)*(x/2 - 1/2))/(exp(2*x) + 2* exp(x) + 1)
Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {3 e^5+2 e^{5+2 x}+e^{5+x} (5-2 x)+e^{5+x} x \log (x)}{4 x+8 e^x x+4 e^{2 x} x+\left (-4 x-8 e^x x-4 e^{2 x} x\right ) \log (x)+\left (x+2 e^x x+e^{2 x} x\right ) \log ^2(x)} \, dx=-\frac {e^{5} \left (e^{x} \mathrm {log}\left (x \right )+\mathrm {log}\left (x \right )+1\right )}{e^{x} \mathrm {log}\left (x \right )-2 e^{x}+\mathrm {log}\left (x \right )-2} \] Input:
int((x*exp(5)*exp(x)*log(x)+2*exp(5)*exp(x)^2+(5-2*x)*exp(5)*exp(x)+3*exp( 5))/((x*exp(x)^2+2*exp(x)*x+x)*log(x)^2+(-4*x*exp(x)^2-8*exp(x)*x-4*x)*log (x)+4*x*exp(x)^2+8*exp(x)*x+4*x),x)
Output:
( - e**5*(e**x*log(x) + log(x) + 1))/(e**x*log(x) - 2*e**x + log(x) - 2)