Integrand size = 73, antiderivative size = 25 \[ \int \frac {2 e^x+e^{5 e^{-x} x} \left (5 x-5 x^2+2 e^x x^2\right )+2 e^x x^2 \log \left (x^2\right )}{e^{x+5 e^{-x} x} x+e^x x \log \left (x^2\right )} \, dx=-5+x^2+\log \left (-e^{5 e^{-x} x}-\log \left (x^2\right )\right ) \] Output:
x^2+ln(-exp(5*x/exp(x))-ln(x^2))-5
Time = 1.38 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {2 e^x+e^{5 e^{-x} x} \left (5 x-5 x^2+2 e^x x^2\right )+2 e^x x^2 \log \left (x^2\right )}{e^{x+5 e^{-x} x} x+e^x x \log \left (x^2\right )} \, dx=x^2+\log \left (e^{5 e^{-x} x}+\log \left (x^2\right )\right ) \] Input:
Integrate[(2*E^x + E^((5*x)/E^x)*(5*x - 5*x^2 + 2*E^x*x^2) + 2*E^x*x^2*Log [x^2])/(E^(x + (5*x)/E^x)*x + E^x*x*Log[x^2]),x]
Output:
x^2 + Log[E^((5*x)/E^x) + Log[x^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.
\(\displaystyle \int \frac {e^{5 e^{-x} x} \left (2 e^x x^2-5 x^2+5 x\right )+2 e^x x^2 \log \left (x^2\right )+2 e^x}{e^x x \log \left (x^2\right )+e^{5 e^{-x} x+x} x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-x} \left (e^{5 e^{-x} x} \left (2 e^x x^2-5 x^2+5 x\right )+2 e^x x^2 \log \left (x^2\right )+2 e^x\right )}{x \left (\log \left (x^2\right )+e^{5 e^{-x} x}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{-x} \left (5 x^2 \log \left (x^2\right )-5 x \log \left (x^2\right )+2 e^x\right )}{x \left (\log \left (x^2\right )+e^{5 e^{-x} x}\right )}-5 e^{-x} x+2 x+5 e^{-x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \int \frac {1}{x \left (\log \left (x^2\right )+e^{5 e^{-x} x}\right )}dx-5 \int \frac {e^{-x} \log \left (x^2\right )}{\log \left (x^2\right )+e^{5 e^{-x} x}}dx+5 \int \frac {e^{-x} x \log \left (x^2\right )}{\log \left (x^2\right )+e^{5 e^{-x} x}}dx+x^2+5 e^{-x} x\) |
Input:
Int[(2*E^x + E^((5*x)/E^x)*(5*x - 5*x^2 + 2*E^x*x^2) + 2*E^x*x^2*Log[x^2]) /(E^(x + (5*x)/E^x)*x + E^x*x*Log[x^2]),x]
Output:
$Aborted
Time = 0.61 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76
method | result | size |
parallelrisch | \(x^{2}+\ln \left ({\mathrm e}^{5 \,{\mathrm e}^{-x} x}+\ln \left (x^{2}\right )\right )\) | \(19\) |
risch | \(x^{2}+\ln \left ({\mathrm e}^{5 \,{\mathrm e}^{-x} x}-\frac {i \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i \ln \left (x \right )\right )}{2}\right )\) | \(68\) |
Input:
int(((2*exp(x)*x^2-5*x^2+5*x)*exp(5*x/exp(x))+2*x^2*exp(x)*ln(x^2)+2*exp(x ))/(x*exp(x)*exp(5*x/exp(x))+x*exp(x)*ln(x^2)),x,method=_RETURNVERBOSE)
Output:
x^2+ln(exp(5*x/exp(x))+ln(x^2))
Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {2 e^x+e^{5 e^{-x} x} \left (5 x-5 x^2+2 e^x x^2\right )+2 e^x x^2 \log \left (x^2\right )}{e^{x+5 e^{-x} x} x+e^x x \log \left (x^2\right )} \, dx=x^{2} - x + \log \left (e^{x} \log \left (x^{2}\right ) + e^{\left ({\left (x e^{x} + 5 \, x\right )} e^{\left (-x\right )}\right )}\right ) \] Input:
integrate(((2*exp(x)*x^2-5*x^2+5*x)*exp(5*x/exp(x))+2*x^2*exp(x)*log(x^2)+ 2*exp(x))/(x*exp(x)*exp(5*x/exp(x))+x*exp(x)*log(x^2)),x, algorithm="frica s")
Output:
x^2 - x + log(e^x*log(x^2) + e^((x*e^x + 5*x)*e^(-x)))
Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {2 e^x+e^{5 e^{-x} x} \left (5 x-5 x^2+2 e^x x^2\right )+2 e^x x^2 \log \left (x^2\right )}{e^{x+5 e^{-x} x} x+e^x x \log \left (x^2\right )} \, dx=x^{2} + \log {\left (e^{5 x e^{- x}} + \log {\left (x^{2} \right )} \right )} \] Input:
integrate(((2*exp(x)*x**2-5*x**2+5*x)*exp(5*x/exp(x))+2*x**2*exp(x)*ln(x** 2)+2*exp(x))/(x*exp(x)*exp(5*x/exp(x))+x*exp(x)*ln(x**2)),x)
Output:
x**2 + log(exp(5*x*exp(-x)) + log(x**2))
Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {2 e^x+e^{5 e^{-x} x} \left (5 x-5 x^2+2 e^x x^2\right )+2 e^x x^2 \log \left (x^2\right )}{e^{x+5 e^{-x} x} x+e^x x \log \left (x^2\right )} \, dx=x^{2} + \log \left (e^{\left (5 \, x e^{\left (-x\right )}\right )} + 2 \, \log \left (x\right )\right ) \] Input:
integrate(((2*exp(x)*x^2-5*x^2+5*x)*exp(5*x/exp(x))+2*x^2*exp(x)*log(x^2)+ 2*exp(x))/(x*exp(x)*exp(5*x/exp(x))+x*exp(x)*log(x^2)),x, algorithm="maxim a")
Output:
x^2 + log(e^(5*x*e^(-x)) + 2*log(x))
\[ \int \frac {2 e^x+e^{5 e^{-x} x} \left (5 x-5 x^2+2 e^x x^2\right )+2 e^x x^2 \log \left (x^2\right )}{e^{x+5 e^{-x} x} x+e^x x \log \left (x^2\right )} \, dx=\int { \frac {2 \, x^{2} e^{x} \log \left (x^{2}\right ) + {\left (2 \, x^{2} e^{x} - 5 \, x^{2} + 5 \, x\right )} e^{\left (5 \, x e^{\left (-x\right )}\right )} + 2 \, e^{x}}{x e^{x} \log \left (x^{2}\right ) + x e^{\left (5 \, x e^{\left (-x\right )} + x\right )}} \,d x } \] Input:
integrate(((2*exp(x)*x^2-5*x^2+5*x)*exp(5*x/exp(x))+2*x^2*exp(x)*log(x^2)+ 2*exp(x))/(x*exp(x)*exp(5*x/exp(x))+x*exp(x)*log(x^2)),x, algorithm="giac" )
Output:
integrate((2*x^2*e^x*log(x^2) + (2*x^2*e^x - 5*x^2 + 5*x)*e^(5*x*e^(-x)) + 2*e^x)/(x*e^x*log(x^2) + x*e^(5*x*e^(-x) + x)), x)
Time = 2.63 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {2 e^x+e^{5 e^{-x} x} \left (5 x-5 x^2+2 e^x x^2\right )+2 e^x x^2 \log \left (x^2\right )}{e^{x+5 e^{-x} x} x+e^x x \log \left (x^2\right )} \, dx=\ln \left (\ln \left (x^2\right )+{\mathrm {e}}^{5\,x\,{\mathrm {e}}^{-x}}\right )+x^2 \] Input:
int((2*exp(x) + exp(5*x*exp(-x))*(5*x + 2*x^2*exp(x) - 5*x^2) + 2*x^2*log( x^2)*exp(x))/(x*exp(5*x*exp(-x))*exp(x) + x*log(x^2)*exp(x)),x)
Output:
log(log(x^2) + exp(5*x*exp(-x))) + x^2
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {2 e^x+e^{5 e^{-x} x} \left (5 x-5 x^2+2 e^x x^2\right )+2 e^x x^2 \log \left (x^2\right )}{e^{x+5 e^{-x} x} x+e^x x \log \left (x^2\right )} \, dx=\mathrm {log}\left (e^{\frac {5 x}{e^{x}}}+\mathrm {log}\left (x^{2}\right )\right )+x^{2} \] Input:
int(((2*exp(x)*x^2-5*x^2+5*x)*exp(5*x/exp(x))+2*x^2*exp(x)*log(x^2)+2*exp( x))/(x*exp(x)*exp(5*x/exp(x))+x*exp(x)*log(x^2)),x)
Output:
log(e**((5*x)/e**x) + log(x**2)) + x**2