Integrand size = 72, antiderivative size = 25 \[ \int \frac {240+e^{2-x} (60+60 x)+e^x \left (240-30 x+e^{2-x} (60+60 x)\right )+\left (15+15 e^x\right ) \log \left (1+2 e^x+e^{2 x}\right )}{x^2+e^x x^2} \, dx=-\frac {15 \left (4 \left (4+e^{2-x}\right )+\log \left (\left (1+e^x\right )^2\right )\right )}{x} \] Output:
-15*(4*exp(2-x)+16+ln((1+exp(x))^2))/x
Time = 5.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {240+e^{2-x} (60+60 x)+e^x \left (240-30 x+e^{2-x} (60+60 x)\right )+\left (15+15 e^x\right ) \log \left (1+2 e^x+e^{2 x}\right )}{x^2+e^x x^2} \, dx=\frac {15 \left (2 \left (-8-2 e^{2-x}+x\right )-\log \left (\left (1+e^x\right )^2\right )\right )}{x} \] Input:
Integrate[(240 + E^(2 - x)*(60 + 60*x) + E^x*(240 - 30*x + E^(2 - x)*(60 + 60*x)) + (15 + 15*E^x)*Log[1 + 2*E^x + E^(2*x)])/(x^2 + E^x*x^2),x]
Output:
(15*(2*(-8 - 2*E^(2 - x) + x) - Log[(1 + E^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.
\(\displaystyle \int \frac {e^{2-x} (60 x+60)+e^x \left (-30 x+e^{2-x} (60 x+60)+240\right )+\left (15 e^x+15\right ) \log \left (2 e^x+e^{2 x}+1\right )+240}{e^x x^2+x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {60 e^{2-x} (x+1)}{x^2}-\frac {15 \left (2 x-\log \left (\left (e^x+1\right )^2\right )-16\right )}{x^2}+\frac {30}{\left (e^x+1\right ) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 30 \int \frac {1}{\left (1+e^x\right ) x}dx+30 \int \frac {e^x}{e^x x+x}dx-\frac {60 e^{2-x}}{x}-\frac {240}{x}-\frac {15 \log \left (\left (e^x+1\right )^2\right )}{x}-30 \log (x)\) |
Input:
Int[(240 + E^(2 - x)*(60 + 60*x) + E^x*(240 - 30*x + E^(2 - x)*(60 + 60*x) ) + (15 + 15*E^x)*Log[1 + 2*E^x + E^(2*x)])/(x^2 + E^x*x^2),x]
Output:
$Aborted
Time = 0.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16
method | result | size |
parallelrisch | \(-\frac {480+120 \,{\mathrm e}^{2-x}+30 \ln \left ({\mathrm e}^{2 x}+2 \,{\mathrm e}^{x}+1\right )}{2 x}\) | \(29\) |
norman | \(\frac {\left (-15 \ln \left ({\mathrm e}^{2 x}+2 \,{\mathrm e}^{x}+1\right ) {\mathrm e}^{x}-60 \,{\mathrm e}^{2}-240 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x}\) | \(33\) |
default | \(\frac {\left (\left (-15 \ln \left (\left ({\mathrm e}^{x}+1\right )^{2}\right )+30 \ln \left ({\mathrm e}^{x}+1\right )-240\right ) {\mathrm e}^{x}-30 \,{\mathrm e}^{x} \ln \left ({\mathrm e}^{x}+1\right )-60 \,{\mathrm e}^{2}\right ) {\mathrm e}^{-x}}{x}\) | \(44\) |
risch | \(\frac {\frac {15 i \operatorname {csgn}\left (i \left ({\mathrm e}^{x}+1\right )^{2}\right ) {\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+1\right )\right )}^{2} \pi }{2}-15 i {\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+1\right )^{2}\right )}^{2} \operatorname {csgn}\left (i \left ({\mathrm e}^{x}+1\right )\right ) \pi +\frac {15 i {\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+1\right )^{2}\right )}^{3} \pi }{2}-60 \,{\mathrm e}^{2-x}-240-30 \ln \left ({\mathrm e}^{x}+1\right )}{x}\) | \(87\) |
Input:
int(((15*exp(x)+15)*ln(exp(x)^2+2*exp(x)+1)+((60*x+60)*exp(2-x)-30*x+240)* exp(x)+(60*x+60)*exp(2-x)+240)/(exp(x)*x^2+x^2),x,method=_RETURNVERBOSE)
Output:
-1/2*(480+120*exp(2-x)+30*ln(exp(x)^2+2*exp(x)+1))/x
Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {240+e^{2-x} (60+60 x)+e^x \left (240-30 x+e^{2-x} (60+60 x)\right )+\left (15+15 e^x\right ) \log \left (1+2 e^x+e^{2 x}\right )}{x^2+e^x x^2} \, dx=-\frac {15 \, {\left (e^{x} \log \left (e^{\left (2 \, x\right )} + 2 \, e^{x} + 1\right ) + 4 \, e^{2} + 16 \, e^{x}\right )} e^{\left (-x\right )}}{x} \] Input:
integrate(((15*exp(x)+15)*log(exp(x)^2+2*exp(x)+1)+((60*x+60)*exp(2-x)-30* x+240)*exp(x)+(60*x+60)*exp(2-x)+240)/(exp(x)*x^2+x^2),x, algorithm="frica s")
Output:
-15*(e^x*log(e^(2*x) + 2*e^x + 1) + 4*e^2 + 16*e^x)*e^(-x)/x
Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {240+e^{2-x} (60+60 x)+e^x \left (240-30 x+e^{2-x} (60+60 x)\right )+\left (15+15 e^x\right ) \log \left (1+2 e^x+e^{2 x}\right )}{x^2+e^x x^2} \, dx=- \frac {15 \log {\left (e^{2 x} + 2 e^{x} + 1 \right )}}{x} - \frac {240}{x} - \frac {60 e^{2} e^{- x}}{x} \] Input:
integrate(((15*exp(x)+15)*ln(exp(x)**2+2*exp(x)+1)+((60*x+60)*exp(2-x)-30* x+240)*exp(x)+(60*x+60)*exp(2-x)+240)/(exp(x)*x**2+x**2),x)
Output:
-15*log(exp(2*x) + 2*exp(x) + 1)/x - 240/x - 60*exp(2)*exp(-x)/x
Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {240+e^{2-x} (60+60 x)+e^x \left (240-30 x+e^{2-x} (60+60 x)\right )+\left (15+15 e^x\right ) \log \left (1+2 e^x+e^{2 x}\right )}{x^2+e^x x^2} \, dx=-\frac {30 \, {\left (2 \, e^{\left (-x + 2\right )} + \log \left (e^{x} + 1\right ) + 8\right )}}{x} \] Input:
integrate(((15*exp(x)+15)*log(exp(x)^2+2*exp(x)+1)+((60*x+60)*exp(2-x)-30* x+240)*exp(x)+(60*x+60)*exp(2-x)+240)/(exp(x)*x^2+x^2),x, algorithm="maxim a")
Output:
-30*(2*e^(-x + 2) + log(e^x + 1) + 8)/x
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {240+e^{2-x} (60+60 x)+e^x \left (240-30 x+e^{2-x} (60+60 x)\right )+\left (15+15 e^x\right ) \log \left (1+2 e^x+e^{2 x}\right )}{x^2+e^x x^2} \, dx=-\frac {30 \, {\left (e^{x} \log \left (e^{x} + 1\right ) + 2 \, e^{2} + 8 \, e^{x}\right )} e^{\left (-x\right )}}{x} \] Input:
integrate(((15*exp(x)+15)*log(exp(x)^2+2*exp(x)+1)+((60*x+60)*exp(2-x)-30* x+240)*exp(x)+(60*x+60)*exp(2-x)+240)/(exp(x)*x^2+x^2),x, algorithm="giac" )
Output:
-30*(e^x*log(e^x + 1) + 2*e^2 + 8*e^x)*e^(-x)/x
Time = 3.82 (sec) , antiderivative size = 102, normalized size of antiderivative = 4.08 \[ \int \frac {240+e^{2-x} (60+60 x)+e^x \left (240-30 x+e^{2-x} (60+60 x)\right )+\left (15+15 e^x\right ) \log \left (1+2 e^x+e^{2 x}\right )}{x^2+e^x x^2} \, dx=-\frac {240\,{\mathrm {e}}^{3\,x}+60\,{\mathrm {e}}^2+{\mathrm {e}}^x\,\left (120\,{\mathrm {e}}^2+240\right )+15\,\ln \left ({\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1\right )\,{\mathrm {e}}^x+{\mathrm {e}}^{2\,x}\,\left (60\,{\mathrm {e}}^2+480\right )+30\,\ln \left ({\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1\right )\,{\mathrm {e}}^{2\,x}+15\,\ln \left ({\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1\right )\,{\mathrm {e}}^{3\,x}}{2\,x\,{\mathrm {e}}^{2\,x}+x\,{\mathrm {e}}^{3\,x}+x\,{\mathrm {e}}^x} \] Input:
int((exp(2 - x)*(60*x + 60) + log(exp(2*x) + 2*exp(x) + 1)*(15*exp(x) + 15 ) + exp(x)*(exp(2 - x)*(60*x + 60) - 30*x + 240) + 240)/(x^2*exp(x) + x^2) ,x)
Output:
-(240*exp(3*x) + 60*exp(2) + exp(x)*(120*exp(2) + 240) + 15*log(exp(2*x) + 2*exp(x) + 1)*exp(x) + exp(2*x)*(60*exp(2) + 480) + 30*log(exp(2*x) + 2*e xp(x) + 1)*exp(2*x) + 15*log(exp(2*x) + 2*exp(x) + 1)*exp(3*x))/(2*x*exp(2 *x) + x*exp(3*x) + x*exp(x))
Time = 0.16 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {240+e^{2-x} (60+60 x)+e^x \left (240-30 x+e^{2-x} (60+60 x)\right )+\left (15+15 e^x\right ) \log \left (1+2 e^x+e^{2 x}\right )}{x^2+e^x x^2} \, dx=\frac {-15 e^{x} \mathrm {log}\left (e^{2 x}+2 e^{x}+1\right )-240 e^{x}-60 e^{2}}{e^{x} x} \] Input:
int(((15*exp(x)+15)*log(exp(x)^2+2*exp(x)+1)+((60*x+60)*exp(2-x)-30*x+240) *exp(x)+(60*x+60)*exp(2-x)+240)/(exp(x)*x^2+x^2),x)
Output:
(15*( - e**x*log(e**(2*x) + 2*e**x + 1) - 16*e**x - 4*e**2))/(e**x*x)