Integrand size = 128, antiderivative size = 28 \[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=e^{-\frac {5}{5+e^x}-x+x^2+x (-1+5 \log (x))^2} \] Output:
exp(x^2-x-5/(exp(x)+5)+(5*ln(x)-1)^2*x)
Time = 0.19 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=e^{-\frac {5}{5+e^x}+x^2+25 x \log ^2(x)} x^{-10 x} \] Input:
Integrate[(E^((-5 + 5*x^2 + E^x*x^2 + (-50*x - 10*E^x*x)*Log[x] + (125*x + 25*E^x*x)*Log[x]^2)/(5 + E^x))*(-250 + 50*x + E^(2*x)*(-10 + 2*x) + E^x*( -95 + 20*x) + (1000 + 400*E^x + 40*E^(2*x))*Log[x] + (625 + 250*E^x + 25*E ^(2*x))*Log[x]^2))/(25 + 10*E^x + E^(2*x)),x]
Output:
E^(-5/(5 + E^x) + x^2 + 25*x*Log[x]^2)/x^(10*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.
| \(\displaystyle \int \frac {\left (50 x+e^{2 x} (2 x-10)+e^x (20 x-95)+\left (250 e^x+25 e^{2 x}+625\right ) \log ^2(x)+\left (400 e^x+40 e^{2 x}+1000\right ) \log (x)-250\right ) \exp \left (\frac {e^x x^2+5 x^2+\left (25 e^x x+125 x\right ) \log ^2(x)+\left (-10 e^x x-50 x\right ) \log (x)-5}{e^x+5}\right )}{10 e^x+e^{2 x}+25} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (50 x+e^{2 x} (2 x-10)+e^x (20 x-95)+\left (250 e^x+25 e^{2 x}+625\right ) \log ^2(x)+\left (400 e^x+40 e^{2 x}+1000\right ) \log (x)-250\right ) \exp \left (\frac {e^x x^2+5 x^2+\left (25 e^x x+125 x\right ) \log ^2(x)+\left (-10 e^x x-50 x\right ) \log (x)-5}{e^x+5}\right )}{\left (e^x+5\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (25 \log ^2(x) \exp \left (\frac {e^x x^2+5 x^2+\left (25 e^x x+125 x\right ) \log ^2(x)+\left (-10 e^x x-50 x\right ) \log (x)-5}{e^x+5}\right )+40 \log (x) \exp \left (\frac {e^x x^2+5 x^2+\left (25 e^x x+125 x\right ) \log ^2(x)+\left (-10 e^x x-50 x\right ) \log (x)-5}{e^x+5}\right )-10 \exp \left (\frac {e^x x^2+5 x^2+\left (25 e^x x+125 x\right ) \log ^2(x)+\left (-10 e^x x-50 x\right ) \log (x)-5}{e^x+5}\right )+2 x \exp \left (\frac {e^x x^2+5 x^2+\left (25 e^x x+125 x\right ) \log ^2(x)+\left (-10 e^x x-50 x\right ) \log (x)-5}{e^x+5}\right )+\frac {5 \exp \left (\frac {e^x x^2+5 x^2+\left (25 e^x x+125 x\right ) \log ^2(x)+\left (-10 e^x x-50 x\right ) \log (x)-5}{e^x+5}\right )}{e^x+5}-\frac {25 \exp \left (\frac {e^x x^2+5 x^2+\left (25 e^x x+125 x\right ) \log ^2(x)+\left (-10 e^x x-50 x\right ) \log (x)-5}{e^x+5}\right )}{\left (e^x+5\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \int e^{x^2+25 \log ^2(x) x-\frac {5}{5+e^x}} x^{1-10 x}dx-10 \int e^{x^2+25 \log ^2(x) x-\frac {5}{5+e^x}} x^{-10 x}dx-25 \int \frac {e^{x^2+25 \log ^2(x) x-\frac {5}{5+e^x}} x^{-10 x}}{\left (5+e^x\right )^2}dx+5 \int \frac {e^{x^2+25 \log ^2(x) x-\frac {5}{5+e^x}} x^{-10 x}}{5+e^x}dx+40 \int e^{x^2+25 \log ^2(x) x-\frac {5}{5+e^x}} x^{-10 x} \log (x)dx+25 \int e^{x^2+25 \log ^2(x) x-\frac {5}{5+e^x}} x^{-10 x} \log ^2(x)dx\) |
Input:
Int[(E^((-5 + 5*x^2 + E^x*x^2 + (-50*x - 10*E^x*x)*Log[x] + (125*x + 25*E^ x*x)*Log[x]^2)/(5 + E^x))*(-250 + 50*x + E^(2*x)*(-10 + 2*x) + E^x*(-95 + 20*x) + (1000 + 400*E^x + 40*E^(2*x))*Log[x] + (625 + 250*E^x + 25*E^(2*x) )*Log[x]^2))/(25 + 10*E^x + E^(2*x)),x]
Output:
$Aborted
Time = 31.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.71
| method | result | size |
| parallelrisch | \({\mathrm e}^{\frac {\left (25 \,{\mathrm e}^{x} x +125 x \right ) \ln \left (x \right )^{2}+\left (-10 \,{\mathrm e}^{x} x -50 x \right ) \ln \left (x \right )+{\mathrm e}^{x} x^{2}+5 x^{2}-5}{{\mathrm e}^{x}+5}}\) | \(48\) |
| risch | \({\mathrm e}^{\frac {25 x \,{\mathrm e}^{x} \ln \left (x \right )^{2}-10 x \,{\mathrm e}^{x} \ln \left (x \right )+{\mathrm e}^{x} x^{2}+125 x \ln \left (x \right )^{2}-50 x \ln \left (x \right )+5 x^{2}-5}{{\mathrm e}^{x}+5}}\) | \(50\) |
Input:
int(((25*exp(x)^2+250*exp(x)+625)*ln(x)^2+(40*exp(x)^2+400*exp(x)+1000)*ln (x)+(2*x-10)*exp(x)^2+(20*x-95)*exp(x)+50*x-250)*exp(((25*exp(x)*x+125*x)* ln(x)^2+(-10*exp(x)*x-50*x)*ln(x)+exp(x)*x^2+5*x^2-5)/(exp(x)+5))/(exp(x)^ 2+10*exp(x)+25),x,method=_RETURNVERBOSE)
Output:
exp(((25*exp(x)*x+125*x)*ln(x)^2+(-10*exp(x)*x-50*x)*ln(x)+exp(x)*x^2+5*x^ 2-5)/(exp(x)+5))
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.68 \[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=e^{\left (\frac {x^{2} e^{x} + 25 \, {\left (x e^{x} + 5 \, x\right )} \log \left (x\right )^{2} + 5 \, x^{2} - 10 \, {\left (x e^{x} + 5 \, x\right )} \log \left (x\right ) - 5}{e^{x} + 5}\right )} \] Input:
integrate(((25*exp(x)^2+250*exp(x)+625)*log(x)^2+(40*exp(x)^2+400*exp(x)+1 000)*log(x)+(2*x-10)*exp(x)^2+(20*x-95)*exp(x)+50*x-250)*exp(((25*exp(x)*x +125*x)*log(x)^2+(-10*exp(x)*x-50*x)*log(x)+exp(x)*x^2+5*x^2-5)/(exp(x)+5) )/(exp(x)^2+10*exp(x)+25),x, algorithm="fricas")
Output:
e^((x^2*e^x + 25*(x*e^x + 5*x)*log(x)^2 + 5*x^2 - 10*(x*e^x + 5*x)*log(x) - 5)/(e^x + 5))
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (22) = 44\).
Time = 0.43 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=e^{\frac {x^{2} e^{x} + 5 x^{2} + \left (- 10 x e^{x} - 50 x\right ) \log {\left (x \right )} + \left (25 x e^{x} + 125 x\right ) \log {\left (x \right )}^{2} - 5}{e^{x} + 5}} \] Input:
integrate(((25*exp(x)**2+250*exp(x)+625)*ln(x)**2+(40*exp(x)**2+400*exp(x) +1000)*ln(x)+(2*x-10)*exp(x)**2+(20*x-95)*exp(x)+50*x-250)*exp(((25*exp(x) *x+125*x)*ln(x)**2+(-10*exp(x)*x-50*x)*ln(x)+exp(x)*x**2+5*x**2-5)/(exp(x) +5))/(exp(x)**2+10*exp(x)+25),x)
Output:
exp((x**2*exp(x) + 5*x**2 + (-10*x*exp(x) - 50*x)*log(x) + (25*x*exp(x) + 125*x)*log(x)**2 - 5)/(exp(x) + 5))
Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (26) = 52\).
Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.04 \[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=e^{\left (\frac {25 \, x e^{x} \log \left (x\right )^{2}}{e^{x} + 5} + \frac {x^{2} e^{x}}{e^{x} + 5} - \frac {10 \, x e^{x} \log \left (x\right )}{e^{x} + 5} + \frac {125 \, x \log \left (x\right )^{2}}{e^{x} + 5} + \frac {5 \, x^{2}}{e^{x} + 5} - \frac {50 \, x \log \left (x\right )}{e^{x} + 5} - \frac {5}{e^{x} + 5}\right )} \] Input:
integrate(((25*exp(x)^2+250*exp(x)+625)*log(x)^2+(40*exp(x)^2+400*exp(x)+1 000)*log(x)+(2*x-10)*exp(x)^2+(20*x-95)*exp(x)+50*x-250)*exp(((25*exp(x)*x +125*x)*log(x)^2+(-10*exp(x)*x-50*x)*log(x)+exp(x)*x^2+5*x^2-5)/(exp(x)+5) )/(exp(x)^2+10*exp(x)+25),x, algorithm="maxima")
Output:
e^(25*x*e^x*log(x)^2/(e^x + 5) + x^2*e^x/(e^x + 5) - 10*x*e^x*log(x)/(e^x + 5) + 125*x*log(x)^2/(e^x + 5) + 5*x^2/(e^x + 5) - 50*x*log(x)/(e^x + 5) - 5/(e^x + 5))
\[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=\int { \frac {{\left (25 \, {\left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right )} \log \left (x\right )^{2} + 2 \, {\left (x - 5\right )} e^{\left (2 \, x\right )} + 5 \, {\left (4 \, x - 19\right )} e^{x} + 40 \, {\left (e^{\left (2 \, x\right )} + 10 \, e^{x} + 25\right )} \log \left (x\right ) + 50 \, x - 250\right )} e^{\left (\frac {x^{2} e^{x} + 25 \, {\left (x e^{x} + 5 \, x\right )} \log \left (x\right )^{2} + 5 \, x^{2} - 10 \, {\left (x e^{x} + 5 \, x\right )} \log \left (x\right ) - 5}{e^{x} + 5}\right )}}{e^{\left (2 \, x\right )} + 10 \, e^{x} + 25} \,d x } \] Input:
integrate(((25*exp(x)^2+250*exp(x)+625)*log(x)^2+(40*exp(x)^2+400*exp(x)+1 000)*log(x)+(2*x-10)*exp(x)^2+(20*x-95)*exp(x)+50*x-250)*exp(((25*exp(x)*x +125*x)*log(x)^2+(-10*exp(x)*x-50*x)*log(x)+exp(x)*x^2+5*x^2-5)/(exp(x)+5) )/(exp(x)^2+10*exp(x)+25),x, algorithm="giac")
Output:
integrate((25*(e^(2*x) + 10*e^x + 25)*log(x)^2 + 2*(x - 5)*e^(2*x) + 5*(4* x - 19)*e^x + 40*(e^(2*x) + 10*e^x + 25)*log(x) + 50*x - 250)*e^((x^2*e^x + 25*(x*e^x + 5*x)*log(x)^2 + 5*x^2 - 10*(x*e^x + 5*x)*log(x) - 5)/(e^x + 5))/(e^(2*x) + 10*e^x + 25), x)
Time = 3.96 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.57 \[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=\frac {{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^x}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{\frac {125\,x\,{\ln \left (x\right )}^2}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{\frac {5\,x^2}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{-\frac {5}{{\mathrm {e}}^x+5}}\,{\mathrm {e}}^{\frac {25\,x\,{\mathrm {e}}^x\,{\ln \left (x\right )}^2}{{\mathrm {e}}^x+5}}}{x^{10\,x}} \] Input:
int((exp((x^2*exp(x) - log(x)*(50*x + 10*x*exp(x)) + log(x)^2*(125*x + 25* x*exp(x)) + 5*x^2 - 5)/(exp(x) + 5))*(50*x + exp(x)*(20*x - 95) + log(x)*( 40*exp(2*x) + 400*exp(x) + 1000) + exp(2*x)*(2*x - 10) + log(x)^2*(25*exp( 2*x) + 250*exp(x) + 625) - 250))/(exp(2*x) + 10*exp(x) + 25),x)
Output:
(exp((x^2*exp(x))/(exp(x) + 5))*exp((125*x*log(x)^2)/(exp(x) + 5))*exp((5* x^2)/(exp(x) + 5))*exp(-5/(exp(x) + 5))*exp((25*x*exp(x)*log(x)^2)/(exp(x) + 5)))/x^(10*x)
\[ \int \frac {e^{\frac {-5+5 x^2+e^x x^2+\left (-50 x-10 e^x x\right ) \log (x)+\left (125 x+25 e^x x\right ) \log ^2(x)}{5+e^x}} \left (-250+50 x+e^{2 x} (-10+2 x)+e^x (-95+20 x)+\left (1000+400 e^x+40 e^{2 x}\right ) \log (x)+\left (625+250 e^x+25 e^{2 x}\right ) \log ^2(x)\right )}{25+10 e^x+e^{2 x}} \, dx=\int \frac {\left (\left (25 \left ({\mathrm e}^{x}\right )^{2}+250 \,{\mathrm e}^{x}+625\right ) \mathrm {log}\left (x \right )^{2}+\left (40 \left ({\mathrm e}^{x}\right )^{2}+400 \,{\mathrm e}^{x}+1000\right ) \mathrm {log}\left (x \right )+\left (2 x -10\right ) \left ({\mathrm e}^{x}\right )^{2}+\left (20 x -95\right ) {\mathrm e}^{x}+50 x -250\right ) {\mathrm e}^{\frac {\left (25 \,{\mathrm e}^{x} x +125 x \right ) \mathrm {log}\left (x \right )^{2}+\left (-10 \,{\mathrm e}^{x} x -50 x \right ) \mathrm {log}\left (x \right )+{\mathrm e}^{x} x^{2}+5 x^{2}-5}{{\mathrm e}^{x}+5}}}{\left ({\mathrm e}^{x}\right )^{2}+10 \,{\mathrm e}^{x}+25}d x \] Input:
int(((25*exp(x)^2+250*exp(x)+625)*log(x)^2+(40*exp(x)^2+400*exp(x)+1000)*l og(x)+(2*x-10)*exp(x)^2+(20*x-95)*exp(x)+50*x-250)*exp(((25*exp(x)*x+125*x )*log(x)^2+(-10*exp(x)*x-50*x)*log(x)+exp(x)*x^2+5*x^2-5)/(exp(x)+5))/(exp (x)^2+10*exp(x)+25),x)
Output:
int(((25*exp(x)^2+250*exp(x)+625)*log(x)^2+(40*exp(x)^2+400*exp(x)+1000)*l og(x)+(2*x-10)*exp(x)^2+(20*x-95)*exp(x)+50*x-250)*exp(((25*exp(x)*x+125*x )*log(x)^2+(-10*exp(x)*x-50*x)*log(x)+exp(x)*x^2+5*x^2-5)/(exp(x)+5))/(exp (x)^2+10*exp(x)+25),x)