Integrand size = 54, antiderivative size = 21 \[ \int e^{-x-e^{-x} \left (10 x^2+2 x^3\right ) \log ^2(5)} \left (e^x+\left (-20 x^2+4 x^3+2 x^4\right ) \log ^2(5)\right ) \, dx=e^{-2 e^{-x} x^2 (5+x) \log ^2(5)} x \] Output:
x/exp(2*x^2*ln(5)^2/exp(x)*(5+x))
Time = 0.55 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int e^{-x-e^{-x} \left (10 x^2+2 x^3\right ) \log ^2(5)} \left (e^x+\left (-20 x^2+4 x^3+2 x^4\right ) \log ^2(5)\right ) \, dx=e^{-2 e^{-x} x^2 (5+x) \log ^2(5)} x \] Input:
Integrate[E^(-x - ((10*x^2 + 2*x^3)*Log[5]^2)/E^x)*(E^x + (-20*x^2 + 4*x^3 + 2*x^4)*Log[5]^2),x]
Output:
x/E^((2*x^2*(5 + x)*Log[5]^2)/E^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.
\(\displaystyle \int e^{-e^{-x} \left (2 x^3+10 x^2\right ) \log ^2(5)-x} \left (\left (2 x^4+4 x^3-20 x^2\right ) \log ^2(5)+e^x\right ) \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \left (\left (2 x^4+4 x^3-20 x^2\right ) \log ^2(5)+e^x\right ) \exp \left (-e^{-x} x \left (2 x^2 \log ^2(5)+e^x+10 x \log ^2(5)\right )\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (2 \left (x^2+2 x-10\right ) x^2 \log ^2(5) \exp \left (-e^{-x} x \left (2 x^2 \log ^2(5)+e^x+10 x \log ^2(5)\right )\right )+\exp \left (x-e^{-x} x \left (2 x^2 \log ^2(5)+e^x+10 x \log ^2(5)\right )\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -20 \log ^2(5) \int \exp \left (-e^{-x} x \left (2 \log ^2(5) x^2+10 \log ^2(5) x+e^x\right )\right ) x^2dx+2 \log ^2(5) \int \exp \left (-e^{-x} x \left (2 \log ^2(5) x^2+10 \log ^2(5) x+e^x\right )\right ) x^4dx+4 \log ^2(5) \int \exp \left (-e^{-x} x \left (2 \log ^2(5) x^2+10 \log ^2(5) x+e^x\right )\right ) x^3dx+\int e^{-2 e^{-x} x^2 (x+5) \log ^2(5)}dx\) |
Input:
Int[E^(-x - ((10*x^2 + 2*x^3)*Log[5]^2)/E^x)*(E^x + (-20*x^2 + 4*x^3 + 2*x ^4)*Log[5]^2),x]
Output:
$Aborted
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95
method | result | size |
risch | \(x \,{\mathrm e}^{-2 x^{2} \left (5+x \right ) \ln \left (5\right )^{2} {\mathrm e}^{-x}}\) | \(20\) |
parallelrisch | \(x \,{\mathrm e}^{-2 x^{2} \left (5+x \right ) \ln \left (5\right )^{2} {\mathrm e}^{-x}}\) | \(22\) |
norman | \(x \,{\mathrm e}^{-\left (2 x^{3}+10 x^{2}\right ) \ln \left (5\right )^{2} {\mathrm e}^{-x}}\) | \(26\) |
Input:
int((exp(x)+(2*x^4+4*x^3-20*x^2)*ln(5)^2)/exp(x)/exp((2*x^3+10*x^2)*ln(5)^ 2/exp(x)),x,method=_RETURNVERBOSE)
Output:
x*exp(-2*x^2*(5+x)*ln(5)^2*exp(-x))
Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int e^{-x-e^{-x} \left (10 x^2+2 x^3\right ) \log ^2(5)} \left (e^x+\left (-20 x^2+4 x^3+2 x^4\right ) \log ^2(5)\right ) \, dx=x e^{\left (-{\left (2 \, {\left (x^{3} + 5 \, x^{2}\right )} \log \left (5\right )^{2} + x e^{x}\right )} e^{\left (-x\right )} + x\right )} \] Input:
integrate((exp(x)+(2*x^4+4*x^3-20*x^2)*log(5)^2)/exp(x)/exp((2*x^3+10*x^2) *log(5)^2/exp(x)),x, algorithm="fricas")
Output:
x*e^(-(2*(x^3 + 5*x^2)*log(5)^2 + x*e^x)*e^(-x) + x)
Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int e^{-x-e^{-x} \left (10 x^2+2 x^3\right ) \log ^2(5)} \left (e^x+\left (-20 x^2+4 x^3+2 x^4\right ) \log ^2(5)\right ) \, dx=x e^{- \left (2 x^{3} + 10 x^{2}\right ) e^{- x} \log {\left (5 \right )}^{2}} \] Input:
integrate((exp(x)+(2*x**4+4*x**3-20*x**2)*ln(5)**2)/exp(x)/exp((2*x**3+10* x**2)*ln(5)**2/exp(x)),x)
Output:
x*exp(-(2*x**3 + 10*x**2)*exp(-x)*log(5)**2)
\[ \int e^{-x-e^{-x} \left (10 x^2+2 x^3\right ) \log ^2(5)} \left (e^x+\left (-20 x^2+4 x^3+2 x^4\right ) \log ^2(5)\right ) \, dx=\int { {\left (2 \, {\left (x^{4} + 2 \, x^{3} - 10 \, x^{2}\right )} \log \left (5\right )^{2} + e^{x}\right )} e^{\left (-2 \, {\left (x^{3} + 5 \, x^{2}\right )} e^{\left (-x\right )} \log \left (5\right )^{2} - x\right )} \,d x } \] Input:
integrate((exp(x)+(2*x^4+4*x^3-20*x^2)*log(5)^2)/exp(x)/exp((2*x^3+10*x^2) *log(5)^2/exp(x)),x, algorithm="maxima")
Output:
integrate((2*(x^4 + 2*x^3 - 10*x^2)*log(5)^2 + e^x)*e^(-2*(x^3 + 5*x^2)*e^ (-x)*log(5)^2 - x), x)
\[ \int e^{-x-e^{-x} \left (10 x^2+2 x^3\right ) \log ^2(5)} \left (e^x+\left (-20 x^2+4 x^3+2 x^4\right ) \log ^2(5)\right ) \, dx=\int { {\left (2 \, {\left (x^{4} + 2 \, x^{3} - 10 \, x^{2}\right )} \log \left (5\right )^{2} + e^{x}\right )} e^{\left (-2 \, {\left (x^{3} + 5 \, x^{2}\right )} e^{\left (-x\right )} \log \left (5\right )^{2} - x\right )} \,d x } \] Input:
integrate((exp(x)+(2*x^4+4*x^3-20*x^2)*log(5)^2)/exp(x)/exp((2*x^3+10*x^2) *log(5)^2/exp(x)),x, algorithm="giac")
Output:
integrate((2*(x^4 + 2*x^3 - 10*x^2)*log(5)^2 + e^x)*e^(-2*(x^3 + 5*x^2)*e^ (-x)*log(5)^2 - x), x)
Time = 3.79 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int e^{-x-e^{-x} \left (10 x^2+2 x^3\right ) \log ^2(5)} \left (e^x+\left (-20 x^2+4 x^3+2 x^4\right ) \log ^2(5)\right ) \, dx=x\,{\mathrm {e}}^{-2\,x^3\,{\mathrm {e}}^{-x}\,{\ln \left (5\right )}^2}\,{\mathrm {e}}^{-10\,x^2\,{\mathrm {e}}^{-x}\,{\ln \left (5\right )}^2} \] Input:
int(exp(-x)*exp(-exp(-x)*log(5)^2*(10*x^2 + 2*x^3))*(exp(x) + log(5)^2*(4* x^3 - 20*x^2 + 2*x^4)),x)
Output:
x*exp(-2*x^3*exp(-x)*log(5)^2)*exp(-10*x^2*exp(-x)*log(5)^2)
Time = 0.19 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48 \[ \int e^{-x-e^{-x} \left (10 x^2+2 x^3\right ) \log ^2(5)} \left (e^x+\left (-20 x^2+4 x^3+2 x^4\right ) \log ^2(5)\right ) \, dx=\frac {x}{e^{\frac {2 \mathrm {log}\left (5\right )^{2} x^{3}+10 \mathrm {log}\left (5\right )^{2} x^{2}}{e^{x}}}} \] Input:
int((exp(x)+(2*x^4+4*x^3-20*x^2)*log(5)^2)/exp(x)/exp((2*x^3+10*x^2)*log(5 )^2/exp(x)),x)
Output:
x/e**((2*log(5)**2*x**3 + 10*log(5)**2*x**2)/e**x)