Integrand size = 49, antiderivative size = 24 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=-2 e^{-x}+e^{9 e^{-2 x} x^2 \log ^2(x)} \] Output:
-2/exp(x)+exp(9*x^2*ln(x)^2/exp(x)^2)
Time = 0.69 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=2 \left (-e^{-x}+\frac {1}{2} e^{9 e^{-2 x} x^2 \log ^2(x)}\right ) \] Input:
Integrate[(2*E^x + E^((9*x^2*Log[x]^2)/E^(2*x))*(18*x*Log[x] + (18*x - 18* x^2)*Log[x]^2))/E^(2*x),x]
Output:
2*(-E^(-x) + E^((9*x^2*Log[x]^2)/E^(2*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.
\(\displaystyle \int e^{-2 x} \left (e^{9 e^{-2 x} x^2 \log ^2(x)} \left (\left (18 x-18 x^2\right ) \log ^2(x)+18 x \log (x)\right )+2 e^x\right ) \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (2 e^{-x}-18 x e^{9 e^{-2 x} x^2 \log ^2(x)-2 x} \log (x) (x \log (x)-\log (x)-1)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 18 \int e^{9 e^{-2 x} x^2 \log ^2(x)-2 x} x \log (x)dx+18 \int e^{9 e^{-2 x} x^2 \log ^2(x)-2 x} x \log ^2(x)dx-18 \int e^{9 e^{-2 x} x^2 \log ^2(x)-2 x} x^2 \log ^2(x)dx-2 e^{-x}\) |
Input:
Int[(2*E^x + E^((9*x^2*Log[x]^2)/E^(2*x))*(18*x*Log[x] + (18*x - 18*x^2)*L og[x]^2))/E^(2*x),x]
Output:
$Aborted
Time = 0.58 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
risch | \(-2 \,{\mathrm e}^{-x}+{\mathrm e}^{9 x^{2} \ln \left (x \right )^{2} {\mathrm e}^{-2 x}}\) | \(22\) |
parallelrisch | \(-{\mathrm e}^{-x} \left (2-{\mathrm e}^{9 x^{2} \ln \left (x \right )^{2} {\mathrm e}^{-2 x}} {\mathrm e}^{x}\right )\) | \(27\) |
Input:
int((((-18*x^2+18*x)*ln(x)^2+18*x*ln(x))*exp(9*x^2*ln(x)^2/exp(x)^2)+2*exp (x))/exp(x)^2,x,method=_RETURNVERBOSE)
Output:
-2*exp(-x)+exp(9*x^2*ln(x)^2*exp(-2*x))
Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx={\left (e^{\left (9 \, x^{2} e^{\left (-2 \, x\right )} \log \left (x\right )^{2} + x\right )} - 2\right )} e^{\left (-x\right )} \] Input:
integrate((((-18*x^2+18*x)*log(x)^2+18*x*log(x))*exp(9*x^2*log(x)^2/exp(x) ^2)+2*exp(x))/exp(x)^2,x, algorithm="fricas")
Output:
(e^(9*x^2*e^(-2*x)*log(x)^2 + x) - 2)*e^(-x)
Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=e^{9 x^{2} e^{- 2 x} \log {\left (x \right )}^{2}} - 2 e^{- x} \] Input:
integrate((((-18*x**2+18*x)*ln(x)**2+18*x*ln(x))*exp(9*x**2*ln(x)**2/exp(x )**2)+2*exp(x))/exp(x)**2,x)
Output:
exp(9*x**2*exp(-2*x)*log(x)**2) - 2*exp(-x)
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=e^{\left (9 \, x^{2} e^{\left (-2 \, x\right )} \log \left (x\right )^{2}\right )} - 2 \, e^{\left (-x\right )} \] Input:
integrate((((-18*x^2+18*x)*log(x)^2+18*x*log(x))*exp(9*x^2*log(x)^2/exp(x) ^2)+2*exp(x))/exp(x)^2,x, algorithm="maxima")
Output:
e^(9*x^2*e^(-2*x)*log(x)^2) - 2*e^(-x)
\[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=\int { -2 \, {\left (9 \, {\left ({\left (x^{2} - x\right )} \log \left (x\right )^{2} - x \log \left (x\right )\right )} e^{\left (9 \, x^{2} e^{\left (-2 \, x\right )} \log \left (x\right )^{2}\right )} - e^{x}\right )} e^{\left (-2 \, x\right )} \,d x } \] Input:
integrate((((-18*x^2+18*x)*log(x)^2+18*x*log(x))*exp(9*x^2*log(x)^2/exp(x) ^2)+2*exp(x))/exp(x)^2,x, algorithm="giac")
Output:
integrate(-2*(9*((x^2 - x)*log(x)^2 - x*log(x))*e^(9*x^2*e^(-2*x)*log(x)^2 ) - e^x)*e^(-2*x), x)
Time = 3.41 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx={\mathrm {e}}^{9\,x^2\,{\mathrm {e}}^{-2\,x}\,{\ln \left (x\right )}^2}-2\,{\mathrm {e}}^{-x} \] Input:
int(exp(-2*x)*(2*exp(x) + exp(9*x^2*exp(-2*x)*log(x)^2)*(log(x)^2*(18*x - 18*x^2) + 18*x*log(x))),x)
Output:
exp(9*x^2*exp(-2*x)*log(x)^2) - 2*exp(-x)
\[ \int e^{-2 x} \left (2 e^x+e^{9 e^{-2 x} x^2 \log ^2(x)} \left (18 x \log (x)+\left (18 x-18 x^2\right ) \log ^2(x)\right )\right ) \, dx=\frac {-18 e^{x} \left (\int \frac {e^{\frac {9 \mathrm {log}\left (x \right )^{2} x^{2}}{e^{2 x}}} \mathrm {log}\left (x \right )^{2} x^{2}}{e^{2 x}}d x \right )+18 e^{x} \left (\int \frac {e^{\frac {9 \mathrm {log}\left (x \right )^{2} x^{2}}{e^{2 x}}} \mathrm {log}\left (x \right )^{2} x}{e^{2 x}}d x \right )+18 e^{x} \left (\int \frac {e^{\frac {9 \mathrm {log}\left (x \right )^{2} x^{2}}{e^{2 x}}} \mathrm {log}\left (x \right ) x}{e^{2 x}}d x \right )-2}{e^{x}} \] Input:
int((((-18*x^2+18*x)*log(x)^2+18*x*log(x))*exp(9*x^2*log(x)^2/exp(x)^2)+2* exp(x))/exp(x)^2,x)
Output:
(2*( - 9*e**x*int((e**((9*log(x)**2*x**2)/e**(2*x))*log(x)**2*x**2)/e**(2* x),x) + 9*e**x*int((e**((9*log(x)**2*x**2)/e**(2*x))*log(x)**2*x)/e**(2*x) ,x) + 9*e**x*int((e**((9*log(x)**2*x**2)/e**(2*x))*log(x)*x)/e**(2*x),x) - 1))/e**x