Integrand size = 40, antiderivative size = 24 \[ \int e^{25-19 x+(25-19 x) \log \left (81 x^4\right )} \left (102 x-95 x^2-19 x^2 \log \left (81 x^4\right )\right ) \, dx=e^{(5 (5-4 x)+x) \left (1+\log \left (81 x^4\right )\right )} x^2 \] Output:
exp((-19*x+25)*(ln(81*x^4)+1))*x^2
\[ \int e^{25-19 x+(25-19 x) \log \left (81 x^4\right )} \left (102 x-95 x^2-19 x^2 \log \left (81 x^4\right )\right ) \, dx=\int e^{25-19 x+(25-19 x) \log \left (81 x^4\right )} \left (102 x-95 x^2-19 x^2 \log \left (81 x^4\right )\right ) \, dx \] Input:
Integrate[E^(25 - 19*x + (25 - 19*x)*Log[81*x^4])*(102*x - 95*x^2 - 19*x^2 *Log[81*x^4]),x]
Output:
Integrate[E^(25 - 19*x + (25 - 19*x)*Log[81*x^4])*(102*x - 95*x^2 - 19*x^2 *Log[81*x^4]), 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.
\(\displaystyle \int e^{(25-19 x) \log \left (81 x^4\right )-19 x+25} \left (-95 x^2-19 x^2 \log \left (81 x^4\right )+102 x\right ) \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int e^{-\left ((19 x-25) \left (\log \left (81 x^4\right )+1\right )\right )} \left (-95 x^2-19 x^2 \log \left (81 x^4\right )+102 x\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (102 x e^{-\left ((19 x-25) \left (\log \left (81 x^4\right )+1\right )\right )}-95 x^2 e^{-\left ((19 x-25) \left (\log \left (81 x^4\right )+1\right )\right )}-19 x^2 e^{-\left ((19 x-25) \left (\log \left (81 x^4\right )+1\right )\right )} \log \left (81 x^4\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 102 \int e^{-\left ((19 x-25) \left (\log \left (81 x^4\right )+1\right )\right )} xdx-95 \int e^{-\left ((19 x-25) \left (\log \left (81 x^4\right )+1\right )\right )} x^2dx-19 \int e^{-\left ((19 x-25) \left (\log \left (81 x^4\right )+1\right )\right )} x^2 \log \left (81 x^4\right )dx\) |
Input:
Int[E^(25 - 19*x + (25 - 19*x)*Log[81*x^4])*(102*x - 95*x^2 - 19*x^2*Log[8 1*x^4]),x]
Output:
$Aborted
Time = 0.35 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
risch | \(x^{2} \left (81 x^{4}\right )^{-19 x +25} {\mathrm e}^{-19 x +25}\) | \(22\) |
default | \(x^{2} {\mathrm e}^{\left (-19 x +25\right ) \ln \left (81 x^{4}\right )-19 x +25}\) | \(23\) |
norman | \(x^{2} {\mathrm e}^{\left (-19 x +25\right ) \ln \left (81 x^{4}\right )-19 x +25}\) | \(23\) |
parallelrisch | \(x^{2} {\mathrm e}^{\left (-19 x +25\right ) \ln \left (81 x^{4}\right )-19 x +25}\) | \(23\) |
Input:
int((-19*x^2*ln(81*x^4)-95*x^2+102*x)*exp((-19*x+25)*ln(81*x^4)-19*x+25),x ,method=_RETURNVERBOSE)
Output:
x^2*(81*x^4)^(-19*x+25)*exp(-19*x+25)
Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int e^{25-19 x+(25-19 x) \log \left (81 x^4\right )} \left (102 x-95 x^2-19 x^2 \log \left (81 x^4\right )\right ) \, dx=x^{2} e^{\left (-{\left (19 \, x - 25\right )} \log \left (81 \, x^{4}\right ) - 19 \, x + 25\right )} \] Input:
integrate((-19*x^2*log(81*x^4)-95*x^2+102*x)*exp((-19*x+25)*log(81*x^4)-19 *x+25),x, algorithm="fricas")
Output:
x^2*e^(-(19*x - 25)*log(81*x^4) - 19*x + 25)
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int e^{25-19 x+(25-19 x) \log \left (81 x^4\right )} \left (102 x-95 x^2-19 x^2 \log \left (81 x^4\right )\right ) \, dx=x^{2} e^{- 19 x + \left (25 - 19 x\right ) \log {\left (81 x^{4} \right )} + 25} \] Input:
integrate((-19*x**2*ln(81*x**4)-95*x**2+102*x)*exp((-19*x+25)*ln(81*x**4)- 19*x+25),x)
Output:
x**2*exp(-19*x + (25 - 19*x)*log(81*x**4) + 25)
Time = 0.17 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int e^{25-19 x+(25-19 x) \log \left (81 x^4\right )} \left (102 x-95 x^2-19 x^2 \log \left (81 x^4\right )\right ) \, dx=515377520732011331036461129765621272702107522001 \, x^{102} e^{\left (-76 \, x \log \left (3\right ) - 76 \, x \log \left (x\right ) - 19 \, x + 25\right )} \] Input:
integrate((-19*x^2*log(81*x^4)-95*x^2+102*x)*exp((-19*x+25)*log(81*x^4)-19 *x+25),x, algorithm="maxima")
Output:
515377520732011331036461129765621272702107522001*x^102*e^(-76*x*log(3) - 7 6*x*log(x) - 19*x + 25)
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int e^{25-19 x+(25-19 x) \log \left (81 x^4\right )} \left (102 x-95 x^2-19 x^2 \log \left (81 x^4\right )\right ) \, dx=x^{2} e^{\left (-19 \, x \log \left (81 \, x^{4}\right ) - 19 \, x + 25 \, \log \left (81 \, x^{4}\right ) + 25\right )} \] Input:
integrate((-19*x^2*log(81*x^4)-95*x^2+102*x)*exp((-19*x+25)*log(81*x^4)-19 *x+25),x, algorithm="giac")
Output:
x^2*e^(-19*x*log(81*x^4) - 19*x + 25*log(81*x^4) + 25)
Time = 3.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int e^{25-19 x+(25-19 x) \log \left (81 x^4\right )} \left (102 x-95 x^2-19 x^2 \log \left (81 x^4\right )\right ) \, dx=\frac {515377520732011331036461129765621272702107522001\,x^{102}\,{\mathrm {e}}^{-19\,x}\,{\mathrm {e}}^{25}}{3^{76\,x}\,{\left (x^4\right )}^{19\,x}} \] Input:
int(-exp(25 - log(81*x^4)*(19*x - 25) - 19*x)*(95*x^2 - 102*x + 19*x^2*log (81*x^4)),x)
Output:
(515377520732011331036461129765621272702107522001*x^102*exp(-19*x)*exp(25) )/(3^(76*x)*(x^4)^(19*x))
\[ \int e^{25-19 x+(25-19 x) \log \left (81 x^4\right )} \left (102 x-95 x^2-19 x^2 \log \left (81 x^4\right )\right ) \, dx=\int \left (-19 x^{2} \mathrm {log}\left (81 x^{4}\right )-95 x^{2}+102 x \right ) {\mathrm e}^{\left (-19 x +25\right ) \mathrm {log}\left (81 x^{4}\right )-19 x +25}d x \] Input:
int((-19*x^2*log(81*x^4)-95*x^2+102*x)*exp((-19*x+25)*log(81*x^4)-19*x+25) ,x)
Output:
int((-19*x^2*log(81*x^4)-95*x^2+102*x)*exp((-19*x+25)*log(81*x^4)-19*x+25) ,x)