Integrand size = 44, antiderivative size = 30 \[ \int \left (-3 x^2+5^x \left (e^{-x^2} x\right )^x \left (1-2 x^2+\log \left (5 e^{-x^2} x\right )\right )+2 \log (\log (4))\right ) \, dx=1+5^x \left (e^{-x^2} x\right )^x+x \left (-x^2+2 \log (\log (4))\right ) \] Output:
1+exp(x*ln(5*x/exp(x^2)))+x*(ln(4*ln(2)^2)-x^2)
\[ \int \left (-3 x^2+5^x \left (e^{-x^2} x\right )^x \left (1-2 x^2+\log \left (5 e^{-x^2} x\right )\right )+2 \log (\log (4))\right ) \, dx=\int \left (-3 x^2+5^x \left (e^{-x^2} x\right )^x \left (1-2 x^2+\log \left (5 e^{-x^2} x\right )\right )+2 \log (\log (4))\right ) \, dx \] Input:
Integrate[-3*x^2 + 5^x*(x/E^x^2)^x*(1 - 2*x^2 + Log[(5*x)/E^x^2]) + 2*Log[ Log[4]],x]
Output:
Integrate[-3*x^2 + 5^x*(x/E^x^2)^x*(1 - 2*x^2 + Log[(5*x)/E^x^2]) + 2*Log[ Log[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.
\(\displaystyle \int \left (-3 x^2+5^x \left (e^{-x^2} x\right )^x \left (-2 x^2+\log \left (5 e^{-x^2} x\right )+1\right )+2 \log (\log (4))\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int 5^x \left (e^{-x^2} x\right )^xdx-2 \int 5^x x^2 \left (e^{-x^2} x\right )^xdx-\int \frac {\int 5^x \left (e^{-x^2} x\right )^xdx}{x}dx+2 \int x \int 5^x \left (e^{-x^2} x\right )^xdxdx+\log \left (5 e^{-x^2} x\right ) \int 5^x \left (e^{-x^2} x\right )^xdx-x^3+2 x \log (\log (4))\) |
Input:
Int[-3*x^2 + 5^x*(x/E^x^2)^x*(1 - 2*x^2 + Log[(5*x)/E^x^2]) + 2*Log[Log[4] ],x]
Output:
$Aborted
Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97
method | result | size |
default | \({\mathrm e}^{x \ln \left (5 x \,{\mathrm e}^{-x^{2}}\right )}-x^{3}+x \ln \left (4 \ln \left (2\right )^{2}\right )\) | \(29\) |
parallelrisch | \({\mathrm e}^{x \ln \left (5 x \,{\mathrm e}^{-x^{2}}\right )}-x^{3}+x \ln \left (4 \ln \left (2\right )^{2}\right )\) | \(29\) |
parts | \({\mathrm e}^{x \ln \left (5 x \,{\mathrm e}^{-x^{2}}\right )}-x^{3}+x \ln \left (4 \ln \left (2\right )^{2}\right )\) | \(29\) |
risch | \(5^{x} \left ({\mathrm e}^{x^{2}}\right )^{-x} x^{x} {\mathrm e}^{-\frac {i \operatorname {csgn}\left (i x \,{\mathrm e}^{-x^{2}}\right ) \pi x \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{-x^{2}}\right )+\operatorname {csgn}\left (i {\mathrm e}^{-x^{2}}\right )\right ) \left (-\operatorname {csgn}\left (i x \,{\mathrm e}^{-x^{2}}\right )+\operatorname {csgn}\left (i x \right )\right )}{2}}-x^{3}+2 x \ln \left (\ln \left (2\right )\right )+2 x \ln \left (2\right )\) | \(93\) |
Input:
int((ln(5*x/exp(x^2))-2*x^2+1)*exp(x*ln(5*x/exp(x^2)))+ln(4*ln(2)^2)-3*x^2 ,x,method=_RETURNVERBOSE)
Output:
exp(x*ln(5*x/exp(x^2)))-x^3+x*ln(4*ln(2)^2)
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \left (-3 x^2+5^x \left (e^{-x^2} x\right )^x \left (1-2 x^2+\log \left (5 e^{-x^2} x\right )\right )+2 \log (\log (4))\right ) \, dx=-x^{3} + x \log \left (4 \, \log \left (2\right )^{2}\right ) + \left (5 \, x e^{\left (-x^{2}\right )}\right )^{x} \] Input:
integrate((log(5*x/exp(x^2))-2*x^2+1)*exp(x*log(5*x/exp(x^2)))+log(4*log(2 )^2)-3*x^2,x, algorithm="fricas")
Output:
-x^3 + x*log(4*log(2)^2) + (5*x*e^(-x^2))^x
Time = 115.72 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \left (-3 x^2+5^x \left (e^{-x^2} x\right )^x \left (1-2 x^2+\log \left (5 e^{-x^2} x\right )\right )+2 \log (\log (4))\right ) \, dx=- x^{3} + x \log {\left (4 \log {\left (2 \right )}^{2} \right )} + e^{x \log {\left (5 x e^{- x^{2}} \right )}} \] Input:
integrate((ln(5*x/exp(x**2))-2*x**2+1)*exp(x*ln(5*x/exp(x**2)))+ln(4*ln(2) **2)-3*x**2,x)
Output:
-x**3 + x*log(4*log(2)**2) + exp(x*log(5*x*exp(-x**2)))
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \left (-3 x^2+5^x \left (e^{-x^2} x\right )^x \left (1-2 x^2+\log \left (5 e^{-x^2} x\right )\right )+2 \log (\log (4))\right ) \, dx=-x^{3} + x \log \left (4 \, \log \left (2\right )^{2}\right ) + e^{\left (-x^{3} + x \log \left (5\right ) + x \log \left (x\right )\right )} \] Input:
integrate((log(5*x/exp(x^2))-2*x^2+1)*exp(x*log(5*x/exp(x^2)))+log(4*log(2 )^2)-3*x^2,x, algorithm="maxima")
Output:
-x^3 + x*log(4*log(2)^2) + e^(-x^3 + x*log(5) + x*log(x))
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \left (-3 x^2+5^x \left (e^{-x^2} x\right )^x \left (1-2 x^2+\log \left (5 e^{-x^2} x\right )\right )+2 \log (\log (4))\right ) \, dx=-x^{3} + x \log \left (4 \, \log \left (2\right )^{2}\right ) + e^{\left (-x^{3} + x \log \left (5 \, x\right )\right )} \] Input:
integrate((log(5*x/exp(x^2))-2*x^2+1)*exp(x*log(5*x/exp(x^2)))+log(4*log(2 )^2)-3*x^2,x, algorithm="giac")
Output:
-x^3 + x*log(4*log(2)^2) + e^(-x^3 + x*log(5*x))
Time = 3.50 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97 \[ \int \left (-3 x^2+5^x \left (e^{-x^2} x\right )^x \left (1-2 x^2+\log \left (5 e^{-x^2} x\right )\right )+2 \log (\log (4))\right ) \, dx=2\,x\,\ln \left (2\right )+{\mathrm {e}}^{-x^3}\,{\left (5\,x\right )}^x+2\,x\,\ln \left (\ln \left (2\right )\right )-x^3 \] Input:
int(log(4*log(2)^2) - 3*x^2 + exp(x*log(5*x*exp(-x^2)))*(log(5*x*exp(-x^2) ) - 2*x^2 + 1),x)
Output:
2*x*log(2) + exp(-x^3)*(5*x)^x + 2*x*log(log(2)) - x^3
\[ \int \left (-3 x^2+5^x \left (e^{-x^2} x\right )^x \left (1-2 x^2+\log \left (5 e^{-x^2} x\right )\right )+2 \log (\log (4))\right ) \, dx=\int \left (\left (\mathrm {log}\left (\frac {5 x}{{\mathrm e}^{x^{2}}}\right )-2 x^{2}+1\right ) {\mathrm e}^{x \,\mathrm {log}\left (\frac {5 x}{{\mathrm e}^{x^{2}}}\right )}+\mathrm {log}\left (4 \mathrm {log}\left (2\right )^{2}\right )-3 x^{2}\right )d x \] Input:
int((log(5*x/exp(x^2))-2*x^2+1)*exp(x*log(5*x/exp(x^2)))+log(4*log(2)^2)-3 *x^2,x)
Output:
int((log(5*x/exp(x^2))-2*x^2+1)*exp(x*log(5*x/exp(x^2)))+log(4*log(2)^2)-3 *x^2,x)