Integrand size = 50, antiderivative size = 33 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=-e^{3+x}+x-x \left (2 x-\frac {e^{-x^2} x \log (\log (4))}{\log (5)}\right ) \] Output:
x-exp(3+x)-x*(2*x-ln(2*ln(2))/ln(5)/exp(x^2)*x)
Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=-e^{3+x}+x-2 x^2+\frac {e^{-x^2} x^2 \log (\log (4))}{\log (5)} \] Input:
Integrate[(E^x^2*(-(E^(3 + x)*Log[5]) + (1 - 4*x)*Log[5]) + (2*x - 2*x^3)* Log[Log[4]])/(E^x^2*Log[5]),x]
Output:
-E^(3 + x) + x - 2*x^2 + (x^2*Log[Log[4]])/(E^x^2*Log[5])
Time = 0.38 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.27, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {27, 25, 7293, 2009}
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.
\(\displaystyle \int \frac {e^{-x^2} \left (\left (2 x-2 x^3\right ) \log (\log (4))+e^{x^2} \left ((1-4 x) \log (5)-e^{x+3} \log (5)\right )\right )}{\log (5)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -e^{-x^2} \left (e^{x^2} \left (e^{x+3} \log (5)-(1-4 x) \log (5)\right )-2 \left (x-x^3\right ) \log (\log (4))\right )dx}{\log (5)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int e^{-x^2} \left (e^{x^2} \left (e^{x+3} \log (5)-(1-4 x) \log (5)\right )-2 \left (x-x^3\right ) \log (\log (4))\right )dx}{\log (5)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (\log (5) \left (4 x+e^{x+3}-1\right )+2 e^{-x^2} x \left (x^2-1\right ) \log (\log (4))\right )dx}{\log (5)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-e^{-x^2} x^2 \log (\log (4))+2 x^2 \log (5)-x \log (5)+e^{x+3} \log (5)}{\log (5)}\) |
Input:
Int[(E^x^2*(-(E^(3 + x)*Log[5]) + (1 - 4*x)*Log[5]) + (2*x - 2*x^3)*Log[Lo g[4]])/(E^x^2*Log[5]),x]
Output:
-((E^(3 + x)*Log[5] - x*Log[5] + 2*x^2*Log[5] - (x^2*Log[Log[4]])/E^x^2)/L og[5])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.49 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00
method | result | size |
parts | \(-2 x^{2}+x +\frac {\ln \left (2 \ln \left (2\right )\right ) {\mathrm e}^{-x^{2}} x^{2}}{\ln \left (5\right )}-{\mathrm e}^{3+x}\) | \(33\) |
risch | \(-2 x^{2}+x +\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) x^{2} {\mathrm e}^{-x^{2}}}{\ln \left (5\right )}-{\mathrm e}^{3+x}\) | \(34\) |
norman | \(\left ({\mathrm e}^{x^{2}} x +\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) x^{2}}{\ln \left (5\right )}-2 x^{2} {\mathrm e}^{x^{2}}-{\mathrm e}^{x^{2}} {\mathrm e}^{3+x}\right ) {\mathrm e}^{-x^{2}}\) | \(48\) |
parallelrisch | \(\frac {\left (-2 x^{2} {\mathrm e}^{x^{2}} \ln \left (5\right )+x^{2} \ln \left (2 \ln \left (2\right )\right )+\ln \left (5\right ) {\mathrm e}^{x^{2}} x -\ln \left (5\right ) {\mathrm e}^{3+x} {\mathrm e}^{x^{2}}\right ) {\mathrm e}^{-x^{2}}}{\ln \left (5\right )}\) | \(53\) |
default | \(\frac {-2 x^{2} \ln \left (5\right )-\ln \left (\ln \left (2\right )\right ) {\mathrm e}^{-x^{2}}+x^{2} {\mathrm e}^{-x^{2}} \ln \left (2\right )-2 \ln \left (\ln \left (2\right )\right ) \left (-\frac {{\mathrm e}^{-x^{2}} x^{2}}{2}-\frac {{\mathrm e}^{-x^{2}}}{2}\right )-\ln \left (5\right ) {\mathrm e}^{x} {\mathrm e}^{3}+x \ln \left (5\right )}{\ln \left (5\right )}\) | \(74\) |
orering | \(\frac {\left (16 x^{7}-72 x^{5}+34 x^{4}+40 x^{3}-69 x^{2}+10 x +9\right ) \left (\left (-2 x^{3}+2 x \right ) \ln \left (2 \ln \left (2\right )\right )+\left (-\ln \left (5\right ) {\mathrm e}^{3+x}+\left (-4 x +1\right ) \ln \left (5\right )\right ) {\mathrm e}^{x^{2}}\right ) {\mathrm e}^{-x^{2}}}{2 \left (16 x^{6}-12 x^{5}-74 x^{4}+74 x^{3}+53 x^{2}-60 x -1\right ) \ln \left (5\right )}-\frac {x \left (16 x^{6}-40 x^{5}-32 x^{4}+182 x^{3}-159 x^{2}-121 x +141\right ) \left (\frac {\left (\left (-6 x^{2}+2\right ) \ln \left (2 \ln \left (2\right )\right )+\left (-\ln \left (5\right ) {\mathrm e}^{3+x}-4 \ln \left (5\right )\right ) {\mathrm e}^{x^{2}}+2 \left (-\ln \left (5\right ) {\mathrm e}^{3+x}+\left (-4 x +1\right ) \ln \left (5\right )\right ) {\mathrm e}^{x^{2}} x \right ) {\mathrm e}^{-x^{2}}}{\ln \left (5\right )}-\frac {2 \left (\left (-2 x^{3}+2 x \right ) \ln \left (2 \ln \left (2\right )\right )+\left (-\ln \left (5\right ) {\mathrm e}^{3+x}+\left (-4 x +1\right ) \ln \left (5\right )\right ) {\mathrm e}^{x^{2}}\right ) {\mathrm e}^{-x^{2}} x}{\ln \left (5\right )}\right )}{2 \left (16 x^{6}-12 x^{5}-74 x^{4}+74 x^{3}+53 x^{2}-60 x -1\right )}-\frac {\left (8 x^{6}-16 x^{5}+51 x^{3}-54 x^{2}-11 x +11\right ) \left (\frac {\left (-12 x \ln \left (2 \ln \left (2\right )\right )-\ln \left (5\right ) {\mathrm e}^{3+x} {\mathrm e}^{x^{2}}+4 \left (-\ln \left (5\right ) {\mathrm e}^{3+x}-4 \ln \left (5\right )\right ) {\mathrm e}^{x^{2}} x +4 \left (-\ln \left (5\right ) {\mathrm e}^{3+x}+\left (-4 x +1\right ) \ln \left (5\right )\right ) {\mathrm e}^{x^{2}} x^{2}+2 \left (-\ln \left (5\right ) {\mathrm e}^{3+x}+\left (-4 x +1\right ) \ln \left (5\right )\right ) {\mathrm e}^{x^{2}}\right ) {\mathrm e}^{-x^{2}}}{\ln \left (5\right )}-\frac {4 \left (\left (-6 x^{2}+2\right ) \ln \left (2 \ln \left (2\right )\right )+\left (-\ln \left (5\right ) {\mathrm e}^{3+x}-4 \ln \left (5\right )\right ) {\mathrm e}^{x^{2}}+2 \left (-\ln \left (5\right ) {\mathrm e}^{3+x}+\left (-4 x +1\right ) \ln \left (5\right )\right ) {\mathrm e}^{x^{2}} x \right ) {\mathrm e}^{-x^{2}} x}{\ln \left (5\right )}+\frac {4 \left (\left (-2 x^{3}+2 x \right ) \ln \left (2 \ln \left (2\right )\right )+\left (-\ln \left (5\right ) {\mathrm e}^{3+x}+\left (-4 x +1\right ) \ln \left (5\right )\right ) {\mathrm e}^{x^{2}}\right ) {\mathrm e}^{-x^{2}} x^{2}}{\ln \left (5\right )}-\frac {2 \left (\left (-2 x^{3}+2 x \right ) \ln \left (2 \ln \left (2\right )\right )+\left (-\ln \left (5\right ) {\mathrm e}^{3+x}+\left (-4 x +1\right ) \ln \left (5\right )\right ) {\mathrm e}^{x^{2}}\right ) {\mathrm e}^{-x^{2}}}{\ln \left (5\right )}\right )}{2 \left (16 x^{6}-12 x^{5}-74 x^{4}+74 x^{3}+53 x^{2}-60 x -1\right )}\) | \(631\) |
Input:
int(((-2*x^3+2*x)*ln(2*ln(2))+(-ln(5)*exp(3+x)+(-4*x+1)*ln(5))*exp(x^2))/l n(5)/exp(x^2),x,method=_RETURNVERBOSE)
Output:
-2*x^2+x+ln(2*ln(2))/ln(5)/exp(x^2)*x^2-exp(3+x)
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=\frac {{\left (x^{2} \log \left (2 \, \log \left (2\right )\right ) - {\left ({\left (2 \, x^{2} - x\right )} \log \left (5\right ) + e^{\left (x + 3\right )} \log \left (5\right )\right )} e^{\left (x^{2}\right )}\right )} e^{\left (-x^{2}\right )}}{\log \left (5\right )} \] Input:
integrate(((-2*x^3+2*x)*log(2*log(2))+(-log(5)*exp(3+x)+(-4*x+1)*log(5))*e xp(x^2))/log(5)/exp(x^2),x, algorithm="fricas")
Output:
(x^2*log(2*log(2)) - ((2*x^2 - x)*log(5) + e^(x + 3)*log(5))*e^(x^2))*e^(- x^2)/log(5)
Time = 0.16 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=- 2 x^{2} + x + \frac {\left (x^{2} \log {\left (\log {\left (2 \right )} \right )} + x^{2} \log {\left (2 \right )}\right ) e^{- x^{2}}}{\log {\left (5 \right )}} - e^{x + 3} \] Input:
integrate(((-2*x**3+2*x)*ln(2*ln(2))+(-ln(5)*exp(3+x)+(-4*x+1)*ln(5))*exp( x**2))/ln(5)/exp(x**2),x)
Output:
-2*x**2 + x + (x**2*log(log(2)) + x**2*log(2))*exp(-x**2)/log(5) - exp(x + 3)
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.70 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=-\frac {2 \, x^{2} \log \left (5\right ) - {\left (x^{2} + 1\right )} e^{\left (-x^{2}\right )} \log \left (2 \, \log \left (2\right )\right ) - x \log \left (5\right ) + e^{\left (x + 3\right )} \log \left (5\right ) + e^{\left (-x^{2}\right )} \log \left (2 \, \log \left (2\right )\right )}{\log \left (5\right )} \] Input:
integrate(((-2*x^3+2*x)*log(2*log(2))+(-log(5)*exp(3+x)+(-4*x+1)*log(5))*e xp(x^2))/log(5)/exp(x^2),x, algorithm="maxima")
Output:
-(2*x^2*log(5) - (x^2 + 1)*e^(-x^2)*log(2*log(2)) - x*log(5) + e^(x + 3)*l og(5) + e^(-x^2)*log(2*log(2)))/log(5)
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.45 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=-\frac {2 \, x^{2} \log \left (5\right ) - {\left (x^{2} \log \left (2\right ) + x^{2} \log \left (\log \left (2\right )\right )\right )} e^{\left (-x^{2}\right )} - x \log \left (5\right ) + e^{\left (x + 3\right )} \log \left (5\right )}{\log \left (5\right )} \] Input:
integrate(((-2*x^3+2*x)*log(2*log(2))+(-log(5)*exp(3+x)+(-4*x+1)*log(5))*e xp(x^2))/log(5)/exp(x^2),x, algorithm="giac")
Output:
-(2*x^2*log(5) - (x^2*log(2) + x^2*log(log(2)))*e^(-x^2) - x*log(5) + e^(x + 3)*log(5))/log(5)
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=x-{\mathrm {e}}^{x+3}-2\,x^2+\frac {x^2\,{\mathrm {e}}^{-x^2}\,\ln \left (2\right )}{\ln \left (5\right )}+\frac {x^2\,{\mathrm {e}}^{-x^2}\,\ln \left (\ln \left (2\right )\right )}{\ln \left (5\right )} \] Input:
int((exp(-x^2)*(log(2*log(2))*(2*x - 2*x^3) - exp(x^2)*(log(5)*(4*x - 1) + exp(x + 3)*log(5))))/log(5),x)
Output:
x - exp(x + 3) - 2*x^2 + (x^2*exp(-x^2)*log(2))/log(5) + (x^2*exp(-x^2)*lo g(log(2)))/log(5)
Time = 0.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.73 \[ \int \frac {e^{-x^2} \left (e^{x^2} \left (-e^{3+x} \log (5)+(1-4 x) \log (5)\right )+\left (2 x-2 x^3\right ) \log (\log (4))\right )}{\log (5)} \, dx=\frac {-e^{x^{2}+x} \mathrm {log}\left (5\right ) e^{3}-2 e^{x^{2}} \mathrm {log}\left (5\right ) x^{2}+e^{x^{2}} \mathrm {log}\left (5\right ) x +\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right ) x^{2}}{e^{x^{2}} \mathrm {log}\left (5\right )} \] Input:
int(((-2*x^3+2*x)*log(2*log(2))+(-log(5)*exp(3+x)+(-4*x+1)*log(5))*exp(x^2 ))/log(5)/exp(x^2),x)
Output:
( - e**(x**2 + x)*log(5)*e**3 - 2*e**(x**2)*log(5)*x**2 + e**(x**2)*log(5) *x + log(2*log(2))*x**2)/(e**(x**2)*log(5))