Integrand size = 66, antiderivative size = 32 \[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\frac {1}{4} e^{-\frac {e^x}{2}-x \left (1-e^x x\right ) \log (3)} (-5+4 x) \] Output:
1/4*(-5+4*x)/exp(ln(3)*x*(-exp(x)*x+1)+1/2*exp(x))
\[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx \] Input:
Integrate[(E^((-2*x*Log[3] - E^x*(1 - 2*x^2*Log[3]))/2)*(8 + (10 - 8*x)*Lo g[3] + E^x*(5 - 4*x + (-20*x + 6*x^2 + 8*x^3)*Log[3])))/8,x]
Output:
Integrate[E^((-2*x*Log[3] - E^x*(1 - 2*x^2*Log[3]))/2)*(8 + (10 - 8*x)*Log [3] + E^x*(5 - 4*x + (-20*x + 6*x^2 + 8*x^3)*Log[3])), x]/8
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.
\(\displaystyle \int \frac {1}{8} \left (e^x \left (\left (8 x^3+6 x^2-20 x\right ) \log (3)-4 x+5\right )+(10-8 x) \log (3)+8\right ) \exp \left (\frac {1}{2} \left (-e^x \left (1-2 x^2 \log (3)\right )-2 x \log (3)\right )\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \int 3^{e^x x^2-x} e^{-\frac {e^x}{2}} \left (2 \log (3) (5-4 x)+e^x \left (-4 x-2 \left (-4 x^3-3 x^2+10 x\right ) \log (3)+5\right )+8\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{8} \int \left (3^{e^x x^2-x} e^{x-\frac {e^x}{2}} \left (\log (9) x^2+\log (81) x-1\right ) (4 x-5)-2\ 3^{e^x x^2-x} e^{-\frac {e^x}{2}} \log (3) (4 x-5)+8\ 3^{e^x x^2-x} e^{-\frac {e^x}{2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{8} \left (8 \int 3^{e^x x^2-x} e^{-\frac {e^x}{2}}dx+5 \int 3^{e^x x^2-x} e^{x-\frac {e^x}{2}}dx+10 \log (3) \int 3^{e^x x^2-x} e^{-\frac {e^x}{2}}dx-8 \log (3) \int 3^{e^x x^2-x} e^{-\frac {e^x}{2}} xdx-(4+5 \log (81)) \int 3^{e^x x^2-x} e^{x-\frac {e^x}{2}} xdx+\log (729) \int 3^{e^x x^2-x} e^{x-\frac {e^x}{2}} x^2dx+4 \log (9) \int 3^{e^x x^2-x} e^{x-\frac {e^x}{2}} x^3dx\right )\) |
Input:
Int[(E^((-2*x*Log[3] - E^x*(1 - 2*x^2*Log[3]))/2)*(8 + (10 - 8*x)*Log[3] + E^x*(5 - 4*x + (-20*x + 6*x^2 + 8*x^3)*Log[3])))/8,x]
Output:
$Aborted
Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81
method | result | size |
norman | \(\left (x -\frac {5}{4}\right ) {\mathrm e}^{-\frac {\left (-2 x^{2} \ln \left (3\right )+1\right ) {\mathrm e}^{x}}{2}-x \ln \left (3\right )}\) | \(26\) |
risch | \(\frac {\left (-10+8 x \right ) 3^{{\mathrm e}^{x} x^{2}} 3^{-x} {\mathrm e}^{-\frac {{\mathrm e}^{x}}{2}}}{8}\) | \(28\) |
parallelrisch | \(\frac {\left (16 x -20\right ) {\mathrm e}^{x^{2} \ln \left (3\right ) {\mathrm e}^{x}-\frac {{\mathrm e}^{x}}{2}-x \ln \left (3\right )}}{16}\) | \(29\) |
Input:
int(1/8*(((8*x^3+6*x^2-20*x)*ln(3)-4*x+5)*exp(x)+(-8*x+10)*ln(3)+8)/exp(1/ 2*(-2*x^2*ln(3)+1)*exp(x)+x*ln(3)),x,method=_RETURNVERBOSE)
Output:
(x-5/4)/exp(1/2*(-2*x^2*ln(3)+1)*exp(x)+x*ln(3))
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\frac {1}{4} \, {\left (4 \, x - 5\right )} e^{\left (\frac {1}{2} \, {\left (2 \, x^{2} \log \left (3\right ) - 1\right )} e^{x} - x \log \left (3\right )\right )} \] Input:
integrate(1/8*(((8*x^3+6*x^2-20*x)*log(3)-4*x+5)*exp(x)+(-8*x+10)*log(3)+8 )/exp(1/2*(-2*x^2*log(3)+1)*exp(x)+x*log(3)),x, algorithm="fricas")
Output:
1/4*(4*x - 5)*e^(1/2*(2*x^2*log(3) - 1)*e^x - x*log(3))
Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\frac {\left (4 x - 5\right ) e^{- x \log {\left (3 \right )} - \left (- x^{2} \log {\left (3 \right )} + \frac {1}{2}\right ) e^{x}}}{4} \] Input:
integrate(1/8*(((8*x**3+6*x**2-20*x)*ln(3)-4*x+5)*exp(x)+(-8*x+10)*ln(3)+8 )/exp(1/2*(-2*x**2*ln(3)+1)*exp(x)+x*ln(3)),x)
Output:
(4*x - 5)*exp(-x*log(3) - (-x**2*log(3) + 1/2)*exp(x))/4
Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\frac {1}{4} \, {\left (4 \, x - 5\right )} e^{\left (x^{2} e^{x} \log \left (3\right ) - x \log \left (3\right ) - \frac {1}{2} \, e^{x}\right )} \] Input:
integrate(1/8*(((8*x^3+6*x^2-20*x)*log(3)-4*x+5)*exp(x)+(-8*x+10)*log(3)+8 )/exp(1/2*(-2*x^2*log(3)+1)*exp(x)+x*log(3)),x, algorithm="maxima")
Output:
1/4*(4*x - 5)*e^(x^2*e^x*log(3) - x*log(3) - 1/2*e^x)
\[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\int { \frac {1}{8} \, {\left ({\left (2 \, {\left (4 \, x^{3} + 3 \, x^{2} - 10 \, x\right )} \log \left (3\right ) - 4 \, x + 5\right )} e^{x} - 2 \, {\left (4 \, x - 5\right )} \log \left (3\right ) + 8\right )} e^{\left (\frac {1}{2} \, {\left (2 \, x^{2} \log \left (3\right ) - 1\right )} e^{x} - x \log \left (3\right )\right )} \,d x } \] Input:
integrate(1/8*(((8*x^3+6*x^2-20*x)*log(3)-4*x+5)*exp(x)+(-8*x+10)*log(3)+8 )/exp(1/2*(-2*x^2*log(3)+1)*exp(x)+x*log(3)),x, algorithm="giac")
Output:
integrate(1/8*((2*(4*x^3 + 3*x^2 - 10*x)*log(3) - 4*x + 5)*e^x - 2*(4*x - 5)*log(3) + 8)*e^(1/2*(2*x^2*log(3) - 1)*e^x - x*log(3)), x)
Time = 2.61 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\frac {3^{x^2\,{\mathrm {e}}^x}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{2}}\,\left (4\,x-5\right )}{4\,3^x} \] Input:
int(exp((exp(x)*(2*x^2*log(3) - 1))/2 - x*log(3))*((exp(x)*(log(3)*(6*x^2 - 20*x + 8*x^3) - 4*x + 5))/8 - (log(3)*(8*x - 10))/8 + 1),x)
Output:
(3^(x^2*exp(x))*exp(-exp(x)/2)*(4*x - 5))/(4*3^x)
Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {1}{8} e^{\frac {1}{2} \left (-2 x \log (3)-e^x \left (1-2 x^2 \log (3)\right )\right )} \left (8+(10-8 x) \log (3)+e^x \left (5-4 x+\left (-20 x+6 x^2+8 x^3\right ) \log (3)\right )\right ) \, dx=\frac {e^{e^{x} \mathrm {log}\left (3\right ) x^{2}} \left (4 x -5\right )}{4 e^{\frac {e^{x}}{2}} 3^{x}} \] Input:
int(1/8*(((8*x^3+6*x^2-20*x)*log(3)-4*x+5)*exp(x)+(-8*x+10)*log(3)+8)/exp( 1/2*(-2*x^2*log(3)+1)*exp(x)+x*log(3)),x)
Output:
(e**(e**x*log(3)*x**2)*(4*x - 5))/(4*e**(e**x/2)*3**x)