Integrand size = 46, antiderivative size = 33 \[ \int \frac {1}{4} e^{-x} \left (-8+12 e^x-2 x-3 x^2+2 x^3+\left (-4 x-x^2+x^3\right ) \log (3 x)\right ) \, dx=3 x-e^{-x} \left (2 x+\frac {1}{4} x \left (2 x+x^2\right ) (2+\log (3 x))\right ) \] Output:
3*x-(1/4*(x^2+2*x)*x*(2+ln(3*x))+2*x)/exp(x)
Time = 0.35 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \[ \int \frac {1}{4} e^{-x} \left (-8+12 e^x-2 x-3 x^2+2 x^3+\left (-4 x-x^2+x^3\right ) \log (3 x)\right ) \, dx=-\frac {1}{4} e^{-x} x \left (2 \left (4-6 e^x+2 x+x^2\right )+x (2+x) \log (3 x)\right ) \] Input:
Integrate[(-8 + 12*E^x - 2*x - 3*x^2 + 2*x^3 + (-4*x - x^2 + x^3)*Log[3*x] )/(4*E^x),x]
Output:
-1/4*(x*(2*(4 - 6*E^x + 2*x + x^2) + x*(2 + x)*Log[3*x]))/E^x
Time = 0.56 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.94, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, 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.
\(\displaystyle \int \frac {1}{4} e^{-x} \left (2 x^3-3 x^2+\left (x^3-x^2-4 x\right ) \log (3 x)-2 x+12 e^x-8\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -e^{-x} \left (-2 x^3+3 x^2+2 x-12 e^x+\left (-x^3+x^2+4 x\right ) \log (3 x)+8\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int e^{-x} \left (-2 x^3+3 x^2+2 x-12 e^x+\left (-x^3+x^2+4 x\right ) \log (3 x)+8\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{4} \int \left (-2 e^{-x} x^3+3 e^{-x} x^2+2 e^{-x} x-e^{-x} \left (x^2-x-4\right ) \log (3 x) x+8 e^{-x}-12\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (-2 e^{-x} x^3-e^{-x} x^3 \log (3 x)-4 e^{-x} x^2-2 e^{-x} x^2 \log (3 x)-8 e^{-x} x+12 x\right )\) |
Input:
Int[(-8 + 12*E^x - 2*x - 3*x^2 + 2*x^3 + (-4*x - x^2 + x^3)*Log[3*x])/(4*E ^x),x]
Output:
(12*x - (8*x)/E^x - (4*x^2)/E^x - (2*x^3)/E^x - (2*x^2*Log[3*x])/E^x - (x^ 3*Log[3*x])/E^x)/4
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.65 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12
method | result | size |
risch | \(-\frac {x^{2} \left (2+x \right ) {\mathrm e}^{-x} \ln \left (3 x \right )}{4}-\frac {x \left (x^{2}+2 x -6 \,{\mathrm e}^{x}+4\right ) {\mathrm e}^{-x}}{2}\) | \(37\) |
parts | \(3 x +\left (-2 x -x^{2}-\frac {x^{3}}{2}-\frac {\ln \left (3 x \right ) x^{2}}{2}-\frac {\ln \left (3 x \right ) x^{3}}{4}\right ) {\mathrm e}^{-x}\) | \(42\) |
default | \(3 x +\frac {\left (-8 x -4 x^{2}-2 x^{3}-2 \ln \left (3 x \right ) x^{2}-\ln \left (3 x \right ) x^{3}\right ) {\mathrm e}^{-x}}{4}\) | \(43\) |
norman | \(\left (-2 x -x^{2}-\frac {x^{3}}{2}+3 \,{\mathrm e}^{x} x -\frac {\ln \left (3 x \right ) x^{2}}{2}-\frac {\ln \left (3 x \right ) x^{3}}{4}\right ) {\mathrm e}^{-x}\) | \(43\) |
parallelrisch | \(-\frac {\left (\ln \left (3 x \right ) x^{3}+2 x^{3}+2 \ln \left (3 x \right ) x^{2}-12 \ln \left ({\mathrm e}^{x}\right ) {\mathrm e}^{x}+4 x^{2}+8 x \right ) {\mathrm e}^{-x}}{4}\) | \(45\) |
orering | \(\frac {\left (-1+x \right ) \left (\left (x^{3}-x^{2}-4 x \right ) \ln \left (3 x \right )+12 \,{\mathrm e}^{x}+2 x^{3}-3 x^{2}-2 x -8\right ) {\mathrm e}^{-x}}{4}+\frac {x \left (2 x^{6}-6 x^{5}-30 x^{4}+102 x^{3}-139 x^{2}+76 x +64\right ) \left (\frac {\left (\left (3 x^{2}-2 x -4\right ) \ln \left (3 x \right )+\frac {x^{3}-x^{2}-4 x}{x}+12 \,{\mathrm e}^{x}+6 x^{2}-6 x -2\right ) {\mathrm e}^{-x}}{4}-\frac {\left (\left (x^{3}-x^{2}-4 x \right ) \ln \left (3 x \right )+12 \,{\mathrm e}^{x}+2 x^{3}-3 x^{2}-2 x -8\right ) {\mathrm e}^{-x}}{4}\right )}{x^{6}-6 x^{5}-6 x^{4}+94 x^{3}-124 x^{2}+16}+\frac {\left (x^{6}-15 x^{4}+21 x^{3}-38 x^{2}+32\right ) x \left (\frac {\left (\left (6 x -2\right ) \ln \left (3 x \right )+\frac {6 x^{2}-4 x -8}{x}-\frac {x^{3}-x^{2}-4 x}{x^{2}}+12 \,{\mathrm e}^{x}+12 x -6\right ) {\mathrm e}^{-x}}{4}-\frac {\left (\left (3 x^{2}-2 x -4\right ) \ln \left (3 x \right )+\frac {x^{3}-x^{2}-4 x}{x}+12 \,{\mathrm e}^{x}+6 x^{2}-6 x -2\right ) {\mathrm e}^{-x}}{2}+\frac {\left (\left (x^{3}-x^{2}-4 x \right ) \ln \left (3 x \right )+12 \,{\mathrm e}^{x}+2 x^{3}-3 x^{2}-2 x -8\right ) {\mathrm e}^{-x}}{4}\right )}{x^{6}-6 x^{5}-6 x^{4}+94 x^{3}-124 x^{2}+16}\) | \(400\) |
Input:
int(1/4*((x^3-x^2-4*x)*ln(3*x)+12*exp(x)+2*x^3-3*x^2-2*x-8)/exp(x),x,metho d=_RETURNVERBOSE)
Output:
-1/4*x^2*(2+x)/exp(x)*ln(3*x)-1/2*x*(x^2+2*x-6*exp(x)+4)/exp(x)
Time = 0.07 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {1}{4} e^{-x} \left (-8+12 e^x-2 x-3 x^2+2 x^3+\left (-4 x-x^2+x^3\right ) \log (3 x)\right ) \, dx=-\frac {1}{4} \, {\left (2 \, x^{3} + 4 \, x^{2} - 12 \, x e^{x} + {\left (x^{3} + 2 \, x^{2}\right )} \log \left (3 \, x\right ) + 8 \, x\right )} e^{\left (-x\right )} \] Input:
integrate(1/4*((x^3-x^2-4*x)*log(3*x)+12*exp(x)+2*x^3-3*x^2-2*x-8)/exp(x), x, algorithm="fricas")
Output:
-1/4*(2*x^3 + 4*x^2 - 12*x*e^x + (x^3 + 2*x^2)*log(3*x) + 8*x)*e^(-x)
Time = 0.12 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {1}{4} e^{-x} \left (-8+12 e^x-2 x-3 x^2+2 x^3+\left (-4 x-x^2+x^3\right ) \log (3 x)\right ) \, dx=3 x + \frac {\left (- x^{3} \log {\left (3 x \right )} - 2 x^{3} - 2 x^{2} \log {\left (3 x \right )} - 4 x^{2} - 8 x\right ) e^{- x}}{4} \] Input:
integrate(1/4*((x**3-x**2-4*x)*ln(3*x)+12*exp(x)+2*x**3-3*x**2-2*x-8)/exp( x),x)
Output:
3*x + (-x**3*log(3*x) - 2*x**3 - 2*x**2*log(3*x) - 4*x**2 - 8*x)*exp(-x)/4
Leaf count of result is larger than twice the leaf count of optimal. 91 vs. \(2 (29) = 58\).
Time = 0.16 (sec) , antiderivative size = 91, normalized size of antiderivative = 2.76 \[ \int \frac {1}{4} e^{-x} \left (-8+12 e^x-2 x-3 x^2+2 x^3+\left (-4 x-x^2+x^3\right ) \log (3 x)\right ) \, dx=-\frac {1}{4} \, {\left (x^{3} \log \left (3\right ) + x^{2} {\left (2 \, \log \left (3\right ) + 1\right )} + {\left (x^{3} + 2 \, x^{2}\right )} \log \left (x\right ) + 4 \, x + 4\right )} e^{\left (-x\right )} - \frac {1}{2} \, {\left (x^{3} + 3 \, x^{2} + 6 \, x + 6\right )} e^{\left (-x\right )} + \frac {3}{4} \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x\right )} + \frac {1}{2} \, {\left (x + 1\right )} e^{\left (-x\right )} + 3 \, x + 2 \, e^{\left (-x\right )} \] Input:
integrate(1/4*((x^3-x^2-4*x)*log(3*x)+12*exp(x)+2*x^3-3*x^2-2*x-8)/exp(x), x, algorithm="maxima")
Output:
-1/4*(x^3*log(3) + x^2*(2*log(3) + 1) + (x^3 + 2*x^2)*log(x) + 4*x + 4)*e^ (-x) - 1/2*(x^3 + 3*x^2 + 6*x + 6)*e^(-x) + 3/4*(x^2 + 2*x + 2)*e^(-x) + 1 /2*(x + 1)*e^(-x) + 3*x + 2*e^(-x)
Time = 0.12 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.67 \[ \int \frac {1}{4} e^{-x} \left (-8+12 e^x-2 x-3 x^2+2 x^3+\left (-4 x-x^2+x^3\right ) \log (3 x)\right ) \, dx=-\frac {1}{4} \, x^{3} e^{\left (-x\right )} \log \left (3 \, x\right ) - \frac {1}{2} \, x^{3} e^{\left (-x\right )} - \frac {1}{2} \, x^{2} e^{\left (-x\right )} \log \left (3 \, x\right ) - x^{2} e^{\left (-x\right )} - 2 \, x e^{\left (-x\right )} + 3 \, x \] Input:
integrate(1/4*((x^3-x^2-4*x)*log(3*x)+12*exp(x)+2*x^3-3*x^2-2*x-8)/exp(x), x, algorithm="giac")
Output:
-1/4*x^3*e^(-x)*log(3*x) - 1/2*x^3*e^(-x) - 1/2*x^2*e^(-x)*log(3*x) - x^2* e^(-x) - 2*x*e^(-x) + 3*x
Time = 1.65 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int \frac {1}{4} e^{-x} \left (-8+12 e^x-2 x-3 x^2+2 x^3+\left (-4 x-x^2+x^3\right ) \log (3 x)\right ) \, dx=\frac {x\,{\mathrm {e}}^{-x}\,\left (12\,{\mathrm {e}}^x-8\right )}{4}-\frac {x^2\,{\mathrm {e}}^{-x}\,\left (2\,\ln \left (3\,x\right )+4\right )}{4}-\frac {x^3\,{\mathrm {e}}^{-x}\,\left (\ln \left (3\,x\right )+2\right )}{4} \] Input:
int(-exp(-x)*(x/2 - 3*exp(x) + (3*x^2)/4 - x^3/2 + (log(3*x)*(4*x + x^2 - x^3))/4 + 2),x)
Output:
(x*exp(-x)*(12*exp(x) - 8))/4 - (x^2*exp(-x)*(2*log(3*x) + 4))/4 - (x^3*ex p(-x)*(log(3*x) + 2))/4
Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {1}{4} e^{-x} \left (-8+12 e^x-2 x-3 x^2+2 x^3+\left (-4 x-x^2+x^3\right ) \log (3 x)\right ) \, dx=\frac {x \left (12 e^{x}-\mathrm {log}\left (3 x \right ) x^{2}-2 \,\mathrm {log}\left (3 x \right ) x -2 x^{2}-4 x -8\right )}{4 e^{x}} \] Input:
int(1/4*((x^3-x^2-4*x)*log(3*x)+12*exp(x)+2*x^3-3*x^2-2*x-8)/exp(x),x)
Output:
(x*(12*e**x - log(3*x)*x**2 - 2*log(3*x)*x - 2*x**2 - 4*x - 8))/(4*e**x)