Integrand size = 48, antiderivative size = 36 \[ \int \frac {1}{8} \left (10+9 x-18 x^2+e^2 (-2+4 x)+e^{-4+x} \left (-2-14 x-6 x^2+e^2 (2+2 x)\right )\right ) \, dx=x+\frac {1}{4} x \left (\frac {x}{4}+\left (1-e^{-4+x}-x\right ) \left (1-e^2+3 x\right )\right ) \] Output:
1/4*(1/4*x+(1-x-exp(-4+x))*(1+3*x-exp(2)))*x+x
Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.61 \[ \int \frac {1}{8} \left (10+9 x-18 x^2+e^2 (-2+4 x)+e^{-4+x} \left (-2-14 x-6 x^2+e^2 (2+2 x)\right )\right ) \, dx=\frac {1}{8} \left (10 x-2 e^2 x+\frac {9 x^2}{2}+2 e^2 x^2-6 x^3+2 e^x \left (\frac {\left (-1+e^2\right ) x}{e^4}-\frac {3 x^2}{e^4}\right )\right ) \] Input:
Integrate[(10 + 9*x - 18*x^2 + E^2*(-2 + 4*x) + E^(-4 + x)*(-2 - 14*x - 6* x^2 + E^2*(2 + 2*x)))/8,x]
Output:
(10*x - 2*E^2*x + (9*x^2)/2 + 2*E^2*x^2 - 6*x^3 + 2*E^x*(((-1 + E^2)*x)/E^ 4 - (3*x^2)/E^4))/8
Time = 0.24 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.92, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {27, 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.
\(\displaystyle \int \frac {1}{8} \left (-18 x^2+e^{x-4} \left (-6 x^2-14 x+e^2 (2 x+2)-2\right )+9 x+e^2 (4 x-2)+10\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{8} \int \left (-18 x^2+9 x-2 e^2 (1-2 x)-2 e^{x-4} \left (3 x^2+7 x-e^2 (x+1)+1\right )+10\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{8} \left (-6 x^3-6 e^{x-4} x^2+\frac {9 x^2}{2}-2 e^{x-4} x+10 x-2 e^{x-2}+\frac {1}{2} e^2 (1-2 x)^2+2 e^{x-2} (x+1)\right )\) |
Input:
Int[(10 + 9*x - 18*x^2 + E^2*(-2 + 4*x) + E^(-4 + x)*(-2 - 14*x - 6*x^2 + E^2*(2 + 2*x)))/8,x]
Output:
(-2*E^(-2 + x) + (E^2*(1 - 2*x)^2)/2 + 10*x - 2*E^(-4 + x)*x + (9*x^2)/2 - 6*E^(-4 + x)*x^2 - 6*x^3 + 2*E^(-2 + x)*(1 + x))/8
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.28
method | result | size |
norman | \(\left (\frac {5}{4}-\frac {{\mathrm e}^{2}}{4}\right ) x +\left (\frac {{\mathrm e}^{2}}{4}+\frac {9}{16}\right ) x^{2}+\left (\frac {{\mathrm e}^{2}}{4}-\frac {1}{4}\right ) x \,{\mathrm e}^{x -4}-\frac {3 x^{3}}{4}-\frac {3 x^{2} {\mathrm e}^{x -4}}{4}\) | \(46\) |
risch | \(\frac {\left (2 \,{\mathrm e}^{2} x -6 x^{2}-2 x \right ) {\mathrm e}^{x -4}}{8}+\frac {x^{2} {\mathrm e}^{2}}{4}-\frac {{\mathrm e}^{2} x}{4}-\frac {3 x^{3}}{4}+\frac {9 x^{2}}{16}+\frac {5 x}{4}\) | \(47\) |
parallelrisch | \(\frac {{\mathrm e}^{2} {\mathrm e}^{x -4} x}{4}-\frac {3 x^{2} {\mathrm e}^{x -4}}{4}-\frac {x \,{\mathrm e}^{x -4}}{4}+\frac {x^{2} {\mathrm e}^{2}}{4}-\frac {{\mathrm e}^{2} x}{4}-\frac {3 x^{3}}{4}+\frac {9 x^{2}}{16}+\frac {5 x}{4}\) | \(52\) |
default | \(\frac {5 x}{4}+\frac {\left (x^{2}-x \right ) {\mathrm e}^{2}}{4}-\frac {25 \,{\mathrm e}^{x -4} \left (x -4\right )}{4}-13 \,{\mathrm e}^{x -4}-\frac {3 \,{\mathrm e}^{x -4} \left (x -4\right )^{2}}{4}+\frac {5 \,{\mathrm e}^{x -4} {\mathrm e}^{2}}{4}+\frac {{\mathrm e}^{2} \left ({\mathrm e}^{x -4} \left (x -4\right )-{\mathrm e}^{x -4}\right )}{4}+\frac {9 x^{2}}{16}-\frac {3 x^{3}}{4}\) | \(79\) |
parts | \(\frac {5 x}{4}+\frac {{\mathrm e}^{2} \left ({\mathrm e}^{x -4} \left (x -4\right )-{\mathrm e}^{x -4}\right )}{4}-\frac {25 \,{\mathrm e}^{x -4} \left (x -4\right )}{4}-13 \,{\mathrm e}^{x -4}-\frac {3 \,{\mathrm e}^{x -4} \left (x -4\right )^{2}}{4}+\frac {5 \,{\mathrm e}^{x -4} {\mathrm e}^{2}}{4}+\frac {9 x^{2}}{16}-\frac {3 x^{3}}{4}+\frac {x^{2} {\mathrm e}^{2}}{4}-\frac {{\mathrm e}^{2} x}{4}\) | \(80\) |
derivativedivides | \(\frac {23 x}{4}-23+\frac {9 \left (x -4\right )^{2}}{16}-\frac {3 x^{3}}{4}+\frac {{\mathrm e}^{2} \left (\left (x -4\right )^{2}+7 x -28\right )}{4}-\frac {25 \,{\mathrm e}^{x -4} \left (x -4\right )}{4}-13 \,{\mathrm e}^{x -4}-\frac {3 \,{\mathrm e}^{x -4} \left (x -4\right )^{2}}{4}+\frac {5 \,{\mathrm e}^{x -4} {\mathrm e}^{2}}{4}+\frac {{\mathrm e}^{2} \left ({\mathrm e}^{x -4} \left (x -4\right )-{\mathrm e}^{x -4}\right )}{4}\) | \(85\) |
Input:
int(1/8*((2+2*x)*exp(2)-6*x^2-14*x-2)*exp(x-4)+1/8*(4*x-2)*exp(2)-9/4*x^2+ 9/8*x+5/4,x,method=_RETURNVERBOSE)
Output:
(5/4-1/4*exp(2))*x+(1/4*exp(2)+9/16)*x^2+(1/4*exp(2)-1/4)*x*exp(x-4)-3/4*x ^3-3/4*x^2*exp(x-4)
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.19 \[ \int \frac {1}{8} \left (10+9 x-18 x^2+e^2 (-2+4 x)+e^{-4+x} \left (-2-14 x-6 x^2+e^2 (2+2 x)\right )\right ) \, dx=-\frac {3}{4} \, x^{3} + \frac {9}{16} \, x^{2} + \frac {1}{4} \, {\left (x^{2} - x\right )} e^{2} - \frac {1}{4} \, {\left (3 \, x^{2} - x e^{2} + x\right )} e^{\left (x - 4\right )} + \frac {5}{4} \, x \] Input:
integrate(1/8*((2+2*x)*exp(2)-6*x^2-14*x-2)*exp(-4+x)+1/8*(4*x-2)*exp(2)-9 /4*x^2+9/8*x+5/4,x, algorithm="fricas")
Output:
-3/4*x^3 + 9/16*x^2 + 1/4*(x^2 - x)*e^2 - 1/4*(3*x^2 - x*e^2 + x)*e^(x - 4 ) + 5/4*x
Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.28 \[ \int \frac {1}{8} \left (10+9 x-18 x^2+e^2 (-2+4 x)+e^{-4+x} \left (-2-14 x-6 x^2+e^2 (2+2 x)\right )\right ) \, dx=- \frac {3 x^{3}}{4} + x^{2} \cdot \left (\frac {9}{16} + \frac {e^{2}}{4}\right ) + x \left (\frac {5}{4} - \frac {e^{2}}{4}\right ) + \frac {\left (- 3 x^{2} - x + x e^{2}\right ) e^{x - 4}}{4} \] Input:
integrate(1/8*((2+2*x)*exp(2)-6*x**2-14*x-2)*exp(-4+x)+1/8*(4*x-2)*exp(2)- 9/4*x**2+9/8*x+5/4,x)
Output:
-3*x**3/4 + x**2*(9/16 + exp(2)/4) + x*(5/4 - exp(2)/4) + (-3*x**2 - x + x *exp(2))*exp(x - 4)/4
Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.22 \[ \int \frac {1}{8} \left (10+9 x-18 x^2+e^2 (-2+4 x)+e^{-4+x} \left (-2-14 x-6 x^2+e^2 (2+2 x)\right )\right ) \, dx=-\frac {3}{4} \, x^{3} + \frac {9}{16} \, x^{2} + \frac {1}{4} \, {\left (x^{2} - x\right )} e^{2} - \frac {1}{4} \, {\left (3 \, x^{2} - x {\left (e^{2} - 1\right )}\right )} e^{\left (x - 4\right )} + \frac {5}{4} \, x \] Input:
integrate(1/8*((2+2*x)*exp(2)-6*x^2-14*x-2)*exp(-4+x)+1/8*(4*x-2)*exp(2)-9 /4*x^2+9/8*x+5/4,x, algorithm="maxima")
Output:
-3/4*x^3 + 9/16*x^2 + 1/4*(x^2 - x)*e^2 - 1/4*(3*x^2 - x*(e^2 - 1))*e^(x - 4) + 5/4*x
Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.25 \[ \int \frac {1}{8} \left (10+9 x-18 x^2+e^2 (-2+4 x)+e^{-4+x} \left (-2-14 x-6 x^2+e^2 (2+2 x)\right )\right ) \, dx=-\frac {3}{4} \, x^{3} + \frac {9}{16} \, x^{2} + \frac {1}{4} \, {\left (x^{2} - x\right )} e^{2} + \frac {1}{4} \, x e^{\left (x - 2\right )} - \frac {1}{4} \, {\left (3 \, x^{2} + x\right )} e^{\left (x - 4\right )} + \frac {5}{4} \, x \] Input:
integrate(1/8*((2+2*x)*exp(2)-6*x^2-14*x-2)*exp(-4+x)+1/8*(4*x-2)*exp(2)-9 /4*x^2+9/8*x+5/4,x, algorithm="giac")
Output:
-3/4*x^3 + 9/16*x^2 + 1/4*(x^2 - x)*e^2 + 1/4*x*e^(x - 2) - 1/4*(3*x^2 + x )*e^(x - 4) + 5/4*x
Time = 0.07 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.31 \[ \int \frac {1}{8} \left (10+9 x-18 x^2+e^2 (-2+4 x)+e^{-4+x} \left (-2-14 x-6 x^2+e^2 (2+2 x)\right )\right ) \, dx=x^2\,\left (\frac {{\mathrm {e}}^2}{4}+\frac {9}{16}\right )-\frac {3\,x^2\,{\mathrm {e}}^{x-4}}{4}-\frac {3\,x^3}{4}-x\,\left (\frac {{\mathrm {e}}^2}{4}-\frac {5}{4}\right )+\frac {x\,{\mathrm {e}}^{x-4}\,\left (2\,{\mathrm {e}}^2-2\right )}{8} \] Input:
int((9*x)/8 - (exp(x - 4)*(14*x + 6*x^2 - exp(2)*(2*x + 2) + 2))/8 - (9*x^ 2)/4 + (exp(2)*(4*x - 2))/8 + 5/4,x)
Output:
x^2*(exp(2)/4 + 9/16) - (3*x^2*exp(x - 4))/4 - (3*x^3)/4 - x*(exp(2)/4 - 5 /4) + (x*exp(x - 4)*(2*exp(2) - 2))/8
Time = 0.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.56 \[ \int \frac {1}{8} \left (10+9 x-18 x^2+e^2 (-2+4 x)+e^{-4+x} \left (-2-14 x-6 x^2+e^2 (2+2 x)\right )\right ) \, dx=\frac {x \left (4 e^{x} e^{2}-12 e^{x} x -4 e^{x}+4 e^{6} x -4 e^{6}-12 e^{4} x^{2}+9 e^{4} x +20 e^{4}\right )}{16 e^{4}} \] Input:
int(1/8*((2+2*x)*exp(2)-6*x^2-14*x-2)*exp(-4+x)+1/8*(4*x-2)*exp(2)-9/4*x^2 +9/8*x+5/4,x)
Output:
(x*(4*e**x*e**2 - 12*e**x*x - 4*e**x + 4*e**6*x - 4*e**6 - 12*e**4*x**2 + 9*e**4*x + 20*e**4))/(16*e**4)