Integrand size = 97, antiderivative size = 21 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2 e^{\frac {1}{5} x \left (5+2 e^{-3+x}+x\right )^2} \] Output:
2*exp(1/5*x*(5+2*exp(x)/exp(3)+x)^2)
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2 e^{\frac {x \left (2 e^x+e^3 (5+x)\right )^2}{5 e^6}} \] Input:
Integrate[(E^(-6 + (4*E^(2*x)*x + E^(3 + x)*(20*x + 4*x^2) + E^6*(25*x + 1 0*x^2 + x^3))/(5*E^6))*(E^(2*x)*(8 + 16*x) + E^6*(50 + 40*x + 6*x^2) + E^( 3 + x)*(40 + 56*x + 8*x^2)))/5,x]
Output:
2*E^((x*(2*E^x + E^3*(5 + x))^2)/(5*E^6))
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.
\(\displaystyle \int \frac {1}{5} \left (e^6 \left (6 x^2+40 x+50\right )+e^{x+3} \left (8 x^2+56 x+40\right )+e^{2 x} (16 x+8)\right ) \exp \left (\frac {e^{x+3} \left (4 x^2+20 x\right )+e^6 \left (x^3+10 x^2+25 x\right )+4 e^{2 x} x}{5 e^6}-6\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int 2 \exp \left (\frac {4 e^{2 x} x+4 e^{x+3} \left (x^2+5 x\right )+e^6 \left (x^3+10 x^2+25 x\right )}{5 e^6}-6\right ) \left (4 e^{2 x} (2 x+1)+4 e^{x+3} \left (x^2+7 x+5\right )+e^6 \left (3 x^2+20 x+25\right )\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{5} \int \exp \left (\frac {4 e^{2 x} x+4 e^{x+3} \left (x^2+5 x\right )+e^6 \left (x^3+10 x^2+25 x\right )}{5 e^6}-6\right ) \left (4 e^{2 x} (2 x+1)+4 e^{x+3} \left (x^2+7 x+5\right )+e^6 \left (3 x^2+20 x+25\right )\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {2}{5} \int \left (4 \exp \left (2 x+\frac {4 e^{2 x} x+4 e^{x+3} \left (x^2+5 x\right )+e^6 \left (x^3+10 x^2+25 x\right )}{5 e^6}-6\right ) (2 x+1)+4 \exp \left (x+\frac {4 e^{2 x} x+4 e^{x+3} \left (x^2+5 x\right )+e^6 \left (x^3+10 x^2+25 x\right )}{5 e^6}-3\right ) \left (x^2+7 x+5\right )+\exp \left (\frac {4 e^{2 x} x+4 e^{x+3} \left (x^2+5 x\right )+e^6 \left (x^3+10 x^2+25 x\right )}{5 e^6}\right ) \left (3 x^2+20 x+25\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{5} \left (20 \int \exp \left (x+\frac {4 e^{2 x} x+4 e^{x+3} \left (x^2+5 x\right )+e^6 \left (x^3+10 x^2+25 x\right )}{5 e^6}-3\right )dx+4 \int \exp \left (2 x+\frac {4 e^{2 x} x+4 e^{x+3} \left (x^2+5 x\right )+e^6 \left (x^3+10 x^2+25 x\right )}{5 e^6}-6\right )dx+28 \int \exp \left (x+\frac {4 e^{2 x} x+4 e^{x+3} \left (x^2+5 x\right )+e^6 \left (x^3+10 x^2+25 x\right )}{5 e^6}-3\right ) xdx+8 \int \exp \left (2 x+\frac {4 e^{2 x} x+4 e^{x+3} \left (x^2+5 x\right )+e^6 \left (x^3+10 x^2+25 x\right )}{5 e^6}-6\right ) xdx+4 \int \exp \left (x+\frac {4 e^{2 x} x+4 e^{x+3} \left (x^2+5 x\right )+e^6 \left (x^3+10 x^2+25 x\right )}{5 e^6}-3\right ) x^2dx+3 \int e^{\frac {x \left (e^3 (x+5)+2 e^x\right )^2}{5 e^6}} x^2dx+25 \int e^{\frac {x \left (e^3 (x+5)+2 e^x\right )^2}{5 e^6}}dx+20 \int e^{\frac {x \left (e^3 (x+5)+2 e^x\right )^2}{5 e^6}} xdx\right )\) |
Input:
Int[(E^(-6 + (4*E^(2*x)*x + E^(3 + x)*(20*x + 4*x^2) + E^6*(25*x + 10*x^2 + x^3))/(5*E^6))*(E^(2*x)*(8 + 16*x) + E^6*(50 + 40*x + 6*x^2) + E^(3 + x) *(40 + 56*x + 8*x^2)))/5,x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(43\) vs. \(2(19)=38\).
Time = 0.59 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.10
method | result | size |
risch | \(2 \,{\mathrm e}^{\frac {x \left (x^{2} {\mathrm e}^{6}+4 \,{\mathrm e}^{3+x} x +10 x \,{\mathrm e}^{6}+4 \,{\mathrm e}^{2 x}+20 \,{\mathrm e}^{3+x}+25 \,{\mathrm e}^{6}\right ) {\mathrm e}^{-6}}{5}}\) | \(44\) |
norman | \(2 \,{\mathrm e}^{\frac {\left (4 x \,{\mathrm e}^{2 x}+\left (4 x^{2}+20 x \right ) {\mathrm e}^{3} {\mathrm e}^{x}+\left (x^{3}+10 x^{2}+25 x \right ) {\mathrm e}^{6}\right ) {\mathrm e}^{-6}}{5}}\) | \(49\) |
parallelrisch | \(2 \,{\mathrm e}^{\frac {\left (4 x \,{\mathrm e}^{2 x}+\left (4 x^{2}+20 x \right ) {\mathrm e}^{3} {\mathrm e}^{x}+\left (x^{3}+10 x^{2}+25 x \right ) {\mathrm e}^{6}\right ) {\mathrm e}^{-6}}{5}}\) | \(49\) |
Input:
int(1/5*((16*x+8)*exp(x)^2+(8*x^2+56*x+40)*exp(3)*exp(x)+(6*x^2+40*x+50)*e xp(3)^2)*exp(1/5*(4*x*exp(x)^2+(4*x^2+20*x)*exp(3)*exp(x)+(x^3+10*x^2+25*x )*exp(3)^2)/exp(3)^2)/exp(3)^2,x,method=_RETURNVERBOSE)
Output:
2*exp(1/5*x*(x^2*exp(6)+4*exp(3+x)*x+10*x*exp(6)+4*exp(2*x)+20*exp(3+x)+25 *exp(6))*exp(-6))
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (17) = 34\).
Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.29 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2 \, e^{\left (\frac {1}{5} \, {\left ({\left (x^{3} + 10 \, x^{2} + 25 \, x - 30\right )} e^{12} + 4 \, x e^{\left (2 \, x + 6\right )} + 4 \, {\left (x^{2} + 5 \, x\right )} e^{\left (x + 9\right )}\right )} e^{\left (-12\right )} + 6\right )} \] Input:
integrate(1/5*((16*x+8)*exp(x)^2+(8*x^2+56*x+40)*exp(3)*exp(x)+(6*x^2+40*x +50)*exp(3)^2)*exp(1/5*(4*x*exp(x)^2+(4*x^2+20*x)*exp(3)*exp(x)+(x^3+10*x^ 2+25*x)*exp(3)^2)/exp(3)^2)/exp(3)^2,x, algorithm="fricas")
Output:
2*e^(1/5*((x^3 + 10*x^2 + 25*x - 30)*e^12 + 4*x*e^(2*x + 6) + 4*(x^2 + 5*x )*e^(x + 9))*e^(-12) + 6)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (19) = 38\).
Time = 0.23 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.33 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2 e^{\frac {\frac {4 x e^{2 x}}{5} + \frac {\left (4 x^{2} + 20 x\right ) e^{3} e^{x}}{5} + \frac {\left (x^{3} + 10 x^{2} + 25 x\right ) e^{6}}{5}}{e^{6}}} \] Input:
integrate(1/5*((16*x+8)*exp(x)**2+(8*x**2+56*x+40)*exp(3)*exp(x)+(6*x**2+4 0*x+50)*exp(3)**2)*exp(1/5*(4*x*exp(x)**2+(4*x**2+20*x)*exp(3)*exp(x)+(x** 3+10*x**2+25*x)*exp(3)**2)/exp(3)**2)/exp(3)**2,x)
Output:
2*exp((4*x*exp(2*x)/5 + (4*x**2 + 20*x)*exp(3)*exp(x)/5 + (x**3 + 10*x**2 + 25*x)*exp(6)/5)*exp(-6))
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (17) = 34\).
Time = 0.24 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.00 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2 \, e^{\left (\frac {1}{5} \, x^{3} + \frac {4}{5} \, x^{2} e^{\left (x - 3\right )} + 2 \, x^{2} + \frac {4}{5} \, x e^{\left (2 \, x - 6\right )} + 4 \, x e^{\left (x - 3\right )} + 5 \, x\right )} \] Input:
integrate(1/5*((16*x+8)*exp(x)^2+(8*x^2+56*x+40)*exp(3)*exp(x)+(6*x^2+40*x +50)*exp(3)^2)*exp(1/5*(4*x*exp(x)^2+(4*x^2+20*x)*exp(3)*exp(x)+(x^3+10*x^ 2+25*x)*exp(3)^2)/exp(3)^2)/exp(3)^2,x, algorithm="maxima")
Output:
2*e^(1/5*x^3 + 4/5*x^2*e^(x - 3) + 2*x^2 + 4/5*x*e^(2*x - 6) + 4*x*e^(x - 3) + 5*x)
\[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=\int { \frac {2}{5} \, {\left ({\left (3 \, x^{2} + 20 \, x + 25\right )} e^{6} + 4 \, {\left (2 \, x + 1\right )} e^{\left (2 \, x\right )} + 4 \, {\left (x^{2} + 7 \, x + 5\right )} e^{\left (x + 3\right )}\right )} e^{\left (\frac {1}{5} \, {\left ({\left (x^{3} + 10 \, x^{2} + 25 \, x\right )} e^{6} + 4 \, x e^{\left (2 \, x\right )} + 4 \, {\left (x^{2} + 5 \, x\right )} e^{\left (x + 3\right )}\right )} e^{\left (-6\right )} - 6\right )} \,d x } \] Input:
integrate(1/5*((16*x+8)*exp(x)^2+(8*x^2+56*x+40)*exp(3)*exp(x)+(6*x^2+40*x +50)*exp(3)^2)*exp(1/5*(4*x*exp(x)^2+(4*x^2+20*x)*exp(3)*exp(x)+(x^3+10*x^ 2+25*x)*exp(3)^2)/exp(3)^2)/exp(3)^2,x, algorithm="giac")
Output:
integrate(2/5*((3*x^2 + 20*x + 25)*e^6 + 4*(2*x + 1)*e^(2*x) + 4*(x^2 + 7* x + 5)*e^(x + 3))*e^(1/5*((x^3 + 10*x^2 + 25*x)*e^6 + 4*x*e^(2*x) + 4*(x^2 + 5*x)*e^(x + 3))*e^(-6) - 6), x)
Time = 3.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.19 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2\,{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^{\frac {x^3}{5}}\,{\mathrm {e}}^{\frac {4\,x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-6}}{5}}\,{\mathrm {e}}^{\frac {4\,x^2\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^x}{5}} \] Input:
int((exp(-6)*exp(exp(-6)*((exp(6)*(25*x + 10*x^2 + x^3))/5 + (4*x*exp(2*x) )/5 + (exp(3)*exp(x)*(20*x + 4*x^2))/5))*(exp(6)*(40*x + 6*x^2 + 50) + exp (2*x)*(16*x + 8) + exp(3)*exp(x)*(56*x + 8*x^2 + 40)))/5,x)
Output:
2*exp(5*x)*exp(4*x*exp(-3)*exp(x))*exp(2*x^2)*exp(x^3/5)*exp((4*x*exp(2*x) *exp(-6))/5)*exp((4*x^2*exp(-3)*exp(x))/5)
Time = 0.77 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.81 \[ \int \frac {1}{5} e^{-6+\frac {4 e^{2 x} x+e^{3+x} \left (20 x+4 x^2\right )+e^6 \left (25 x+10 x^2+x^3\right )}{5 e^6}} \left (e^{2 x} (8+16 x)+e^6 \left (50+40 x+6 x^2\right )+e^{3+x} \left (40+56 x+8 x^2\right )\right ) \, dx=2 e^{\frac {4 e^{2 x} x +4 e^{x} e^{3} x^{2}+20 e^{x} e^{3} x +e^{6} x^{3}+10 e^{6} x^{2}+25 e^{6} x}{5 e^{6}}} \] Input:
int(1/5*((16*x+8)*exp(x)^2+(8*x^2+56*x+40)*exp(3)*exp(x)+(6*x^2+40*x+50)*e xp(3)^2)*exp(1/5*(4*x*exp(x)^2+(4*x^2+20*x)*exp(3)*exp(x)+(x^3+10*x^2+25*x )*exp(3)^2)/exp(3)^2)/exp(3)^2,x)
Output:
2*e**((4*e**(2*x)*x + 4*e**x*e**3*x**2 + 20*e**x*e**3*x + e**6*x**3 + 10*e **6*x**2 + 25*e**6*x)/(5*e**6))