Integrand size = 59, antiderivative size = 20 \[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {1}{5} e^{11-4 e^{-2 x} x} (-3+x) x \] Output:
1/5*exp(3-4*x/exp(x)^2)*exp(8)*x*(-3+x)
Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {1}{5} e^{11-4 e^{-2 x} x} (-3+x) x \] Input:
Integrate[(E^((3*E^(2*x) - 4*x)/E^(2*x) - 2*x)*(E^(8 + 2*x)*(-3 + 2*x) + E ^8*(12*x - 28*x^2 + 8*x^3)))/5,x]
Output:
(E^(11 - (4*x)/E^(2*x))*(-3 + x)*x)/5
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.
\(\displaystyle \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^8 \left (8 x^3-28 x^2+12 x\right )+e^{2 x+8} (2 x-3)\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{2 x+8} (3-2 x)-4 e^8 \left (2 x^3-7 x^2+3 x\right )\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{2 x+8} (3-2 x)-4 e^8 \left (2 x^3-7 x^2+3 x\right )\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {1}{5} \int e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x+8} \left (-8 x^3+28 x^2-2 e^{2 x} x-12 x+3 e^{2 x}\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{5} \int \left (-8 e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x+8} x^3+28 e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x+8} x^2-2 e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )+8} x-12 e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x+8} x+3 e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )+8}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (8 \int e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x+8} x^3dx-28 \int e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x+8} x^2dx-3 \int e^{11-4 e^{-2 x} x}dx+12 \int e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x+8} xdx+2 \int e^{11-4 e^{-2 x} x} xdx\right )\) |
Input:
Int[(E^((3*E^(2*x) - 4*x)/E^(2*x) - 2*x)*(E^(8 + 2*x)*(-3 + 2*x) + E^8*(12 *x - 28*x^2 + 8*x^3)))/5,x]
Output:
$Aborted
Time = 0.33 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35
method | result | size |
risch | \(\frac {x \left (-3+x \right ) {\mathrm e}^{3 \,{\mathrm e}^{-2 x} {\mathrm e}^{2 x}-4 \,{\mathrm e}^{-2 x} x +8}}{5}\) | \(27\) |
parallelrisch | \(\frac {{\mathrm e}^{8} x^{2} {\mathrm e}^{\left (3 \,{\mathrm e}^{2 x}-4 x \right ) {\mathrm e}^{-2 x}}}{5}-\frac {3 \,{\mathrm e}^{8} x \,{\mathrm e}^{\left (3 \,{\mathrm e}^{2 x}-4 x \right ) {\mathrm e}^{-2 x}}}{5}\) | \(46\) |
norman | \(\left (\frac {x^{2} {\mathrm e}^{8} {\mathrm e}^{2 x} {\mathrm e}^{\left (3 \,{\mathrm e}^{2 x}-4 x \right ) {\mathrm e}^{-2 x}}}{5}-\frac {3 \,{\mathrm e}^{8} {\mathrm e}^{2 x} x \,{\mathrm e}^{\left (3 \,{\mathrm e}^{2 x}-4 x \right ) {\mathrm e}^{-2 x}}}{5}\right ) {\mathrm e}^{-2 x}\) | \(59\) |
Input:
int(1/5*((-3+2*x)*exp(8)*exp(x)^2+(8*x^3-28*x^2+12*x)*exp(8))*exp((3*exp(x )^2-4*x)/exp(x)^2)/exp(x)^2,x,method=_RETURNVERBOSE)
Output:
1/5*x*(-3+x)*exp(3*exp(-2*x)*exp(2*x)-4*exp(-2*x)*x+8)
Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (16) = 32\).
Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.05 \[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {1}{5} \, {\left (x^{2} - 3 \, x\right )} e^{\left (-{\left (4 \, x e^{8} + {\left (2 \, x - 3\right )} e^{\left (2 \, x + 8\right )}\right )} e^{\left (-2 \, x - 8\right )} + 2 \, x + 8\right )} \] Input:
integrate(1/5*((-3+2*x)*exp(8)*exp(x)^2+(8*x^3-28*x^2+12*x)*exp(8))*exp((3 *exp(x)^2-4*x)/exp(x)^2)/exp(x)^2,x, algorithm="fricas")
Output:
1/5*(x^2 - 3*x)*e^(-(4*x*e^8 + (2*x - 3)*e^(2*x + 8))*e^(-2*x - 8) + 2*x + 8)
Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {\left (x^{2} e^{8} - 3 x e^{8}\right ) e^{\left (- 4 x + 3 e^{2 x}\right ) e^{- 2 x}}}{5} \] Input:
integrate(1/5*((-3+2*x)*exp(8)*exp(x)**2+(8*x**3-28*x**2+12*x)*exp(8))*exp ((3*exp(x)**2-4*x)/exp(x)**2)/exp(x)**2,x)
Output:
(x**2*exp(8) - 3*x*exp(8))*exp((-4*x + 3*exp(2*x))*exp(-2*x))/5
\[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\int { \frac {1}{5} \, {\left (4 \, {\left (2 \, x^{3} - 7 \, x^{2} + 3 \, x\right )} e^{8} + {\left (2 \, x - 3\right )} e^{\left (2 \, x + 8\right )}\right )} e^{\left (-{\left (4 \, x - 3 \, e^{\left (2 \, x\right )}\right )} e^{\left (-2 \, x\right )} - 2 \, x\right )} \,d x } \] Input:
integrate(1/5*((-3+2*x)*exp(8)*exp(x)^2+(8*x^3-28*x^2+12*x)*exp(8))*exp((3 *exp(x)^2-4*x)/exp(x)^2)/exp(x)^2,x, algorithm="maxima")
Output:
1/5*integrate((4*(2*x^3 - 7*x^2 + 3*x)*e^8 + (2*x - 3)*e^(2*x + 8))*e^(-(4 *x - 3*e^(2*x))*e^(-2*x) - 2*x), x)
Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (16) = 32\).
Time = 0.12 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.00 \[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {1}{5} \, {\left (x^{2} e^{\left (-4 \, x e^{\left (-2 \, x\right )} - 2 \, x + 11\right )} - 3 \, x e^{\left (-4 \, x e^{\left (-2 \, x\right )} - 2 \, x + 11\right )}\right )} e^{\left (2 \, x\right )} \] Input:
integrate(1/5*((-3+2*x)*exp(8)*exp(x)^2+(8*x^3-28*x^2+12*x)*exp(8))*exp((3 *exp(x)^2-4*x)/exp(x)^2)/exp(x)^2,x, algorithm="giac")
Output:
1/5*(x^2*e^(-4*x*e^(-2*x) - 2*x + 11) - 3*x*e^(-4*x*e^(-2*x) - 2*x + 11))* e^(2*x)
Time = 3.49 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {x\,{\mathrm {e}}^{11}\,{\mathrm {e}}^{-4\,x\,{\mathrm {e}}^{-2\,x}}\,\left (x-3\right )}{5} \] Input:
int((exp(-2*x)*exp(-exp(-2*x)*(4*x - 3*exp(2*x)))*(exp(8)*(12*x - 28*x^2 + 8*x^3) + exp(2*x)*exp(8)*(2*x - 3)))/5,x)
Output:
(x*exp(11)*exp(-4*x*exp(-2*x))*(x - 3))/5
\[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {e^{11} \left (8 \left (\int \frac {x^{3}}{e^{\frac {2 e^{2 x} x +4 x}{e^{2 x}}}}d x \right )-28 \left (\int \frac {x^{2}}{e^{\frac {2 e^{2 x} x +4 x}{e^{2 x}}}}d x \right )+12 \left (\int \frac {x}{e^{\frac {2 e^{2 x} x +4 x}{e^{2 x}}}}d x \right )+2 \left (\int \frac {x}{e^{\frac {4 x}{e^{2 x}}}}d x \right )-3 \left (\int \frac {1}{e^{\frac {4 x}{e^{2 x}}}}d x \right )\right )}{5} \] Input:
int(1/5*((-3+2*x)*exp(8)*exp(x)^2+(8*x^3-28*x^2+12*x)*exp(8))*exp((3*exp(x )^2-4*x)/exp(x)^2)/exp(x)^2,x)
Output:
(e**11*(8*int(x**3/e**((2*e**(2*x)*x + 4*x)/e**(2*x)),x) - 28*int(x**2/e** ((2*e**(2*x)*x + 4*x)/e**(2*x)),x) + 12*int(x/e**((2*e**(2*x)*x + 4*x)/e** (2*x)),x) + 2*int(x/e**((4*x)/e**(2*x)),x) - 3*int(1/e**((4*x)/e**(2*x)),x )))/5