\(\int \frac {1}{5} e^{e^{-2 x} (3 e^{2 x}-4 x)-2 x} (e^{8+2 x} (-3+2 x)+e^8 (12 x-28 x^2+8 x^3)) \, dx\) [4544]

Optimal result

Integrand size = 59, antiderivative size = 20 \[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {1}{5} e^{11-4 e^{-2 x} x} (-3+x) x \]

[Out]

1/5*exp(3-4*x/exp(x)^2)*exp(8)*x*(-3+x)

Rubi [F]

\[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx \]

[In]

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]

[Out]

(-3*Defer[Int][E^(11 - (4*x)/E^(2*x)), x])/5 + (2*Defer[Int][E^(11 - (4*x)/E^(2*x))*x, x])/5 + (12*Defer[Int][
E^(11 - 2*x - (4*x)/E^(2*x))*x, x])/5 - (28*Defer[Int][E^(11 - 2*x - (4*x)/E^(2*x))*x^2, x])/5 + (8*Defer[Int]
[E^(11 - 2*x - (4*x)/E^(2*x))*x^3, x])/5

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \int 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} \int e^{11-2 x-4 e^{-2 x} x} \left (e^{2 x} (-3+2 x)+4 x \left (3-7 x+2 x^2\right )\right ) \, dx \\ & = \frac {1}{5} \int \left (e^{11-4 e^{-2 x} x} (-3+2 x)+4 e^{11-2 x-4 e^{-2 x} x} x \left (3-7 x+2 x^2\right )\right ) \, dx \\ & = \frac {1}{5} \int e^{11-4 e^{-2 x} x} (-3+2 x) \, dx+\frac {4}{5} \int e^{11-2 x-4 e^{-2 x} x} x \left (3-7 x+2 x^2\right ) \, dx \\ & = \frac {1}{5} \int \left (-3 e^{11-4 e^{-2 x} x}+2 e^{11-4 e^{-2 x} x} x\right ) \, dx+\frac {4}{5} \int \left (3 e^{11-2 x-4 e^{-2 x} x} x-7 e^{11-2 x-4 e^{-2 x} x} x^2+2 e^{11-2 x-4 e^{-2 x} x} x^3\right ) \, dx \\ & = \frac {2}{5} \int e^{11-4 e^{-2 x} x} x \, dx-\frac {3}{5} \int e^{11-4 e^{-2 x} x} \, dx+\frac {8}{5} \int e^{11-2 x-4 e^{-2 x} x} x^3 \, dx+\frac {12}{5} \int e^{11-2 x-4 e^{-2 x} x} x \, dx-\frac {28}{5} \int e^{11-2 x-4 e^{-2 x} x} x^2 \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {1}{5} e^{11-4 e^{-2 x} x} (-3+x) x \]

[In]

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]

[Out]

(E^(11 - (4*x)/E^(2*x))*(-3 + x)*x)/5

Maple [A] (verified)

Time = 0.31 (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	\(\left (\frac {x^{2} {\mathrm e}^{8} {\mathrm e}^{2 x} {\mathrm e}^{\left (3 \,{\mathrm e}^{2 x}-4 x \right ) {\mathrm e}^{-2 x}}}{5}-\frac {3 \,{\mathrm e}^{8} {\mathrm e}^{2 x} x \,{\mathrm e}^{\left (3 \,{\mathrm e}^{2 x}-4 x \right ) {\mathrm e}^{-2 x}}}{5}\right ) {\mathrm e}^{-2 x}\)	\(59\)

[In]

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)

[Out]

1/5*x*(-3+x)*exp(3*exp(-2*x)*exp(2*x)-4*exp(-2*x)*x+8)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (16) = 32\).

Time = 0.24 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.05 \[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {1}{5} \, {\left (x^{2} - 3 \, x\right )} e^{\left (-{\left (4 \, x e^{8} + {\left (2 \, x - 3\right )} e^{\left (2 \, x + 8\right )}\right )} e^{\left (-2 \, x - 8\right )} + 2 \, x + 8\right )} \]

[In]

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")

[Out]

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)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55 \[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {\left (x^{2} e^{8} - 3 x e^{8}\right ) e^{\left (- 4 x + 3 e^{2 x}\right ) e^{- 2 x}}}{5} \]

[In]

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)

[Out]

(x**2*exp(8) - 3*x*exp(8))*exp((-4*x + 3*exp(2*x))*exp(-2*x))/5

Maxima [F]

\[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\int { \frac {1}{5} \, {\left (4 \, {\left (2 \, x^{3} - 7 \, x^{2} + 3 \, x\right )} e^{8} + {\left (2 \, x - 3\right )} e^{\left (2 \, x + 8\right )}\right )} e^{\left (-{\left (4 \, x - 3 \, e^{\left (2 \, x\right )}\right )} e^{\left (-2 \, x\right )} - 2 \, x\right )} \,d x } \]

[In]

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")

[Out]

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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 40 vs. \(2 (16) = 32\).

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 2.00 \[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {1}{5} \, {\left (x^{2} e^{\left (-4 \, x e^{\left (-2 \, x\right )} - 2 \, x + 11\right )} - 3 \, x e^{\left (-4 \, x e^{\left (-2 \, x\right )} - 2 \, x + 11\right )}\right )} e^{\left (2 \, x\right )} \]

[In]

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")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 12.93 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {1}{5} e^{e^{-2 x} \left (3 e^{2 x}-4 x\right )-2 x} \left (e^{8+2 x} (-3+2 x)+e^8 \left (12 x-28 x^2+8 x^3\right )\right ) \, dx=\frac {x\,{\mathrm {e}}^{11}\,{\mathrm {e}}^{-4\,x\,{\mathrm {e}}^{-2\,x}}\,\left (x-3\right )}{5} \]

[In]

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)

[Out]

(x*exp(11)*exp(-4*x*exp(-2*x))*(x - 3))/5