\(\int \frac {e^{2-e^x-4 x} (-e^{2 x} x^2-4 x^3+e^x (2 x-6 x^2-x^3))}{5 e^{3 x}+15 e^{2 x} x+15 e^x x^2+5 x^3} \, dx\) [3128]

Optimal result

Integrand size = 78, antiderivative size = 26 \[ \int \frac {e^{2-e^x-4 x} \left (-e^{2 x} x^2-4 x^3+e^x \left (2 x-6 x^2-x^3\right )\right )}{5 e^{3 x}+15 e^{2 x} x+15 e^x x^2+5 x^3} \, dx=\frac {e^{2-e^x-4 x} x^2}{5 \left (e^x+x\right )^2} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{2-e^x-4 x} \left (-e^{2 x} x^2-4 x^3+e^x \left (2 x-6 x^2-x^3\right )\right )}{5 e^{3 x}+15 e^{2 x} x+15 e^x x^2+5 x^3} \, dx=\int \frac {e^{2-e^x-4 x} \left (-e^{2 x} x^2-4 x^3+e^x \left (2 x-6 x^2-x^3\right )\right )}{5 e^{3 x}+15 e^{2 x} x+15 e^x x^2+5 x^3} \, dx \]

[In]

Int[(E^(2 - E^x - 4*x)*(-(E^(2*x)*x^2) - 4*x^3 + E^x*(2*x - 6*x^2 - x^3)))/(5*E^(3*x) + 15*E^(2*x)*x + 15*E^x*
x^2 + 5*x^3),x]

[Out]

(-2*Defer[Int][(E^(2 - E^x - 4*x)*x^2)/(E^x + x)^3, x])/5 + (2*Defer[Int][(E^(2 - E^x - 4*x)*x^3)/(E^x + x)^3,
x])/5 + (2*Defer[Int][(E^(2 - E^x - 4*x)*x)/(E^x + x)^2, x])/5 - (6*Defer[Int][(E^(2 - E^x - 4*x)*x^2)/(E^x +
x)^2, x])/5 + Defer[Int][(E^(2 - E^x - 4*x)*x^3)/(E^x + x)^2, x]/5 - Defer[Int][(E^(2 - E^x - 4*x)*x^2)/(E^x
+ x), x]/5

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2-e^x-4 x} x \left (-e^{2 x} x-4 x^2-e^x \left (-2+6 x+x^2\right )\right )}{5 \left (e^x+x\right )^3} \, dx \\ & = \frac {1}{5} \int \frac {e^{2-e^x-4 x} x \left (-e^{2 x} x-4 x^2-e^x \left (-2+6 x+x^2\right )\right )}{\left (e^x+x\right )^3} \, dx \\ & = \frac {1}{5} \int \left (\frac {2 e^{2-e^x-4 x} (-1+x) x^2}{\left (e^x+x\right )^3}-\frac {e^{2-e^x-4 x} x^2}{e^x+x}+\frac {e^{2-e^x-4 x} x \left (2-6 x+x^2\right )}{\left (e^x+x\right )^2}\right ) \, dx \\ & = -\left (\frac {1}{5} \int \frac {e^{2-e^x-4 x} x^2}{e^x+x} \, dx\right )+\frac {1}{5} \int \frac {e^{2-e^x-4 x} x \left (2-6 x+x^2\right )}{\left (e^x+x\right )^2} \, dx+\frac {2}{5} \int \frac {e^{2-e^x-4 x} (-1+x) x^2}{\left (e^x+x\right )^3} \, dx \\ & = -\left (\frac {1}{5} \int \frac {e^{2-e^x-4 x} x^2}{e^x+x} \, dx\right )+\frac {1}{5} \int \left (\frac {2 e^{2-e^x-4 x} x}{\left (e^x+x\right )^2}-\frac {6 e^{2-e^x-4 x} x^2}{\left (e^x+x\right )^2}+\frac {e^{2-e^x-4 x} x^3}{\left (e^x+x\right )^2}\right ) \, dx+\frac {2}{5} \int \left (-\frac {e^{2-e^x-4 x} x^2}{\left (e^x+x\right )^3}+\frac {e^{2-e^x-4 x} x^3}{\left (e^x+x\right )^3}\right ) \, dx \\ & = \frac {1}{5} \int \frac {e^{2-e^x-4 x} x^3}{\left (e^x+x\right )^2} \, dx-\frac {1}{5} \int \frac {e^{2-e^x-4 x} x^2}{e^x+x} \, dx-\frac {2}{5} \int \frac {e^{2-e^x-4 x} x^2}{\left (e^x+x\right )^3} \, dx+\frac {2}{5} \int \frac {e^{2-e^x-4 x} x^3}{\left (e^x+x\right )^3} \, dx+\frac {2}{5} \int \frac {e^{2-e^x-4 x} x}{\left (e^x+x\right )^2} \, dx-\frac {6}{5} \int \frac {e^{2-e^x-4 x} x^2}{\left (e^x+x\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2-e^x-4 x} \left (-e^{2 x} x^2-4 x^3+e^x \left (2 x-6 x^2-x^3\right )\right )}{5 e^{3 x}+15 e^{2 x} x+15 e^x x^2+5 x^3} \, dx=\frac {e^{2-e^x-4 x} x^2}{5 \left (e^x+x\right )^2} \]

[In]

Integrate[(E^(2 - E^x - 4*x)*(-(E^(2*x)*x^2) - 4*x^3 + E^x*(2*x - 6*x^2 - x^3)))/(5*E^(3*x) + 15*E^(2*x)*x + 1
5*E^x*x^2 + 5*x^3),x]

[Out]

(E^(2 - E^x - 4*x)*x^2)/(5*(E^x + x)^2)

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {x^{2} {\mathrm e}^{-4 x +2-{\mathrm e}^{x}}}{5 \left ({\mathrm e}^{x}+x \right )^{2}}\)	\(22\)
parallelrisch	\(\frac {x^{2} {\mathrm e}^{-4 x +2} {\mathrm e}^{-{\mathrm e}^{x}}}{5 \,{\mathrm e}^{2 x}+10 \,{\mathrm e}^{x} x +5 x^{2}}\)	\(34\)

[In]

int((-exp(x)^2*x^2+(-x^3-6*x^2+2*x)*exp(x)-4*x^3)/(5*exp(x)^3+15*x*exp(x)^2+15*exp(x)*x^2+5*x^3)/exp(-1+2*x)^2
/exp(exp(x)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {e^{2-e^x-4 x} \left (-e^{2 x} x^2-4 x^3+e^x \left (2 x-6 x^2-x^3\right )\right )}{5 e^{3 x}+15 e^{2 x} x+15 e^x x^2+5 x^3} \, dx=\frac {x^{2} e^{\left (-4 \, x - e^{x} + 2\right )}}{5 \, {\left (x^{2} + 2 \, x e^{x} + e^{\left (2 \, x\right )}\right )}} \]

[In]

integrate((-exp(x)^2*x^2+(-x^3-6*x^2+2*x)*exp(x)-4*x^3)/(5*exp(x)^3+15*x*exp(x)^2+15*exp(x)*x^2+5*x^3)/exp(-1+
2*x)^2/exp(exp(x)),x, algorithm="fricas")

[Out]

1/5*x^2*e^(-4*x - e^x + 2)/(x^2 + 2*x*e^x + e^(2*x))

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {e^{2-e^x-4 x} \left (-e^{2 x} x^2-4 x^3+e^x \left (2 x-6 x^2-x^3\right )\right )}{5 e^{3 x}+15 e^{2 x} x+15 e^x x^2+5 x^3} \, dx=\frac {x^{2} e^{2} e^{- e^{x}}}{5 x^{2} e^{4 x} + 10 x e^{5 x} + 5 e^{6 x}} \]

[In]

integrate((-exp(x)**2*x**2+(-x**3-6*x**2+2*x)*exp(x)-4*x**3)/(5*exp(x)**3+15*x*exp(x)**2+15*exp(x)*x**2+5*x**3
)/exp(-1+2*x)**2/exp(exp(x)),x)

[Out]

x**2*exp(2)*exp(-exp(x))/(5*x**2*exp(4*x) + 10*x*exp(5*x) + 5*exp(6*x))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {e^{2-e^x-4 x} \left (-e^{2 x} x^2-4 x^3+e^x \left (2 x-6 x^2-x^3\right )\right )}{5 e^{3 x}+15 e^{2 x} x+15 e^x x^2+5 x^3} \, dx=\frac {x^{2} e^{\left (-e^{x} + 2\right )}}{5 \, {\left (x^{2} e^{\left (4 \, x\right )} + 2 \, x e^{\left (5 \, x\right )} + e^{\left (6 \, x\right )}\right )}} \]

[In]

integrate((-exp(x)^2*x^2+(-x^3-6*x^2+2*x)*exp(x)-4*x^3)/(5*exp(x)^3+15*x*exp(x)^2+15*exp(x)*x^2+5*x^3)/exp(-1+
2*x)^2/exp(exp(x)),x, algorithm="maxima")

[Out]

1/5*x^2*e^(-e^x + 2)/(x^2*e^(4*x) + 2*x*e^(5*x) + e^(6*x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.46 \[ \int \frac {e^{2-e^x-4 x} \left (-e^{2 x} x^2-4 x^3+e^x \left (2 x-6 x^2-x^3\right )\right )}{5 e^{3 x}+15 e^{2 x} x+15 e^x x^2+5 x^3} \, dx=\frac {x^{2} e^{2}}{5 \, {\left (x^{2} e^{\left (4 \, x + e^{x}\right )} + 2 \, x e^{\left (5 \, x + e^{x}\right )} + e^{\left (6 \, x + e^{x}\right )}\right )}} \]

[In]

integrate((-exp(x)^2*x^2+(-x^3-6*x^2+2*x)*exp(x)-4*x^3)/(5*exp(x)^3+15*x*exp(x)^2+15*exp(x)*x^2+5*x^3)/exp(-1+
2*x)^2/exp(exp(x)),x, algorithm="giac")

[Out]

1/5*x^2*e^2/(x^2*e^(4*x + e^x) + 2*x*e^(5*x + e^x) + e^(6*x + e^x))

Mupad [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \frac {e^{2-e^x-4 x} \left (-e^{2 x} x^2-4 x^3+e^x \left (2 x-6 x^2-x^3\right )\right )}{5 e^{3 x}+15 e^{2 x} x+15 e^x x^2+5 x^3} \, dx=\frac {x^2\,{\mathrm {e}}^2\,{\mathrm {e}}^{-{\mathrm {e}}^x}}{5\,\left ({\mathrm {e}}^{6\,x}+2\,x\,{\mathrm {e}}^{5\,x}+x^2\,{\mathrm {e}}^{4\,x}\right )} \]

[In]

int(-(exp(2 - 4*x)*exp(-exp(x))*(x^2*exp(2*x) + exp(x)*(6*x^2 - 2*x + x^3) + 4*x^3))/(5*exp(3*x) + 15*x*exp(2*
x) + 15*x^2*exp(x) + 5*x^3),x)

[Out]

(x^2*exp(2)*exp(-exp(x)))/(5*(exp(6*x) + 2*x*exp(5*x) + x^2*exp(4*x)))