\(\int e^{8-2 x} x^3 \, dx\) [708]

Optimal result

Integrand size = 11, antiderivative size = 52 \[ \int e^{8-2 x} x^3 \, dx=-\frac {3}{8} e^{8-2 x}-\frac {3}{4} e^{8-2 x} x-\frac {3}{4} e^{8-2 x} x^2-\frac {1}{2} e^{8-2 x} x^3 \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2207, 2225} \[ \int e^{8-2 x} x^3 \, dx=-\frac {1}{2} e^{8-2 x} x^3-\frac {3}{4} e^{8-2 x} x^2-\frac {3}{4} e^{8-2 x} x-\frac {3}{8} e^{8-2 x} \]

[In]

Int[E^(8 - 2*x)*x^3,x]

[Out]

(-3*E^(8 - 2*x))/8 - (3*E^(8 - 2*x)*x)/4 - (3*E^(8 - 2*x)*x^2)/4 - (E^(8 - 2*x)*x^3)/2

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2} e^{8-2 x} x^3+\frac {3}{2} \int e^{8-2 x} x^2 \, dx \\ & = -\frac {3}{4} e^{8-2 x} x^2-\frac {1}{2} e^{8-2 x} x^3+\frac {3}{2} \int e^{8-2 x} x \, dx \\ & = -\frac {3}{4} e^{8-2 x} x-\frac {3}{4} e^{8-2 x} x^2-\frac {1}{2} e^{8-2 x} x^3+\frac {3}{4} \int e^{8-2 x} \, dx \\ & = -\frac {3}{8} e^{8-2 x}-\frac {3}{4} e^{8-2 x} x-\frac {3}{4} e^{8-2 x} x^2-\frac {1}{2} e^{8-2 x} x^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.50 \[ \int e^{8-2 x} x^3 \, dx=-\frac {1}{8} e^{8-2 x} \left (3+6 x+6 x^2+4 x^3\right ) \]

[In]

Integrate[E^(8 - 2*x)*x^3,x]

[Out]

-1/8*(E^(8 - 2*x)*(3 + 6*x + 6*x^2 + 4*x^3))

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.44

method	result	size
risch	\(\left (-\frac {1}{2} x^{3}-\frac {3}{4} x^{2}-\frac {3}{4} x -\frac {3}{8}\right ) {\mathrm e}^{8-2 x}\)	\(23\)
gosper	\(-\frac {\left (4 x^{3}+6 x^{2}+6 x +3\right ) {\mathrm e}^{8-2 x}}{8}\)	\(24\)
meijerg	\(\frac {{\mathrm e}^{8} \left (6-\frac {\left (32 x^{3}+48 x^{2}+48 x +24\right ) {\mathrm e}^{-2 x}}{4}\right )}{16}\)	\(28\)
norman	\(-\frac {3 \,{\mathrm e}^{8-2 x}}{8}-\frac {3 \,{\mathrm e}^{8-2 x} x}{4}-\frac {3 \,{\mathrm e}^{8-2 x} x^{2}}{4}-\frac {{\mathrm e}^{8-2 x} x^{3}}{2}\)	\(41\)
parallelrisch	\(-\frac {3 \,{\mathrm e}^{8-2 x}}{8}-\frac {3 \,{\mathrm e}^{8-2 x} x}{4}-\frac {3 \,{\mathrm e}^{8-2 x} x^{2}}{4}-\frac {{\mathrm e}^{8-2 x} x^{3}}{2}\)	\(41\)
parts	\(-\frac {{\mathrm e}^{8-2 x} x^{3}}{2}-\frac {3 \,{\mathrm e}^{8-2 x} \left (8-2 x \right )^{2}}{16}+\frac {27 \,{\mathrm e}^{8-2 x} \left (8-2 x \right )}{8}-\frac {123 \,{\mathrm e}^{8-2 x}}{8}\)	\(49\)
derivativedivides	\(\frac {123 \,{\mathrm e}^{8-2 x} \left (8-2 x \right )}{8}-\frac {379 \,{\mathrm e}^{8-2 x}}{8}-\frac {27 \,{\mathrm e}^{8-2 x} \left (8-2 x \right )^{2}}{16}+\frac {{\mathrm e}^{8-2 x} \left (8-2 x \right )^{3}}{16}\)	\(53\)
default	\(\frac {123 \,{\mathrm e}^{8-2 x} \left (8-2 x \right )}{8}-\frac {379 \,{\mathrm e}^{8-2 x}}{8}-\frac {27 \,{\mathrm e}^{8-2 x} \left (8-2 x \right )^{2}}{16}+\frac {{\mathrm e}^{8-2 x} \left (8-2 x \right )^{3}}{16}\)	\(53\)

[In]

int(exp(8-2*x)*x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.44 \[ \int e^{8-2 x} x^3 \, dx=-\frac {1}{8} \, {\left (4 \, x^{3} + 6 \, x^{2} + 6 \, x + 3\right )} e^{\left (-2 \, x + 8\right )} \]

[In]

integrate(exp(8-2*x)*x^3,x, algorithm="fricas")

[Out]

-1/8*(4*x^3 + 6*x^2 + 6*x + 3)*e^(-2*x + 8)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.46 \[ \int e^{8-2 x} x^3 \, dx=\frac {\left (- 4 x^{3} - 6 x^{2} - 6 x - 3\right ) e^{8 - 2 x}}{8} \]

[In]

integrate(exp(8-2*x)*x**3,x)

[Out]

(-4*x**3 - 6*x**2 - 6*x - 3)*exp(8 - 2*x)/8

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.58 \[ \int e^{8-2 x} x^3 \, dx=-\frac {1}{8} \, {\left (4 \, x^{3} e^{8} + 6 \, x^{2} e^{8} + 6 \, x e^{8} + 3 \, e^{8}\right )} e^{\left (-2 \, x\right )} \]

[In]

integrate(exp(8-2*x)*x^3,x, algorithm="maxima")

[Out]

-1/8*(4*x^3*e^8 + 6*x^2*e^8 + 6*x*e^8 + 3*e^8)*e^(-2*x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.44 \[ \int e^{8-2 x} x^3 \, dx=-\frac {1}{8} \, {\left (4 \, x^{3} + 6 \, x^{2} + 6 \, x + 3\right )} e^{\left (-2 \, x + 8\right )} \]

[In]

integrate(exp(8-2*x)*x^3,x, algorithm="giac")

[Out]

-1/8*(4*x^3 + 6*x^2 + 6*x + 3)*e^(-2*x + 8)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.44 \[ \int e^{8-2 x} x^3 \, dx=-{\mathrm {e}}^{8-2\,x}\,\left (\frac {x^3}{2}+\frac {3\,x^2}{4}+\frac {3\,x}{4}+\frac {3}{8}\right ) \]

[In]

int(x^3*exp(8 - 2*x),x)

[Out]

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