\(\int e^{-t} t^3 \, dt\) [168]

Optimal result

Integrand size = 9, antiderivative size = 36 \[ \int e^{-t} t^3 \, dt=-6 e^{-t}-6 e^{-t} t-3 e^{-t} t^2-e^{-t} t^3 \]

[Out]

-6/exp(t)-6*t/exp(t)-3*t^2/exp(t)-t^3/exp(t)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2207, 2225} \[ \int e^{-t} t^3 \, dt=-e^{-t} t^3-3 e^{-t} t^2-6 e^{-t} t-6 e^{-t} \]

[In]

Int[t^3/E^t,t]

[Out]

-6/E^t - (6*t)/E^t - (3*t^2)/E^t - t^3/E^t

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}& = -e^{-t} t^3+3 \int e^{-t} t^2 \, dt \\ & = -3 e^{-t} t^2-e^{-t} t^3+6 \int e^{-t} t \, dt \\ & = -6 e^{-t} t-3 e^{-t} t^2-e^{-t} t^3+6 \int e^{-t} \, dt \\ & = -6 e^{-t}-6 e^{-t} t-3 e^{-t} t^2-e^{-t} t^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.58 \[ \int e^{-t} t^3 \, dt=e^{-t} \left (-6-6 t-3 t^2-t^3\right ) \]

[In]

Integrate[t^3/E^t,t]

[Out]

(-6 - 6*t - 3*t^2 - t^3)/E^t

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.56

method	result	size
gosper	\(-\left (t^{3}+3 t^{2}+6 t +6\right ) {\mathrm e}^{-t}\)	\(20\)
norman	\(\left (-t^{3}-3 t^{2}-6 t -6\right ) {\mathrm e}^{-t}\)	\(21\)
risch	\(\left (-t^{3}-3 t^{2}-6 t -6\right ) {\mathrm e}^{-t}\)	\(21\)
parallelrisch	\(\left (-t^{3}-3 t^{2}-6 t -6\right ) {\mathrm e}^{-t}\)	\(21\)
meijerg	\(6-\frac {\left (4 t^{3}+12 t^{2}+24 t +24\right ) {\mathrm e}^{-t}}{4}\)	\(24\)
default	\(-6 \,{\mathrm e}^{-t}-6 t \,{\mathrm e}^{-t}-3 t^{2} {\mathrm e}^{-t}-t^{3} {\mathrm e}^{-t}\)	\(33\)

[In]

int(t^3/exp(t),t,method=_RETURNVERBOSE)

[Out]

-(t^3+3*t^2+6*t+6)/exp(t)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.53 \[ \int e^{-t} t^3 \, dt=-{\left (t^{3} + 3 \, t^{2} + 6 \, t + 6\right )} e^{\left (-t\right )} \]

[In]

integrate(t^3/exp(t),t, algorithm="fricas")

[Out]

-(t^3 + 3*t^2 + 6*t + 6)*e^(-t)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.47 \[ \int e^{-t} t^3 \, dt=\left (- t^{3} - 3 t^{2} - 6 t - 6\right ) e^{- t} \]

[In]

integrate(t**3/exp(t),t)

[Out]

(-t**3 - 3*t**2 - 6*t - 6)*exp(-t)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.53 \[ \int e^{-t} t^3 \, dt=-{\left (t^{3} + 3 \, t^{2} + 6 \, t + 6\right )} e^{\left (-t\right )} \]

[In]

integrate(t^3/exp(t),t, algorithm="maxima")

[Out]

-(t^3 + 3*t^2 + 6*t + 6)*e^(-t)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.53 \[ \int e^{-t} t^3 \, dt=-{\left (t^{3} + 3 \, t^{2} + 6 \, t + 6\right )} e^{\left (-t\right )} \]

[In]

integrate(t^3/exp(t),t, algorithm="giac")

[Out]

-(t^3 + 3*t^2 + 6*t + 6)*e^(-t)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.53 \[ \int e^{-t} t^3 \, dt=-{\mathrm {e}}^{-t}\,\left (t^3+3\,t^2+6\,t+6\right ) \]

[In]

int(t^3*exp(-t),t)

[Out]

-exp(-t)*(6*t + 3*t^2 + t^3 + 6)