\(\int -\frac {4}{5} e^{-x^2} x \, dx\) [8613]

Optimal result

Integrand size = 12, antiderivative size = 11 \[ \int -\frac {4}{5} e^{-x^2} x \, dx=\frac {2 e^{-x^2}}{5} \]

[Out]

2/5/exp(x^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 2240} \[ \int -\frac {4}{5} e^{-x^2} x \, dx=\frac {2 e^{-x^2}}{5} \]

[In]

Int[(-4*x)/(5*E^x^2),x]

[Out]

2/(5*E^x^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int -\frac {4}{5} e^{-x^2} x \, dx=\frac {2 e^{-x^2}}{5} \]

[In]

Integrate[(-4*x)/(5*E^x^2),x]

[Out]

2/(5*E^x^2)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {2 \,{\mathrm e}^{-x^{2}}}{5}\)	\(9\)
derivativedivides	\(\frac {2 \,{\mathrm e}^{-x^{2}}}{5}\)	\(9\)
default	\(\frac {2 \,{\mathrm e}^{-x^{2}}}{5}\)	\(9\)
norman	\(\frac {2 \,{\mathrm e}^{-x^{2}}}{5}\)	\(9\)
risch	\(\frac {2 \,{\mathrm e}^{-x^{2}}}{5}\)	\(9\)
parallelrisch	\(\frac {2 \,{\mathrm e}^{-x^{2}}}{5}\)	\(9\)
meijerg	\(-\frac {2}{5}+\frac {2 \,{\mathrm e}^{-x^{2}}}{5}\)	\(11\)

[In]

int(-4/5*x/exp(x^2),x,method=_RETURNVERBOSE)

[Out]

2/5/exp(x^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int -\frac {4}{5} e^{-x^2} x \, dx=\frac {2}{5} \, e^{\left (-x^{2}\right )} \]

[In]

integrate(-4/5*x/exp(x^2),x, algorithm="fricas")

[Out]

2/5*e^(-x^2)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int -\frac {4}{5} e^{-x^2} x \, dx=\frac {2 e^{- x^{2}}}{5} \]

[In]

integrate(-4/5*x/exp(x**2),x)

[Out]

2*exp(-x**2)/5

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int -\frac {4}{5} e^{-x^2} x \, dx=\frac {2}{5} \, e^{\left (-x^{2}\right )} \]

[In]

integrate(-4/5*x/exp(x^2),x, algorithm="maxima")

[Out]

2/5*e^(-x^2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int -\frac {4}{5} e^{-x^2} x \, dx=\frac {2}{5} \, e^{\left (-x^{2}\right )} \]

[In]

integrate(-4/5*x/exp(x^2),x, algorithm="giac")

[Out]

2/5*e^(-x^2)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int -\frac {4}{5} e^{-x^2} x \, dx=\frac {2\,{\mathrm {e}}^{-x^2}}{5} \]

[In]

int(-(4*x*exp(-x^2))/5,x)

[Out]

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