\(\int e^{-x^3} x^5 \, dx\) [286]

Optimal result

Integrand size = 11, antiderivative size = 26 \[ \int e^{-x^3} x^5 \, dx=-\frac {e^{-x^3}}{3}-\frac {1}{3} e^{-x^3} x^3 \]

[Out]

-1/3/exp(x^3)-1/3*x^3/exp(x^3)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2243, 2240} \[ \int e^{-x^3} x^5 \, dx=-\frac {1}{3} e^{-x^3} x^3-\frac {e^{-x^3}}{3} \]

[In]

Int[x^5/E^x^3,x]

[Out]

-1/3*1/E^x^3 - x^3/(3*E^x^3)

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]

Rule 2243

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^(m
- n + 1)*(F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2*((m + 1)/n)] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.62 \[ \int e^{-x^3} x^5 \, dx=-\frac {1}{3} e^{-x^3} \left (1+x^3\right ) \]

[In]

Integrate[x^5/E^x^3,x]

[Out]

-1/3*(1 + x^3)/E^x^3

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.54

method	result	size
gosper	\(-\frac {\left (x^{3}+1\right ) {\mathrm e}^{-x^{3}}}{3}\)	\(14\)
norman	\(\left (-\frac {x^{3}}{3}-\frac {1}{3}\right ) {\mathrm e}^{-x^{3}}\)	\(15\)
risch	\(\left (-\frac {x^{3}}{3}-\frac {1}{3}\right ) {\mathrm e}^{-x^{3}}\)	\(15\)
parallelrisch	\(\frac {\left (-x^{3}-1\right ) {\mathrm e}^{-x^{3}}}{3}\)	\(16\)
meijerg	\(\frac {1}{3}-\frac {\left (2 x^{3}+2\right ) {\mathrm e}^{-x^{3}}}{6}\)	\(18\)
derivativedivides	\(-\frac {{\mathrm e}^{-x^{3}}}{3}-\frac {x^{3} {\mathrm e}^{-x^{3}}}{3}\)	\(21\)
default	\(-\frac {{\mathrm e}^{-x^{3}}}{3}-\frac {x^{3} {\mathrm e}^{-x^{3}}}{3}\)	\(21\)

[In]

int(x^5/exp(x^3),x,method=_RETURNVERBOSE)

[Out]

-1/3*(x^3+1)/exp(x^3)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.50 \[ \int e^{-x^3} x^5 \, dx=-\frac {1}{3} \, {\left (x^{3} + 1\right )} e^{\left (-x^{3}\right )} \]

[In]

integrate(x^5/exp(x^3),x, algorithm="fricas")

[Out]

-1/3*(x^3 + 1)*e^(-x^3)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.46 \[ \int e^{-x^3} x^5 \, dx=\frac {\left (- x^{3} - 1\right ) e^{- x^{3}}}{3} \]

[In]

integrate(x**5/exp(x**3),x)

[Out]

(-x**3 - 1)*exp(-x**3)/3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.50 \[ \int e^{-x^3} x^5 \, dx=-\frac {1}{3} \, {\left (x^{3} + 1\right )} e^{\left (-x^{3}\right )} \]

[In]

integrate(x^5/exp(x^3),x, algorithm="maxima")

[Out]

-1/3*(x^3 + 1)*e^(-x^3)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.50 \[ \int e^{-x^3} x^5 \, dx=-\frac {1}{3} \, {\left (x^{3} + 1\right )} e^{\left (-x^{3}\right )} \]

[In]

integrate(x^5/exp(x^3),x, algorithm="giac")

[Out]

-1/3*(x^3 + 1)*e^(-x^3)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.50 \[ \int e^{-x^3} x^5 \, dx=-\frac {{\mathrm {e}}^{-x^3}\,\left (x^3+1\right )}{3} \]

[In]

int(x^5*exp(-x^3),x)

[Out]

-(exp(-x^3)*(x^3 + 1))/3