\(\int e^{4 x^3} x^2 \, dx\) [114]

Optimal result

Integrand size = 11, antiderivative size = 11 \[ \int e^{4 x^3} x^2 \, dx=\frac {e^{4 x^3}}{12} \]

[Out]

1/12*exp(4*x^3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2240} \[ \int e^{4 x^3} x^2 \, dx=\frac {e^{4 x^3}}{12} \]

[In]

Int[E^(4*x^3)*x^2,x]

[Out]

E^(4*x^3)/12

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}& = \frac {e^{4 x^3}}{12} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int e^{4 x^3} x^2 \, dx=\frac {e^{4 x^3}}{12} \]

[In]

Integrate[E^(4*x^3)*x^2,x]

[Out]

E^(4*x^3)/12

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.82

method	result	size
gosper	\(\frac {{\mathrm e}^{4 x^{3}}}{12}\)	\(9\)
derivativedivides	\(\frac {{\mathrm e}^{4 x^{3}}}{12}\)	\(9\)
default	\(\frac {{\mathrm e}^{4 x^{3}}}{12}\)	\(9\)
norman	\(\frac {{\mathrm e}^{4 x^{3}}}{12}\)	\(9\)
risch	\(\frac {{\mathrm e}^{4 x^{3}}}{12}\)	\(9\)
parallelrisch	\(\frac {{\mathrm e}^{4 x^{3}}}{12}\)	\(9\)
meijerg	\(-\frac {1}{12}+\frac {{\mathrm e}^{4 x^{3}}}{12}\)	\(11\)

[In]

int(exp(4*x^3)*x^2,x,method=_RETURNVERBOSE)

[Out]

1/12*exp(4*x^3)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int e^{4 x^3} x^2 \, dx=\frac {1}{12} \, e^{\left (4 \, x^{3}\right )} \]

[In]

integrate(exp(4*x^3)*x^2,x, algorithm="fricas")

[Out]

1/12*e^(4*x^3)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.64 \[ \int e^{4 x^3} x^2 \, dx=\frac {e^{4 x^{3}}}{12} \]

[In]

integrate(exp(4*x**3)*x**2,x)

[Out]

exp(4*x**3)/12

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int e^{4 x^3} x^2 \, dx=\frac {1}{12} \, e^{\left (4 \, x^{3}\right )} \]

[In]

integrate(exp(4*x^3)*x^2,x, algorithm="maxima")

[Out]

1/12*e^(4*x^3)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int e^{4 x^3} x^2 \, dx=\frac {1}{12} \, e^{\left (4 \, x^{3}\right )} \]

[In]

integrate(exp(4*x^3)*x^2,x, algorithm="giac")

[Out]

1/12*e^(4*x^3)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.73 \[ \int e^{4 x^3} x^2 \, dx=\frac {{\mathrm {e}}^{4\,x^3}}{12} \]

[In]

int(x^2*exp(4*x^3),x)

[Out]

exp(4*x^3)/12