\(\int e^{x^3} (1-e^{4 x^3})^2 x^2 \, dx\) [714]

Optimal result

Integrand size = 22, antiderivative size = 32 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {e^{x^3}}{3}-\frac {2 e^{5 x^3}}{15}+\frac {e^{9 x^3}}{27} \]

[Out]

1/3*exp(x^3)-2/15*exp(5*x^3)+1/27*exp(9*x^3)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6847, 2281, 200} \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {e^{x^3}}{3}-\frac {2 e^{5 x^3}}{15}+\frac {e^{9 x^3}}{27} \]

[In]

Int[E^x^3*(1 - E^(4*x^3))^2*x^2,x]

[Out]

E^x^3/3 - (2*E^(5*x^3))/15 + E^(9*x^3)/27

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 2281

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_.)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Wit
h[{m = FullSimplify[d*e*(Log[F]/(g*h*Log[G]))]}, Dist[Denominator[m]/(g*h*Log[G]), Subst[Int[x^(Denominator[m]
- 1)*(a + b*F^(c*e - d*e*(f/g))*x^Numerator[m])^p, x], x, G^(h*((f + g*x)/Denominator[m]))], x] /; LtQ[m, -1]
|| GtQ[m, 1]] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x]

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int e^x \left (1-e^{4 x}\right )^2 \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (1-x^4\right )^2 \, dx,x,e^{x^3}\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (1-2 x^4+x^8\right ) \, dx,x,e^{x^3}\right ) \\ & = \frac {e^{x^3}}{3}-\frac {2 e^{5 x^3}}{15}+\frac {e^{9 x^3}}{27} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {1}{135} e^{x^3} \left (45-18 e^{4 x^3}+5 e^{8 x^3}\right ) \]

[In]

Integrate[E^x^3*(1 - E^(4*x^3))^2*x^2,x]

[Out]

(E^x^3*(45 - 18*E^(4*x^3) + 5*E^(8*x^3)))/135

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {{\mathrm e}^{x^{3}}}{3}-\frac {2 \,{\mathrm e}^{5 x^{3}}}{15}+\frac {{\mathrm e}^{9 x^{3}}}{27}\)	\(24\)
default	\(\frac {{\mathrm e}^{x^{3}}}{3}-\frac {2 \,{\mathrm e}^{5 x^{3}}}{15}+\frac {{\mathrm e}^{9 x^{3}}}{27}\)	\(24\)
risch	\(\frac {{\mathrm e}^{x^{3}}}{3}-\frac {2 \,{\mathrm e}^{5 x^{3}}}{15}+\frac {{\mathrm e}^{9 x^{3}}}{27}\)	\(24\)
meijerg	\(-\frac {32}{135}+\frac {{\mathrm e}^{x^{3}}}{3}-\frac {2 \,{\mathrm e}^{5 x^{3}}}{15}+\frac {{\mathrm e}^{9 x^{3}}}{27}\)	\(25\)
parallelrisch	\(\frac {{\mathrm e}^{x^{3}}}{3}+\frac {{\mathrm e}^{8 x^{3}} {\mathrm e}^{x^{3}}}{27}-\frac {2 \,{\mathrm e}^{x^{3}} {\mathrm e}^{4 x^{3}}}{15}\)	\(34\)

[In]

int(exp(x^3)*(1-exp(4*x^3))^2*x^2,x,method=_RETURNVERBOSE)

[Out]

1/27*exp(x^3)^9-2/15*exp(x^3)^5+1/3*exp(x^3)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {1}{27} \, e^{\left (9 \, x^{3}\right )} - \frac {2}{15} \, e^{\left (5 \, x^{3}\right )} + \frac {1}{3} \, e^{\left (x^{3}\right )} \]

[In]

integrate(exp(x^3)*(1-exp(4*x^3))^2*x^2,x, algorithm="fricas")

[Out]

1/27*e^(9*x^3) - 2/15*e^(5*x^3) + 1/3*e^(x^3)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {e^{9 x^{3}}}{27} - \frac {2 e^{5 x^{3}}}{15} + \frac {e^{x^{3}}}{3} \]

[In]

integrate(exp(x**3)*(1-exp(4*x**3))**2*x**2,x)

[Out]

exp(9*x**3)/27 - 2*exp(5*x**3)/15 + exp(x**3)/3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {1}{27} \, e^{\left (9 \, x^{3}\right )} - \frac {2}{15} \, e^{\left (5 \, x^{3}\right )} + \frac {1}{3} \, e^{\left (x^{3}\right )} \]

[In]

integrate(exp(x^3)*(1-exp(4*x^3))^2*x^2,x, algorithm="maxima")

[Out]

1/27*e^(9*x^3) - 2/15*e^(5*x^3) + 1/3*e^(x^3)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {1}{27} \, e^{\left (9 \, x^{3}\right )} - \frac {2}{15} \, e^{\left (5 \, x^{3}\right )} + \frac {1}{3} \, e^{\left (x^{3}\right )} \]

[In]

integrate(exp(x^3)*(1-exp(4*x^3))^2*x^2,x, algorithm="giac")

[Out]

1/27*e^(9*x^3) - 2/15*e^(5*x^3) + 1/3*e^(x^3)

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int e^{x^3} \left (1-e^{4 x^3}\right )^2 x^2 \, dx=\frac {{\mathrm {e}}^{x^3}\,\left (5\,{\mathrm {e}}^{8\,x^3}-18\,{\mathrm {e}}^{4\,x^3}+45\right )}{135} \]

[In]

int(x^2*exp(x^3)*(exp(4*x^3) - 1)^2,x)

[Out]

(exp(x^3)*(5*exp(8*x^3) - 18*exp(4*x^3) + 45))/135