\(\int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^2 \, dx\) [210]

Optimal result

Integrand size = 33, antiderivative size = 99 \[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^2 \, dx=\frac {e^{(a+b x)^3}}{3 b^3}-\frac {a^2 (a+b x) \Gamma \left (\frac {1}{3},-(a+b x)^3\right )}{3 b^3 \sqrt [3]{-(a+b x)^3}}+\frac {2 a (a+b x)^2 \Gamma \left (\frac {2}{3},-(a+b x)^3\right )}{3 b^3 \left (-(a+b x)^3\right )^{2/3}} \]

[Out]

1/3*exp((b*x+a)^3)/b^3-1/3*a^2*(b*x+a)*GAMMA(1/3,-(b*x+a)^3)/b^3/(-(b*x+a)^3)^(1/3)+2/3*a*(b*x+a)^2*GAMMA(2/3,
-(b*x+a)^3)/b^3/(-(b*x+a)^3)^(2/3)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2259, 2258, 2239, 2250, 2240} \[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^2 \, dx=-\frac {a^2 (a+b x) \Gamma \left (\frac {1}{3},-(a+b x)^3\right )}{3 b^3 \sqrt [3]{-(a+b x)^3}}+\frac {e^{(a+b x)^3}}{3 b^3}+\frac {2 a (a+b x)^2 \Gamma \left (\frac {2}{3},-(a+b x)^3\right )}{3 b^3 \left (-(a+b x)^3\right )^{2/3}} \]

[In]

Int[E^(a^3 + 3*a^2*b*x + 3*a*b^2*x^2 + b^3*x^3)*x^2,x]

[Out]

E^(a + b*x)^3/(3*b^3) - (a^2*(a + b*x)*Gamma[1/3, -(a + b*x)^3])/(3*b^3*(-(a + b*x)^3)^(1/3)) + (2*a*(a + b*x)
^2*Gamma[2/3, -(a + b*x)^3])/(3*b^3*(-(a + b*x)^3)^(2/3))

Rule 2239

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

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 2250

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

Rule 2258

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rule 2259

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rubi steps \begin{align*} \text {integral}& = \int e^{(a+b x)^3} x^2 \, dx \\ & = \int \left (\frac {a^2 e^{(a+b x)^3}}{b^2}-\frac {2 a e^{(a+b x)^3} (a+b x)}{b^2}+\frac {e^{(a+b x)^3} (a+b x)^2}{b^2}\right ) \, dx \\ & = \frac {\int e^{(a+b x)^3} (a+b x)^2 \, dx}{b^2}-\frac {(2 a) \int e^{(a+b x)^3} (a+b x) \, dx}{b^2}+\frac {a^2 \int e^{(a+b x)^3} \, dx}{b^2} \\ & = \frac {e^{(a+b x)^3}}{3 b^3}-\frac {a^2 (a+b x) \Gamma \left (\frac {1}{3},-(a+b x)^3\right )}{3 b^3 \sqrt [3]{-(a+b x)^3}}+\frac {2 a (a+b x)^2 \Gamma \left (\frac {2}{3},-(a+b x)^3\right )}{3 b^3 \left (-(a+b x)^3\right )^{2/3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.90 \[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^2 \, dx=\frac {e^{(a+b x)^3}-\frac {a^2 (a+b x) \Gamma \left (\frac {1}{3},-(a+b x)^3\right )}{\sqrt [3]{-(a+b x)^3}}+\frac {2 a (a+b x)^2 \Gamma \left (\frac {2}{3},-(a+b x)^3\right )}{\left (-(a+b x)^3\right )^{2/3}}}{3 b^3} \]

[In]

Integrate[E^(a^3 + 3*a^2*b*x + 3*a*b^2*x^2 + b^3*x^3)*x^2,x]

[Out]

(E^(a + b*x)^3 - (a^2*(a + b*x)*Gamma[1/3, -(a + b*x)^3])/(-(a + b*x)^3)^(1/3) + (2*a*(a + b*x)^2*Gamma[2/3, -
(a + b*x)^3])/(-(a + b*x)^3)^(2/3))/(3*b^3)

Maple [F]

\[\int {\mathrm e}^{b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}} x^{2}d x\]

[In]

int(exp(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)*x^2,x)

[Out]

int(exp(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)*x^2,x)

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.25 \[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^2 \, dx=\frac {\left (-b^{3}\right )^{\frac {2}{3}} a^{2} \Gamma \left (\frac {1}{3}, -b^{3} x^{3} - 3 \, a b^{2} x^{2} - 3 \, a^{2} b x - a^{3}\right ) - 2 \, \left (-b^{3}\right )^{\frac {1}{3}} a b \Gamma \left (\frac {2}{3}, -b^{3} x^{3} - 3 \, a b^{2} x^{2} - 3 \, a^{2} b x - a^{3}\right ) + b^{2} e^{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}}{3 \, b^{5}} \]

[In]

integrate(exp(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)*x^2,x, algorithm="fricas")

[Out]

1/3*((-b^3)^(2/3)*a^2*gamma(1/3, -b^3*x^3 - 3*a*b^2*x^2 - 3*a^2*b*x - a^3) - 2*(-b^3)^(1/3)*a*b*gamma(2/3, -b^
3*x^3 - 3*a*b^2*x^2 - 3*a^2*b*x - a^3) + b^2*e^(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3))/b^5

Sympy [F]

\[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^2 \, dx=e^{a^{3}} \int x^{2} e^{b^{3} x^{3}} e^{3 a b^{2} x^{2}} e^{3 a^{2} b x}\, dx \]

[In]

integrate(exp(b**3*x**3+3*a*b**2*x**2+3*a**2*b*x+a**3)*x**2,x)

[Out]

exp(a**3)*Integral(x**2*exp(b**3*x**3)*exp(3*a*b**2*x**2)*exp(3*a**2*b*x), x)

Maxima [F]

\[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^2 \, dx=\int { x^{2} e^{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \,d x } \]

[In]

integrate(exp(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)*x^2,x, algorithm="maxima")

[Out]

integrate(x^2*e^(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3), x)

Giac [F]

\[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^2 \, dx=\int { x^{2} e^{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \,d x } \]

[In]

integrate(exp(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)*x^2,x, algorithm="giac")

[Out]

integrate(x^2*e^(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3), x)

Mupad [F(-1)]

Timed out. \[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^2 \, dx=\int x^2\,{\mathrm {e}}^{a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3} \,d x \]

[In]

int(x^2*exp(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x),x)

[Out]

int(x^2*exp(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x), x)