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

Optimal. Leaf size=99 \[ -\frac{a^2 (a+b x) \text{Gamma}\left (\frac{1}{3},-(a+b x)^3\right )}{3 b^3 \sqrt [3]{-(a+b x)^3}}+\frac{2 a (a+b x)^2 \text{Gamma}\left (\frac{2}{3},-(a+b x)^3\right )}{3 b^3 \left (-(a+b x)^3\right )^{2/3}}+\frac{e^{(a+b x)^3}}{3 b^3} \]

[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))

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Rubi [A]  time = 0.120953, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {2227, 2226, 2208, 2218, 2209} \[ -\frac{a^2 (a+b x) \text{Gamma}\left (\frac{1}{3},-(a+b x)^3\right )}{3 b^3 \sqrt [3]{-(a+b x)^3}}+\frac{2 a (a+b x)^2 \text{Gamma}\left (\frac{2}{3},-(a+b x)^3\right )}{3 b^3 \left (-(a+b x)^3\right )^{2/3}}+\frac{e^{(a+b x)^3}}{3 b^3} \]

Antiderivative was successfully verified.

[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 2227

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]

Rule 2226

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 2208

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 2218

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

Rule 2209

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*} \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^2 \, dx &=\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]  time = 0.0809934, size = 89, normalized size = 0.9 \[ \frac{-\frac{a^2 (a+b x) \text{Gamma}\left (\frac{1}{3},-(a+b x)^3\right )}{\sqrt [3]{-(a+b x)^3}}+\frac{2 a (a+b x)^2 \text{Gamma}\left (\frac{2}{3},-(a+b x)^3\right )}{\left (-(a+b x)^3\right )^{2/3}}+e^{(a+b x)^3}}{3 b^3} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 0.017, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{{b}^{3}{x}^{3}+3\,a{b}^{2}{x}^{2}+3\,{a}^{2}bx+{a}^{3}}}{x}^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.55053, size = 277, normalized size = 2.8 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)