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

Optimal. Leaf size=80 \[ \frac{a (a+b x) \text{Gamma}\left (\frac{1}{3},-(a+b x)^3\right )}{3 b^2 \sqrt [3]{-(a+b x)^3}}-\frac{(a+b x)^2 \text{Gamma}\left (\frac{2}{3},-(a+b x)^3\right )}{3 b^2 \left (-(a+b x)^3\right )^{2/3}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0684645, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2227, 2226, 2208, 2218} \[ \frac{a (a+b x) \text{Gamma}\left (\frac{1}{3},-(a+b x)^3\right )}{3 b^2 \sqrt [3]{-(a+b x)^3}}-\frac{(a+b x)^2 \text{Gamma}\left (\frac{2}{3},-(a+b x)^3\right )}{3 b^2 \left (-(a+b x)^3\right )^{2/3}} \]

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,x]

[Out]

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

Rubi steps

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

Mathematica [A]  time = 0.0309232, size = 74, normalized size = 0.92 \[ \frac{(a+b x) \left (a \sqrt [3]{-(a+b x)^3} \text{Gamma}\left (\frac{1}{3},-(a+b x)^3\right )-(a+b x) \text{Gamma}\left (\frac{2}{3},-(a+b x)^3\right )\right )}{3 b^2 \left (-(a+b x)^3\right )^{2/3}} \]

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,x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.015, 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\, 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,x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x 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,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.46661, size = 203, normalized size = 2.54 \begin{align*} -\frac{\left (-b^{3}\right )^{\frac{2}{3}} a \Gamma \left (\frac{1}{3}, -b^{3} x^{3} - 3 \, a b^{2} x^{2} - 3 \, a^{2} b x - a^{3}\right ) - \left (-b^{3}\right )^{\frac{1}{3}} b \Gamma \left (\frac{2}{3}, -b^{3} x^{3} - 3 \, a b^{2} x^{2} - 3 \, a^{2} b x - a^{3}\right )}{3 \, b^{4}} \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,x, algorithm="fricas")

[Out]

-1/3*((-b^3)^(2/3)*a*gamma(1/3, -b^3*x^3 - 3*a*b^2*x^2 - 3*a^2*b*x - a^3) - (-b^3)^(1/3)*b*gamma(2/3, -b^3*x^3
- 3*a*b^2*x^2 - 3*a^2*b*x - a^3))/b^4

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} e^{a^{3}} \int x e^{b^{3} x^{3}} e^{3 a b^{2} x^{2}} e^{3 a^{2} b x}\, dx \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,x)

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x 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,x, algorithm="giac")

[Out]

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