3.208 \(\int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^4 \, dx\)

Optimal. Leaf size=183 \[ \frac{4 a^3 (a+b x)^2 \text{Gamma}\left (\frac{2}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{2/3}}-\frac{a^4 (a+b x) \text{Gamma}\left (\frac{1}{3},-(a+b x)^3\right )}{3 b^5 \sqrt [3]{-(a+b x)^3}}-\frac{(a+b x)^5 \text{Gamma}\left (\frac{5}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{5/3}}+\frac{4 a (a+b x)^4 \text{Gamma}\left (\frac{4}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{4/3}}+\frac{2 a^2 e^{(a+b x)^3}}{b^5} \]

[Out]

(2*a^2*E^(a + b*x)^3)/b^5 - (a^4*(a + b*x)*Gamma[1/3, -(a + b*x)^3])/(3*b^5*(-(a + b*x)^3)^(1/3)) + (4*a^3*(a
+ b*x)^2*Gamma[2/3, -(a + b*x)^3])/(3*b^5*(-(a + b*x)^3)^(2/3)) + (4*a*(a + b*x)^4*Gamma[4/3, -(a + b*x)^3])/(
3*b^5*(-(a + b*x)^3)^(4/3)) - ((a + b*x)^5*Gamma[5/3, -(a + b*x)^3])/(3*b^5*(-(a + b*x)^3)^(5/3))

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Rubi [A]  time = 0.183204, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{4 a^3 (a+b x)^2 \text{Gamma}\left (\frac{2}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{2/3}}-\frac{a^4 (a+b x) \text{Gamma}\left (\frac{1}{3},-(a+b x)^3\right )}{3 b^5 \sqrt [3]{-(a+b x)^3}}-\frac{(a+b x)^5 \text{Gamma}\left (\frac{5}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{5/3}}+\frac{4 a (a+b x)^4 \text{Gamma}\left (\frac{4}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{4/3}}+\frac{2 a^2 e^{(a+b x)^3}}{b^5} \]

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^4,x]

[Out]

(2*a^2*E^(a + b*x)^3)/b^5 - (a^4*(a + b*x)*Gamma[1/3, -(a + b*x)^3])/(3*b^5*(-(a + b*x)^3)^(1/3)) + (4*a^3*(a
+ b*x)^2*Gamma[2/3, -(a + b*x)^3])/(3*b^5*(-(a + b*x)^3)^(2/3)) + (4*a*(a + b*x)^4*Gamma[4/3, -(a + b*x)^3])/(
3*b^5*(-(a + b*x)^3)^(4/3)) - ((a + b*x)^5*Gamma[5/3, -(a + b*x)^3])/(3*b^5*(-(a + b*x)^3)^(5/3))

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^4 \, dx &=\int e^{(a+b x)^3} x^4 \, dx\\ &=\int \left (\frac{a^4 e^{(a+b x)^3}}{b^4}-\frac{4 a^3 e^{(a+b x)^3} (a+b x)}{b^4}+\frac{6 a^2 e^{(a+b x)^3} (a+b x)^2}{b^4}-\frac{4 a e^{(a+b x)^3} (a+b x)^3}{b^4}+\frac{e^{(a+b x)^3} (a+b x)^4}{b^4}\right ) \, dx\\ &=\frac{\int e^{(a+b x)^3} (a+b x)^4 \, dx}{b^4}-\frac{(4 a) \int e^{(a+b x)^3} (a+b x)^3 \, dx}{b^4}+\frac{\left (6 a^2\right ) \int e^{(a+b x)^3} (a+b x)^2 \, dx}{b^4}-\frac{\left (4 a^3\right ) \int e^{(a+b x)^3} (a+b x) \, dx}{b^4}+\frac{a^4 \int e^{(a+b x)^3} \, dx}{b^4}\\ &=\frac{2 a^2 e^{(a+b x)^3}}{b^5}-\frac{a^4 (a+b x) \Gamma \left (\frac{1}{3},-(a+b x)^3\right )}{3 b^5 \sqrt [3]{-(a+b x)^3}}+\frac{4 a^3 (a+b x)^2 \Gamma \left (\frac{2}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{2/3}}+\frac{4 a (a+b x)^4 \Gamma \left (\frac{4}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{4/3}}-\frac{(a+b x)^5 \Gamma \left (\frac{5}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{5/3}}\\ \end{align*}

Mathematica [A]  time = 0.179857, size = 164, normalized size = 0.9 \[ \frac{a^4 (-(a+b x)) \sqrt [3]{-(a+b x)^3} \text{Gamma}\left (\frac{1}{3},-(a+b x)^3\right )+4 a^3 (a+b x)^2 \text{Gamma}\left (\frac{2}{3},-(a+b x)^3\right )-4 a (a+b x) \sqrt [3]{-(a+b x)^3} \text{Gamma}\left (\frac{4}{3},-(a+b x)^3\right )+(a+b x)^2 \text{Gamma}\left (\frac{5}{3},-(a+b x)^3\right )+6 a^2 e^{(a+b x)^3} \left (-(a+b x)^3\right )^{2/3}}{3 b^5 \left (-(a+b x)^3\right )^{2/3}} \]

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^4,x]

[Out]

(6*a^2*E^(a + b*x)^3*(-(a + b*x)^3)^(2/3) - a^4*(a + b*x)*(-(a + b*x)^3)^(1/3)*Gamma[1/3, -(a + b*x)^3] + 4*a^
3*(a + b*x)^2*Gamma[2/3, -(a + b*x)^3] - 4*a*(a + b*x)*(-(a + b*x)^3)^(1/3)*Gamma[4/3, -(a + b*x)^3] + (a + b*
x)^2*Gamma[5/3, -(a + b*x)^3])/(3*b^5*(-(a + b*x)^3)^(2/3))

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Maple [F]  time = 0.025, 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}^{4}\, 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^4,x)

[Out]

int(exp(b^3*x^3+3*a*b^2*x^2+3*a^2*b*x+a^3)*x^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} 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^4,x, algorithm="maxima")

[Out]

integrate(x^4*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 = 1.51148, size = 348, normalized size = 1.9 \begin{align*} -\frac{2 \,{\left (6 \, a^{3} + 1\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 ) -{\left (3 \, a^{4} + 4 \, a\right )} \left (-b^{3}\right )^{\frac{2}{3}} \Gamma \left (\frac{1}{3}, -b^{3} x^{3} - 3 \, a b^{2} x^{2} - 3 \, a^{2} b x - a^{3}\right ) - 3 \,{\left (b^{4} x^{2} - 2 \, a b^{3} x + 3 \, a^{2} b^{2}\right )} e^{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}}{9 \, b^{7}} \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^4,x, algorithm="fricas")

[Out]

-1/9*(2*(6*a^3 + 1)*(-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) - (3*a^4 + 4*a)*(-b^3)
^(2/3)*gamma(1/3, -b^3*x^3 - 3*a*b^2*x^2 - 3*a^2*b*x - a^3) - 3*(b^4*x^2 - 2*a*b^3*x + 3*a^2*b^2)*e^(b^3*x^3 +
3*a*b^2*x^2 + 3*a^2*b*x + a^3))/b^7

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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**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} 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^4,x, algorithm="giac")

[Out]

integrate(x^4*e^(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3), x)