3.391 \(\int e^{e (c+d x)^3} (a+b x)^3 \, dx\)

Optimal. Leaf size=177 \[ -\frac{b (c+d x)^2 (b c-a d)^2 \text{Gamma}\left (\frac{2}{3},-e (c+d x)^3\right )}{d^4 \left (-e (c+d x)^3\right )^{2/3}}+\frac{(c+d x) (b c-a d)^3 \text{Gamma}\left (\frac{1}{3},-e (c+d x)^3\right )}{3 d^4 \sqrt [3]{-e (c+d x)^3}}-\frac{b^3 (c+d x)^4 \text{Gamma}\left (\frac{4}{3},-e (c+d x)^3\right )}{3 d^4 \left (-e (c+d x)^3\right )^{4/3}}-\frac{b^2 (b c-a d) e^{e (c+d x)^3}}{d^4 e} \]

[Out]

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

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Rubi [A]  time = 0.155975, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2226, 2208, 2218, 2209} \[ -\frac{b (c+d x)^2 (b c-a d)^2 \text{Gamma}\left (\frac{2}{3},-e (c+d x)^3\right )}{d^4 \left (-e (c+d x)^3\right )^{2/3}}+\frac{(c+d x) (b c-a d)^3 \text{Gamma}\left (\frac{1}{3},-e (c+d x)^3\right )}{3 d^4 \sqrt [3]{-e (c+d x)^3}}-\frac{b^3 (c+d x)^4 \text{Gamma}\left (\frac{4}{3},-e (c+d x)^3\right )}{3 d^4 \left (-e (c+d x)^3\right )^{4/3}}-\frac{b^2 (b c-a d) e^{e (c+d x)^3}}{d^4 e} \]

Antiderivative was successfully verified.

[In]

Int[E^(e*(c + d*x)^3)*(a + b*x)^3,x]

[Out]

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

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

Mathematica [A]  time = 0.213214, size = 167, normalized size = 0.94 \[ \frac{-\frac{3 b (c+d x)^2 (b c-a d)^2 \text{Gamma}\left (\frac{2}{3},-e (c+d x)^3\right )}{\left (-e (c+d x)^3\right )^{2/3}}+\frac{(c+d x) (b c-a d)^3 \text{Gamma}\left (\frac{1}{3},-e (c+d x)^3\right )}{\sqrt [3]{-e (c+d x)^3}}+\frac{b^3 (c+d x) \text{Gamma}\left (\frac{4}{3},-e (c+d x)^3\right )}{e \sqrt [3]{-e (c+d x)^3}}-\frac{3 b^2 (b c-a d) e^{e (c+d x)^3}}{e}}{3 d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(e*(c + d*x)^3)*(a + b*x)^3,x]

[Out]

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

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{e \left ( dx+c \right ) ^{3}}} \left ( bx+a \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(e*(d*x+c)^3)*(b*x+a)^3,x)

[Out]

int(exp(e*(d*x+c)^3)*(b*x+a)^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{3} e^{\left ({\left (d x + c\right )}^{3} e\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.53813, size = 514, normalized size = 2.9 \begin{align*} \frac{9 \,{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} \left (-d^{3} e\right )^{\frac{1}{3}} e \Gamma \left (\frac{2}{3}, -d^{3} e x^{3} - 3 \, c d^{2} e x^{2} - 3 \, c^{2} d e x - c^{3} e\right ) - \left (-d^{3} e\right )^{\frac{2}{3}}{\left (b^{3} + 3 \,{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} e\right )} \Gamma \left (\frac{1}{3}, -d^{3} e x^{3} - 3 \, c d^{2} e x^{2} - 3 \, c^{2} d e x - c^{3} e\right ) + 3 \,{\left (b^{3} d^{3} e x -{\left (2 \, b^{3} c d^{2} - 3 \, a b^{2} d^{3}\right )} e\right )} e^{\left (d^{3} e x^{3} + 3 \, c d^{2} e x^{2} + 3 \, c^{2} d e x + c^{3} e\right )}}{9 \, d^{6} e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(e*(d*x+c)**3)*(b*x+a)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{3} e^{\left ({\left (d x + c\right )}^{3} e\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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