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

Optimal. Leaf size=126 \[ \frac{2 b (c+d x)^2 (b c-a d) \text{Gamma}\left (\frac{2}{3},-e (c+d x)^3\right )}{3 d^3 \left (-e (c+d x)^3\right )^{2/3}}-\frac{(c+d x) (b c-a d)^2 \text{Gamma}\left (\frac{1}{3},-e (c+d x)^3\right )}{3 d^3 \sqrt [3]{-e (c+d x)^3}}+\frac{b^2 e^{e (c+d x)^3}}{3 d^3 e} \]

[Out]

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

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Rubi [A]  time = 0.106701, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {2226, 2208, 2218, 2209} \[ \frac{2 b (c+d x)^2 (b c-a d) \text{Gamma}\left (\frac{2}{3},-e (c+d x)^3\right )}{3 d^3 \left (-e (c+d x)^3\right )^{2/3}}-\frac{(c+d x) (b c-a d)^2 \text{Gamma}\left (\frac{1}{3},-e (c+d x)^3\right )}{3 d^3 \sqrt [3]{-e (c+d x)^3}}+\frac{b^2 e^{e (c+d x)^3}}{3 d^3 e} \]

Antiderivative was successfully verified.

[In]

Int[E^(e*(c + d*x)^3)*(a + b*x)^2,x]

[Out]

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

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

Mathematica [A]  time = 0.0854033, size = 117, normalized size = 0.93 \[ \frac{\frac{2 b (c+d x)^2 (b c-a d) \text{Gamma}\left (\frac{2}{3},-e (c+d x)^3\right )}{\left (-e (c+d x)^3\right )^{2/3}}-\frac{(c+d x) (b c-a d)^2 \text{Gamma}\left (\frac{1}{3},-e (c+d x)^3\right )}{\sqrt [3]{-e (c+d x)^3}}+\frac{b^2 e^{e (c+d x)^3}}{e}}{3 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{e \left ( dx+c \right ) ^{3}}} \left ( bx+a \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{2} 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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.53158, size = 385, normalized size = 3.06 \begin{align*} \frac{b^{2} d^{2} e^{\left (d^{3} e x^{3} + 3 \, c d^{2} e x^{2} + 3 \, c^{2} d e x + c^{3} e\right )} +{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \left (-d^{3} e\right )^{\frac{2}{3}} \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 ) - 2 \,{\left (b^{2} c d - a b d^{2}\right )} \left (-d^{3} e\right )^{\frac{1}{3}} \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 )}{3 \, d^{5} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**2,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 )}^{2} 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)^2,x, algorithm="giac")

[Out]

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