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

Optimal. Leaf size=151 \[ -\frac{2 b (c+d x)^2 (b c-a d) \left (-\frac{e}{(c+d x)^3}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},-\frac{e}{(c+d x)^3}\right )}{3 d^3}+\frac{(c+d x) (b c-a d)^2 \sqrt [3]{-\frac{e}{(c+d x)^3}} \text{Gamma}\left (-\frac{1}{3},-\frac{e}{(c+d x)^3}\right )}{3 d^3}-\frac{b^2 e \text{Ei}\left (\frac{e}{(c+d x)^3}\right )}{3 d^3}+\frac{b^2 (c+d x)^3 e^{\frac{e}{(c+d x)^3}}}{3 d^3} \]

[Out]

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

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Rubi [A]  time = 0.133174, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {2226, 2208, 2218, 2214, 2210} \[ -\frac{2 b (c+d x)^2 (b c-a d) \left (-\frac{e}{(c+d x)^3}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},-\frac{e}{(c+d x)^3}\right )}{3 d^3}+\frac{(c+d x) (b c-a d)^2 \sqrt [3]{-\frac{e}{(c+d x)^3}} \text{Gamma}\left (-\frac{1}{3},-\frac{e}{(c+d x)^3}\right )}{3 d^3}-\frac{b^2 e \text{Ei}\left (\frac{e}{(c+d x)^3}\right )}{3 d^3}+\frac{b^2 (c+d x)^3 e^{\frac{e}{(c+d x)^3}}}{3 d^3} \]

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)*(c + d*x)^3)/(3*d^3) - (b^2*e*ExpIntegralEi[e/(c + d*x)^3])/(3*d^3) - (2*b*(b*c - a*d)*
(-(e/(c + d*x)^3))^(2/3)*(c + d*x)^2*Gamma[-2/3, -(e/(c + d*x)^3)])/(3*d^3) + ((b*c - a*d)^2*(-(e/(c + d*x)^3)
)^(1/3)*(c + d*x)*Gamma[-1/3, -(e/(c + d*x)^3)])/(3*d^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 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), Int[(c + d*x)^(m + n)*F^(a + b*(c +
d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 2210

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

Rubi steps

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

Mathematica [A]  time = 0.0759187, size = 136, normalized size = 0.9 \[ \frac{-2 b (c+d x)^2 (b c-a d) \left (-\frac{e}{(c+d x)^3}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},-\frac{e}{(c+d x)^3}\right )+(c+d x) (b c-a d)^2 \sqrt [3]{-\frac{e}{(c+d x)^3}} \text{Gamma}\left (-\frac{1}{3},-\frac{e}{(c+d x)^3}\right )-b^2 e \text{Ei}\left (\frac{e}{(c+d x)^3}\right )+b^2 (c+d x)^3 e^{\frac{e}{(c+d x)^3}}}{3 d^3} \]

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

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Maple [F]  time = 0.076, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{{\frac{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*} \frac{1}{3} \,{\left (b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} x\right )} e^{\left (\frac{e}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right )} + \int \frac{{\left (b^{2} d e x^{3} + 3 \, a b d e x^{2} + 3 \, a^{2} d e x\right )} e^{\left (\frac{e}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right )}}{d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4}}\,{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]

1/3*(b^2*x^3 + 3*a*b*x^2 + 3*a^2*x)*e^(e/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + integrate((b^2*d*e*x^3 +
3*a*b*d*e*x^2 + 3*a^2*d*e*x)*e^(e/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3))/(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d
^2*x^2 + 4*c^3*d*x + c^4), x)

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Fricas [A]  time = 1.57348, size = 551, normalized size = 3.65 \begin{align*} -\frac{b^{2} e{\rm Ei}\left (\frac{e}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) - 3 \,{\left (b^{2} c d^{2} - a b d^{3}\right )} \left (-\frac{e}{d^{3}}\right )^{\frac{2}{3}} \Gamma \left (\frac{1}{3}, -\frac{e}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) + 3 \,{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \left (-\frac{e}{d^{3}}\right )^{\frac{1}{3}} \Gamma \left (\frac{2}{3}, -\frac{e}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) -{\left (b^{2} d^{3} x^{3} + 3 \, a b d^{3} x^{2} + 3 \, a^{2} d^{3} x + b^{2} c^{3} - 3 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} e^{\left (\frac{e}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right )}}{3 \, d^{3}} \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*e*Ei(e/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) - 3*(b^2*c*d^2 - a*b*d^3)*(-e/d^3)^(2/3)*gamma(1/3
, -e/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*(-e/d^3)^(1/3)*gamma(2
/3, -e/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) - (b^2*d^3*x^3 + 3*a*b*d^3*x^2 + 3*a^2*d^3*x + b^2*c^3 - 3*a
*b*c^2*d + 3*a^2*c*d^2)*e^(e/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)))/d^3

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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 (\frac{e}{{\left (d x + c\right )}^{3}}\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^(e/(d*x + c)^3), x)