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

Optimal. Leaf size=92 \[ \frac{b (c+d x)^2 \left (-\frac{e}{(c+d x)^3}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},-\frac{e}{(c+d x)^3}\right )}{3 d^2}-\frac{(c+d x) (b c-a d) \sqrt [3]{-\frac{e}{(c+d x)^3}} \text{Gamma}\left (-\frac{1}{3},-\frac{e}{(c+d x)^3}\right )}{3 d^2} \]

[Out]

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

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Rubi [A]  time = 0.0567096, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2226, 2208, 2218} \[ \frac{b (c+d x)^2 \left (-\frac{e}{(c+d x)^3}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},-\frac{e}{(c+d x)^3}\right )}{3 d^2}-\frac{(c+d x) (b c-a d) \sqrt [3]{-\frac{e}{(c+d x)^3}} \text{Gamma}\left (-\frac{1}{3},-\frac{e}{(c+d x)^3}\right )}{3 d^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(e/(c + d*x)^3)*(a + b*x),x]

[Out]

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

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

Mathematica [A]  time = 0.0847917, size = 85, normalized size = 0.92 \[ \frac{(c+d x) \left ((a d-b c) \sqrt [3]{-\frac{e}{(c+d x)^3}} \text{Gamma}\left (-\frac{1}{3},-\frac{e}{(c+d x)^3}\right )+b (c+d x) \left (-\frac{e}{(c+d x)^3}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},-\frac{e}{(c+d x)^3}\right )\right )}{3 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.059, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{{\frac{e}{ \left ( dx+c \right ) ^{3}}}}} \left ( bx+a \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \,{\left (b x^{2} + 2 \, a 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{3 \,{\left (b d e x^{2} + 2 \, a 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 )}}{2 \,{\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4}\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),x, algorithm="maxima")

[Out]

1/2*(b*x^2 + 2*a*x)*e^(e/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)) + integrate(3/2*(b*d*e*x^2 + 2*a*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 [B]  time = 1.53108, size = 370, normalized size = 4.02 \begin{align*} -\frac{b d^{2} \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 ) - 2 \,{\left (b c d - a d^{2}\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 d^{2} x^{2} + 2 \, a d^{2} x - b c^{2} + 2 \, a c d\right )} e^{\left (\frac{e}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right )}}{2 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*d^2*(-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)) - 2*(b*c*d - a*d^2)*(-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*d^2*x^2 + 2*a*d^2*x - b*c^2 + 2*a*c*d)*
e^(e/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)))/d^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),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 )} 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),x, algorithm="giac")

[Out]

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