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

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

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 92, normalized size of antiderivative = 1.00, 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} \[ \frac {(c+d x) (b c-a d) \text {Gamma}\left (\frac {1}{3},-e (c+d x)^3\right )}{3 d^2 \sqrt [3]{-e (c+d x)^3}}-\frac {b (c+d x)^2 \text {Gamma}\left (\frac {2}{3},-e (c+d x)^3\right )}{3 d^2 \left (-e (c+d x)^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[E^(e*(c + d*x)^3)*(a + b*x),x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 86, normalized size = 0.93 \[ -\frac {(c+d x) \left (b (c+d x) \Gamma \left (\frac {2}{3},-e (c+d x)^3\right )-(b c-a d) \sqrt [3]{-e (c+d x)^3} \Gamma \left (\frac {1}{3},-e (c+d x)^3\right )\right )}{3 d^2 \left (-e (c+d x)^3\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 110, normalized size = 1.20 \[ \frac {\left (-d^{3} e\right )^{\frac {1}{3}} b d \Gamma \left (\frac {2}{3}, -d^{3} e x^{3} - 3 \, c d^{2} e x^{2} - 3 \, c^{2} d e x - c^{3} e\right ) - \left (-d^{3} e\right )^{\frac {2}{3}} {\left (b c - a d\right )} \Gamma \left (\frac {1}{3}, -d^{3} e x^{3} - 3 \, c d^{2} e x^{2} - 3 \, c^{2} d e x - c^{3} e\right )}{3 \, d^{4} e} \]

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]

1/3*((-d^3*e)^(1/3)*b*d*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*c - a
*d)*gamma(1/3, -d^3*e*x^3 - 3*c*d^2*e*x^2 - 3*c^2*d*e*x - c^3*e))/(d^4*e)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )} e^{\left ({\left (d x + c\right )}^{3} e\right )}\,{d x} \]

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^((d*x + c)^3*e), x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \left (b x +a \right ) {\mathrm e}^{\left (d x +c \right )^{3} e}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )} e^{\left ({\left (d x + c\right )}^{3} e\right )}\,{d x} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {e}}^{e\,{\left (c+d\,x\right )}^3}\,\left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(e*(c + d*x)^3)*(a + b*x),x)

[Out]

int(exp(e*(c + d*x)^3)*(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \left (\int a e^{d^{3} e x^{3}} e^{3 c d^{2} e x^{2}} e^{3 c^{2} d e x}\, dx + \int b x e^{d^{3} e x^{3}} e^{3 c d^{2} e x^{2}} e^{3 c^{2} d e x}\, dx\right ) e^{c^{3} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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