3.4.91 \(\int e^{e (c+d x)^3} (a+b x)^3 \, dx\) [391]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2258, 2239, 2250, 2240} \begin {gather*} -\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} \end {gather*}

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 2239

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 2240

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]

Rule 2250

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(-F^a)*((e +
f*x)^(m + 1)/(f*n*((-b)*(c + d*x)^n*Log[F])^((m + 1)/n)))*Gamma[(m + 1)/n, (-b)*(c + d*x)^n*Log[F]], x] /; Fre
eQ[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rule 2258

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)^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*}

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Mathematica [A]
time = 0.45, size = 167, normalized size = 0.94 \begin {gather*} \frac {-\frac {3 b^2 (b c-a d) e^{e (c+d x)^3}}{e}+\frac {(b c-a d)^3 (c+d x) \Gamma \left (\frac {1}{3},-e (c+d x)^3\right )}{\sqrt [3]{-e (c+d x)^3}}-\frac {3 b (b c-a d)^2 (c+d x)^2 \Gamma \left (\frac {2}{3},-e (c+d x)^3\right )}{\left (-e (c+d x)^3\right )^{2/3}}+\frac {b^3 (c+d x) \Gamma \left (\frac {4}{3},-e (c+d x)^3\right )}{e \sqrt [3]{-e (c+d x)^3}}}{3 d^4} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{e \left (d x +c \right )^{3}} \left (b x +a \right )^{3}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.08, size = 231, normalized size = 1.31 \begin {gather*} \frac {{\left (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}, -{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} 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}, -{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} e\right ) + 3 \, {\left (b^{3} d^{3} x - 2 \, b^{3} c d^{2} + 3 \, a b^{2} d^{3}\right )} e^{\left ({\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} e + 1\right )}\right )} e^{\left (-2\right )}}{9 \, d^{6}} \end {gather*}

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \left (\int a^{3} e^{d^{3} e x^{3}} e^{3 c d^{2} e x^{2}} e^{3 c^{2} d e x}\, dx + \int b^{3} x^{3} e^{d^{3} e x^{3}} e^{3 c d^{2} e x^{2}} e^{3 c^{2} d e x}\, dx + \int 3 a b^{2} x^{2} e^{d^{3} e x^{3}} e^{3 c d^{2} e x^{2}} e^{3 c^{2} d e x}\, dx + \int 3 a^{2} 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**3*exp(d**3*e*x**3)*exp(3*c*d**2*e*x**2)*exp(3*c**2*d*e*x), x) + Integral(b**3*x**3*exp(d**3*e*x**
3)*exp(3*c*d**2*e*x**2)*exp(3*c**2*d*e*x), x) + Integral(3*a*b**2*x**2*exp(d**3*e*x**3)*exp(3*c*d**2*e*x**2)*e
xp(3*c**2*d*e*x), x) + Integral(3*a**2*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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(e*(c + d*x)^3)*(a + b*x)^3,x)

[Out]

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

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