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

Optimal. Leaf size=216 \[ \frac {\sqrt {\pi } e^{-\frac {(b c-a d)^2}{4 b d}} (a d+b c)^2 \text {erfi}\left (\frac {a d+b c+2 b d x}{2 \sqrt {b} \sqrt {d}}\right )}{8 b^{5/2} d^{5/2}}-\frac {\sqrt {\pi } e^{-\frac {(b c-a d)^2}{4 b d}} \text {erfi}\left (\frac {a d+b c+2 b d x}{2 \sqrt {b} \sqrt {d}}\right )}{4 b^{3/2} d^{3/2}}-\frac {(a d+b c) e^{x (a d+b c)+a c+b d x^2}}{4 b^2 d^2}+\frac {x e^{x (a d+b c)+a c+b d x^2}}{2 b d} \]

[Out]

-1/4*(a*d+b*c)*exp(a*c+(a*d+b*c)*x+b*d*x^2)/b^2/d^2+1/2*exp(a*c+(a*d+b*c)*x+b*d*x^2)*x/b/d-1/4*erfi(1/2*(2*b*d
*x+a*d+b*c)/b^(1/2)/d^(1/2))*Pi^(1/2)/b^(3/2)/d^(3/2)/exp(1/4*(-a*d+b*c)^2/b/d)+1/8*(a*d+b*c)^2*erfi(1/2*(2*b*
d*x+a*d+b*c)/b^(1/2)/d^(1/2))*Pi^(1/2)/b^(5/2)/d^(5/2)/exp(1/4*(-a*d+b*c)^2/b/d)

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Rubi [A]  time = 0.27, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {2244, 2241, 2240, 2234, 2204} \[ \frac {\sqrt {\pi } e^{-\frac {(b c-a d)^2}{4 b d}} (a d+b c)^2 \text {Erfi}\left (\frac {a d+b c+2 b d x}{2 \sqrt {b} \sqrt {d}}\right )}{8 b^{5/2} d^{5/2}}-\frac {\sqrt {\pi } e^{-\frac {(b c-a d)^2}{4 b d}} \text {Erfi}\left (\frac {a d+b c+2 b d x}{2 \sqrt {b} \sqrt {d}}\right )}{4 b^{3/2} d^{3/2}}-\frac {(a d+b c) e^{x (a d+b c)+a c+b d x^2}}{4 b^2 d^2}+\frac {x e^{x (a d+b c)+a c+b d x^2}}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[E^((a + b*x)*(c + d*x))*x^2,x]

[Out]

-((b*c + a*d)*E^(a*c + (b*c + a*d)*x + b*d*x^2))/(4*b^2*d^2) + (E^(a*c + (b*c + a*d)*x + b*d*x^2)*x)/(2*b*d) -
(Sqrt[Pi]*Erfi[(b*c + a*d + 2*b*d*x)/(2*Sqrt[b]*Sqrt[d])])/(4*b^(3/2)*d^(3/2)*E^((b*c - a*d)^2/(4*b*d))) + ((
b*c + a*d)^2*Sqrt[Pi]*Erfi[(b*c + a*d + 2*b*d*x)/(2*Sqrt[b]*Sqrt[d])])/(8*b^(5/2)*d^(5/2)*E^((b*c - a*d)^2/(4*
b*d)))

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2234

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2240

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(e*F^(a + b*x + c*x^2))/(
2*c*Log[F]), x] - Dist[(b*e - 2*c*d)/(2*c), Int[F^(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
NeQ[b*e - 2*c*d, 0]

Rule 2241

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*(d + e*x)^(m - 1)
*F^(a + b*x + c*x^2))/(2*c*Log[F]), x] + (-Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*F^(a + b*x + c*x^2)
, x], x] - Dist[((m - 1)*e^2)/(2*c*Log[F]), Int[(d + e*x)^(m - 2)*F^(a + b*x + c*x^2), x], x]) /; FreeQ[{F, a,
b, c, d, e}, x] && NeQ[b*e - 2*c*d, 0] && GtQ[m, 1]

Rule 2244

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && QuadraticQ[v, x] &&  !(LinearMatchQ[u, x] && QuadraticMatchQ[v, x])

Rubi steps

\begin {align*} \int e^{(a+b x) (c+d x)} x^2 \, dx &=\int e^{a c+(b c+a d) x+b d x^2} x^2 \, dx\\ &=\frac {e^{a c+(b c+a d) x+b d x^2} x}{2 b d}-\frac {\int e^{a c+(b c+a d) x+b d x^2} \, dx}{2 b d}-\frac {(b c+a d) \int e^{a c+(b c+a d) x+b d x^2} x \, dx}{2 b d}\\ &=-\frac {(b c+a d) e^{a c+(b c+a d) x+b d x^2}}{4 b^2 d^2}+\frac {e^{a c+(b c+a d) x+b d x^2} x}{2 b d}+\frac {(b c+a d)^2 \int e^{a c+(b c+a d) x+b d x^2} \, dx}{4 b^2 d^2}-\frac {e^{-\frac {(b c-a d)^2}{4 b d}} \int e^{\frac {(b c+a d+2 b d x)^2}{4 b d}} \, dx}{2 b d}\\ &=-\frac {(b c+a d) e^{a c+(b c+a d) x+b d x^2}}{4 b^2 d^2}+\frac {e^{a c+(b c+a d) x+b d x^2} x}{2 b d}-\frac {e^{-\frac {(b c-a d)^2}{4 b d}} \sqrt {\pi } \text {erfi}\left (\frac {b c+a d+2 b d x}{2 \sqrt {b} \sqrt {d}}\right )}{4 b^{3/2} d^{3/2}}+\frac {\left ((b c+a d)^2 e^{-\frac {(b c-a d)^2}{4 b d}}\right ) \int e^{\frac {(b c+a d+2 b d x)^2}{4 b d}} \, dx}{4 b^2 d^2}\\ &=-\frac {(b c+a d) e^{a c+(b c+a d) x+b d x^2}}{4 b^2 d^2}+\frac {e^{a c+(b c+a d) x+b d x^2} x}{2 b d}-\frac {e^{-\frac {(b c-a d)^2}{4 b d}} \sqrt {\pi } \text {erfi}\left (\frac {b c+a d+2 b d x}{2 \sqrt {b} \sqrt {d}}\right )}{4 b^{3/2} d^{3/2}}+\frac {(b c+a d)^2 e^{-\frac {(b c-a d)^2}{4 b d}} \sqrt {\pi } \text {erfi}\left (\frac {b c+a d+2 b d x}{2 \sqrt {b} \sqrt {d}}\right )}{8 b^{5/2} d^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 144, normalized size = 0.67 \[ \frac {e^{-\frac {(b c-a d)^2}{4 b d}} \left (\sqrt {\pi } \left (a^2 d^2+2 b d (a c-1)+b^2 c^2\right ) \text {erfi}\left (\frac {a d+b (c+2 d x)}{2 \sqrt {b} \sqrt {d}}\right )-2 \sqrt {b} \sqrt {d} e^{\frac {(a d+b (c+2 d x))^2}{4 b d}} (a d+b (c-2 d x))\right )}{8 b^{5/2} d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[b]*Sqrt[d]*E^((a*d + b*(c + 2*d*x))^2/(4*b*d))*(a*d + b*(c - 2*d*x)) + (b^2*c^2 + 2*b*(-1 + a*c)*d +
a^2*d^2)*Sqrt[Pi]*Erfi[(a*d + b*(c + 2*d*x))/(2*Sqrt[b]*Sqrt[d])])/(8*b^(5/2)*d^(5/2)*E^((b*c - a*d)^2/(4*b*d)
))

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fricas [A]  time = 0.41, size = 148, normalized size = 0.69 \[ -\frac {\sqrt {\pi } {\left (b^{2} c^{2} + a^{2} d^{2} + 2 \, {\left (a b c - b\right )} d\right )} \sqrt {-b d} \operatorname {erf}\left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d}}{2 \, b d}\right ) e^{\left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{4 \, b d}\right )} - 2 \, {\left (2 \, b^{2} d^{2} x - b^{2} c d - a b d^{2}\right )} e^{\left (b d x^{2} + a c + {\left (b c + a d\right )} x\right )}}{8 \, b^{3} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(sqrt(pi)*(b^2*c^2 + a^2*d^2 + 2*(a*b*c - b)*d)*sqrt(-b*d)*erf(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)/(b*d)
)*e^(-1/4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/(b*d)) - 2*(2*b^2*d^2*x - b^2*c*d - a*b*d^2)*e^(b*d*x^2 + a*c + (b*c
+ a*d)*x))/(b^3*d^3)

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giac [A]  time = 0.44, size = 152, normalized size = 0.70 \[ -\frac {\frac {\sqrt {\pi } {\left (b^{2} c^{2} + 2 \, a b c d + a^{2} d^{2} - 2 \, b d\right )} \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {-b d} {\left (2 \, x + \frac {b c + a d}{b d}\right )}\right ) e^{\left (-\frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{4 \, b d}\right )}}{\sqrt {-b d}} - 2 \, {\left (b d {\left (2 \, x + \frac {b c + a d}{b d}\right )} - 2 \, b c - 2 \, a d\right )} e^{\left (b d x^{2} + b c x + a d x + a c\right )}}{8 \, b^{2} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(sqrt(pi)*(b^2*c^2 + 2*a*b*c*d + a^2*d^2 - 2*b*d)*erf(-1/2*sqrt(-b*d)*(2*x + (b*c + a*d)/(b*d)))*e^(-1/4*
(b^2*c^2 - 2*a*b*c*d + a^2*d^2)/(b*d))/sqrt(-b*d) - 2*(b*d*(2*x + (b*c + a*d)/(b*d)) - 2*b*c - 2*a*d)*e^(b*d*x
^2 + b*c*x + a*d*x + a*c))/(b^2*d^2)

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maple [A]  time = 0.02, size = 212, normalized size = 0.98 \[ \frac {x \,{\mathrm e}^{b d \,x^{2}+a c +\left (a d +b c \right ) x}}{2 b d}+\frac {\sqrt {\pi }\, \erf \left (-\sqrt {-b d}\, x +\frac {a d +b c}{2 \sqrt {-b d}}\right ) {\mathrm e}^{a c -\frac {\left (a d +b c \right )^{2}}{4 b d}}}{4 \sqrt {-b d}\, b d}-\frac {\left (a d +b c \right ) \left (\frac {\left (a d +b c \right ) \sqrt {\pi }\, \erf \left (-\sqrt {-b d}\, x +\frac {a d +b c}{2 \sqrt {-b d}}\right ) {\mathrm e}^{a c -\frac {\left (a d +b c \right )^{2}}{4 b d}}}{4 \sqrt {-b d}\, b d}+\frac {{\mathrm e}^{b d \,x^{2}+a c +\left (a d +b c \right ) x}}{2 b d}\right )}{2 b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b/d*x*exp(b*d*x^2+a*c+(a*d+b*c)*x)-1/2*(a*d+b*c)/b/d*(1/2/b/d*exp(b*d*x^2+a*c+(a*d+b*c)*x)+1/4*(a*d+b*c)/b
/d*Pi^(1/2)*exp(a*c-1/4*(a*d+b*c)^2/b/d)/(-b*d)^(1/2)*erf(-(-b*d)^(1/2)*x+1/2*(a*d+b*c)/(-b*d)^(1/2)))+1/4/b/d
*Pi^(1/2)*exp(a*c-1/4*(a*d+b*c)^2/b/d)/(-b*d)^(1/2)*erf(-(-b*d)^(1/2)*x+1/2*(a*d+b*c)/(-b*d)^(1/2))

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maxima [A]  time = 1.90, size = 221, normalized size = 1.02 \[ \frac {{\left (\frac {\sqrt {\pi } {\left (2 \, b d x + b c + a d\right )} {\left (b c + a d\right )}^{2} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-\frac {{\left (2 \, b d x + b c + a d\right )}^{2}}{b d}}\right ) - 1\right )}}{\left (b d\right )^{\frac {5}{2}} \sqrt {-\frac {{\left (2 \, b d x + b c + a d\right )}^{2}}{b d}}} - \frac {4 \, {\left (b c + a d\right )} b d e^{\left (\frac {{\left (2 \, b d x + b c + a d\right )}^{2}}{4 \, b d}\right )}}{\left (b d\right )^{\frac {5}{2}}} - \frac {4 \, {\left (2 \, b d x + b c + a d\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {{\left (2 \, b d x + b c + a d\right )}^{2}}{4 \, b d}\right )}{\left (b d\right )^{\frac {5}{2}} \left (-\frac {{\left (2 \, b d x + b c + a d\right )}^{2}}{b d}\right )^{\frac {3}{2}}}\right )} e^{\left (a c - \frac {{\left (b c + a d\right )}^{2}}{4 \, b d}\right )}}{8 \, \sqrt {b d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(sqrt(pi)*(2*b*d*x + b*c + a*d)*(b*c + a*d)^2*(erf(1/2*sqrt(-(2*b*d*x + b*c + a*d)^2/(b*d))) - 1)/((b*d)^(
5/2)*sqrt(-(2*b*d*x + b*c + a*d)^2/(b*d))) - 4*(b*c + a*d)*b*d*e^(1/4*(2*b*d*x + b*c + a*d)^2/(b*d))/(b*d)^(5/
2) - 4*(2*b*d*x + b*c + a*d)^3*gamma(3/2, -1/4*(2*b*d*x + b*c + a*d)^2/(b*d))/((b*d)^(5/2)*(-(2*b*d*x + b*c +
a*d)^2/(b*d))^(3/2)))*e^(a*c - 1/4*(b*c + a*d)^2/(b*d))/sqrt(b*d)

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mupad [B]  time = 0.30, size = 150, normalized size = 0.69 \[ \frac {x\,{\mathrm {e}}^{a\,c+a\,d\,x+b\,c\,x+b\,d\,x^2}}{2\,b\,d}-\frac {{\mathrm {e}}^{a\,c+a\,d\,x+b\,c\,x+b\,d\,x^2}\,\left (\frac {a\,d}{4}+\frac {b\,c}{4}\right )}{b^2\,d^2}+\frac {\sqrt {\pi }\,{\mathrm {e}}^{\frac {a\,c}{2}-\frac {a^2\,d}{4\,b}-\frac {b\,c^2}{4\,d}}\,\mathrm {erfi}\left (\frac {\frac {a\,d}{2}+\frac {b\,c}{2}+b\,d\,x}{\sqrt {b\,d}}\right )\,\left (a^2\,d^2+2\,a\,b\,c\,d+b^2\,c^2-2\,b\,d\right )}{8\,b^2\,d^2\,\sqrt {b\,d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*exp((a + b*x)*(c + d*x)),x)

[Out]

(x*exp(a*c + a*d*x + b*c*x + b*d*x^2))/(2*b*d) - (exp(a*c + a*d*x + b*c*x + b*d*x^2)*((a*d)/4 + (b*c)/4))/(b^2
*d^2) + (pi^(1/2)*exp((a*c)/2 - (a^2*d)/(4*b) - (b*c^2)/(4*d))*erfi(((a*d)/2 + (b*c)/2 + b*d*x)/(b*d)^(1/2))*(
a^2*d^2 - 2*b*d + b^2*c^2 + 2*a*b*c*d))/(8*b^2*d^2*(b*d)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp((b*x+a)*(d*x+c))*x**2,x)

[Out]

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

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