3.75 \(\int e^x \sin (a+b x+c x^2) \, dx\)

Optimal. Leaf size=144 \[ \frac{(-1)^{3/4} \sqrt{\pi } e^{\frac{1}{4} i \left (4 a+\frac{(1+i b)^2}{c}\right )} \text{Erf}\left (\frac{\sqrt [4]{-1} (i b+2 i c x+1)}{2 \sqrt{c}}\right )}{4 \sqrt{c}}+\frac{(-1)^{3/4} \sqrt{\pi } e^{\frac{i (b+i)^2}{4 c}-i a} \text{Erfi}\left (\frac{\sqrt [4]{-1} (-i b-2 i c x+1)}{2 \sqrt{c}}\right )}{4 \sqrt{c}} \]

[Out]

((-1)^(3/4)*E^((I/4)*(4*a + (1 + I*b)^2/c))*Sqrt[Pi]*Erf[((-1)^(1/4)*(1 + I*b + (2*I)*c*x))/(2*Sqrt[c])])/(4*S
qrt[c]) + ((-1)^(3/4)*E^((-I)*a + ((I/4)*(I + b)^2)/c)*Sqrt[Pi]*Erfi[((-1)^(1/4)*(1 - I*b - (2*I)*c*x))/(2*Sqr
t[c])])/(4*Sqrt[c])

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Rubi [A]  time = 0.216196, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {4472, 2234, 2204, 2205} \[ \frac{(-1)^{3/4} \sqrt{\pi } e^{\frac{1}{4} i \left (4 a+\frac{(1+i b)^2}{c}\right )} \text{Erf}\left (\frac{\sqrt [4]{-1} (i b+2 i c x+1)}{2 \sqrt{c}}\right )}{4 \sqrt{c}}+\frac{(-1)^{3/4} \sqrt{\pi } e^{\frac{i (b+i)^2}{4 c}-i a} \text{Erfi}\left (\frac{\sqrt [4]{-1} (-i b-2 i c x+1)}{2 \sqrt{c}}\right )}{4 \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sin[a + b*x + c*x^2],x]

[Out]

((-1)^(3/4)*E^((I/4)*(4*a + (1 + I*b)^2/c))*Sqrt[Pi]*Erf[((-1)^(1/4)*(1 + I*b + (2*I)*c*x))/(2*Sqrt[c])])/(4*S
qrt[c]) + ((-1)^(3/4)*E^((-I)*a + ((I/4)*(I + b)^2)/c)*Sqrt[Pi]*Erfi[((-1)^(1/4)*(1 - I*b - (2*I)*c*x))/(2*Sqr
t[c])])/(4*Sqrt[c])

Rule 4472

Int[(F_)^(u_)*Sin[v_]^(n_.), x_Symbol] :> Int[ExpandTrigToExp[F^u, Sin[v]^n, x], x] /; FreeQ[F, x] && (LinearQ
[u, x] || PolyQ[u, x, 2]) && (LinearQ[v, x] || PolyQ[v, x, 2]) && IGtQ[n, 0]

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 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 2205

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

Rubi steps

\begin{align*} \int e^x \sin \left (a+b x+c x^2\right ) \, dx &=\int \left (\frac{1}{2} i e^{-i a+(1-i b) x-i c x^2}-\frac{1}{2} i e^{i a+(1+i b) x+i c x^2}\right ) \, dx\\ &=\frac{1}{2} i \int e^{-i a+(1-i b) x-i c x^2} \, dx-\frac{1}{2} i \int e^{i a+(1+i b) x+i c x^2} \, dx\\ &=-\left (\frac{1}{2} \left (i e^{\frac{1}{4} i \left (4 a+\frac{(1+i b)^2}{c}\right )}\right ) \int e^{-\frac{i (1+i b+2 i c x)^2}{4 c}} \, dx\right )+\frac{1}{2} \left (i e^{-\frac{i \left (1-2 i b-b^2+4 a c\right )}{4 c}}\right ) \int e^{\frac{i (1-i b-2 i c x)^2}{4 c}} \, dx\\ &=\frac{(-1)^{3/4} e^{\frac{1}{4} i \left (4 a+\frac{(1+i b)^2}{c}\right )} \sqrt{\pi } \text{erf}\left (\frac{\sqrt [4]{-1} (1+i b+2 i c x)}{2 \sqrt{c}}\right )}{4 \sqrt{c}}+\frac{(-1)^{3/4} e^{-\frac{i \left (1-2 i b-b^2+4 a c\right )}{4 c}} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt [4]{-1} (1-i b-2 i c x)}{2 \sqrt{c}}\right )}{4 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.255904, size = 134, normalized size = 0.93 \[ -\frac{\sqrt [4]{-1} \sqrt{\pi } e^{-\frac{i \left (b^2-2 i b+1\right )}{4 c}} \left (e^{\frac{i b^2}{2 c}} (\sin (a)+i \cos (a)) \text{Erfi}\left (\frac{(-1)^{3/4} (b+2 c x+i)}{2 \sqrt{c}}\right )+e^{\left .\frac{i}{2}\right /c} (\cos (a)+i \sin (a)) \text{Erfi}\left (\frac{\sqrt [4]{-1} (b+2 c x-i)}{2 \sqrt{c}}\right )\right )}{4 \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sin[a + b*x + c*x^2],x]

[Out]

-((-1)^(1/4)*Sqrt[Pi]*(E^((I/2)/c)*Erfi[((-1)^(1/4)*(-I + b + 2*c*x))/(2*Sqrt[c])]*(Cos[a] + I*Sin[a]) + E^(((
I/2)*b^2)/c)*Erfi[((-1)^(3/4)*(I + b + 2*c*x))/(2*Sqrt[c])]*(I*Cos[a] + Sin[a])))/(4*Sqrt[c]*E^(((I/4)*(1 - (2
*I)*b + b^2))/c))

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Maple [A]  time = 0.109, size = 117, normalized size = 0.8 \begin{align*}{{\frac{i}{4}}\sqrt{\pi }{{\rm e}^{{\frac{{\frac{i}{4}} \left ( -{b}^{2}+2\,ib+4\,ac+1 \right ) }{c}}}}{\it Erf} \left ( -\sqrt{-ic}x+{\frac{1+ib}{2}{\frac{1}{\sqrt{-ic}}}} \right ){\frac{1}{\sqrt{-ic}}}}+{{\frac{i}{4}}\sqrt{\pi }{{\rm e}^{{\frac{{\frac{i}{4}} \left ( 2\,ib-4\,ac+{b}^{2}-1 \right ) }{c}}}}{\it Erf} \left ( \sqrt{ic}x-{\frac{-ib+1}{2}{\frac{1}{\sqrt{ic}}}} \right ){\frac{1}{\sqrt{ic}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.08344, size = 417, normalized size = 2.9 \begin{align*} -\frac{\sqrt{\pi }{\left ({\left ({\left (i \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right ) + i \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right ) + \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right ) - \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right )\right )} \cos \left (-\frac{b^{2} - 4 \, a c - 1}{4 \, c}\right ) +{\left (\cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right ) + \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right ) - i \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right ) + i \, \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right )\right )} \sin \left (-\frac{b^{2} - 4 \, a c - 1}{4 \, c}\right )\right )} \operatorname{erf}\left (\frac{i \,{\left (2 i \, c x + i \, b - 1\right )} \sqrt{i \, c}}{2 \, c}\right ) +{\left ({\left (-i \, \cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right ) - i \, \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right ) + \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right ) - \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right )\right )} \cos \left (-\frac{b^{2} - 4 \, a c - 1}{4 \, c}\right ) +{\left (\cos \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right ) + \cos \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right ) + i \, \sin \left (\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right ) - i \, \sin \left (-\frac{1}{4} \, \pi + \frac{1}{2} \, \arctan \left (0, c\right )\right )\right )} \sin \left (-\frac{b^{2} - 4 \, a c - 1}{4 \, c}\right )\right )} \operatorname{erf}\left (\frac{i \,{\left (2 i \, c x + i \, b + 1\right )} \sqrt{-i \, c}}{2 \, c}\right )\right )} e^{\left (-\frac{b}{2 \, c}\right )}}{8 \, \sqrt{{\left | c \right |}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(pi)*(((I*cos(1/4*pi + 1/2*arctan2(0, c)) + I*cos(-1/4*pi + 1/2*arctan2(0, c)) + sin(1/4*pi + 1/2*arc
tan2(0, c)) - sin(-1/4*pi + 1/2*arctan2(0, c)))*cos(-1/4*(b^2 - 4*a*c - 1)/c) + (cos(1/4*pi + 1/2*arctan2(0, c
)) + cos(-1/4*pi + 1/2*arctan2(0, c)) - I*sin(1/4*pi + 1/2*arctan2(0, c)) + I*sin(-1/4*pi + 1/2*arctan2(0, c))
)*sin(-1/4*(b^2 - 4*a*c - 1)/c))*erf(1/2*I*(2*I*c*x + I*b - 1)*sqrt(I*c)/c) + ((-I*cos(1/4*pi + 1/2*arctan2(0,
c)) - I*cos(-1/4*pi + 1/2*arctan2(0, c)) + sin(1/4*pi + 1/2*arctan2(0, c)) - sin(-1/4*pi + 1/2*arctan2(0, c))
)*cos(-1/4*(b^2 - 4*a*c - 1)/c) + (cos(1/4*pi + 1/2*arctan2(0, c)) + cos(-1/4*pi + 1/2*arctan2(0, c)) + I*sin(
1/4*pi + 1/2*arctan2(0, c)) - I*sin(-1/4*pi + 1/2*arctan2(0, c)))*sin(-1/4*(b^2 - 4*a*c - 1)/c))*erf(1/2*I*(2*
I*c*x + I*b + 1)*sqrt(-I*c)/c))*e^(-1/2*b/c)/sqrt(abs(c))

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Fricas [B]  time = 0.500768, size = 647, normalized size = 4.49 \begin{align*} \frac{i \, \sqrt{2} \pi \sqrt{\frac{c}{\pi }} e^{\left (\frac{i \, b^{2} - 4 i \, a c - 2 \, b - i}{4 \, c}\right )} \operatorname{C}\left (\frac{\sqrt{2}{\left (2 \, c x + b + i\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) + i \, \sqrt{2} \pi \sqrt{\frac{c}{\pi }} e^{\left (\frac{-i \, b^{2} + 4 i \, a c - 2 \, b + i}{4 \, c}\right )} \operatorname{C}\left (-\frac{\sqrt{2}{\left (2 \, c x + b - i\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) + \sqrt{2} \pi \sqrt{\frac{c}{\pi }} e^{\left (\frac{i \, b^{2} - 4 i \, a c - 2 \, b - i}{4 \, c}\right )} \operatorname{S}\left (\frac{\sqrt{2}{\left (2 \, c x + b + i\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) - \sqrt{2} \pi \sqrt{\frac{c}{\pi }} e^{\left (\frac{-i \, b^{2} + 4 i \, a c - 2 \, b + i}{4 \, c}\right )} \operatorname{S}\left (-\frac{\sqrt{2}{\left (2 \, c x + b - i\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right )}{4 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(I*sqrt(2)*pi*sqrt(c/pi)*e^(1/4*(I*b^2 - 4*I*a*c - 2*b - I)/c)*fresnel_cos(1/2*sqrt(2)*(2*c*x + b + I)*sqr
t(c/pi)/c) + I*sqrt(2)*pi*sqrt(c/pi)*e^(1/4*(-I*b^2 + 4*I*a*c - 2*b + I)/c)*fresnel_cos(-1/2*sqrt(2)*(2*c*x +
b - I)*sqrt(c/pi)/c) + sqrt(2)*pi*sqrt(c/pi)*e^(1/4*(I*b^2 - 4*I*a*c - 2*b - I)/c)*fresnel_sin(1/2*sqrt(2)*(2*
c*x + b + I)*sqrt(c/pi)/c) - sqrt(2)*pi*sqrt(c/pi)*e^(1/4*(-I*b^2 + 4*I*a*c - 2*b + I)/c)*fresnel_sin(-1/2*sqr
t(2)*(2*c*x + b - I)*sqrt(c/pi)/c))/c

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{x} \sin{\left (a + b x + c x^{2} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(exp(x)*sin(a + b*x + c*x**2), x)

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Giac [A]  time = 1.16922, size = 198, normalized size = 1.38 \begin{align*} \frac{i \, \sqrt{2} \sqrt{\pi } \operatorname{erf}\left (-\frac{1}{4} \, \sqrt{2}{\left (2 \, x + \frac{b - i}{c}\right )}{\left (-\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}\right ) e^{\left (-\frac{i \, b^{2} - 4 i \, a c + 2 \, b - i}{4 \, c}\right )}}{4 \,{\left (-\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}} - \frac{i \, \sqrt{2} \sqrt{\pi } \operatorname{erf}\left (-\frac{1}{4} \, \sqrt{2}{\left (2 \, x + \frac{b + i}{c}\right )}{\left (\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}\right ) e^{\left (-\frac{-i \, b^{2} + 4 i \, a c + 2 \, b + i}{4 \, c}\right )}}{4 \,{\left (\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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