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

Optimal. Leaf size=115 \[ \frac{(-1)^{3/4} \sqrt{\pi } e^{\frac{1}{4} i \left (4 a+\frac{1}{c}\right )} \text{Erf}\left (\frac{\sqrt [4]{-1} (1+2 i c x)}{2 \sqrt{c}}\right )}{4 \sqrt{c}}+\frac{(-1)^{3/4} \sqrt{\pi } e^{-\frac{1}{4} i \left (4 a+\frac{1}{c}\right )} \text{Erfi}\left (\frac{\sqrt [4]{-1} (1-2 i c x)}{2 \sqrt{c}}\right )}{4 \sqrt{c}} \]

[Out]

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

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Rubi [A]  time = 0.123032, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4472, 2234, 2204, 2205} \[ \frac{(-1)^{3/4} \sqrt{\pi } e^{\frac{1}{4} i \left (4 a+\frac{1}{c}\right )} \text{Erf}\left (\frac{\sqrt [4]{-1} (1+2 i c x)}{2 \sqrt{c}}\right )}{4 \sqrt{c}}+\frac{(-1)^{3/4} \sqrt{\pi } e^{-\frac{1}{4} i \left (4 a+\frac{1}{c}\right )} \text{Erfi}\left (\frac{\sqrt [4]{-1} (1-2 i c x)}{2 \sqrt{c}}\right )}{4 \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sin[a + c*x^2],x]

[Out]

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

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

Mathematica [A]  time = 0.156724, size = 108, normalized size = 0.94 \[ -\frac{\sqrt [4]{-1} \sqrt{\pi } e^{\left .-\frac{i}{4}\right /c} \left (e^{\left .\frac{i}{2}\right /c} (\cos (a)+i \sin (a)) \text{Erfi}\left (\frac{\sqrt [4]{-1} (2 c x-i)}{2 \sqrt{c}}\right )+(\sin (a)+i \cos (a)) \text{Erfi}\left (\frac{(-1)^{3/4} (2 c x+i)}{2 \sqrt{c}}\right )\right )}{4 \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sin[a + c*x^2],x]

[Out]

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

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Maple [A]  time = 0.124, size = 88, normalized size = 0.8 \begin{align*}{-{\frac{i}{4}}\sqrt{\pi }{{\rm e}^{{\frac{{\frac{i}{4}} \left ( 4\,ac+1 \right ) }{c}}}}{\it Erf} \left ( \sqrt{-ic}x-{\frac{1}{2}{\frac{1}{\sqrt{-ic}}}} \right ){\frac{1}{\sqrt{-ic}}}}+{{\frac{i}{4}}\sqrt{\pi }{{\rm e}^{{\frac{-{\frac{i}{4}} \left ( 4\,ac+1 \right ) }{c}}}}{\it Erf} \left ( \sqrt{ic}x-{\frac{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+a),x)

[Out]

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

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Maxima [B]  time = 1.92927, size = 377, normalized size = 3.28 \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{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{4 \, a c + 1}{4 \, c}\right )\right )} \operatorname{erf}\left (\frac{2 i \, c x - 1}{2 \, \sqrt{i \, 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{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{4 \, a c + 1}{4 \, c}\right )\right )} \operatorname{erf}\left (\frac{2 i \, c x + 1}{2 \, \sqrt{-i \, c}}\right )\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+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*ar
ctan2(0, c)) + sin(-1/4*pi + 1/2*arctan2(0, c)))*cos(1/4*(4*a*c + 1)/c) - (cos(1/4*pi + 1/2*arctan2(0, c)) + c
os(-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*(4*a*c + 1)/c))*erf(1/2*(2*I*c*x - 1)/sqrt(I*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*(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*(4*a*c + 1)/c))*erf(1/2*(2*I*c*x + 1)/sqrt(-I*c)))/sqrt(abs(c))

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Fricas [B]  time = 0.492526, size = 549, normalized size = 4.77 \begin{align*} \frac{i \, \sqrt{2} \pi \sqrt{\frac{c}{\pi }} e^{\left (\frac{-4 i \, a c - i}{4 \, c}\right )} \operatorname{C}\left (\frac{\sqrt{2}{\left (2 \, c x + i\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) + i \, \sqrt{2} \pi \sqrt{\frac{c}{\pi }} e^{\left (\frac{4 i \, a c + i}{4 \, c}\right )} \operatorname{C}\left (-\frac{\sqrt{2}{\left (2 \, c x - i\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) + \sqrt{2} \pi \sqrt{\frac{c}{\pi }} e^{\left (\frac{-4 i \, a c - i}{4 \, c}\right )} \operatorname{S}\left (\frac{\sqrt{2}{\left (2 \, c x + i\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) - \sqrt{2} \pi \sqrt{\frac{c}{\pi }} e^{\left (\frac{4 i \, a c + i}{4 \, c}\right )} \operatorname{S}\left (-\frac{\sqrt{2}{\left (2 \, c x - 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+a),x, algorithm="fricas")

[Out]

1/4*(I*sqrt(2)*pi*sqrt(c/pi)*e^(1/4*(-4*I*a*c - I)/c)*fresnel_cos(1/2*sqrt(2)*(2*c*x + I)*sqrt(c/pi)/c) + I*sq
rt(2)*pi*sqrt(c/pi)*e^(1/4*(4*I*a*c + I)/c)*fresnel_cos(-1/2*sqrt(2)*(2*c*x - I)*sqrt(c/pi)/c) + sqrt(2)*pi*sq
rt(c/pi)*e^(1/4*(-4*I*a*c - I)/c)*fresnel_sin(1/2*sqrt(2)*(2*c*x + I)*sqrt(c/pi)/c) - sqrt(2)*pi*sqrt(c/pi)*e^
(1/4*(4*I*a*c + I)/c)*fresnel_sin(-1/2*sqrt(2)*(2*c*x - 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 + 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+a),x)

[Out]

Integral(exp(x)*sin(a + c*x**2), x)

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Giac [A]  time = 1.14007, size = 171, normalized size = 1.49 \begin{align*} -\frac{i \, \sqrt{2} \sqrt{\pi } \operatorname{erf}\left (-\frac{1}{4} \, \sqrt{2}{\left (2 \, x + \frac{i}{c}\right )}{\left (\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}\right ) e^{\left (-\frac{4 i \, a c + 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{i}{c}\right )}{\left (-\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}\right ) e^{\left (-\frac{-4 i \, a c - 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+a),x, algorithm="giac")

[Out]

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