\(\int e^x \sin (a+b x+c x^2) \, dx\) [75]

Optimal result

Integrand size = 15, antiderivative size = 144 \[ \int e^x \sin \left (a+b x+c x^2\right ) \, dx=\frac {(-1)^{3/4} e^{\frac {1}{4} i \left (4 a+\frac {(1+i b)^2}{c}\right )} \sqrt {\pi } \text {erf}\left (\frac {\sqrt [4]{-1} (1+i b+2 i c x)}{2 \sqrt {c}}\right )}{4 \sqrt {c}}+\frac {(-1)^{3/4} e^{-i a+\frac {i (i+b)^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {\sqrt [4]{-1} (1-i b-2 i c x)}{2 \sqrt {c}}\right )}{4 \sqrt {c}} \]

[Out]

1/4*(-1)^(3/4)*exp(1/4*I*(4*a+(1+I*b)^2/c))*erf(1/2*(-1)^(1/4)*(1+I*b+2*I*c*x)/c^(1/2))*Pi^(1/2)/c^(1/2)+1/4*(
-1)^(3/4)*exp(-I*a+1/4*I*(I+b)^2/c)*erfi(1/2*(-1)^(1/4)*(1-I*b-2*I*c*x)/c^(1/2))*Pi^(1/2)/c^(1/2)

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {4560, 2266, 2235, 2236} \[ \int e^x \sin \left (a+b x+c x^2\right ) \, dx=\frac {(-1)^{3/4} \sqrt {\pi } e^{\frac {1}{4} i \left (4 a+\frac {(1+i b)^2}{c}\right )} \text {erf}\left (\frac {\sqrt [4]{-1} (i b+2 i c x+1)}{2 \sqrt {c}}\right )}{4 \sqrt {c}}+\frac {(-1)^{3/4} \sqrt {\pi } e^{\frac {i (b+i)^2}{4 c}-i a} \text {erfi}\left (\frac {\sqrt [4]{-1} (-i b-2 i c x+1)}{2 \sqrt {c}}\right )}{4 \sqrt {c}} \]

[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 2235

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 2236

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]

Rule 2266

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 4560

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.19 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.93 \[ \int e^x \sin \left (a+b x+c x^2\right ) \, dx=-\frac {\sqrt [4]{-1} e^{-\frac {i \left (1-2 i b+b^2\right )}{4 c}} \sqrt {\pi } \left (e^{\left .\frac {i}{2}\right /c} \text {erfi}\left (\frac {\sqrt [4]{-1} (-i+b+2 c x)}{2 \sqrt {c}}\right ) (\cos (a)+i \sin (a))+e^{\frac {i b^2}{2 c}} \text {erfi}\left (\frac {(-1)^{3/4} (i+b+2 c x)}{2 \sqrt {c}}\right ) (i \cos (a)+\sin (a))\right )}{4 \sqrt {c}} \]

[In]

Integrate[E^x*Sin[a + b*x + c*x^2],x]

[Out]

-1/4*((-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])))/(Sqrt[c]*E^(((I/4)*(1 -
(2*I)*b + b^2))/c))

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {i \sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c -b^{2}+2 i b +1\right )}{4 c}} \operatorname {erf}\left (-\sqrt {-i c}\, x +\frac {i b +1}{2 \sqrt {-i c}}\right )}{4 \sqrt {-i c}}+\frac {i \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c -b^{2}-2 i b +1\right )}{4 c}} \operatorname {erf}\left (\sqrt {i c}\, x -\frac {-i b +1}{2 \sqrt {i c}}\right )}{4 \sqrt {i c}}\)	\(119\)

[In]

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

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (91) = 182\).

Time = 0.25 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.59 \[ \int e^x \sin \left (a+b x+c x^2\right ) \, dx=\frac {i \, \sqrt {2} \pi \sqrt {\frac {c}{\pi }} e^{\left (\frac {i \, b^{2} - 4 i \, a c - 2 \, b - i}{4 \, c}\right )} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x + b + i\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) + i \, \sqrt {2} \pi \sqrt {\frac {c}{\pi }} e^{\left (\frac {-i \, b^{2} + 4 i \, a c - 2 \, b + i}{4 \, c}\right )} \operatorname {C}\left (-\frac {\sqrt {2} {\left (2 \, c x + b - i\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) + \sqrt {2} \pi \sqrt {\frac {c}{\pi }} e^{\left (\frac {i \, b^{2} - 4 i \, a c - 2 \, b - i}{4 \, c}\right )} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x + b + i\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - \sqrt {2} \pi \sqrt {\frac {c}{\pi }} e^{\left (\frac {-i \, b^{2} + 4 i \, a c - 2 \, b + i}{4 \, c}\right )} \operatorname {S}\left (-\frac {\sqrt {2} {\left (2 \, c x + b - i\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right )}{4 \, c} \]

[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

Sympy [F]

\[ \int e^x \sin \left (a+b x+c x^2\right ) \, dx=\int e^{x} \sin {\left (a + b x + c x^{2} \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.91 \[ \int e^x \sin \left (a+b x+c x^2\right ) \, dx=-\frac {\sqrt {2} \sqrt {\pi } {\left ({\left (\left (i + 1\right ) \, \cos \left (-\frac {b^{2} - 4 \, a c - 1}{4 \, c}\right ) - \left (i - 1\right ) \, \sin \left (-\frac {b^{2} - 4 \, a c - 1}{4 \, c}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, c x + i \, b - 1\right )} \sqrt {i \, c}}{2 \, c}\right ) + {\left (-\left (i - 1\right ) \, \cos \left (-\frac {b^{2} - 4 \, a c - 1}{4 \, c}\right ) + \left (i + 1\right ) \, \sin \left (-\frac {b^{2} - 4 \, a c - 1}{4 \, c}\right )\right )} \operatorname {erf}\left (\frac {i \, {\left (2 i \, c x + i \, b + 1\right )} \sqrt {-i \, c}}{2 \, c}\right )\right )} e^{\left (-\frac {b}{2 \, c}\right )}}{8 \, \sqrt {c}} \]

[In]

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

[Out]

-1/8*sqrt(2)*sqrt(pi)*(((I + 1)*cos(-1/4*(b^2 - 4*a*c - 1)/c) - (I - 1)*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 - 1)*cos(-1/4*(b^2 - 4*a*c - 1)/c) + (I + 1)*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(c)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.02 \[ \int e^x \sin \left (a+b x+c x^2\right ) \, dx=\frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{4} i \, \sqrt {2} {\left (2 \, x + \frac {b - i}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c + 2 \, b - i}{4 \, c}\right )}}{4 \, {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} + \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\frac {1}{4} i \, \sqrt {2} {\left (2 \, x + \frac {b + i}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c + 2 \, b + i}{4 \, c}\right )}}{4 \, {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} \]

[In]

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

[Out]

1/4*sqrt(2)*sqrt(pi)*erf(-1/4*I*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*sqrt(2)*sqrt(pi)*erf(1/4*I*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)))

Mupad [F(-1)]

Timed out. \[ \int e^x \sin \left (a+b x+c x^2\right ) \, dx=\int \sin \left (c\,x^2+b\,x+a\right )\,{\mathrm {e}}^x \,d x \]

[In]

int(sin(a + b*x + c*x^2)*exp(x),x)

[Out]

int(sin(a + b*x + c*x^2)*exp(x), x)