\(\int e^{x^2} \sin (a+b x+c x^2) \, dx\) [90]

Optimal result

Integrand size = 17, antiderivative size = 155 \[ \int e^{x^2} \sin \left (a+b x+c x^2\right ) \, dx=-\frac {i e^{-i \left (a-\frac {b^2}{4 i+4 c}\right )} \sqrt {\pi } \text {erfi}\left (\frac {i b-2 (1-i c) x}{2 \sqrt {1-i c}}\right )}{4 \sqrt {1-i c}}-\frac {i e^{i a+\frac {b^2}{4 (1+i c)}} \sqrt {\pi } \text {erfi}\left (\frac {i b+2 (1+i c) x}{2 \sqrt {1+i c}}\right )}{4 \sqrt {1+i c}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.40 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.06 \[ \int e^{x^2} \sin \left (a+b x+c x^2\right ) \, dx=-\frac {(-1)^{3/4} e^{\frac {i b^2}{4 i-4 c}} \sqrt {\pi } \left ((-i+c) \sqrt {i+c} e^{\frac {i b^2 c}{2+2 c^2}} \text {erfi}\left (\frac {(-1)^{3/4} (b+2 (i+c) x)}{2 \sqrt {i+c}}\right ) (\cos (a)-i \sin (a))+\sqrt {-i+c} (i+c) \text {erfi}\left (\frac {\sqrt [4]{-1} (b+2 (-i+c) x)}{2 \sqrt {-i+c}}\right ) (-i \cos (a)+\sin (a))\right )}{4 \left (1+c^2\right )} \] Input:

Integrate[E^x^2*Sin[a + b*x + c*x^2],x]
 

Output:

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

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4975, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 4975

\(\displaystyle \int \left (\frac {1}{2} i e^{-i a-i b x+(1-i c) x^2}-\frac {1}{2} i e^{i a+i b x+(1+i c) x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {i \sqrt {\pi } e^{-i \left (a-\frac {b^2}{4 c+4 i}\right )} \text {erfi}\left (\frac {i b-2 (1-i c) x}{2 \sqrt {1-i c}}\right )}{4 \sqrt {1-i c}}-\frac {i \sqrt {\pi } e^{i a+\frac {b^2}{4 (1+i c)}} \text {erfi}\left (\frac {i b+2 (1+i c) x}{2 \sqrt {1+i c}}\right )}{4 \sqrt {1+i c}}\)

Input:

Int[E^x^2*Sin[a + b*x + c*x^2],x]
 

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4975

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]
 

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.83

method	result	size
risch	\(\frac {i \sqrt {\pi }\, {\mathrm e}^{-\frac {4 a c -4 i a -b^{2}}{4 \left (i c +1\right )}} \operatorname {erf}\left (-\sqrt {-i c -1}\, x +\frac {i b}{2 \sqrt {-i c -1}}\right )}{4 \sqrt {-i c -1}}+\frac {i \sqrt {\pi }\, {\mathrm e}^{\frac {4 a c +4 i a -b^{2}}{4 i c -4}} \operatorname {erf}\left (\sqrt {i c -1}\, x +\frac {i b}{2 \sqrt {i c -1}}\right )}{4 \sqrt {i c -1}}\)	\(129\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.04 \[ \int e^{x^2} \sin \left (a+b x+c x^2\right ) \, dx=-\frac {\sqrt {\pi } {\left (c - i\right )} \sqrt {i \, c - 1} \operatorname {erf}\left (-\frac {{\left (b c + 2 \, {\left (c^{2} + 1\right )} x - i \, b\right )} \sqrt {i \, c - 1}}{2 \, {\left (c^{2} + 1\right )}}\right ) e^{\left (\frac {i \, b^{2} c - 4 i \, a c^{2} + b^{2} - 4 i \, a}{4 \, {\left (c^{2} + 1\right )}}\right )} - \sqrt {\pi } {\left (c + i\right )} \sqrt {-i \, c - 1} \operatorname {erf}\left (\frac {{\left (b c + 2 \, {\left (c^{2} + 1\right )} x + i \, b\right )} \sqrt {-i \, c - 1}}{2 \, {\left (c^{2} + 1\right )}}\right ) e^{\left (\frac {-i \, b^{2} c + 4 i \, a c^{2} + b^{2} + 4 i \, a}{4 \, {\left (c^{2} + 1\right )}}\right )}}{4 \, {\left (c^{2} + 1\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 475 vs. \(2 (100) = 200\).

Time = 0.05 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.06 \[ \int e^{x^2} \sin \left (a+b x+c x^2\right ) \, dx=\frac {\sqrt {\pi } \sqrt {2 \, c^{2} + 2} {\left ({\left (\cos \left (-\frac {b^{2} c - 4 \, a c^{2} - 4 \, a}{4 \, {\left (c^{2} + 1\right )}}\right ) e^{\left (\frac {b^{2}}{4 \, {\left (c^{2} + 1\right )}}\right )} - i \, e^{\left (\frac {b^{2}}{4 \, {\left (c^{2} + 1\right )}}\right )} \sin \left (-\frac {b^{2} c - 4 \, a c^{2} - 4 \, a}{4 \, {\left (c^{2} + 1\right )}}\right )\right )} \operatorname {erf}\left (-\frac {2 \, {\left (-i \, c + 1\right )} x - i \, b}{2 \, \sqrt {i \, c - 1}}\right ) - {\left (\cos \left (-\frac {b^{2} c - 4 \, a c^{2} - 4 \, a}{4 \, {\left (c^{2} + 1\right )}}\right ) e^{\left (\frac {b^{2}}{4 \, {\left (c^{2} + 1\right )}}\right )} + i \, e^{\left (\frac {b^{2}}{4 \, {\left (c^{2} + 1\right )}}\right )} \sin \left (-\frac {b^{2} c - 4 \, a c^{2} - 4 \, a}{4 \, {\left (c^{2} + 1\right )}}\right )\right )} \operatorname {erf}\left (-\frac {2 \, {\left (-i \, c - 1\right )} x - i \, b}{2 \, \sqrt {-i \, c - 1}}\right )\right )} \sqrt {\sqrt {c^{2} + 1} + 1} - \sqrt {\pi } \sqrt {2 \, c^{2} + 2} {\left ({\left (-i \, \cos \left (-\frac {b^{2} c - 4 \, a c^{2} - 4 \, a}{4 \, {\left (c^{2} + 1\right )}}\right ) e^{\left (\frac {b^{2}}{4 \, {\left (c^{2} + 1\right )}}\right )} - e^{\left (\frac {b^{2}}{4 \, {\left (c^{2} + 1\right )}}\right )} \sin \left (-\frac {b^{2} c - 4 \, a c^{2} - 4 \, a}{4 \, {\left (c^{2} + 1\right )}}\right )\right )} \operatorname {erf}\left (-\frac {2 \, {\left (-i \, c + 1\right )} x - i \, b}{2 \, \sqrt {i \, c - 1}}\right ) + {\left (-i \, \cos \left (-\frac {b^{2} c - 4 \, a c^{2} - 4 \, a}{4 \, {\left (c^{2} + 1\right )}}\right ) e^{\left (\frac {b^{2}}{4 \, {\left (c^{2} + 1\right )}}\right )} + e^{\left (\frac {b^{2}}{4 \, {\left (c^{2} + 1\right )}}\right )} \sin \left (-\frac {b^{2} c - 4 \, a c^{2} - 4 \, a}{4 \, {\left (c^{2} + 1\right )}}\right )\right )} \operatorname {erf}\left (-\frac {2 \, {\left (-i \, c - 1\right )} x - i \, b}{2 \, \sqrt {-i \, c - 1}}\right )\right )} \sqrt {\sqrt {c^{2} + 1} - 1}}{8 \, {\left (c^{2} + 1\right )}} \] Input:

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

Output:

1/8*(sqrt(pi)*sqrt(2*c^2 + 2)*((cos(-1/4*(b^2*c - 4*a*c^2 - 4*a)/(c^2 + 1) 
)*e^(1/4*b^2/(c^2 + 1)) - I*e^(1/4*b^2/(c^2 + 1))*sin(-1/4*(b^2*c - 4*a*c^ 
2 - 4*a)/(c^2 + 1)))*erf(-1/2*(2*(-I*c + 1)*x - I*b)/sqrt(I*c - 1)) - (cos 
(-1/4*(b^2*c - 4*a*c^2 - 4*a)/(c^2 + 1))*e^(1/4*b^2/(c^2 + 1)) + I*e^(1/4* 
b^2/(c^2 + 1))*sin(-1/4*(b^2*c - 4*a*c^2 - 4*a)/(c^2 + 1)))*erf(-1/2*(2*(- 
I*c - 1)*x - I*b)/sqrt(-I*c - 1)))*sqrt(sqrt(c^2 + 1) + 1) - sqrt(pi)*sqrt 
(2*c^2 + 2)*((-I*cos(-1/4*(b^2*c - 4*a*c^2 - 4*a)/(c^2 + 1))*e^(1/4*b^2/(c 
^2 + 1)) - e^(1/4*b^2/(c^2 + 1))*sin(-1/4*(b^2*c - 4*a*c^2 - 4*a)/(c^2 + 1 
)))*erf(-1/2*(2*(-I*c + 1)*x - I*b)/sqrt(I*c - 1)) + (-I*cos(-1/4*(b^2*c - 
 4*a*c^2 - 4*a)/(c^2 + 1))*e^(1/4*b^2/(c^2 + 1)) + e^(1/4*b^2/(c^2 + 1))*s 
in(-1/4*(b^2*c - 4*a*c^2 - 4*a)/(c^2 + 1)))*erf(-1/2*(2*(-I*c - 1)*x - I*b 
)/sqrt(-I*c - 1)))*sqrt(sqrt(c^2 + 1) - 1))/(c^2 + 1)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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