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

Optimal. Leaf size=97 \[ \frac{\sqrt{\frac{\pi }{2}} \sin \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )}{\sqrt{c}}+\frac{\sqrt{\frac{\pi }{2}} \cos \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{\sqrt{c}} \]

[Out]

(Sqrt[Pi/2]*Cos[a - b^2/(4*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])])/Sqrt[c] + (Sqrt[Pi/2]*FresnelC[(b +
2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)])/Sqrt[c]

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Rubi [A]  time = 0.0212291, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3447, 3351, 3352} \[ \frac{\sqrt{\frac{\pi }{2}} \sin \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )}{\sqrt{c}}+\frac{\sqrt{\frac{\pi }{2}} \cos \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{\sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x + c*x^2],x]

[Out]

(Sqrt[Pi/2]*Cos[a - b^2/(4*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])])/Sqrt[c] + (Sqrt[Pi/2]*FresnelC[(b +
2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)])/Sqrt[c]

Rule 3447

Int[Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[Cos[(b^2 - 4*a*c)/(4*c)], Int[Sin[(b + 2*c*x)^2/
(4*c)], x], x] - Dist[Sin[(b^2 - 4*a*c)/(4*c)], Int[Cos[(b + 2*c*x)^2/(4*c)], x], x] /; FreeQ[{a, b, c}, x] &&
NeQ[b^2 - 4*a*c, 0]

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rubi steps

\begin{align*} \int \sin \left (a+b x+c x^2\right ) \, dx &=\cos \left (a-\frac{b^2}{4 c}\right ) \int \sin \left (\frac{(b+2 c x)^2}{4 c}\right ) \, dx+\sin \left (a-\frac{b^2}{4 c}\right ) \int \cos \left (\frac{(b+2 c x)^2}{4 c}\right ) \, dx\\ &=\frac{\sqrt{\frac{\pi }{2}} \cos \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{\sqrt{c}}+\frac{\sqrt{\frac{\pi }{2}} C\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right ) \sin \left (a-\frac{b^2}{4 c}\right )}{\sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.172663, size = 84, normalized size = 0.87 \[ \frac{\sqrt{\frac{\pi }{2}} \left (\sin \left (a-\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )+\cos \left (a-\frac{b^2}{4 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )\right )}{\sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x + c*x^2],x]

[Out]

(Sqrt[Pi/2]*(Cos[a - b^2/(4*c)]*FresnelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])] + FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqr
t[2*Pi])]*Sin[a - b^2/(4*c)]))/Sqrt[c]

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Maple [A]  time = 0.007, size = 82, normalized size = 0.9 \begin{align*}{\frac{\sqrt{2}\sqrt{\pi }}{2} \left ( \cos \left ({\frac{1}{c} \left ({\frac{{b}^{2}}{4}}-ca \right ) } \right ){\it FresnelS} \left ({\frac{\sqrt{2}}{\sqrt{\pi }} \left ( cx+{\frac{b}{2}} \right ){\frac{1}{\sqrt{c}}}} \right ) -\sin \left ({\frac{1}{c} \left ({\frac{{b}^{2}}{4}}-ca \right ) } \right ){\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi }} \left ( cx+{\frac{b}{2}} \right ){\frac{1}{\sqrt{c}}}} \right ) \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*2^(1/2)*Pi^(1/2)/c^(1/2)*(cos((1/4*b^2-c*a)/c)*FresnelS(2^(1/2)/Pi^(1/2)/c^(1/2)*(c*x+1/2*b))-sin((1/4*b^2
-c*a)/c)*FresnelC(2^(1/2)/Pi^(1/2)/c^(1/2)*(c*x+1/2*b)))

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Maxima [C]  time = 2.31587, size = 393, normalized size = 4.05 \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}{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}{4 \, c}\right )\right )} \operatorname{erf}\left (\frac{2 i \, c x + i \, b}{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{b^{2} - 4 \, a c}{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}{4 \, c}\right )\right )} \operatorname{erf}\left (\frac{2 i \, c x + i \, b}{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(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*ar
ctan2(0, c)) + sin(-1/4*pi + 1/2*arctan2(0, c)))*cos(-1/4*(b^2 - 4*a*c)/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)))*s
in(-1/4*(b^2 - 4*a*c)/c))*erf(1/2*(2*I*c*x + I*b)/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*(b^2
- 4*a*c)/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*arctan
2(0, c)) - I*sin(-1/4*pi + 1/2*arctan2(0, c)))*sin(-1/4*(b^2 - 4*a*c)/c))*erf(1/2*(2*I*c*x + I*b)/sqrt(-I*c)))
/sqrt(abs(c))

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Fricas [A]  time = 1.34997, size = 279, normalized size = 2.88 \begin{align*} \frac{\sqrt{2} \pi \sqrt{\frac{c}{\pi }} \cos \left (-\frac{b^{2} - 4 \, a c}{4 \, c}\right ) \operatorname{S}\left (\frac{\sqrt{2}{\left (2 \, c x + b\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) + \sqrt{2} \pi \sqrt{\frac{c}{\pi }} \operatorname{C}\left (\frac{\sqrt{2}{\left (2 \, c x + b\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) \sin \left (-\frac{b^{2} - 4 \, a c}{4 \, c}\right )}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(sqrt(2)*pi*sqrt(c/pi)*cos(-1/4*(b^2 - 4*a*c)/c)*fresnel_sin(1/2*sqrt(2)*(2*c*x + b)*sqrt(c/pi)/c) + sqrt(
2)*pi*sqrt(c/pi)*fresnel_cos(1/2*sqrt(2)*(2*c*x + b)*sqrt(c/pi)/c)*sin(-1/4*(b^2 - 4*a*c)/c))/c

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Sympy [A]  time = 0.547151, size = 88, normalized size = 0.91 \begin{align*} \frac{\sqrt{2} \sqrt{\pi } \left (\sin{\left (a - \frac{b^{2}}{4 c} \right )} C\left (\frac{\sqrt{2} \left (b + 2 c x\right )}{2 \sqrt{\pi } \sqrt{c}}\right ) + \cos{\left (a - \frac{b^{2}}{4 c} \right )} S\left (\frac{\sqrt{2} \left (b + 2 c x\right )}{2 \sqrt{\pi } \sqrt{c}}\right )\right ) \sqrt{\frac{1}{c}}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(2)*sqrt(pi)*(sin(a - b**2/(4*c))*fresnelc(sqrt(2)*(b + 2*c*x)/(2*sqrt(pi)*sqrt(c))) + cos(a - b**2/(4*c))
*fresnels(sqrt(2)*(b + 2*c*x)/(2*sqrt(pi)*sqrt(c))))*sqrt(1/c)/2

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Giac [C]  time = 1.21517, size = 182, normalized size = 1.88 \begin{align*} \frac{i \, \sqrt{2} \sqrt{\pi } \operatorname{erf}\left (-\frac{1}{4} \, \sqrt{2}{\left (2 \, x + \frac{b}{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}{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}{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}{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(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/c)*(-I*c/abs(c) + 1)*sqrt(abs(c)))*e^(-1/4*(I*b^2 - 4*I*a*c)/
c)/((-I*c/abs(c) + 1)*sqrt(abs(c))) - 1/4*I*sqrt(2)*sqrt(pi)*erf(-1/4*sqrt(2)*(2*x + b/c)*(I*c/abs(c) + 1)*sqr
t(abs(c)))*e^(-1/4*(-I*b^2 + 4*I*a*c)/c)/((I*c/abs(c) + 1)*sqrt(abs(c)))