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

Optimal. Leaf size=100 \[ -\frac{\sqrt{\pi } \cos \left (2 a-\frac{b^2}{2 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{\pi } \sqrt{c}}\right )}{4 \sqrt{c}}+\frac{\sqrt{\pi } \sin \left (2 a-\frac{b^2}{2 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{\pi }}\right )}{4 \sqrt{c}}+\frac{x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0484573, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {3449, 3448, 3352, 3351} \[ -\frac{\sqrt{\pi } \cos \left (2 a-\frac{b^2}{2 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{\pi } \sqrt{c}}\right )}{4 \sqrt{c}}+\frac{\sqrt{\pi } \sin \left (2 a-\frac{b^2}{2 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{\pi }}\right )}{4 \sqrt{c}}+\frac{x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x + c*x^2]^2,x]

[Out]

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

Rule 3449

Int[Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]^(n_), x_Symbol] :> Int[ExpandTrigReduce[Sin[a + b*x + c*x^2]^n, x],
x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 1]

Rule 3448

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

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]

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]

Rubi steps

\begin{align*} \int \sin ^2\left (a+b x+c x^2\right ) \, dx &=\int \left (\frac{1}{2}-\frac{1}{2} \cos \left (2 a+2 b x+2 c x^2\right )\right ) \, dx\\ &=\frac{x}{2}-\frac{1}{2} \int \cos \left (2 a+2 b x+2 c x^2\right ) \, dx\\ &=\frac{x}{2}-\frac{1}{2} \cos \left (2 a-\frac{b^2}{2 c}\right ) \int \cos \left (\frac{(2 b+4 c x)^2}{8 c}\right ) \, dx+\frac{1}{2} \sin \left (2 a-\frac{b^2}{2 c}\right ) \int \sin \left (\frac{(2 b+4 c x)^2}{8 c}\right ) \, dx\\ &=\frac{x}{2}-\frac{\sqrt{\pi } \cos \left (2 a-\frac{b^2}{2 c}\right ) C\left (\frac{b+2 c x}{\sqrt{c} \sqrt{\pi }}\right )}{4 \sqrt{c}}+\frac{\sqrt{\pi } S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{\pi }}\right ) \sin \left (2 a-\frac{b^2}{2 c}\right )}{4 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.0728906, size = 97, normalized size = 0.97 \[ \frac{-\sqrt{\pi } \cos \left (2 a-\frac{b^2}{2 c}\right ) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{\pi } \sqrt{c}}\right )+\sqrt{\pi } \sin \left (2 a-\frac{b^2}{2 c}\right ) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{\pi }}\right )+2 \sqrt{c} x}{4 \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x + c*x^2]^2,x]

[Out]

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

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Maple [A]  time = 0.013, size = 72, normalized size = 0.7 \begin{align*}{\frac{x}{2}}-{\frac{\sqrt{\pi }}{4} \left ( \cos \left ({\frac{-4\,ca+{b}^{2}}{2\,c}} \right ){\it FresnelC} \left ({\frac{2\,cx+b}{\sqrt{\pi }}{\frac{1}{\sqrt{c}}}} \right ) +\sin \left ({\frac{-4\,ca+{b}^{2}}{2\,c}} \right ){\it FresnelS} \left ({\frac{2\,cx+b}{\sqrt{\pi }}{\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)^2,x)

[Out]

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

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Maxima [C]  time = 2.16694, size = 410, normalized size = 4.1 \begin{align*} -\frac{\sqrt{2} \sqrt{\pi }{\left ({\left ({\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 )} \cos \left (-\frac{b^{2} - 4 \, a c}{2 \, c}\right ) -{\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 )} \sin \left (-\frac{b^{2} - 4 \, a c}{2 \, c}\right )\right )} \operatorname{erf}\left (\frac{2 i \, c x + i \, b}{\sqrt{2 i \, c}}\right ) -{\left ({\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 )} \cos \left (-\frac{b^{2} - 4 \, a c}{2 \, c}\right ) +{\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 )} \sin \left (-\frac{b^{2} - 4 \, a c}{2 \, c}\right )\right )} \operatorname{erf}\left (\frac{2 i \, c x + i \, b}{\sqrt{-2 i \, c}}\right )\right )} \sqrt{{\left | c \right |}} - 16 \, x{\left | c \right |}}{32 \,{\left | c \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(sqrt(2)*sqrt(pi)*(((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)))*cos(-1/2*(b^2 - 4*a*c)/c) - (I*cos(1/4*pi + 1/2*arct
an2(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)))*sin(-1/2*(b^2 - 4*a*c)/c))*erf((2*I*c*x + I*b)/sqrt(2*I*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)))*cos(-1/
2*(b^2 - 4*a*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)))*sin(-1/2*(b^2 - 4*a*c)/c))*erf((2*I*c*x + I*b)/sqrt(-2*I*
c)))*sqrt(abs(c)) - 16*x*abs(c))/abs(c)

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Fricas [A]  time = 1.46734, size = 238, normalized size = 2.38 \begin{align*} -\frac{\pi \sqrt{\frac{c}{\pi }} \cos \left (-\frac{b^{2} - 4 \, a c}{2 \, c}\right ) \operatorname{C}\left (\frac{{\left (2 \, c x + b\right )} \sqrt{\frac{c}{\pi }}}{c}\right ) - \pi \sqrt{\frac{c}{\pi }} \operatorname{S}\left (\frac{{\left (2 \, c x + b\right )} \sqrt{\frac{c}{\pi }}}{c}\right ) \sin \left (-\frac{b^{2} - 4 \, a c}{2 \, c}\right ) - 2 \, c x}{4 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.20047, size = 83, normalized size = 0.83 \begin{align*} \frac{x}{2} - \frac{\sqrt{\pi } \left (- \sin{\left (2 a - \frac{b^{2}}{2 c} \right )} S\left (\frac{2 b + 4 c x}{2 \sqrt{\pi } \sqrt{c}}\right ) + \cos{\left (2 a - \frac{b^{2}}{2 c} \right )} C\left (\frac{2 b + 4 c x}{2 \sqrt{\pi } \sqrt{c}}\right )\right ) \sqrt{\frac{1}{c}}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.32185, size = 165, normalized size = 1.65 \begin{align*} \frac{1}{2} \, x + \frac{\sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{c}{\left (2 \, x + \frac{b}{c}\right )}{\left (-\frac{i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac{i \, b^{2} - 4 i \, a c}{2 \, c}\right )}}{8 \, \sqrt{c}{\left (-\frac{i \, c}{{\left | c \right |}} + 1\right )}} + \frac{\sqrt{\pi } \operatorname{erf}\left (-\frac{1}{2} \, \sqrt{c}{\left (2 \, x + \frac{b}{c}\right )}{\left (\frac{i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac{-i \, b^{2} + 4 i \, a c}{2 \, c}\right )}}{8 \, \sqrt{c}{\left (\frac{i \, c}{{\left | c \right |}} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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