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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 100, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.07, size = 97, normalized size = 0.97 \[ \frac {-\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )+\sqrt {\pi } \sin \left (2 a-\frac {b^2}{2 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )+2 \sqrt {c} x}{4 \sqrt {c}} \]

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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fricas [A]  time = 0.44, size = 93, normalized size = 0.93 \[ -\frac {\pi \sqrt {\frac {c}{\pi }} \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) \operatorname {C}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) - \pi \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) - 2 \, c x}{4 \, c} \]

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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giac [C]  time = 0.27, size = 122, normalized size = 1.22 \[ \frac {1}{2} \, x + \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c}{2 \, c}\right )}}{8 \, \sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}} + \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c}{2 \, c}\right )}}{8 \, \sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}} \]

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))

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maple [A]  time = 0.06, size = 72, normalized size = 0.72 \[ \frac {x}{2}-\frac {\sqrt {\pi }\, \left (\cos \left (\frac {-4 c a +b^{2}}{2 c}\right ) \FresnelC \left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 c a +b^{2}}{2 c}\right ) \mathrm {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 \sqrt {c}} \]

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 = 0.42, size = 124, normalized size = 1.24 \[ \frac {4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i - 1\right ) \, \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) + \left (i + 1\right ) \, \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right )\right )} \operatorname {erf}\left (\frac {2 i \, c x + i \, b}{\sqrt {2 i \, c}}\right ) + {\left (\left (i + 1\right ) \, \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) + \left (i - 1\right ) \, \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right )\right )} \operatorname {erf}\left (\frac {2 i \, c x + i \, b}{\sqrt {-2 i \, c}}\right )\right )} c^{\frac {3}{2}} + 16 \, c^{2} x}{32 \, c^{2}} \]

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*(4^(1/4)*sqrt(2)*sqrt(pi)*(((I - 1)*cos(-1/2*(b^2 - 4*a*c)/c) + (I + 1)*sin(-1/2*(b^2 - 4*a*c)/c))*erf((2
*I*c*x + I*b)/sqrt(2*I*c)) + ((I + 1)*cos(-1/2*(b^2 - 4*a*c)/c) + (I - 1)*sin(-1/2*(b^2 - 4*a*c)/c))*erf((2*I*
c*x + I*b)/sqrt(-2*I*c)))*c^(3/2) + 16*c^2*x)/c^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (c\,x^2+b\,x+a\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x + c*x^2)^2,x)

[Out]

int(sin(a + b*x + c*x^2)^2, x)

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sympy [A]  time = 1.18, size = 83, normalized size = 0.83 \[ \frac {x}{2} - \frac {\sqrt {\pi } \left (- \sin {\left (2 a - \frac {b^{2}}{2 c} \right )} S\left (\frac {2 b + 4 c x}{2 \sqrt {\pi } \sqrt {c}}\right ) + \cos {\left (2 a - \frac {b^{2}}{2 c} \right )} C\left (\frac {2 b + 4 c x}{2 \sqrt {\pi } \sqrt {c}}\right )\right ) \sqrt {\frac {1}{c}}}{4} \]

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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