3.22 \(\int \cos ^2(a+b x-c x^2) \, dx\)

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*b^2/c)*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*b^2/c)*Pi^(1/2)/c^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3450, 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[Cos[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 3450

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

Rubi steps

\begin {align*} \int \cos ^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 = 0.11, size = 100, normalized size = 1.00 \[ \frac {\sqrt {\pi } \cos \left (2 a+\frac {b^2}{2 c}\right ) C\left (\frac {2 c x-b}{\sqrt {c} \sqrt {\pi }}\right )+\sqrt {\pi } \sin \left (2 a+\frac {b^2}{2 c}\right ) S\left (\frac {2 c x-b}{\sqrt {c} \sqrt {\pi }}\right )+2 \sqrt {c} x}{4 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[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 = 1.20, size = 96, normalized size = 0.96 \[ \frac {\pi \sqrt {\frac {c}{\pi }} \cos \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right ) \operatorname {C}\left (\frac {{\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) + \pi \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {{\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) \sin \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right ) + 2 \, c x}{4 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(-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_sin(
(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 = 0.77, size = 124, normalized size = 1.24 \[ \frac {1}{2} \, x - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x - \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {i \, b^{2} + 4 i \, a c}{2 \, c}\right )}}{8 \, \sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x - \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {-i \, b^{2} - 4 i \, a c}{2 \, c}\right )}}{8 \, \sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(-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 = 0.04, size = 76, normalized size = 0.76 \[ \frac {x}{2}+\frac {\sqrt {\pi }\, \left (\cos \left (\frac {4 c a +b^{2}}{2 c}\right ) \FresnelC \left (\frac {2 c x -b}{\sqrt {\pi }\, \sqrt {c}}\right )+\sin \left (\frac {4 c a +b^{2}}{2 c}\right ) \mathrm {S}\left (\frac {2 c x -b}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [C]  time = 1.16, 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(cos(-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 \[ \int {\cos \left (-c\,x^2+b\,x+a\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x - c*x^2)^2,x)

[Out]

int(cos(a + b*x - c*x^2)^2, x)

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sympy [A]  time = 0.97, size = 88, normalized size = 0.88 \[ \frac {x}{2} + \frac {\sqrt {\pi } \sqrt {- \frac {1}{c}} \left (- \sin {\left (2 a + \frac {b^{2}}{2 c} \right )} S\left (\frac {2 b - 4 c x}{2 \sqrt {\pi } \sqrt {- c}}\right ) + \cos {\left (2 a + \frac {b^{2}}{2 c} \right )} C\left (\frac {2 b - 4 c x}{2 \sqrt {\pi } \sqrt {- c}}\right )\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(-c*x**2+b*x+a)**2,x)

[Out]

x/2 + sqrt(pi)*sqrt(-1/c)*(-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))*fresnelc((2*b - 4*c*x)/(2*sqrt(pi)*sqrt(-c))))/4

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