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

Optimal. Leaf size=126 \[ -\frac {\sqrt {\pi } b \cos \left (2 a+\frac {b^2}{2 c}\right ) C\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{8 c^{3/2}}-\frac {\sqrt {\pi } b \sin \left (2 a+\frac {b^2}{2 c}\right ) S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{8 c^{3/2}}-\frac {\sin \left (2 a+2 b x-2 c x^2\right )}{8 c}+\frac {x^2}{4} \]

[Out]

1/4*x^2-1/8*sin(-2*c*x^2+2*b*x+2*a)/c-1/8*b*cos(2*a+1/2*b^2/c)*FresnelC((-2*c*x+b)/c^(1/2)/Pi^(1/2))*Pi^(1/2)/
c^(3/2)-1/8*b*FresnelS((-2*c*x+b)/c^(1/2)/Pi^(1/2))*sin(2*a+1/2*b^2/c)*Pi^(1/2)/c^(3/2)

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Rubi [A]  time = 0.08, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3468, 3462, 3448, 3352, 3351} \[ -\frac {\sqrt {\pi } b \cos \left (2 a+\frac {b^2}{2 c}\right ) \text {FresnelC}\left (\frac {b-2 c x}{\sqrt {\pi } \sqrt {c}}\right )}{8 c^{3/2}}-\frac {\sqrt {\pi } b \sin \left (2 a+\frac {b^2}{2 c}\right ) S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{8 c^{3/2}}-\frac {\sin \left (2 a+2 b x-2 c x^2\right )}{8 c}+\frac {x^2}{4} \]

Antiderivative was successfully verified.

[In]

Int[x*Cos[a + b*x - c*x^2]^2,x]

[Out]

x^2/4 - (b*Sqrt[Pi]*Cos[2*a + b^2/(2*c)]*FresnelC[(b - 2*c*x)/(Sqrt[c]*Sqrt[Pi])])/(8*c^(3/2)) - (b*Sqrt[Pi]*F
resnelS[(b - 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*Sin[2*a + b^2/(2*c)])/(8*c^(3/2)) - Sin[2*a + 2*b*x - 2*c*x^2]/(8*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 3462

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

Rule 3468

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

Rubi steps

\begin {align*} \int x \cos ^2\left (a+b x-c x^2\right ) \, dx &=\int \left (\frac {x}{2}+\frac {1}{2} x \cos \left (2 a+2 b x-2 c x^2\right )\right ) \, dx\\ &=\frac {x^2}{4}+\frac {1}{2} \int x \cos \left (2 a+2 b x-2 c x^2\right ) \, dx\\ &=\frac {x^2}{4}-\frac {\sin \left (2 a+2 b x-2 c x^2\right )}{8 c}+\frac {b \int \cos \left (2 a+2 b x-2 c x^2\right ) \, dx}{4 c}\\ &=\frac {x^2}{4}-\frac {\sin \left (2 a+2 b x-2 c x^2\right )}{8 c}+\frac {\left (b \cos \left (2 a+\frac {b^2}{2 c}\right )\right ) \int \cos \left (\frac {(2 b-4 c x)^2}{8 c}\right ) \, dx}{4 c}+\frac {\left (b \sin \left (2 a+\frac {b^2}{2 c}\right )\right ) \int \sin \left (\frac {(2 b-4 c x)^2}{8 c}\right ) \, dx}{4 c}\\ &=\frac {x^2}{4}-\frac {b \sqrt {\pi } \cos \left (2 a+\frac {b^2}{2 c}\right ) C\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{8 c^{3/2}}-\frac {b \sqrt {\pi } S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a+\frac {b^2}{2 c}\right )}{8 c^{3/2}}-\frac {\sin \left (2 a+2 b x-2 c x^2\right )}{8 c}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 122, normalized size = 0.97 \[ \frac {\sqrt {\pi } b \cos \left (2 a+\frac {b^2}{2 c}\right ) C\left (\frac {2 c x-b}{\sqrt {c} \sqrt {\pi }}\right )+\sqrt {\pi } b \sin \left (2 a+\frac {b^2}{2 c}\right ) S\left (\frac {2 c x-b}{\sqrt {c} \sqrt {\pi }}\right )+\sqrt {c} \left (2 c x^2-\sin (2 (a+x (b-c x)))\right )}{8 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cos[a + b*x - c*x^2]^2,x]

[Out]

(b*Sqrt[Pi]*Cos[2*a + b^2/(2*c)]*FresnelC[(-b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])] + b*Sqrt[Pi]*FresnelS[(-b + 2*c*x)/
(Sqrt[c]*Sqrt[Pi])]*Sin[2*a + b^2/(2*c)] + Sqrt[c]*(2*c*x^2 - Sin[2*(a + x*(b - c*x))]))/(8*c^(3/2))

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fricas [A]  time = 0.89, size = 133, normalized size = 1.06 \[ \frac {\pi b \sqrt {\frac {c}{\pi }} \cos \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right ) \operatorname {C}\left (\frac {{\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) + \pi b \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {{\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) \sin \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right ) + 2 \, c^{2} x^{2} + 2 \, c \cos \left (c x^{2} - b x - a\right ) \sin \left (c x^{2} - b x - a\right )}{8 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(pi*b*sqrt(c/pi)*cos(1/2*(b^2 + 4*a*c)/c)*fresnel_cos((2*c*x - b)*sqrt(c/pi)/c) + pi*b*sqrt(c/pi)*fresnel_
sin((2*c*x - b)*sqrt(c/pi)/c)*sin(1/2*(b^2 + 4*a*c)/c) + 2*c^2*x^2 + 2*c*cos(c*x^2 - b*x - a)*sin(c*x^2 - b*x
- a))/c^2

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giac [C]  time = 0.63, size = 172, normalized size = 1.37 \[ \frac {1}{4} \, x^{2} - \frac {\frac {\sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x - \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {i \, b^{2} + 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}} + i \, e^{\left (2 i \, c x^{2} - 2 i \, b x - 2 i \, a\right )}}{16 \, c} - \frac {\frac {\sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{2} \, \sqrt {c} {\left (2 \, x - \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {-i \, b^{2} - 4 i \, a c}{2 \, c}\right )}}{\sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}} - i \, e^{\left (-2 i \, c x^{2} + 2 i \, b x + 2 i \, a\right )}}{16 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^2 - 1/16*(sqrt(pi)*b*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)) + I*e^(2*I*c*x^2 - 2*I*b*x - 2*I*a))/c - 1/16*(sqrt(pi)*b*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)) - I*e^(-2*I*c*x^2 + 2*I*b*x + 2*I*a)
)/c

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maple [A]  time = 0.04, size = 99, normalized size = 0.79 \[ \frac {x^{2}}{4}-\frac {\sin \left (-2 c \,x^{2}+2 b x +2 a \right )}{8 c}+\frac {b \sqrt {\pi }\, \left (\cos \left (\frac {4 c a +b^{2}}{2 c}\right ) \FresnelC \left (\frac {2 c x -b}{\sqrt {\pi }\, \sqrt {c}}\right )+\sin \left (\frac {4 c a +b^{2}}{2 c}\right ) \mathrm {S}\left (\frac {2 c x -b}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{8 c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^2-1/8*sin(-2*c*x^2+2*b*x+2*a)/c+1/8*b/c^(3/2)*Pi^(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 = 2.61, size = 608, normalized size = 4.83 \[ \frac {\sqrt {2} {\left ({\left (\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2}} \sqrt {\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )} - \left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2}} \sqrt {-\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )}\right )} b^{2} \cos \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right ) + {\left (-\left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2}} \sqrt {\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )} + \left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2}} \sqrt {-\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )}\right )} b^{2} \sin \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right ) + {\left ({\left (-\left (2 i - 2\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2}} \sqrt {\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )} + \left (2 i + 2\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2}} \sqrt {-\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )}\right )} b c \cos \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right ) + {\left (\left (2 i + 2\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2}} \sqrt {\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )} - \left (2 i - 2\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {1}{2}} \sqrt {-\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{c}}\right ) - 1\right )}\right )} b c \sin \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right )\right )} x + \sqrt {2} {\left (8 \, c^{2} x^{2} + c {\left (-2 i \, e^{\left (\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{2 \, c}\right )} + 2 i \, e^{\left (-\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{2 \, c}\right )}\right )} \cos \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right ) - 2 \, c {\left (e^{\left (\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{2 \, c}\right )} + e^{\left (-\frac {4 i \, c^{2} x^{2} - 4 i \, b c x + i \, b^{2}}{2 \, c}\right )}\right )} \sin \left (\frac {b^{2} + 4 \, a c}{2 \, c}\right )\right )} \sqrt {\frac {4 \, c^{2} x^{2} - 4 \, b c x + b^{2}}{c}}\right )}}{64 \, c^{2} \sqrt {\frac {4 \, c^{2} x^{2} - 4 \, b c x + b^{2}}{c}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*sqrt(2)*(((I - 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) - (I +
1)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*cos(1/2*(b^2 + 4*a*c)
/c) + (-(I + 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + (I - 1)*sqrt
(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*sin(1/2*(b^2 + 4*a*c)/c) + (
(-(2*I - 2)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + (2*I + 2)*sqrt(2
)*sqrt(pi)*(erf(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b*c*cos(1/2*(b^2 + 4*a*c)/c) + ((2
*I + 2)*sqrt(2)*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) - (2*I - 2)*sqrt(2)*sq
rt(pi)*(erf(sqrt(1/2)*sqrt(-(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1))*b*c*sin(1/2*(b^2 + 4*a*c)/c))*x + sqrt
(2)*(8*c^2*x^2 + c*(-2*I*e^(1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c) + 2*I*e^(-1/2*(4*I*c^2*x^2 - 4*I*b*c*x +
I*b^2)/c))*cos(1/2*(b^2 + 4*a*c)/c) - 2*c*(e^(1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c) + e^(-1/2*(4*I*c^2*x^2
- 4*I*b*c*x + I*b^2)/c))*sin(1/2*(b^2 + 4*a*c)/c))*sqrt((4*c^2*x^2 - 4*b*c*x + b^2)/c))/(c^2*sqrt((4*c^2*x^2 -
4*b*c*x + b^2)/c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cos(a + b*x - c*x^2)^2,x)

[Out]

int(x*cos(a + b*x - c*x^2)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \cos ^{2}{\left (a + b x - c x^{2} \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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