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

Optimal. Leaf size=123 \[ -\frac {\sqrt {\frac {\pi }{2}} b \cos \left (a-\frac {b^2}{4 c}\right ) C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} b \sin \left (a-\frac {b^2}{4 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}+\frac {\sin \left (a+b x+c x^2\right )}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3462, 3448, 3352, 3351} \[ -\frac {\sqrt {\frac {\pi }{2}} b \cos \left (a-\frac {b^2}{4 c}\right ) \text {FresnelC}\left (\frac {b+2 c x}{\sqrt {2 \pi } \sqrt {c}}\right )}{2 c^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} b \sin \left (a-\frac {b^2}{4 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}+\frac {\sin \left (a+b x+c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x*Cos[a + b*x + c*x^2],x]

[Out]

-(b*Sqrt[Pi/2]*Cos[a - b^2/(4*c)]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])])/(2*c^(3/2)) + (b*Sqrt[Pi/2]*Fres
nelS[(b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]*Sin[a - b^2/(4*c)])/(2*c^(3/2)) + Sin[a + b*x + c*x^2]/(2*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]

Rubi steps

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

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Mathematica [A]  time = 0.30, size = 112, normalized size = 0.91 \[ \frac {\sqrt {2 \pi } (-b) \cos \left (a-\frac {b^2}{4 c}\right ) C\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )+\sqrt {2 \pi } b \sin \left (a-\frac {b^2}{4 c}\right ) S\left (\frac {b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )+2 \sqrt {c} \sin (a+x (b+c x))}{4 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cos[a + b*x + c*x^2],x]

[Out]

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

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fricas [A]  time = 0.99, size = 119, normalized size = 0.97 \[ -\frac {\sqrt {2} \pi b \sqrt {\frac {c}{\pi }} \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - \sqrt {2} \pi b \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) - 2 \, c \sin \left (c x^{2} + b x + a\right )}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.62, size = 181, normalized size = 1.47 \[ \frac {\frac {\sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x + \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c}{4 \, c}\right )}}{{\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - 2 i \, e^{\left (i \, c x^{2} + i \, b x + i \, a\right )}}{8 \, c} + \frac {\frac {\sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x + \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c}{4 \, c}\right )}}{{\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} + 2 i \, e^{\left (-i \, c x^{2} - i \, b x - i \, a\right )}}{8 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 99, normalized size = 0.80 \[ \frac {\sin \left (c \,x^{2}+b x +a \right )}{2 c}-\frac {b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )+\sin \left (\frac {\frac {b^{2}}{4}-c a}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 2.65, size = 578, normalized size = 4.70 \[ \frac {{\left (\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\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 (\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}{4 \, c}\right ) + {\left (\left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\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 (\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}{4 \, c}\right ) + {\left ({\left (\left (2 i - 2\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\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 (\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}{4 \, c}\right ) + {\left (\left (2 i + 2\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\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 (\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}{4 \, c}\right )\right )} x + {\left (c {\left (-4 i \, e^{\left (\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{4 \, c}\right )} + 4 i \, e^{\left (-\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{4 \, c}\right )}\right )} \cos \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right ) + 4 \, c {\left (e^{\left (\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{4 \, c}\right )} + e^{\left (-\frac {4 i \, c^{2} x^{2} + 4 i \, b c x + i \, b^{2}}{4 \, c}\right )}\right )} \sin \left (-\frac {b^{2} - 4 \, a c}{4 \, c}\right )\right )} \sqrt {\frac {4 \, c^{2} x^{2} + 4 \, b c x + b^{2}}{c}}}{16 \, 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),x, algorithm="maxima")

[Out]

1/16*(((I - 1)*sqrt(2)*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1) - (I + 1)*sqrt(2)*sqr
t(pi)*(erf(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*cos(-1/4*(b^2 - 4*a*c)/c) + ((I + 1)*sqrt
(2)*sqrt(pi)*(erf(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(1/2*sqrt
(-(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c)) - 1))*b^2*sin(-1/4*(b^2 - 4*a*c)/c) + (((2*I - 2)*sqrt(2)*sqrt(pi)*(er
f(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(1/2*sqrt(-(4*I*c^2*x^2
+ 4*I*b*c*x + I*b^2)/c)) - 1))*b*c*cos(-1/4*(b^2 - 4*a*c)/c) + ((2*I + 2)*sqrt(2)*sqrt(pi)*(erf(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(1/2*sqrt(-(4*I*c^2*x^2 + 4*I*b*c*x +
I*b^2)/c)) - 1))*b*c*sin(-1/4*(b^2 - 4*a*c)/c))*x + (c*(-4*I*e^(1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c) + 4*I
*e^(-1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*cos(-1/4*(b^2 - 4*a*c)/c) + 4*c*(e^(1/4*(4*I*c^2*x^2 + 4*I*b*c*
x + I*b^2)/c) + e^(-1/4*(4*I*c^2*x^2 + 4*I*b*c*x + I*b^2)/c))*sin(-1/4*(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 \[ \int x\,\cos \left (c\,x^2+b\,x+a\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cos(a + b*x + c*x^2),x)

[Out]

int(x*cos(a + b*x + c*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*cos(a + b*x + c*x**2), x)

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