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

Optimal. Leaf size=124 \[ -\frac{\sqrt{\frac{\pi }{2}} b \cos \left (a+\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{b-2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )}{2 c^{3/2}}-\frac{\sqrt{\frac{\pi }{2}} b \sin \left (a+\frac{b^2}{4 c}\right ) S\left (\frac{b-2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{2 c^{3/2}}-\frac{\sin \left (a+b x-c x^2\right )}{2 c} \]

[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)

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Rubi [A]  time = 0.0458486, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, 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 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 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 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*}

Mathematica [A]  time = 0.333524, size = 116, normalized size = 0.94 \[ \frac{\sqrt{2 \pi } b \cos \left (a+\frac{b^2}{4 c}\right ) \text{FresnelC}\left (\frac{2 c x-b}{\sqrt{2 \pi } \sqrt{c}}\right )+\sqrt{2 \pi } b \sin \left (a+\frac{b^2}{4 c}\right ) S\left (\frac{2 c x-b}{\sqrt{c} \sqrt{2 \pi }}\right )-2 \sqrt{c} \sin (a+x (b-c x))}{4 c^{3/2}} \]

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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Maple [A]  time = 0.027, size = 98, normalized size = 0.8 \begin{align*} -{\frac{\sin \left ( -c{x}^{2}+bx+a \right ) }{2\,c}}+{\frac{\sqrt{2}b\sqrt{\pi }}{4} \left ( \cos \left ({\frac{1}{c} \left ({\frac{{b}^{2}}{4}}+ca \right ) } \right ){\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi }} \left ( cx-{\frac{b}{2}} \right ){\frac{1}{\sqrt{c}}}} \right ) +\sin \left ({\frac{1}{c} \left ({\frac{{b}^{2}}{4}}+ca \right ) } \right ){\it FresnelS} \left ({\frac{\sqrt{2}}{\sqrt{\pi }} \left ( cx-{\frac{b}{2}} \right ){\frac{1}{\sqrt{c}}}} \right ) \right ){c}^{-{\frac{3}{2}}}} \end{align*}

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 = 1.82983, size = 1308, normalized size = 10.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(((sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + sqrt(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*sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 - 4*I*
b*c*x + I*b^2)/c)) - 1) - I*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*(sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + sqrt(pi)*(erf(1/2*sqr
t(-(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*sqrt(pi)*(erf(1/2*sqrt((4*
I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + 2*I*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)*cos(1/2*arctan2((4*c^2*x^2 - 4*b*c*x + b^2)/c, 0)) + ((-I*sqrt(pi)*(erf(1
/2*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + I*sqrt(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) + (sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1)
+ 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*sq
rt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) - 2*I*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*(sqrt(pi)*(erf(1/2*sqrt((4*I*c^2*x^2 - 4*I*b*c*x
+ I*b^2)/c)) - 1) + 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)*sin(1/2*arctan2((4*c^2*x^2 - 4*b*c*x + b^2)/c, 0)) + (c*(2*I*e^(1/4*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^
2)/c) - 2*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) + 2*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)/abs(c)))/(c^2*sqrt((4*c^2*x^2 - 4*b*c*x + b^2)/abs(c)))

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Fricas [A]  time = 1.41153, size = 321, normalized size = 2.59 \begin{align*} \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}} \end{align*}

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) + sqrt
(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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \cos{\left (a + b x - c x^{2} \right )}\, dx \end{align*}

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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Giac [C]  time = 1.17293, size = 247, normalized size = 1.99 \begin{align*} -\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} \end{align*}

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*sq
rt(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