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

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

[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)

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Rubi [A]  time = 0.0663787, antiderivative size = 126, normalized size of antiderivative = 1., 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 = {3467, 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*Sin[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 3467

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

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 \sin ^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*}

Mathematica [A]  time = 0.291463, size = 122, normalized size = 0.97 \[ \frac{\sqrt{\pi } (-b) \cos \left (2 a+\frac{b^2}{2 c}\right ) \text{FresnelC}\left (\frac{2 c x-b}{\sqrt{\pi } \sqrt{c}}\right )-\sqrt{\pi } b \sin \left (2 a+\frac{b^2}{2 c}\right ) S\left (\frac{2 c x-b}{\sqrt{c} \sqrt{\pi }}\right )+\sqrt{c} \left (\sin (2 (a+x (b-c x)))+2 c x^2\right )}{8 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sin[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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Maple [A]  time = 0.014, size = 99, normalized size = 0.8 \begin{align*}{\frac{{x}^{2}}{4}}+{\frac{\sin \left ( -2\,c{x}^{2}+2\,bx+2\,a \right ) }{8\,c}}-{\frac{b\sqrt{\pi }}{8} \left ( \cos \left ({\frac{4\,ca+{b}^{2}}{2\,c}} \right ){\it FresnelC} \left ({\frac{2\,cx-b}{\sqrt{\pi }}{\frac{1}{\sqrt{c}}}} \right ) +\sin \left ({\frac{4\,ca+{b}^{2}}{2\,c}} \right ){\it FresnelS} \left ({\frac{2\,cx-b}{\sqrt{\pi }}{\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*sin(-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 = 1.95154, size = 1372, normalized size = 10.89 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*sqrt(2)*(sqrt(2)*(4*c^2*x^2 + c*(I*e^(1/2*(4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c) - 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) + 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)/abs(c)) + ((sqr
t(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + 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*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2
- 4*I*b*c*x + I*b^2)/c)) - 1) - I*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*(sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + 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*cos(1/2*(b^2 + 4*a*c)/c) - (-2*I*s
qrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + 2*I*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*sin(1/2*(b^2 + 4*a*c)/c))*x)*cos(1/2*arctan2(2*(4*c^2*x^2 - 4*b*c
*x + b^2)/c, 0)) + ((-I*sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + I*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) + (sqrt(pi)*(erf(
sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + 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*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*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*(sqrt(pi)*(erf(sqrt(1/2)*sqrt((4*I*c^2*x^2 - 4*I*b*c*x + I*b^2)/c)) - 1) + 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*sin(1/2*(b^2 + 4*a*c)/c))*x)*sin(1/2*arct
an2(2*(4*c^2*x^2 - 4*b*c*x + b^2)/c, 0)))/(c^2*sqrt((4*c^2*x^2 - 4*b*c*x + b^2)/abs(c)))

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Fricas [A]  time = 1.4373, size = 313, normalized size = 2.48 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.40317, size = 232, normalized size = 1.84 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(-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