3.30 \(\int (d+e x) \sin (a+b x+c x^2) \, dx\)

Optimal. Leaf size=140 \[ \frac{\sqrt{\frac{\pi }{2}} \sin \left (a-\frac{b^2}{4 c}\right ) (2 c d-b e) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )}{2 c^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} \cos \left (a-\frac{b^2}{4 c}\right ) (2 c d-b e) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{2 c^{3/2}}-\frac{e \cos \left (a+b x+c x^2\right )}{2 c} \]

[Out]

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

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Rubi [A]  time = 0.0558215, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {3461, 3447, 3351, 3352} \[ \frac{\sqrt{\frac{\pi }{2}} \sin \left (a-\frac{b^2}{4 c}\right ) (2 c d-b e) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )}{2 c^{3/2}}+\frac{\sqrt{\frac{\pi }{2}} \cos \left (a-\frac{b^2}{4 c}\right ) (2 c d-b e) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )}{2 c^{3/2}}-\frac{e \cos \left (a+b x+c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*Sin[a + b*x + c*x^2],x]

[Out]

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

Rule 3461

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

Rule 3447

Int[Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[Cos[(b^2 - 4*a*c)/(4*c)], Int[Sin[(b + 2*c*x)^2/
(4*c)], x], x] - Dist[Sin[(b^2 - 4*a*c)/(4*c)], Int[Cos[(b + 2*c*x)^2/(4*c)], x], x] /; FreeQ[{a, b, c}, x] &&
NeQ[b^2 - 4*a*c, 0]

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]

Rubi steps

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

Mathematica [A]  time = 0.668156, size = 128, normalized size = 0.91 \[ \frac{\sqrt{2 \pi } \sin \left (a-\frac{b^2}{4 c}\right ) (2 c d-b e) \text{FresnelC}\left (\frac{b+2 c x}{\sqrt{2 \pi } \sqrt{c}}\right )+\sqrt{2 \pi } \cos \left (a-\frac{b^2}{4 c}\right ) (2 c d-b e) S\left (\frac{b+2 c x}{\sqrt{c} \sqrt{2 \pi }}\right )-2 \sqrt{c} e \cos (a+x (b+c x))}{4 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*Sin[a + b*x + c*x^2],x]

[Out]

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

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Maple [A]  time = 0.01, size = 184, normalized size = 1.3 \begin{align*} -{\frac{e\cos \left ( c{x}^{2}+bx+a \right ) }{2\,c}}-{\frac{be\sqrt{2}\sqrt{\pi }}{4} \left ( \cos \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 ) -\sin \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 ) \right ){c}^{-{\frac{3}{2}}}}+{\frac{\sqrt{2}\sqrt{\pi }d}{2} \left ( \cos \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 ) -\sin \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 ) \right ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*sin(c*x^2+b*x+a),x)

[Out]

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

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Maxima [C]  time = 2.71412, size = 1702, normalized size = 12.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(pi)*(((-I*cos(1/4*pi + 1/2*arctan2(0, c)) - I*cos(-1/4*pi + 1/2*arctan2(0, c)) - sin(1/4*pi + 1/2*ar
ctan2(0, c)) + sin(-1/4*pi + 1/2*arctan2(0, c)))*cos(-1/4*(b^2 - 4*a*c)/c) - (cos(1/4*pi + 1/2*arctan2(0, c))
+ cos(-1/4*pi + 1/2*arctan2(0, c)) - I*sin(1/4*pi + 1/2*arctan2(0, c)) + I*sin(-1/4*pi + 1/2*arctan2(0, c)))*s
in(-1/4*(b^2 - 4*a*c)/c))*erf(1/2*(2*I*c*x + I*b)/sqrt(I*c)) + ((-I*cos(1/4*pi + 1/2*arctan2(0, c)) - I*cos(-1
/4*pi + 1/2*arctan2(0, c)) + sin(1/4*pi + 1/2*arctan2(0, c)) - sin(-1/4*pi + 1/2*arctan2(0, c)))*cos(-1/4*(b^2
- 4*a*c)/c) + (cos(1/4*pi + 1/2*arctan2(0, c)) + cos(-1/4*pi + 1/2*arctan2(0, c)) + I*sin(1/4*pi + 1/2*arctan
2(0, c)) - I*sin(-1/4*pi + 1/2*arctan2(0, c)))*sin(-1/4*(b^2 - 4*a*c)/c))*erf(1/2*(2*I*c*x + I*b)/sqrt(-I*c)))
*d/sqrt(abs(c)) - 1/8*(((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*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*cos(-1/4*(b^2 - 4*a*c)/c) + 2*(sqr
t(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)*cos(1/2*arctan2((4*c^2*x^2 + 4*b*c*x + b^2)/c, 0)) +
((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*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*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)*sin(1/2*arctan2((4*c^2*x^2 + 4*b*c*x + b^2)/c, 0)) + (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))*cos(-1/4*(b^2 - 4*a*c)/c) + 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))*sin(-1/4*(b^
2 - 4*a*c)/c))*sqrt((4*c^2*x^2 + 4*b*c*x + b^2)/abs(c)))*e/(c^2*sqrt((4*c^2*x^2 + 4*b*c*x + b^2)/abs(c)))

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Fricas [A]  time = 1.74135, size = 359, normalized size = 2.56 \begin{align*} \frac{\sqrt{2} \pi{\left (2 \, c d - b e\right )} \sqrt{\frac{c}{\pi }} \cos \left (-\frac{b^{2} - 4 \, a c}{4 \, c}\right ) \operatorname{S}\left (\frac{\sqrt{2}{\left (2 \, c x + b\right )} \sqrt{\frac{c}{\pi }}}{2 \, c}\right ) + \sqrt{2} \pi{\left (2 \, c d - b e\right )} \sqrt{\frac{c}{\pi }} \operatorname{C}\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 e \cos \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((e*x+d)*sin(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [C]  time = 1.28206, size = 441, normalized size = 3.15 \begin{align*} \frac{i \, \sqrt{2} \sqrt{\pi } d \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 )}}{4 \,{\left (-\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}} - \frac{i \, \sqrt{2} \sqrt{\pi } d \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 )}}{4 \,{\left (\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}} - \frac{\frac{i \, \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}{4 \, c}\right )}}{{\left (-\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}} + 2 \, e^{\left (i \, c x^{2} + i \, b x + i \, a + 1\right )}}{8 \, c} - \frac{-\frac{i \, \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}{4 \, c}\right )}}{{\left (\frac{i \, c}{{\left | c \right |}} + 1\right )} \sqrt{{\left | c \right |}}} + 2 \, e^{\left (-i \, c x^{2} - i \, b x - i \, a + 1\right )}}{8 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I*sqrt(2)*sqrt(pi)*d*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))) - 1/4*I*sqrt(2)*sqrt(pi)*d*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))) - 1/8*(I*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 - 4*c)/c)/((-I*c/abs(c) + 1)*
sqrt(abs(c))) + 2*e^(I*c*x^2 + I*b*x + I*a + 1))/c - 1/8*(-I*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 - 4*c)/c)/((I*c/abs(c) + 1)*sqrt(abs(c))) + 2*e^(-I*c*
x^2 - I*b*x - I*a + 1))/c