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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 124, normalized size of antiderivative = 1.00, 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 = {3461, 3447, 3351, 3352} \[ -\frac {\sqrt {\frac {\pi }{2}} b \sin \left (a+\frac {b^2}{4 c}\right ) \text {FresnelC}\left (\frac {b-2 c x}{\sqrt {2 \pi } \sqrt {c}}\right )}{2 c^{3/2}}+\frac {\sqrt {\frac {\pi }{2}} b \cos \left (a+\frac {b^2}{4 c}\right ) S\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{2 c^{3/2}}+\frac {\cos \left (a+b x-c x^2\right )}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x*Sin[a + b*x - c*x^2],x]

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.33, size = 117, normalized size = 0.94 \[ \frac {\sqrt {2 \pi } b \sin \left (a+\frac {b^2}{4 c}\right ) C\left (\frac {2 c x-b}{\sqrt {c} \sqrt {2 \pi }}\right )-\sqrt {2 \pi } b \cos \left (a+\frac {b^2}{4 c}\right ) S\left (\frac {2 c x-b}{\sqrt {c} \sqrt {2 \pi }}\right )+2 \sqrt {c} \cos (a+x (b-c x))}{4 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sin[a + b*x - c*x^2],x]

[Out]

(2*Sqrt[c]*Cos[a + x*(b - c*x)] - b*Sqrt[2*Pi]*Cos[a + b^2/(4*c)]*FresnelS[(-b + 2*c*x)/(Sqrt[c]*Sqrt[2*Pi])]
+ b*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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fricas [A]  time = 0.42, size = 126, normalized size = 1.02 \[ -\frac {\sqrt {2} \pi b \sqrt {\frac {c}{\pi }} \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - \sqrt {2} \pi b \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) \sin \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) - 2 \, c \cos \left (c x^{2} - b x - a\right )}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*sin(-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_sin(1/2*sqrt(2)*(2*c*x - b)*sqrt(c/pi)/c) - sqr
t(2)*pi*b*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*cos(c*x^
2 - b*x - a))/c^2

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giac [C]  time = 0.21, size = 185, normalized size = 1.49 \[ -\frac {\frac {i \, \sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x - \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {i \, b^{2} + 4 i \, a c}{4 \, c}\right )}}{{\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - 2 \, e^{\left (i \, c x^{2} - i \, b x - i \, a\right )}}{8 \, c} - \frac {-\frac {i \, \sqrt {2} \sqrt {\pi } b \operatorname {erf}\left (-\frac {1}{4} \, \sqrt {2} {\left (2 \, x - \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {-i \, b^{2} - 4 i \, a c}{4 \, c}\right )}}{{\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - 2 \, e^{\left (-i \, c x^{2} + i \, b x + i \, a\right )}}{8 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 99, normalized size = 0.80 \[ \frac {\cos \left (-c \,x^{2}+b x +a \right )}{2 c}-\frac {b \sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}+c a}{c}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \left (c x -\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )-\sin \left (\frac {\frac {b^{2}}{4}+c a}{c}\right ) \FresnelC \left (\frac {\sqrt {2}\, \left (c x -\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {c}}\right )\right )}{4 c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*cos(-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)*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 = 3.14, size = 578, normalized size = 4.66 \[ \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 (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 )} \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) + 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 )} \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*sin(-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)*(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*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 + (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))*cos(1/4*(b^2 + 4*a*c)/c) + 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))*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 \[ \int x\,\sin \left (-c\,x^2+b\,x+a\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(a + b*x - c*x^2),x)

[Out]

int(x*sin(a + b*x - c*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sin(a + b*x - c*x**2), x)

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