\(\int \sin (a+b x-c x^2) \, dx\) [8]

Optimal result

Integrand size = 12, antiderivative size = 98 \[ \int \sin \left (a+b x-c x^2\right ) \, dx=\frac {\sqrt {\frac {\pi }{2}} \cos \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelS}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{\sqrt {c}}-\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a+\frac {b^2}{4 c}\right )}{\sqrt {c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3528, 3432, 3433} \[ \int \sin \left (a+b x-c x^2\right ) \, dx=\frac {\sqrt {\frac {\pi }{2}} \cos \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelS}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{\sqrt {c}}-\frac {\sqrt {\frac {\pi }{2}} \sin \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelC}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{\sqrt {c}} \]

[In]

Int[Sin[a + b*x - c*x^2],x]

[Out]

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

Rule 3432

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelS[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3433

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2
]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]

Rule 3528

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]

Rubi steps \begin{align*} \text {integral}& = -\left (\cos \left (a+\frac {b^2}{4 c}\right ) \int \sin \left (\frac {(b-2 c x)^2}{4 c}\right ) \, dx\right )+\sin \left (a+\frac {b^2}{4 c}\right ) \int \cos \left (\frac {(b-2 c x)^2}{4 c}\right ) \, dx \\ & = \frac {\sqrt {\frac {\pi }{2}} \cos \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelS}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )}{\sqrt {c}}-\frac {\sqrt {\frac {\pi }{2}} \operatorname {FresnelC}\left (\frac {b-2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a+\frac {b^2}{4 c}\right )}{\sqrt {c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.91 \[ \int \sin \left (a+b x-c x^2\right ) \, dx=\frac {\sqrt {\frac {\pi }{2}} \left (-\cos \left (a+\frac {b^2}{4 c}\right ) \operatorname {FresnelS}\left (\frac {-b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right )+\operatorname {FresnelC}\left (\frac {-b+2 c x}{\sqrt {c} \sqrt {2 \pi }}\right ) \sin \left (a+\frac {b^2}{4 c}\right )\right )}{\sqrt {c}} \]

[In]

Integrate[Sin[a + b*x - c*x^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.89

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {\pi }\, \left (\cos \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \operatorname {S}\left (\frac {\sqrt {2}\, \left (-c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {-c}}\right )+\sin \left (\frac {\frac {b^{2}}{4}+a c}{c}\right ) \operatorname {C}\left (\frac {\sqrt {2}\, \left (-c x +\frac {b}{2}\right )}{\sqrt {\pi }\, \sqrt {-c}}\right )\right )}{2 \sqrt {-c}}\)	\(87\)
risch	\(\frac {i \sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c +b^{2}\right )}{4 c}} \operatorname {erf}\left (\sqrt {-i c}\, x +\frac {i b}{2 \sqrt {-i c}}\right )}{4 \sqrt {-i c}}+\frac {i \sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c +b^{2}\right )}{4 c}} \operatorname {erf}\left (-\sqrt {i c}\, x +\frac {i b}{2 \sqrt {i c}}\right )}{4 \sqrt {i c}}\)	\(97\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.09 \[ \int \sin \left (a+b x-c x^2\right ) \, dx=-\frac {\sqrt {2} \pi \sqrt {\frac {c}{\pi }} \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) - \sqrt {2} \pi \sqrt {\frac {c}{\pi }} \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, c x - b\right )} \sqrt {\frac {c}{\pi }}}{2 \, c}\right ) \sin \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right )}{2 \, c} \]

[In]

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

[Out]

-1/2*(sqrt(2)*pi*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*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))/c

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.96 \[ \int \sin \left (a+b x-c x^2\right ) \, dx=\frac {\sqrt {2} \sqrt {\pi } \sqrt {- \frac {1}{c}} \left (\sin {\left (a + \frac {b^{2}}{4 c} \right )} C\left (\frac {\sqrt {2} \left (b - 2 c x\right )}{2 \sqrt {\pi } \sqrt {- c}}\right ) + \cos {\left (a + \frac {b^{2}}{4 c} \right )} S\left (\frac {\sqrt {2} \left (b - 2 c x\right )}{2 \sqrt {\pi } \sqrt {- c}}\right )\right )}{2} \]

[In]

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

[Out]

sqrt(2)*sqrt(pi)*sqrt(-1/c)*(sin(a + b**2/(4*c))*fresnelc(sqrt(2)*(b - 2*c*x)/(2*sqrt(pi)*sqrt(-c))) + cos(a +
b**2/(4*c))*fresnels(sqrt(2)*(b - 2*c*x)/(2*sqrt(pi)*sqrt(-c))))/2

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.14 \[ \int \sin \left (a+b x-c x^2\right ) \, dx=-\frac {\sqrt {2} \sqrt {\pi } {\left ({\left (\left (i + 1\right ) \, \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) + \left (i - 1\right ) \, \sin \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right )\right )} \operatorname {erf}\left (\frac {2 i \, c x - i \, b}{2 \, \sqrt {i \, c}}\right ) + {\left (\left (i - 1\right ) \, \cos \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right ) + \left (i + 1\right ) \, \sin \left (\frac {b^{2} + 4 \, a c}{4 \, c}\right )\right )} \operatorname {erf}\left (\frac {2 i \, c x - i \, b}{2 \, \sqrt {-i \, c}}\right )\right )}}{8 \, \sqrt {c}} \]

[In]

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

[Out]

-1/8*sqrt(2)*sqrt(pi)*(((I + 1)*cos(1/4*(b^2 + 4*a*c)/c) + (I - 1)*sin(1/4*(b^2 + 4*a*c)/c))*erf(1/2*(2*I*c*x
- I*b)/sqrt(I*c)) + ((I - 1)*cos(1/4*(b^2 + 4*a*c)/c) + (I + 1)*sin(1/4*(b^2 + 4*a*c)/c))*erf(1/2*(2*I*c*x - I
*b)/sqrt(-I*c)))/sqrt(c)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.40 \[ \int \sin \left (a+b x-c x^2\right ) \, dx=-\frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{4} i \, \sqrt {2} {\left (2 \, x - \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {i \, b^{2} + 4 i \, a c}{4 \, c}\right )}}{4 \, {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} - \frac {\sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\frac {1}{4} i \, \sqrt {2} {\left (2 \, x - \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}\right ) e^{\left (-\frac {-i \, b^{2} - 4 i \, a c}{4 \, c}\right )}}{4 \, {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )} \sqrt {{\left | c \right |}}} \]

[In]

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

[Out]

-1/4*sqrt(2)*sqrt(pi)*erf(-1/4*I*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*sqrt(2)*sqrt(pi)*erf(1/4*I*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)))

Mupad [B] (verification not implemented)

Time = 5.60 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.07 \[ \int \sin \left (a+b x-c x^2\right ) \, dx=\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {S}\left (\frac {\sqrt {2}\,\left (\frac {b}{2}-c\,x\right )\,\sqrt {-\frac {1}{c}}}{\sqrt {\pi }}\right )\,\cos \left (\frac {b^2+4\,a\,c}{4\,c}\right )\,\sqrt {-\frac {1}{c}}}{2}+\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {C}\left (\frac {\sqrt {2}\,\left (\frac {b}{2}-c\,x\right )\,\sqrt {-\frac {1}{c}}}{\sqrt {\pi }}\right )\,\sin \left (\frac {b^2+4\,a\,c}{4\,c}\right )\,\sqrt {-\frac {1}{c}}}{2} \]

[In]

int(sin(a + b*x - c*x^2),x)

[Out]

(2^(1/2)*pi^(1/2)*fresnels((2^(1/2)*(b/2 - c*x)*(-1/c)^(1/2))/pi^(1/2))*cos((4*a*c + b^2)/(4*c))*(-1/c)^(1/2))
/2 + (2^(1/2)*pi^(1/2)*fresnelc((2^(1/2)*(b/2 - c*x)*(-1/c)^(1/2))/pi^(1/2))*sin((4*a*c + b^2)/(4*c))*(-1/c)^(
1/2))/2