\(\int \sin ^2(a+b x+c x^2) \, dx\) [35]

Optimal result

Integrand size = 13, antiderivative size = 100 \[ \int \sin ^2\left (a+b x+c x^2\right ) \, dx=\frac {x}{2}-\frac {\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{4 \sqrt {c}}+\frac {\sqrt {\pi } \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{4 \sqrt {c}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3530, 3529, 3433, 3432} \[ \int \sin ^2\left (a+b x+c x^2\right ) \, dx=-\frac {\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{4 \sqrt {c}}+\frac {\sqrt {\pi } \sin \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{4 \sqrt {c}}+\frac {x}{2} \]

[In]

Int[Sin[a + b*x + c*x^2]^2,x]

[Out]

x/2 - (Sqrt[Pi]*Cos[2*a - b^2/(2*c)]*FresnelC[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])])/(4*Sqrt[c]) + (Sqrt[Pi]*Fresnel
S[(b + 2*c*x)/(Sqrt[c]*Sqrt[Pi])]*Sin[2*a - b^2/(2*c)])/(4*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 3529

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 3530

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2}-\frac {1}{2} \cos \left (2 a+2 b x+2 c x^2\right )\right ) \, dx \\ & = \frac {x}{2}-\frac {1}{2} \int \cos \left (2 a+2 b x+2 c x^2\right ) \, dx \\ & = \frac {x}{2}-\frac {1}{2} \cos \left (2 a-\frac {b^2}{2 c}\right ) \int \cos \left (\frac {(2 b+4 c x)^2}{8 c}\right ) \, dx+\frac {1}{2} \sin \left (2 a-\frac {b^2}{2 c}\right ) \int \sin \left (\frac {(2 b+4 c x)^2}{8 c}\right ) \, dx \\ & = \frac {x}{2}-\frac {\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )}{4 \sqrt {c}}+\frac {\sqrt {\pi } \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{4 \sqrt {c}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.97 \[ \int \sin ^2\left (a+b x+c x^2\right ) \, dx=\frac {2 \sqrt {c} x-\sqrt {\pi } \cos \left (2 a-\frac {b^2}{2 c}\right ) \operatorname {FresnelC}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right )+\sqrt {\pi } \operatorname {FresnelS}\left (\frac {b+2 c x}{\sqrt {c} \sqrt {\pi }}\right ) \sin \left (2 a-\frac {b^2}{2 c}\right )}{4 \sqrt {c}} \]

[In]

Integrate[Sin[a + b*x + c*x^2]^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.72

method	result	size
default	\(\frac {x}{2}-\frac {\sqrt {\pi }\, \left (\cos \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {C}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )+\sin \left (\frac {-4 a c +b^{2}}{2 c}\right ) \operatorname {S}\left (\frac {2 c x +b}{\sqrt {c}\, \sqrt {\pi }}\right )\right )}{4 \sqrt {c}}\)	\(72\)
risch	\(\frac {x}{2}-\frac {\sqrt {\pi }\, {\mathrm e}^{-\frac {i \left (4 a c -b^{2}\right )}{2 c}} \sqrt {2}\, \operatorname {erf}\left (\sqrt {2}\, \sqrt {i c}\, x +\frac {i b \sqrt {2}}{2 \sqrt {i c}}\right )}{16 \sqrt {i c}}+\frac {\sqrt {\pi }\, {\mathrm e}^{\frac {i \left (4 a c -b^{2}\right )}{2 c}} \operatorname {erf}\left (-\sqrt {-2 i c}\, x +\frac {i b}{\sqrt {-2 i c}}\right )}{8 \sqrt {-2 i c}}\)	\(111\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.93 \[ \int \sin ^2\left (a+b x+c x^2\right ) \, dx=-\frac {\pi \sqrt {\frac {c}{\pi }} \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) \operatorname {C}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) - \pi \sqrt {\frac {c}{\pi }} \operatorname {S}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {\frac {c}{\pi }}}{c}\right ) \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) - 2 \, c x}{4 \, c} \]

[In]

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

[Out]

-1/4*(pi*sqrt(c/pi)*cos(-1/2*(b^2 - 4*a*c)/c)*fresnel_cos((2*c*x + b)*sqrt(c/pi)/c) - pi*sqrt(c/pi)*fresnel_si
n((2*c*x + b)*sqrt(c/pi)/c)*sin(-1/2*(b^2 - 4*a*c)/c) - 2*c*x)/c

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.83 \[ \int \sin ^2\left (a+b x+c x^2\right ) \, dx=\frac {x}{2} - \frac {\sqrt {\pi } \left (- \sin {\left (2 a - \frac {b^{2}}{2 c} \right )} S\left (\frac {2 b + 4 c x}{2 \sqrt {\pi } \sqrt {c}}\right ) + \cos {\left (2 a - \frac {b^{2}}{2 c} \right )} C\left (\frac {2 b + 4 c x}{2 \sqrt {\pi } \sqrt {c}}\right )\right ) \sqrt {\frac {1}{c}}}{4} \]

[In]

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

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.24 \[ \int \sin ^2\left (a+b x+c x^2\right ) \, dx=\frac {4^{\frac {1}{4}} \sqrt {2} \sqrt {\pi } {\left ({\left (\left (i - 1\right ) \, \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) + \left (i + 1\right ) \, \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right )\right )} \operatorname {erf}\left (\frac {2 i \, c x + i \, b}{\sqrt {2 i \, c}}\right ) + {\left (\left (i + 1\right ) \, \cos \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right ) + \left (i - 1\right ) \, \sin \left (-\frac {b^{2} - 4 \, a c}{2 \, c}\right )\right )} \operatorname {erf}\left (\frac {2 i \, c x + i \, b}{\sqrt {-2 i \, c}}\right )\right )} c^{\frac {3}{2}} + 16 \, c^{2} x}{32 \, c^{2}} \]

[In]

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

[Out]

1/32*(4^(1/4)*sqrt(2)*sqrt(pi)*(((I - 1)*cos(-1/2*(b^2 - 4*a*c)/c) + (I + 1)*sin(-1/2*(b^2 - 4*a*c)/c))*erf((2
*I*c*x + I*b)/sqrt(2*I*c)) + ((I + 1)*cos(-1/2*(b^2 - 4*a*c)/c) + (I - 1)*sin(-1/2*(b^2 - 4*a*c)/c))*erf((2*I*
c*x + I*b)/sqrt(-2*I*c)))*c^(3/2) + 16*c^2*x)/c^2

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.22 \[ \int \sin ^2\left (a+b x+c x^2\right ) \, dx=\frac {1}{2} \, x - \frac {i \, \sqrt {\pi } \operatorname {erf}\left (-\frac {1}{2} i \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {i \, b^{2} - 4 i \, a c}{2 \, c}\right )}}{8 \, \sqrt {c} {\left (\frac {i \, c}{{\left | c \right |}} + 1\right )}} + \frac {i \, \sqrt {\pi } \operatorname {erf}\left (\frac {1}{2} i \, \sqrt {c} {\left (2 \, x + \frac {b}{c}\right )} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}\right ) e^{\left (-\frac {-i \, b^{2} + 4 i \, a c}{2 \, c}\right )}}{8 \, \sqrt {c} {\left (-\frac {i \, c}{{\left | c \right |}} + 1\right )}} \]

[In]

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

[Out]

1/2*x - 1/8*I*sqrt(pi)*erf(-1/2*I*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)) + 1/8*I*sqrt(pi)*erf(1/2*I*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))

Mupad [F(-1)]

Timed out. \[ \int \sin ^2\left (a+b x+c x^2\right ) \, dx=\int {\sin \left (c\,x^2+b\,x+a\right )}^2 \,d x \]

[In]

int(sin(a + b*x + c*x^2)^2,x)

[Out]

int(sin(a + b*x + c*x^2)^2, x)