\(\int \sin ^2(\frac {1}{4}+x+x^2) \, dx\) [26]

Optimal result

Integrand size = 11, antiderivative size = 27 \[ \int \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {x}{2}-\frac {1}{4} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {\pi }}\right ) \]

[Out]

1/2*x-1/4*FresnelC((1+2*x)/Pi^(1/2))*Pi^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3530, 3527, 3433} \[ \int \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {x}{2}-\frac {1}{4} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 x+1}{\sqrt {\pi }}\right ) \]

[In]

Int[Sin[1/4 + x + x^2]^2,x]

[Out]

x/2 - (Sqrt[Pi]*FresnelC[(1 + 2*x)/Sqrt[Pi]])/4

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 3527

Int[Cos[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Int[Cos[(b + 2*c*x)^2/(4*c)], x] /; FreeQ[{a, b, c},
x] && EqQ[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 (\frac {1}{2}+2 x+2 x^2\right )\right ) \, dx \\ & = \frac {x}{2}-\frac {1}{2} \int \cos \left (\frac {1}{2}+2 x+2 x^2\right ) \, dx \\ & = \frac {x}{2}-\frac {1}{2} \int \cos \left (\frac {1}{8} (2+4 x)^2\right ) \, dx \\ & = \frac {x}{2}-\frac {1}{4} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {\pi }}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{4} \left (2 x-\sqrt {\pi } \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {\pi }}\right )\right ) \]

[In]

Integrate[Sin[1/4 + x + x^2]^2,x]

[Out]

(2*x - Sqrt[Pi]*FresnelC[(1 + 2*x)/Sqrt[Pi]])/4

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {x}{2}-\frac {\operatorname {C}\left (\frac {1+2 x}{\sqrt {\pi }}\right ) \sqrt {\pi }}{4}\)	\(20\)
risch	\(\frac {x}{2}+\frac {\sqrt {\pi }\, \sqrt {2}\, \left (-1\right )^{\frac {3}{4}} \operatorname {erf}\left (\sqrt {2}\, \left (-1\right )^{\frac {1}{4}} x +\frac {\sqrt {2}\, \left (-1\right )^{\frac {1}{4}}}{2}\right )}{16}-\frac {\sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-2 i}\, x -\frac {i}{\sqrt {-2 i}}\right )}{8 \sqrt {-2 i}}\)	\(58\)

[In]

int(sin(1/4+x+x^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*x-1/4*FresnelC((1+2*x)/Pi^(1/2))*Pi^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=-\frac {1}{4} \, \sqrt {\pi } \operatorname {C}\left (\frac {2 \, x + 1}{\sqrt {\pi }}\right ) + \frac {1}{2} \, x \]

[In]

integrate(sin(1/4+x+x^2)^2,x, algorithm="fricas")

[Out]

-1/4*sqrt(pi)*fresnel_cos((2*x + 1)/sqrt(pi)) + 1/2*x

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {x}{2} - \frac {\sqrt {\pi } C\left (\frac {4 x + 2}{2 \sqrt {\pi }}\right )}{4} \]

[In]

integrate(sin(1/4+x+x**2)**2,x)

[Out]

x/2 - sqrt(pi)*fresnelc((4*x + 2)/(2*sqrt(pi)))/4

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{16} \, \sqrt {\pi } {\left (\left (i - 1\right ) \, \operatorname {erf}\left (\frac {2 i \, x + i}{\sqrt {2 i}}\right ) + \left (i + 1\right ) \, \operatorname {erf}\left (\frac {2 i \, x + i}{\sqrt {-2 i}}\right )\right )} + \frac {1}{2} \, x \]

[In]

integrate(sin(1/4+x+x^2)^2,x, algorithm="maxima")

[Out]

1/16*sqrt(pi)*((I - 1)*erf((2*I*x + I)/sqrt(2*I)) + (I + 1)*erf((2*I*x + I)/sqrt(-2*I))) + 1/2*x

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {\pi } \operatorname {erf}\left (\left (i - 1\right ) \, x + \frac {1}{2} i - \frac {1}{2}\right ) - \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {\pi } \operatorname {erf}\left (-\left (i + 1\right ) \, x - \frac {1}{2} i - \frac {1}{2}\right ) + \frac {1}{2} \, x \]

[In]

integrate(sin(1/4+x+x^2)^2,x, algorithm="giac")

[Out]

(1/16*I + 1/16)*sqrt(pi)*erf((I - 1)*x + 1/2*I - 1/2) - (1/16*I - 1/16)*sqrt(pi)*erf(-(I + 1)*x - 1/2*I - 1/2)
+ 1/2*x

Mupad [F(-1)]

Timed out. \[ \int \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\int {\sin \left (x^2+x+\frac {1}{4}\right )}^2 \,d x \]

[In]

int(sin(x + x^2 + 1/4)^2,x)

[Out]

int(sin(x + x^2 + 1/4)^2, x)