\(\int \sin (\frac {1}{4}+x+x^2) \, dx\) [13]

Optimal result

Integrand size = 9, antiderivative size = 24 \[ \int \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=\sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {1+2 x}{\sqrt {2 \pi }}\right ) \]

[Out]

1/2*FresnelS(1/2*(1+2*x)*2^(1/2)/Pi^(1/2))*2^(1/2)*Pi^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3526, 3432} \[ \int \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=\sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {2 x+1}{\sqrt {2 \pi }}\right ) \]

[In]

Int[Sin[1/4 + x + x^2],x]

[Out]

Sqrt[Pi/2]*FresnelS[(1 + 2*x)/Sqrt[2*Pi]]

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 3526

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

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=\sqrt {\frac {\pi }{2}} \operatorname {FresnelS}\left (\frac {1+2 x}{\sqrt {2 \pi }}\right ) \]

[In]

Integrate[Sin[1/4 + x + x^2],x]

[Out]

Sqrt[Pi/2]*FresnelS[(1 + 2*x)/Sqrt[2*Pi]]

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {\sqrt {2}\, \sqrt {\pi }\, \operatorname {S}\left (\frac {\sqrt {2}\, \left (x +\frac {1}{2}\right )}{\sqrt {\pi }}\right )}{2}\)	\(20\)
risch	\(\frac {\left (-1\right )^{\frac {1}{4}} \sqrt {\pi }\, \operatorname {erf}\left (\left (-1\right )^{\frac {1}{4}} x +\frac {\left (-1\right )^{\frac {1}{4}}}{2}\right )}{4}-\frac {i \sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-i}\, x -\frac {i}{2 \sqrt {-i}}\right )}{4 \sqrt {-i}}\)	\(47\)

[In]

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

[Out]

1/2*2^(1/2)*Pi^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)*(x+1/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{2} \, \sqrt {2} \sqrt {\pi } \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, x + 1\right )}}{2 \, \sqrt {\pi }}\right ) \]

[In]

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

[Out]

1/2*sqrt(2)*sqrt(pi)*fresnel_sin(1/2*sqrt(2)*(2*x + 1)/sqrt(pi))

Sympy [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {\sqrt {2} \sqrt {\pi } S\left (\frac {\sqrt {2} \cdot \left (2 x + 1\right )}{2 \sqrt {\pi }}\right )}{2} \]

[In]

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

[Out]

sqrt(2)*sqrt(pi)*fresnels(sqrt(2)*(2*x + 1)/(2*sqrt(pi)))/2

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.92 \[ \int \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{16} \, \sqrt {\pi } {\left (\left (i + 1\right ) \, \sqrt {2} \operatorname {erf}\left (-\frac {1}{2} \, \left (-1\right )^{\frac {3}{4}} {\left (2 i \, x + i\right )}\right ) + \left (i + 1\right ) \, \sqrt {2} \operatorname {erf}\left (-\left (\frac {1}{4} i - \frac {1}{4}\right ) \, \sqrt {2} {\left (2 i \, x + i\right )}\right ) - \left (i - 1\right ) \, \sqrt {2} \operatorname {erf}\left (-\left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2} {\left (2 i \, x + i\right )}\right ) + \left (i - 1\right ) \, \sqrt {2} \operatorname {erf}\left (\frac {2 i \, x + i}{2 \, \sqrt {-i}}\right )\right )} \]

[In]

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

[Out]

1/16*sqrt(pi)*((I + 1)*sqrt(2)*erf(-1/2*(-1)^(3/4)*(2*I*x + I)) + (I + 1)*sqrt(2)*erf(-(1/4*I - 1/4)*sqrt(2)*(
2*I*x + I)) - (I - 1)*sqrt(2)*erf(-(1/4*I + 1/4)*sqrt(2)*(2*I*x + I)) + (I - 1)*sqrt(2)*erf(1/2*(2*I*x + I)/sq
rt(-I)))

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=\left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{4} i - \frac {1}{4}\right ) \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) - \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) \]

[In]

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

[Out]

(1/8*I - 1/8)*sqrt(2)*sqrt(pi)*erf((1/4*I - 1/4)*sqrt(2)*(2*x + 1)) - (1/8*I + 1/8)*sqrt(2)*sqrt(pi)*erf(-(1/4
*I + 1/4)*sqrt(2)*(2*x + 1))

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \sin \left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {S}\left (\frac {\sqrt {2}\,\left (x+\frac {1}{2}\right )}{\sqrt {\pi }}\right )}{2} \]

[In]

int(sin(x + x^2 + 1/4),x)

[Out]

(2^(1/2)*pi^(1/2)*fresnels((2^(1/2)*(x + 1/2))/pi^(1/2)))/2