\(\int x \sin ^2(\frac {1}{4}+x+x^2) \, dx\) [25]

Optimal result

Integrand size = 13, antiderivative size = 46 \[ \int x \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {x^2}{4}+\frac {1}{8} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {\pi }}\right )-\frac {1}{8} \sin \left (\frac {1}{2}+2 x+2 x^2\right ) \]

[Out]

1/4*x^2-1/8*sin(1/2+2*x+2*x^2)+1/8*FresnelC((1+2*x)/Pi^(1/2))*Pi^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, 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 = {3548, 3543, 3527, 3433} \[ \int x \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{8} \sqrt {\pi } \operatorname {FresnelC}\left (\frac {2 x+1}{\sqrt {\pi }}\right )+\frac {x^2}{4}-\frac {1}{8} \sin \left (2 x^2+2 x+\frac {1}{2}\right ) \]

[In]

Int[x*Sin[1/4 + x + x^2]^2,x]

[Out]

x^2/4 + (Sqrt[Pi]*FresnelC[(1 + 2*x)/Sqrt[Pi]])/8 - Sin[1/2 + 2*x + 2*x^2]/8

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 3543

Int[Cos[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[e*(Sin[a + b*x + c*x^2]/(2*
c)), x] + Dist[(2*c*d - b*e)/(2*c), Int[Cos[a + b*x + c*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d
- b*e, 0]

Rule 3548

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91 \[ \int x \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{8} \left (2 x^2+\sqrt {\pi } \operatorname {FresnelC}\left (\frac {1+2 x}{\sqrt {\pi }}\right )-\sin \left (\frac {1}{2} (1+2 x)^2\right )\right ) \]

[In]

Integrate[x*Sin[1/4 + x + x^2]^2,x]

[Out]

(2*x^2 + Sqrt[Pi]*FresnelC[(1 + 2*x)/Sqrt[Pi]] - Sin[(1 + 2*x)^2/2])/8

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {x^{2}}{4}-\frac {\sin \left (\frac {1}{2}+2 x +2 x^{2}\right )}{8}+\frac {\operatorname {C}\left (\frac {1+2 x}{\sqrt {\pi }}\right ) \sqrt {\pi }}{8}\)	\(35\)
risch	\(-\frac {\sqrt {\pi }\, \sqrt {2}\, \left (-1\right )^{\frac {3}{4}} \operatorname {erf}\left (\sqrt {2}\, \left (-1\right )^{\frac {1}{4}} x +\frac {\sqrt {2}\, \left (-1\right )^{\frac {1}{4}}}{2}\right )}{32}+\frac {\sqrt {\pi }\, \operatorname {erf}\left (\sqrt {-2 i}\, x -\frac {i}{\sqrt {-2 i}}\right )}{16 \sqrt {-2 i}}+\frac {x^{2}}{4}-\frac {\sin \left (\frac {\left (1+2 x \right )^{2}}{2}\right )}{8}\)	\(72\)

[In]

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

[Out]

1/4*x^2-1/8*sin(1/2+2*x+2*x^2)+1/8*FresnelC((1+2*x)/Pi^(1/2))*Pi^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int x \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{4} \, x^{2} - \frac {1}{4} \, \cos \left (x^{2} + x + \frac {1}{4}\right ) \sin \left (x^{2} + x + \frac {1}{4}\right ) + \frac {1}{8} \, \sqrt {\pi } \operatorname {C}\left (\frac {2 \, x + 1}{\sqrt {\pi }}\right ) \]

[In]

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

[Out]

1/4*x^2 - 1/4*cos(x^2 + x + 1/4)*sin(x^2 + x + 1/4) + 1/8*sqrt(pi)*fresnel_cos((2*x + 1)/sqrt(pi))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (37) = 74\).

Time = 0.87 (sec) , antiderivative size = 121, normalized size of antiderivative = 2.63 \[ \int x \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {x^{2}}{4} - \frac {\sqrt {\pi } x C\left (\frac {2 x}{\sqrt {\pi }} + \frac {1}{\sqrt {\pi }}\right )}{4} + \frac {\sqrt {\pi } x C\left (\frac {2 x}{\sqrt {\pi }} + \frac {1}{\sqrt {\pi }}\right ) \Gamma \left (\frac {1}{4}\right )}{16 \Gamma \left (\frac {5}{4}\right )} - \frac {\sin {\left (2 \left (x + \frac {1}{2}\right )^{2} \right )} \Gamma \left (\frac {1}{4}\right )}{32 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {\pi } C\left (\frac {2 x}{\sqrt {\pi }} + \frac {1}{\sqrt {\pi }}\right ) \Gamma \left (\frac {1}{4}\right )}{32 \Gamma \left (\frac {5}{4}\right )} \]

[In]

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

[Out]

x**2/4 - sqrt(pi)*x*fresnelc(2*x/sqrt(pi) + 1/sqrt(pi))/4 + sqrt(pi)*x*fresnelc(2*x/sqrt(pi) + 1/sqrt(pi))*gam
ma(1/4)/(16*gamma(5/4)) - sin(2*(x + 1/2)**2)*gamma(1/4)/(32*gamma(5/4)) + sqrt(pi)*fresnelc(2*x/sqrt(pi) + 1/
sqrt(pi))*gamma(1/4)/(32*gamma(5/4))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.36 (sec) , antiderivative size = 137, normalized size of antiderivative = 2.98 \[ \int x \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {32 \, x^{3} + 16 \, x^{2} - 8 \, x {\left (-i \, e^{\left (2 i \, x^{2} + 2 i \, x + \frac {1}{2} i\right )} + i \, e^{\left (-2 i \, x^{2} - 2 i \, x - \frac {1}{2} i\right )}\right )} - \sqrt {8 \, x^{2} + 8 \, x + 2} {\left (\left (i - 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {2 i \, x^{2} + 2 i \, x + \frac {1}{2} i}\right ) - 1\right )} - \left (i + 1\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-2 i \, x^{2} - 2 i \, x - \frac {1}{2} i}\right ) - 1\right )}\right )} + 4 i \, e^{\left (2 i \, x^{2} + 2 i \, x + \frac {1}{2} i\right )} - 4 i \, e^{\left (-2 i \, x^{2} - 2 i \, x - \frac {1}{2} i\right )}}{64 \, {\left (2 \, x + 1\right )}} \]

[In]

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

[Out]

1/64*(32*x^3 + 16*x^2 - 8*x*(-I*e^(2*I*x^2 + 2*I*x + 1/2*I) + I*e^(-2*I*x^2 - 2*I*x - 1/2*I)) - sqrt(8*x^2 + 8
*x + 2)*((I - 1)*sqrt(2)*sqrt(pi)*(erf(sqrt(2*I*x^2 + 2*I*x + 1/2*I)) - 1) - (I + 1)*sqrt(2)*sqrt(pi)*(erf(sqr
t(-2*I*x^2 - 2*I*x - 1/2*I)) - 1)) + 4*I*e^(2*I*x^2 + 2*I*x + 1/2*I) - 4*I*e^(-2*I*x^2 - 2*I*x - 1/2*I))/(2*x
+ 1)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.31 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.17 \[ \int x \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx=\frac {1}{4} \, x^{2} - \left (\frac {1}{32} i + \frac {1}{32}\right ) \, \sqrt {\pi } \operatorname {erf}\left (\left (i - 1\right ) \, x + \frac {1}{2} i - \frac {1}{2}\right ) + \left (\frac {1}{32} i - \frac {1}{32}\right ) \, \sqrt {\pi } \operatorname {erf}\left (-\left (i + 1\right ) \, x - \frac {1}{2} i - \frac {1}{2}\right ) + \frac {1}{16} i \, e^{\left (2 i \, x^{2} + 2 i \, x + \frac {1}{2} i\right )} - \frac {1}{16} i \, e^{\left (-2 i \, x^{2} - 2 i \, x - \frac {1}{2} i\right )} \]

[In]

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

[Out]

1/4*x^2 - (1/32*I + 1/32)*sqrt(pi)*erf((I - 1)*x + 1/2*I - 1/2) + (1/32*I - 1/32)*sqrt(pi)*erf(-(I + 1)*x - 1/
2*I - 1/2) + 1/16*I*e^(2*I*x^2 + 2*I*x + 1/2*I) - 1/16*I*e^(-2*I*x^2 - 2*I*x - 1/2*I)

Mupad [F(-1)]

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

[In]

int(x*sin(x + x^2 + 1/4)^2,x)

[Out]

int(x*sin(x + x^2 + 1/4)^2, x)