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

Optimal. Leaf size=46 \[ \frac {1}{8} \sqrt {\pi } C\left (\frac {2 x+1}{\sqrt {\pi }}\right )+\frac {x^2}{4}-\frac {1}{8} \sin \left (2 x^2+2 x+\frac {1}{2}\right ) \]

[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)

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Rubi [A]  time = 0.03, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3467, 3462, 3446, 3352} \[ \frac {1}{8} \sqrt {\pi } \text {FresnelC}\left (\frac {2 x+1}{\sqrt {\pi }}\right )+\frac {x^2}{4}-\frac {1}{8} \sin \left (2 x^2+2 x+\frac {1}{2}\right ) \]

Antiderivative was successfully verified.

[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 3352

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

Rule 3446

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 3462

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 3467

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*} \int x \sin ^2\left (\frac {1}{4}+x+x^2\right ) \, dx &=\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 } C\left (\frac {1+2 x}{\sqrt {\pi }}\right )-\frac {1}{8} \sin \left (\frac {1}{2}+2 x+2 x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 42, normalized size = 0.91 \[ \frac {1}{8} \left (\sqrt {\pi } C\left (\frac {2 x+1}{\sqrt {\pi }}\right )+2 x^2-\sin \left (\frac {1}{2} (2 x+1)^2\right )\right ) \]

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.40, size = 37, normalized size = 0.80 \[ \frac {1}{4} \, x^{2} - \frac {1}{4} \, \cos \left (x^{2} + x + \frac {1}{4}\right ) \sin \left (x^{2} + x + \frac {1}{4}\right ) + \frac {1}{8} \, \sqrt {\pi } \operatorname {C}\left (\frac {2 \, x + 1}{\sqrt {\pi }}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [C]  time = 0.19, size = 54, normalized size = 1.17 \[ \frac {1}{4} \, x^{2} - \left (\frac {1}{32} i + \frac {1}{32}\right ) \, \sqrt {\pi } \operatorname {erf}\left (\left (i - 1\right ) \, x + \frac {1}{2} i - \frac {1}{2}\right ) + \left (\frac {1}{32} i - \frac {1}{32}\right ) \, \sqrt {\pi } \operatorname {erf}\left (-\left (i + 1\right ) \, x - \frac {1}{2} i - \frac {1}{2}\right ) + \frac {1}{16} i \, e^{\left (2 i \, x^{2} + 2 i \, x + \frac {1}{2} i\right )} - \frac {1}{16} i \, e^{\left (-2 i \, x^{2} - 2 i \, x - \frac {1}{2} i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.06, size = 35, normalized size = 0.76 \[ \frac {x^{2}}{4}-\frac {\sin \left (\frac {1}{2}+2 x +2 x^{2}\right )}{8}+\frac {\FresnelC \left (\frac {1+2 x}{\sqrt {\pi }}\right ) \sqrt {\pi }}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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maxima [C]  time = 0.54, size = 135, normalized size = 2.93 \[ \frac {65536 \, x^{3} + 32768 \, x^{2} + x {\left (16384 i \, e^{\left (2 i \, x^{2} + 2 i \, x + \frac {1}{2} i\right )} - 16384 i \, e^{\left (-2 i \, x^{2} - 2 i \, x - \frac {1}{2} i\right )}\right )} + \sqrt {8 \, x^{2} + 8 \, x + 2} {\left (-\left (2048 i - 2048\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {2 i \, x^{2} + 2 i \, x + \frac {1}{2} i}\right ) - 1\right )} + \left (2048 i + 2048\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-2 i \, x^{2} - 2 i \, x - \frac {1}{2} i}\right ) - 1\right )}\right )} + 8192 i \, e^{\left (2 i \, x^{2} + 2 i \, x + \frac {1}{2} i\right )} - 8192 i \, e^{\left (-2 i \, x^{2} - 2 i \, x - \frac {1}{2} i\right )}}{131072 \, {\left (2 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/131072*(65536*x^3 + 32768*x^2 + x*(16384*I*e^(2*I*x^2 + 2*I*x + 1/2*I) - 16384*I*e^(-2*I*x^2 - 2*I*x - 1/2*I
)) + sqrt(8*x^2 + 8*x + 2)*(-(2048*I - 2048)*sqrt(2)*sqrt(pi)*(erf(sqrt(2*I*x^2 + 2*I*x + 1/2*I)) - 1) + (2048
*I + 2048)*sqrt(2)*sqrt(pi)*(erf(sqrt(-2*I*x^2 - 2*I*x - 1/2*I)) - 1)) + 8192*I*e^(2*I*x^2 + 2*I*x + 1/2*I) -
8192*I*e^(-2*I*x^2 - 2*I*x - 1/2*I))/(2*x + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,{\sin \left (x^2+x+\frac {1}{4}\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(x + x^2 + 1/4)^2,x)

[Out]

int(x*sin(x + x^2 + 1/4)^2, x)

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sympy [B]  time = 1.91, size = 121, normalized size = 2.63 \[ \frac {x^{2}}{4} - \frac {\sqrt {\pi } x C\left (\frac {2 x}{\sqrt {\pi }} + \frac {1}{\sqrt {\pi }}\right )}{4} + \frac {\sqrt {\pi } x C\left (\frac {2 x}{\sqrt {\pi }} + \frac {1}{\sqrt {\pi }}\right ) \Gamma \left (\frac {1}{4}\right )}{16 \Gamma \left (\frac {5}{4}\right )} - \frac {\sin {\left (2 \left (x + \frac {1}{2}\right )^{2} \right )} \Gamma \left (\frac {1}{4}\right )}{32 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {\pi } C\left (\frac {2 x}{\sqrt {\pi }} + \frac {1}{\sqrt {\pi }}\right ) \Gamma \left (\frac {1}{4}\right )}{32 \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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