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

Optimal. Leaf size=82 \[ \frac{1}{2} \sqrt{\frac{\pi }{2}} \text{FresnelC}\left (\frac{2 x+1}{\sqrt{2 \pi }}\right )+\frac{1}{4} \sqrt{\frac{\pi }{2}} S\left (\frac{2 x+1}{\sqrt{2 \pi }}\right )-\frac{1}{2} x \cos \left (x^2+x+\frac{1}{4}\right )+\frac{1}{4} \cos \left (x^2+x+\frac{1}{4}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0330391, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {3463, 3446, 3352, 3461, 3445, 3351} \[ \frac{1}{2} \sqrt{\frac{\pi }{2}} \text{FresnelC}\left (\frac{2 x+1}{\sqrt{2 \pi }}\right )+\frac{1}{4} \sqrt{\frac{\pi }{2}} S\left (\frac{2 x+1}{\sqrt{2 \pi }}\right )-\frac{1}{2} x \cos \left (x^2+x+\frac{1}{4}\right )+\frac{1}{4} \cos \left (x^2+x+\frac{1}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sin[1/4 + x + x^2],x]

[Out]

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

Rule 3463

Int[((d_.) + (e_.)*(x_))^(m_)*Sin[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> -Simp[(e*(d + e*x)^(m - 1)*
Cos[a + b*x + c*x^2])/(2*c), x] + (Dist[(e^2*(m - 1))/(2*c), Int[(d + e*x)^(m - 2)*Cos[a + b*x + c*x^2], x], x
] - Dist[(b*e - 2*c*d)/(2*c), Int[(d + e*x)^(m - 1)*Sin[a + b*x + c*x^2], x], x]) /; FreeQ[{a, b, c, d, e}, x]
&& NeQ[b*e - 2*c*d, 0] && GtQ[m, 1]

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 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 3461

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

Rule 3445

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]

Rule 3351

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

Rubi steps

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

Mathematica [A]  time = 0.117096, size = 66, normalized size = 0.8 \[ \frac{1}{8} \left (2 \sqrt{2 \pi } \text{FresnelC}\left (\frac{2 x+1}{\sqrt{2 \pi }}\right )+\sqrt{2 \pi } S\left (\frac{2 x+1}{\sqrt{2 \pi }}\right )+2 (1-2 x) \cos \left (x^2+x+\frac{1}{4}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sin[1/4 + x + x^2],x]

[Out]

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

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Maple [A]  time = 0.009, size = 59, normalized size = 0.7 \begin{align*} -{\frac{x}{2}\cos \left ({\frac{1}{4}}+x+{x}^{2} \right ) }+{\frac{1}{4}\cos \left ({\frac{1}{4}}+x+{x}^{2} \right ) }+{\frac{\sqrt{2}\sqrt{\pi }}{8}{\it FresnelS} \left ({\frac{\sqrt{2}}{\sqrt{\pi }} \left ( x+{\frac{1}{2}} \right ) } \right ) }+{\frac{\sqrt{2}\sqrt{\pi }}{4}{\it FresnelC} \left ({\frac{\sqrt{2}}{\sqrt{\pi }} \left ( x+{\frac{1}{2}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x*cos(1/4+x+x^2)+1/4*cos(1/4+x+x^2)+1/8*2^(1/2)*Pi^(1/2)*FresnelS(2^(1/2)/Pi^(1/2)*(x+1/2))+1/4*2^(1/2)*P
i^(1/2)*FresnelC(2^(1/2)/Pi^(1/2)*(x+1/2))

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Maxima [C]  time = 3.97883, size = 211, normalized size = 2.57 \begin{align*} \frac{512 \, x{\left (e^{\left (i \, x^{2} + i \, x + \frac{1}{4} i\right )} + e^{\left (-i \, x^{2} - i \, x - \frac{1}{4} i\right )}\right )} + \sqrt{4 \, x^{2} + 4 \, x + 1}{\left (\left (32 i + 32\right ) \, \sqrt{2} \sqrt{\pi }{\left (\operatorname{erf}\left (\sqrt{i \, x^{2} + i \, x + \frac{1}{4} i}\right ) - 1\right )} - \left (32 i - 32\right ) \, \sqrt{2} \sqrt{\pi }{\left (\operatorname{erf}\left (\sqrt{-i \, x^{2} - i \, x - \frac{1}{4} i}\right ) - 1\right )} + \left (128 i - 128\right ) \, \sqrt{2} \Gamma \left (\frac{3}{2}, i \, x^{2} + i \, x + \frac{1}{4} i\right ) - \left (128 i + 128\right ) \, \sqrt{2} \Gamma \left (\frac{3}{2}, -i \, x^{2} - i \, x - \frac{1}{4} i\right )\right )} + 256 \, e^{\left (i \, x^{2} + i \, x + \frac{1}{4} i\right )} + 256 \, e^{\left (-i \, x^{2} - i \, x - \frac{1}{4} i\right )}}{1024 \,{\left (2 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1024*(512*x*(e^(I*x^2 + I*x + 1/4*I) + e^(-I*x^2 - I*x - 1/4*I)) + sqrt(4*x^2 + 4*x + 1)*((32*I + 32)*sqrt(2
)*sqrt(pi)*(erf(sqrt(I*x^2 + I*x + 1/4*I)) - 1) - (32*I - 32)*sqrt(2)*sqrt(pi)*(erf(sqrt(-I*x^2 - I*x - 1/4*I)
) - 1) + (128*I - 128)*sqrt(2)*gamma(3/2, I*x^2 + I*x + 1/4*I) - (128*I + 128)*sqrt(2)*gamma(3/2, -I*x^2 - I*x
- 1/4*I)) + 256*e^(I*x^2 + I*x + 1/4*I) + 256*e^(-I*x^2 - I*x - 1/4*I))/(2*x + 1)

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Fricas [A]  time = 1.48286, size = 228, normalized size = 2.78 \begin{align*} -\frac{1}{4} \,{\left (2 \, x - 1\right )} \cos \left (x^{2} + x + \frac{1}{4}\right ) + \frac{1}{4} \, \sqrt{2} \sqrt{\pi } \operatorname{C}\left (\frac{\sqrt{2}{\left (2 \, x + 1\right )}}{2 \, \sqrt{\pi }}\right ) + \frac{1}{8} \, \sqrt{2} \sqrt{\pi } \operatorname{S}\left (\frac{\sqrt{2}{\left (2 \, x + 1\right )}}{2 \, \sqrt{\pi }}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sin{\left (x^{2} + x + \frac{1}{4} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*sin(x**2 + x + 1/4), x)

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Giac [C]  time = 2.41814, size = 101, normalized size = 1.23 \begin{align*} -\left (\frac{1}{32} i + \frac{3}{32}\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}{32} i - \frac{3}{32}\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 ) - \frac{1}{8} i \,{\left (-2 i \, x + i\right )} e^{\left (i \, x^{2} + i \, x + \frac{1}{4} i\right )} - \frac{1}{8} i \,{\left (-2 i \, x + i\right )} e^{\left (-i \, x^{2} - i \, x - \frac{1}{4} i\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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