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

Optimal. Leaf size=41 \[ -\frac {1}{2} \sqrt {\frac {\pi }{2}} S\left (\frac {2 x+1}{\sqrt {2 \pi }}\right )-\frac {1}{2} \cos \left (x^2+x+\frac {1}{4}\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3461, 3445, 3351} \[ -\frac {1}{2} \sqrt {\frac {\pi }{2}} S\left (\frac {2 x+1}{\sqrt {2 \pi }}\right )-\frac {1}{2} \cos \left (x^2+x+\frac {1}{4}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Sin[1/4 + x + x^2],x]

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 39, normalized size = 0.95 \[ \frac {1}{4} \left (-\sqrt {2 \pi } S\left (\frac {2 x+1}{\sqrt {2 \pi }}\right )-2 \cos \left (x^2+x+\frac {1}{4}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sin[1/4 + x + x^2],x]

[Out]

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

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fricas [A]  time = 0.45, size = 32, normalized size = 0.78 \[ -\frac {1}{4} \, \sqrt {2} \sqrt {\pi } \operatorname {S}\left (\frac {\sqrt {2} {\left (2 \, x + 1\right )}}{2 \, \sqrt {\pi }}\right ) - \frac {1}{2} \, \cos \left (x^{2} + x + \frac {1}{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*sqrt(pi)*fresnel_sin(1/2*sqrt(2)*(2*x + 1)/sqrt(pi)) - 1/2*cos(x^2 + x + 1/4)

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giac [C]  time = 0.18, size = 65, normalized size = 1.59 \[ -\left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{4} i - \frac {1}{4}\right ) \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) + \left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) - \frac {1}{4} \, e^{\left (i \, x^{2} + i \, x + \frac {1}{4} i\right )} - \frac {1}{4} \, e^{\left (-i \, x^{2} - i \, x - \frac {1}{4} i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 30, normalized size = 0.73 \[ -\frac {\cos \left (\frac {1}{4}+x +x^{2}\right )}{2}-\frac {\sqrt {2}\, \sqrt {\pi }\, \mathrm {S}\left (\frac {\sqrt {2}\, \left (x +\frac {1}{2}\right )}{\sqrt {\pi }}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.70, size = 123, normalized size = 3.00 \[ -\frac {2048 \, x {\left (e^{\left (i \, x^{2} + i \, x + \frac {1}{4} i\right )} + e^{\left (-i \, x^{2} - i \, x - \frac {1}{4} i\right )}\right )} - \sqrt {4 \, x^{2} + 4 \, x + 1} {\left (-\left (256 i + 256\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, x^{2} + i \, x + \frac {1}{4} i}\right ) - 1\right )} + \left (256 i - 256\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, x^{2} - i \, x - \frac {1}{4} i}\right ) - 1\right )}\right )} + 1024 \, e^{\left (i \, x^{2} + i \, x + \frac {1}{4} i\right )} + 1024 \, e^{\left (-i \, x^{2} - i \, x - \frac {1}{4} i\right )}}{4096 \, {\left (2 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4096*(2048*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)*(-(256*I + 256)*s
qrt(2)*sqrt(pi)*(erf(sqrt(I*x^2 + I*x + 1/4*I)) - 1) + (256*I - 256)*sqrt(2)*sqrt(pi)*(erf(sqrt(-I*x^2 - I*x -
1/4*I)) - 1)) + 1024*e^(I*x^2 + I*x + 1/4*I) + 1024*e^(-I*x^2 - I*x - 1/4*I))/(2*x + 1)

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mupad [B]  time = 0.06, size = 32, normalized size = 0.78 \[ -\frac {\cos \left (x^2+x+\frac {1}{4}\right )}{2}-\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {S}\left (\frac {\sqrt {2}\,\left (2\,x+1\right )}{2\,\sqrt {\pi }}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*sin(x + x^2 + 1/4),x)

[Out]

- cos(x + x^2 + 1/4)/2 - (2^(1/2)*pi^(1/2)*fresnels((2^(1/2)*(2*x + 1))/(2*pi^(1/2))))/4

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sympy [B]  time = 1.42, size = 160, normalized size = 3.90 \[ - \frac {3 \sqrt {2} \sqrt {\pi } x S\left (\frac {\sqrt {2} x}{\sqrt {\pi }} + \frac {\sqrt {2}}{2 \sqrt {\pi }}\right ) \Gamma \left (\frac {3}{4}\right )}{8 \Gamma \left (\frac {7}{4}\right )} + \frac {\sqrt {2} \sqrt {\pi } x S\left (\frac {\sqrt {2} x}{\sqrt {\pi }} + \frac {\sqrt {2}}{2 \sqrt {\pi }}\right )}{2} - \frac {3 \cos {\left (\left (x + \frac {1}{2}\right )^{2} \right )} \Gamma \left (\frac {3}{4}\right )}{8 \Gamma \left (\frac {7}{4}\right )} - \frac {3 \sqrt {2} \sqrt {\pi } S\left (\frac {\sqrt {2} x}{\sqrt {\pi }} + \frac {\sqrt {2}}{2 \sqrt {\pi }}\right ) \Gamma \left (\frac {3}{4}\right )}{16 \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*sqrt(2)*sqrt(pi)*x*fresnels(sqrt(2)*x/sqrt(pi) + sqrt(2)/(2*sqrt(pi)))*gamma(3/4)/(8*gamma(7/4)) + sqrt(2)*
sqrt(pi)*x*fresnels(sqrt(2)*x/sqrt(pi) + sqrt(2)/(2*sqrt(pi)))/2 - 3*cos((x + 1/2)**2)*gamma(3/4)/(8*gamma(7/4
)) - 3*sqrt(2)*sqrt(pi)*fresnels(sqrt(2)*x/sqrt(pi) + sqrt(2)/(2*sqrt(pi)))*gamma(3/4)/(16*gamma(7/4))

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