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

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

[Out]

1/2*sin(1/4+x+x^2)-1/4*FresnelC(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 = {3462, 3446, 3352} \[ \frac {1}{2} \sin \left (x^2+x+\frac {1}{4}\right )-\frac {1}{2} \sqrt {\frac {\pi }{2}} \text {FresnelC}\left (\frac {2 x+1}{\sqrt {2 \pi }}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x*Cos[1/4 + x + x^2],x]

[Out]

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

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]

Rubi steps

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

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

Antiderivative was successfully verified.

[In]

Integrate[x*Cos[1/4 + x + x^2],x]

[Out]

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

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fricas [A]  time = 0.79, size = 32, normalized size = 0.78 \[ -\frac {1}{4} \, \sqrt {2} \sqrt {\pi } \operatorname {C}\left (\frac {\sqrt {2} {\left (2 \, x + 1\right )}}{2 \, \sqrt {\pi }}\right ) + \frac {1}{2} \, \sin \left (x^{2} + x + \frac {1}{4}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.50, size = 65, normalized size = 1.59 \[ \left (\frac {1}{16} i + \frac {1}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (\left (\frac {1}{4} i - \frac {1}{4}\right ) \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) - \left (\frac {1}{16} i - \frac {1}{16}\right ) \, \sqrt {2} \sqrt {\pi } \operatorname {erf}\left (-\left (\frac {1}{4} i + \frac {1}{4}\right ) \, \sqrt {2} {\left (2 \, x + 1\right )}\right ) - \frac {1}{4} i \, e^{\left (i \, x^{2} + i \, x + \frac {1}{4} i\right )} + \frac {1}{4} i \, e^{\left (-i \, x^{2} - i \, x - \frac {1}{4} i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cos(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*I*e^(I*x^2 + I*x + 1/4*I) + 1/4*I*e^(-I*x^2 - I*x - 1/4*I)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 1.55, size = 125, normalized size = 3.05 \[ -\frac {x {\left (2048 i \, e^{\left (i \, x^{2} + i \, x + \frac {1}{4} i\right )} - 2048 i \, e^{\left (-i \, x^{2} - i \, x - \frac {1}{4} i\right )}\right )} + \sqrt {4 \, x^{2} + 4 \, x + 1} {\left (-\left (256 i - 256\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {i \, x^{2} + i \, x + \frac {1}{4} i}\right ) - 1\right )} + \left (256 i + 256\right ) \, \sqrt {2} \sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-i \, x^{2} - i \, x - \frac {1}{4} i}\right ) - 1\right )}\right )} + 1024 i \, e^{\left (i \, x^{2} + i \, x + \frac {1}{4} i\right )} - 1024 i \, e^{\left (-i \, x^{2} - i \, x - \frac {1}{4} i\right )}}{4096 \, {\left (2 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.48, size = 32, normalized size = 0.78 \[ \frac {\sin \left (x^2+x+\frac {1}{4}\right )}{2}-\frac {\sqrt {2}\,\sqrt {\pi }\,\mathrm {C}\left (\frac {\sqrt {2}\,\left (2\,x+1\right )}{2\,\sqrt {\pi }}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cos(x + x^2 + 1/4),x)

[Out]

sin(x + x^2 + 1/4)/2 - (2^(1/2)*pi^(1/2)*fresnelc((2^(1/2)*(2*x + 1))/(2*pi^(1/2))))/4

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sympy [B]  time = 1.10, size = 155, normalized size = 3.78 \[ - \frac {\sqrt {2} \sqrt {\pi } x C\left (\frac {\sqrt {2} x}{\sqrt {\pi }} + \frac {\sqrt {2}}{2 \sqrt {\pi }}\right ) \Gamma \left (\frac {1}{4}\right )}{8 \Gamma \left (\frac {5}{4}\right )} + \frac {\sqrt {2} \sqrt {\pi } x C\left (\frac {\sqrt {2} x}{\sqrt {\pi }} + \frac {\sqrt {2}}{2 \sqrt {\pi }}\right )}{2} + \frac {\sin {\left (\left (x + \frac {1}{2}\right )^{2} \right )} \Gamma \left (\frac {1}{4}\right )}{8 \Gamma \left (\frac {5}{4}\right )} - \frac {\sqrt {2} \sqrt {\pi } C\left (\frac {\sqrt {2} x}{\sqrt {\pi }} + \frac {\sqrt {2}}{2 \sqrt {\pi }}\right ) \Gamma \left (\frac {1}{4}\right )}{16 \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*sqrt(pi)*x*fresnelc(sqrt(2)*x/sqrt(pi) + sqrt(2)/(2*sqrt(pi)))*gamma(1/4)/(8*gamma(5/4)) + sqrt(2)*sq
rt(pi)*x*fresnelc(sqrt(2)*x/sqrt(pi) + sqrt(2)/(2*sqrt(pi)))/2 + sin((x + 1/2)**2)*gamma(1/4)/(8*gamma(5/4)) -
sqrt(2)*sqrt(pi)*fresnelc(sqrt(2)*x/sqrt(pi) + sqrt(2)/(2*sqrt(pi)))*gamma(1/4)/(16*gamma(5/4))

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