3.116 \(\int x C(b x) \, dx\)

Optimal. Leaf size=49 \[ \frac {S(b x)}{2 \pi b^2}-\frac {x \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 \pi b}+\frac {1}{2} x^2 C(b x) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6427, 3386, 3351} \[ \frac {S(b x)}{2 \pi b^2}-\frac {x \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 \pi b}+\frac {1}{2} x^2 \text {FresnelC}(b x) \]

Antiderivative was successfully verified.

[In]

Int[x*FresnelC[b*x],x]

[Out]

(x^2*FresnelC[b*x])/2 + FresnelS[b*x]/(2*b^2*Pi) - (x*Sin[(b^2*Pi*x^2)/2])/(2*b*Pi)

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 3386

Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Sin[c + d*
x^n])/(d*n), x] - Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x
] && IGtQ[n, 0] && LtQ[n, m + 1]

Rule 6427

Int[FresnelC[(b_.)*(x_)]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*FresnelC[b*x])/(d*(m + 1)), x] -
Dist[b/(d*(m + 1)), Int[(d*x)^(m + 1)*Cos[(Pi*b^2*x^2)/2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 49, normalized size = 1.00 \[ \frac {S(b x)}{2 \pi b^2}-\frac {x \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 \pi b}+\frac {1}{2} x^2 C(b x) \]

Antiderivative was successfully verified.

[In]

Integrate[x*FresnelC[b*x],x]

[Out]

(x^2*FresnelC[b*x])/2 + FresnelS[b*x]/(2*b^2*Pi) - (x*Sin[(b^2*Pi*x^2)/2])/(2*b*Pi)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x {\rm fresnelc}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnelc(b*x),x, algorithm="fricas")

[Out]

integral(x*fresnelc(b*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x {\rm fresnelc}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnelc(b*x),x, algorithm="giac")

[Out]

integrate(x*fresnelc(b*x), x)

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maple [A]  time = 0.01, size = 44, normalized size = 0.90 \[ \frac {\frac {b^{2} x^{2} \FresnelC \left (b x \right )}{2}-\frac {b x \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 \pi }+\frac {\mathrm {S}\left (b x \right )}{2 \pi }}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*FresnelC(b*x),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x {\rm fresnelc}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnelc(b*x),x, algorithm="maxima")

[Out]

integrate(x*fresnelc(b*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,\mathrm {FresnelC}\left (b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*FresnelC(b*x),x)

[Out]

int(x*FresnelC(b*x), x)

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sympy [A]  time = 0.60, size = 49, normalized size = 1.00 \[ \frac {b x^{3} \Gamma \left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} \\ \frac {1}{2}, \frac {5}{4}, \frac {7}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 \Gamma \left (\frac {5}{4}\right ) \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnelc(b*x),x)

[Out]

b*x**3*gamma(1/4)*gamma(3/4)*hyper((1/4, 3/4), (1/2, 5/4, 7/4), -pi**2*b**4*x**4/16)/(16*gamma(5/4)*gamma(7/4)
)

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