3.146 \(\int x \text{FresnelC}(b x)^2 \, dx\)

Optimal. Leaf size=144 \[ \frac{i x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},-\frac{1}{2} i \pi b^2 x^2\right )}{8 \pi }-\frac{i x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},\frac{1}{2} i \pi b^2 x^2\right )}{8 \pi }+\frac{\text{FresnelC}(b x) S(b x)}{2 \pi b^2}-\frac{x \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b}-\frac{\cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^2}+\frac{1}{2} x^2 \text{FresnelC}(b x)^2 \]

[Out]

-Cos[b^2*Pi*x^2]/(4*b^2*Pi^2) + (x^2*FresnelC[b*x]^2)/2 + (FresnelC[b*x]*FresnelS[b*x])/(2*b^2*Pi) + ((I/8)*x^
2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (-I/2)*b^2*Pi*x^2])/Pi - ((I/8)*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2},
(I/2)*b^2*Pi*x^2])/Pi - (x*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(b*Pi)

________________________________________________________________________________________

Rubi [A]  time = 0.0839067, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {6431, 6455, 6447, 3379, 2638} \[ \frac{i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi }-\frac{i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi }+\frac{\text{FresnelC}(b x) S(b x)}{2 \pi b^2}-\frac{x \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b}-\frac{\cos \left (\pi b^2 x^2\right )}{4 \pi ^2 b^2}+\frac{1}{2} x^2 \text{FresnelC}(b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x*FresnelC[b*x]^2,x]

[Out]

-Cos[b^2*Pi*x^2]/(4*b^2*Pi^2) + (x^2*FresnelC[b*x]^2)/2 + (FresnelC[b*x]*FresnelS[b*x])/(2*b^2*Pi) + ((I/8)*x^
2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (-I/2)*b^2*Pi*x^2])/Pi - ((I/8)*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2},
(I/2)*b^2*Pi*x^2])/Pi - (x*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(b*Pi)

Rule 6431

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

Rule 6455

Int[Cos[(d_.)*(x_)^2]*FresnelC[(b_.)*(x_)]*(x_)^(m_), x_Symbol] :> Simp[(x^(m - 1)*Sin[d*x^2]*FresnelC[b*x])/(
2*d), x] + (-Dist[(m - 1)/(2*d), Int[x^(m - 2)*Sin[d*x^2]*FresnelC[b*x], x], x] - Dist[b/(4*d), Int[x^(m - 1)*
Sin[2*d*x^2], x], x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4] && IGtQ[m, 1]

Rule 6447

Int[FresnelC[(b_.)*(x_)]*Sin[(d_.)*(x_)^2], x_Symbol] :> Simp[(b*Pi*FresnelC[b*x]*FresnelS[b*x])/(4*d), x] + (
Simp[(1*I*b*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, -(I*d*x^2)])/8, x] - Simp[(1*I*b*x^2*HypergeometricPFQ[{1,
1}, {3/2, 2}, I*d*x^2])/8, x]) /; FreeQ[{b, d}, x] && EqQ[d^2, (Pi^2*b^4)/4]

Rule 3379

Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x C(b x)^2 \, dx &=\frac{1}{2} x^2 C(b x)^2-b \int x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x) \, dx\\ &=\frac{1}{2} x^2 C(b x)^2-\frac{x C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b \pi }+\frac{\int x \sin \left (b^2 \pi x^2\right ) \, dx}{2 \pi }+\frac{\int C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b \pi }\\ &=\frac{1}{2} x^2 C(b x)^2+\frac{C(b x) S(b x)}{2 b^2 \pi }+\frac{i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi }-\frac{i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi }-\frac{x C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b \pi }+\frac{\operatorname{Subst}\left (\int \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 \pi }\\ &=-\frac{\cos \left (b^2 \pi x^2\right )}{4 b^2 \pi ^2}+\frac{1}{2} x^2 C(b x)^2+\frac{C(b x) S(b x)}{2 b^2 \pi }+\frac{i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi }-\frac{i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi }-\frac{x C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b \pi }\\ \end{align*}

Mathematica [F]  time = 0.174331, size = 0, normalized size = 0. \[ \int x \text{FresnelC}(b x)^2 \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[x*FresnelC[b*x]^2,x]

[Out]

Integrate[x*FresnelC[b*x]^2, x]

________________________________________________________________________________________

Maple [F]  time = 0.051, size = 0, normalized size = 0. \begin{align*} \int x \left ({\it FresnelC} \left ( bx \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*FresnelC(b*x)^2,x)

[Out]

int(x*FresnelC(b*x)^2,x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm fresnelc}\left (b x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnelc(b*x)^2,x, algorithm="maxima")

[Out]

integrate(x*fresnelc(b*x)^2, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x{\rm fresnelc}\left (b x\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnelc(b*x)^2,x, algorithm="fricas")

[Out]

integral(x*fresnelc(b*x)^2, x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x C^{2}\left (b x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnelc(b*x)**2,x)

[Out]

Integral(x*fresnelc(b*x)**2, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x{\rm fresnelc}\left (b x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*fresnelc(b*x)^2,x, algorithm="giac")

[Out]

integrate(x*fresnelc(b*x)^2, x)