3.204 \(\int x^4 \text{FresnelC}(b x) \sin (\frac{1}{2} b^2 \pi x^2) \, dx\)

Optimal. Leaf size=196 \[ -\frac{3 i x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},-\frac{1}{2} i \pi b^2 x^2\right )}{8 \pi ^2 b^3}+\frac{3 i x^2 \text{HypergeometricPFQ}\left (\{1,1\},\left \{\frac{3}{2},2\right \},\frac{1}{2} i \pi b^2 x^2\right )}{8 \pi ^2 b^3}-\frac{3 \text{FresnelC}(b x) S(b x)}{2 \pi ^2 b^5}+\frac{3 x \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac{x^3 \text{FresnelC}(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac{x^2 \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac{\cos \left (\pi b^2 x^2\right )}{\pi ^3 b^5}+\frac{x^4}{8 \pi b} \]

[Out]

x^4/(8*b*Pi) + Cos[b^2*Pi*x^2]/(b^5*Pi^3) - (x^3*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(b^2*Pi) - (3*FresnelC[b*x
]*FresnelS[b*x])/(2*b^5*Pi^2) - (((3*I)/8)*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (-I/2)*b^2*Pi*x^2])/(b^3*Pi
^2) + (((3*I)/8)*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (I/2)*b^2*Pi*x^2])/(b^3*Pi^2) + (3*x*FresnelC[b*x]*Si
n[(b^2*Pi*x^2)/2])/(b^4*Pi^2) + (x^2*Sin[b^2*Pi*x^2])/(4*b^3*Pi^2)

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Rubi [A]  time = 0.15681, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {6463, 6455, 6447, 3379, 2638, 3380, 3309, 30, 3296} \[ -\frac{3 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi ^2 b^3}+\frac{3 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 \pi ^2 b^3}-\frac{3 \text{FresnelC}(b x) S(b x)}{2 \pi ^2 b^5}+\frac{3 x \text{FresnelC}(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi ^2 b^4}-\frac{x^3 \text{FresnelC}(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{\pi b^2}+\frac{x^2 \sin \left (\pi b^2 x^2\right )}{4 \pi ^2 b^3}+\frac{\cos \left (\pi b^2 x^2\right )}{\pi ^3 b^5}+\frac{x^4}{8 \pi b} \]

Antiderivative was successfully verified.

[In]

Int[x^4*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2],x]

[Out]

x^4/(8*b*Pi) + Cos[b^2*Pi*x^2]/(b^5*Pi^3) - (x^3*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(b^2*Pi) - (3*FresnelC[b*x
]*FresnelS[b*x])/(2*b^5*Pi^2) - (((3*I)/8)*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (-I/2)*b^2*Pi*x^2])/(b^3*Pi
^2) + (((3*I)/8)*x^2*HypergeometricPFQ[{1, 1}, {3/2, 2}, (I/2)*b^2*Pi*x^2])/(b^3*Pi^2) + (3*x*FresnelC[b*x]*Si
n[(b^2*Pi*x^2)/2])/(b^4*Pi^2) + (x^2*Sin[b^2*Pi*x^2])/(4*b^3*Pi^2)

Rule 6463

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

Rule 3380

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[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 3309

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + ((f_.)*(x_))/2]^2, x_Symbol] :> Dist[1/2, Int[(c + d*x)^m, x], x] -
Dist[1/2, Int[(c + d*x)^m*Cos[2*e + f*x], x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rubi steps

\begin{align*} \int x^4 C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx &=-\frac{x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }+\frac{3 \int x^2 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x) \, dx}{b^2 \pi }+\frac{\int x^3 \cos ^2\left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b \pi }\\ &=-\frac{x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }+\frac{3 x C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac{3 \int C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx}{b^4 \pi ^2}-\frac{3 \int x \sin \left (b^2 \pi x^2\right ) \, dx}{2 b^3 \pi ^2}+\frac{\operatorname{Subst}\left (\int x \cos ^2\left (\frac{1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{2 b \pi }\\ &=-\frac{x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }-\frac{3 C(b x) S(b x)}{2 b^5 \pi ^2}-\frac{3 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 b^3 \pi ^2}+\frac{3 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 b^3 \pi ^2}+\frac{3 x C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}-\frac{3 \operatorname{Subst}\left (\int \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^3 \pi ^2}+\frac{\operatorname{Subst}\left (\int x \, dx,x,x^2\right )}{4 b \pi }+\frac{\operatorname{Subst}\left (\int x \cos \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b \pi }\\ &=\frac{x^4}{8 b \pi }+\frac{3 \cos \left (b^2 \pi x^2\right )}{4 b^5 \pi ^3}-\frac{x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }-\frac{3 C(b x) S(b x)}{2 b^5 \pi ^2}-\frac{3 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 b^3 \pi ^2}+\frac{3 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 b^3 \pi ^2}+\frac{3 x C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac{x^2 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}-\frac{\operatorname{Subst}\left (\int \sin \left (b^2 \pi x\right ) \, dx,x,x^2\right )}{4 b^3 \pi ^2}\\ &=\frac{x^4}{8 b \pi }+\frac{\cos \left (b^2 \pi x^2\right )}{b^5 \pi ^3}-\frac{x^3 \cos \left (\frac{1}{2} b^2 \pi x^2\right ) C(b x)}{b^2 \pi }-\frac{3 C(b x) S(b x)}{2 b^5 \pi ^2}-\frac{3 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;-\frac{1}{2} i b^2 \pi x^2\right )}{8 b^3 \pi ^2}+\frac{3 i x^2 \, _2F_2\left (1,1;\frac{3}{2},2;\frac{1}{2} i b^2 \pi x^2\right )}{8 b^3 \pi ^2}+\frac{3 x C(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{b^4 \pi ^2}+\frac{x^2 \sin \left (b^2 \pi x^2\right )}{4 b^3 \pi ^2}\\ \end{align*}

Mathematica [F]  time = 0.0371357, size = 0, normalized size = 0. \[ \int x^4 \text{FresnelC}(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right ) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[x^4*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2],x]

[Out]

Integrate[x^4*FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2], x]

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Maple [F]  time = 0.078, size = 0, normalized size = 0. \begin{align*} \int{x}^{4}{\it FresnelC} \left ( bx \right ) \sin \left ({\frac{{b}^{2}\pi \,{x}^{2}}{2}} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2),x)

[Out]

int(x^4*FresnelC(b*x)*sin(1/2*b^2*Pi*x^2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4}{\rm fresnelc}\left (b x\right ) \sin \left (\frac{1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4*fresnelc(b*x)*sin(1/2*pi*b^2*x^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{4}{\rm fresnelc}\left (b x\right ) \sin \left (\frac{1}{2} \, \pi b^{2} x^{2}\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^4*fresnelc(b*x)*sin(1/2*pi*b^2*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*fresnelc(b*x)*sin(1/2*b**2*pi*x**2),x)

[Out]

Integral(x**4*sin(pi*b**2*x**2/2)*fresnelc(b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4}{\rm fresnelc}\left (b x\right ) \sin \left (\frac{1}{2} \, \pi b^{2} x^{2}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4*fresnelc(b*x)*sin(1/2*pi*b^2*x^2), x)