3.105 \(\int \frac {\cos (\frac {1}{2} b^2 \pi x^2) S(b x)}{x^6} \, dx\)

Optimal. Leaf size=163 \[ \frac {1}{30} \pi ^3 b^5 S(b x)^2+\frac {\pi b^3}{60 x^2}-\frac {S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}+\frac {\pi b^2 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{15 x^3}-\frac {b \sin \left (\pi b^2 x^2\right )}{40 x^4}-\frac {7}{120} \pi ^2 b^5 \text {Si}\left (b^2 \pi x^2\right )+\frac {\pi ^2 b^4 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{15 x}-\frac {\pi b^3 \cos \left (\pi b^2 x^2\right )}{24 x^2} \]

[Out]

1/60*b^3*Pi/x^2-1/24*b^3*Pi*cos(b^2*Pi*x^2)/x^2-1/5*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/x^5+1/15*b^4*Pi^2*cos(1/
2*b^2*Pi*x^2)*FresnelS(b*x)/x+1/30*b^5*Pi^3*FresnelS(b*x)^2-7/120*b^5*Pi^2*Si(b^2*Pi*x^2)+1/15*b^2*Pi*FresnelS
(b*x)*sin(1/2*b^2*Pi*x^2)/x^3-1/40*b*sin(b^2*Pi*x^2)/x^4

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Rubi [A]  time = 0.21, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {6464, 6456, 6440, 30, 3375, 3380, 3297, 3299, 3379} \[ \frac {\pi b^2 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{15 x^3}+\frac {\pi ^2 b^4 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{15 x}-\frac {S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}+\frac {1}{30} \pi ^3 b^5 S(b x)^2-\frac {7}{120} \pi ^2 b^5 \text {Si}\left (b^2 \pi x^2\right )+\frac {\pi b^3}{60 x^2}-\frac {b \sin \left (\pi b^2 x^2\right )}{40 x^4}-\frac {\pi b^3 \cos \left (\pi b^2 x^2\right )}{24 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/x^6,x]

[Out]

(b^3*Pi)/(60*x^2) - (b^3*Pi*Cos[b^2*Pi*x^2])/(24*x^2) - (Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(5*x^5) + (b^4*Pi^
2*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(15*x) + (b^5*Pi^3*FresnelS[b*x]^2)/30 + (b^2*Pi*FresnelS[b*x]*Sin[(b^2*P
i*x^2)/2])/(15*x^3) - (b*Sin[b^2*Pi*x^2])/(40*x^4) - (7*b^5*Pi^2*SinIntegral[b^2*Pi*x^2])/120

Rule 30

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

Rule 3297

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

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3375

Int[Sin[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[SinIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

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 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 6440

Int[FresnelS[(b_.)*(x_)]^(n_.)*Sin[(d_.)*(x_)^2], x_Symbol] :> Dist[(Pi*b)/(2*d), Subst[Int[x^n, x], x, Fresne
lS[b*x]], x] /; FreeQ[{b, d, n}, x] && EqQ[d^2, (Pi^2*b^4)/4]

Rule 6456

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

Rule 6464

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

Rubi steps

\begin {align*} \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{x^6} \, dx &=-\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{5 x^5}+\frac {1}{10} b \int \frac {\sin \left (b^2 \pi x^2\right )}{x^5} \, dx-\frac {1}{5} \left (b^2 \pi \right ) \int \frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4} \, dx\\ &=\frac {b^3 \pi }{60 x^2}-\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{5 x^5}+\frac {b^2 \pi S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{15 x^3}+\frac {1}{20} b \operatorname {Subst}\left (\int \frac {\sin \left (b^2 \pi x\right )}{x^3} \, dx,x,x^2\right )+\frac {1}{30} \left (b^3 \pi \right ) \int \frac {\cos \left (b^2 \pi x^2\right )}{x^3} \, dx-\frac {1}{15} \left (b^4 \pi ^2\right ) \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{x^2} \, dx\\ &=\frac {b^3 \pi }{60 x^2}-\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{5 x^5}+\frac {b^4 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{15 x}+\frac {b^2 \pi S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{15 x^3}-\frac {b \sin \left (b^2 \pi x^2\right )}{40 x^4}+\frac {1}{60} \left (b^3 \pi \right ) \operatorname {Subst}\left (\int \frac {\cos \left (b^2 \pi x\right )}{x^2} \, dx,x,x^2\right )+\frac {1}{40} \left (b^3 \pi \right ) \operatorname {Subst}\left (\int \frac {\cos \left (b^2 \pi x\right )}{x^2} \, dx,x,x^2\right )-\frac {1}{30} \left (b^5 \pi ^2\right ) \int \frac {\sin \left (b^2 \pi x^2\right )}{x} \, dx+\frac {1}{15} \left (b^6 \pi ^3\right ) \int S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx\\ &=\frac {b^3 \pi }{60 x^2}-\frac {b^3 \pi \cos \left (b^2 \pi x^2\right )}{24 x^2}-\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{5 x^5}+\frac {b^4 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{15 x}+\frac {b^2 \pi S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{15 x^3}-\frac {b \sin \left (b^2 \pi x^2\right )}{40 x^4}-\frac {1}{60} b^5 \pi ^2 \text {Si}\left (b^2 \pi x^2\right )-\frac {1}{60} \left (b^5 \pi ^2\right ) \operatorname {Subst}\left (\int \frac {\sin \left (b^2 \pi x\right )}{x} \, dx,x,x^2\right )-\frac {1}{40} \left (b^5 \pi ^2\right ) \operatorname {Subst}\left (\int \frac {\sin \left (b^2 \pi x\right )}{x} \, dx,x,x^2\right )+\frac {1}{15} \left (b^5 \pi ^3\right ) \operatorname {Subst}(\int x \, dx,x,S(b x))\\ &=\frac {b^3 \pi }{60 x^2}-\frac {b^3 \pi \cos \left (b^2 \pi x^2\right )}{24 x^2}-\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{5 x^5}+\frac {b^4 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{15 x}+\frac {1}{30} b^5 \pi ^3 S(b x)^2+\frac {b^2 \pi S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{15 x^3}-\frac {b \sin \left (b^2 \pi x^2\right )}{40 x^4}-\frac {7}{120} b^5 \pi ^2 \text {Si}\left (b^2 \pi x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 163, normalized size = 1.00 \[ \frac {1}{30} \pi ^3 b^5 S(b x)^2+\frac {\pi b^3}{60 x^2}-\frac {S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}+\frac {\pi b^2 S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{15 x^3}-\frac {b \sin \left (\pi b^2 x^2\right )}{40 x^4}-\frac {7}{120} \pi ^2 b^5 \text {Si}\left (b^2 \pi x^2\right )+\frac {\pi ^2 b^4 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{15 x}-\frac {\pi b^3 \cos \left (\pi b^2 x^2\right )}{24 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/x^6,x]

[Out]

(b^3*Pi)/(60*x^2) - (b^3*Pi*Cos[b^2*Pi*x^2])/(24*x^2) - (Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(5*x^5) + (b^4*Pi^
2*Cos[(b^2*Pi*x^2)/2]*FresnelS[b*x])/(15*x) + (b^5*Pi^3*FresnelS[b*x]^2)/30 + (b^2*Pi*FresnelS[b*x]*Sin[(b^2*P
i*x^2)/2])/(15*x^3) - (b*Sin[b^2*Pi*x^2])/(40*x^4) - (7*b^5*Pi^2*SinIntegral[b^2*Pi*x^2])/120

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) {\rm fresnels}\left (b x\right )}{x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/2*b^2*pi*x^2)*fresnels(b*x)/x^6,x, algorithm="fricas")

[Out]

integral(cos(1/2*pi*b^2*x^2)*fresnels(b*x)/x^6, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) {\rm fresnels}\left (b x\right )}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/2*b^2*pi*x^2)*fresnels(b*x)/x^6,x, algorithm="giac")

[Out]

integrate(cos(1/2*pi*b^2*x^2)*fresnels(b*x)/x^6, x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right ) \mathrm {S}\left (b x \right )}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/x^6,x)

[Out]

int(cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/x^6,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) {\rm fresnels}\left (b x\right )}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/2*b^2*pi*x^2)*fresnels(b*x)/x^6,x, algorithm="maxima")

[Out]

integrate(cos(1/2*pi*b^2*x^2)*fresnels(b*x)/x^6, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {FresnelS}\left (b\,x\right )\,\cos \left (\frac {\Pi \,b^2\,x^2}{2}\right )}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((FresnelS(b*x)*cos((Pi*b^2*x^2)/2))/x^6,x)

[Out]

int((FresnelS(b*x)*cos((Pi*b^2*x^2)/2))/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(1/2*b**2*pi*x**2)*fresnels(b*x)/x**6,x)

[Out]

Integral(cos(pi*b**2*x**2/2)*fresnels(b*x)/x**6, x)

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