3.3.14 \(\int \frac {\text {FresnelC}(b x) \sin (\frac {1}{2} b^2 \pi x^2)}{x^6} \, dx\) [214]

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

[Out]

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

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Rubi [A]
time = 0.15, 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 = {6600, 6592, 6576, 30, 3456, 3461, 3378, 3380, 3460} \begin {gather*} -\frac {1}{30} \pi ^3 b^5 \text {FresnelC}(b x)^2-\frac {\pi b^3}{60 x^2}-\frac {\text {FresnelC}(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 x^5}-\frac {\pi b^2 \text {FresnelC}(b x) \cos \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 \text {FresnelC}(b x) \sin \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/x^6,x]

[Out]

-1/60*(b^3*Pi)/x^2 - (b^3*Pi*Cos[b^2*Pi*x^2])/(24*x^2) - (b^2*Pi*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(15*x^3) -
(b^5*Pi^3*FresnelC[b*x]^2)/30 - (FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(5*x^5) + (b^4*Pi^2*FresnelC[b*x]*Sin[(b^
2*Pi*x^2)/2])/(15*x) - (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 3378

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 3380

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 3456

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

Rule 3460

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 3461

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 6576

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

Rule 6592

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

Rule 6600

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

Rubi steps

\begin {align*} \int \frac {C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx &=-\frac {C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{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 {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{x^4} \, dx\\ &=-\frac {b^3 \pi }{60 x^2}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{15 x^3}-\frac {C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 x^5}+\frac {1}{20} b \text {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 {C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2} \, dx\\ &=-\frac {b^3 \pi }{60 x^2}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{15 x^3}-\frac {C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 x^5}+\frac {b^4 \pi ^2 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{15 x}-\frac {b \sin \left (b^2 \pi x^2\right )}{40 x^4}+\frac {1}{60} \left (b^3 \pi \right ) \text {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 ) \text {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 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x) \, dx\\ &=-\frac {b^3 \pi }{60 x^2}-\frac {b^3 \pi \cos \left (b^2 \pi x^2\right )}{24 x^2}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{15 x^3}-\frac {C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 x^5}+\frac {b^4 \pi ^2 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{15 x}-\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 ) \text {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 ) \text {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 ) \text {Subst}(\int x \, dx,x,C(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 {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) C(b x)}{15 x^3}-\frac {1}{30} b^5 \pi ^3 C(b x)^2-\frac {C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 x^5}+\frac {b^4 \pi ^2 C(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{15 x}-\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 \begin {gather*} -\frac {b^3 \pi }{60 x^2}-\frac {b^3 \pi \cos \left (b^2 \pi x^2\right )}{24 x^2}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \text {FresnelC}(b x)}{15 x^3}-\frac {1}{30} b^5 \pi ^3 \text {FresnelC}(b x)^2-\frac {\text {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 x^5}+\frac {b^4 \pi ^2 \text {FresnelC}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{15 x}-\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 {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/x^6,x]

[Out]

-1/60*(b^3*Pi)/x^2 - (b^3*Pi*Cos[b^2*Pi*x^2])/(24*x^2) - (b^2*Pi*Cos[(b^2*Pi*x^2)/2]*FresnelC[b*x])/(15*x^3) -
(b^5*Pi^3*FresnelC[b*x]^2)/30 - (FresnelC[b*x]*Sin[(b^2*Pi*x^2)/2])/(5*x^5) + (b^4*Pi^2*FresnelC[b*x]*Sin[(b^
2*Pi*x^2)/2])/(15*x) - (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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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\FresnelC \left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{x^{6}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelC(b*x)*sin(1/2*b^2*Pi*x^2)/x^6,x)

[Out]

int(FresnelC(b*x)*sin(1/2*b^2*Pi*x^2)/x^6,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnel_cos(b*x)*sin(1/2*pi*b^2*x^2)/x^6, x)

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Fricas [A]
time = 0.34, size = 141, normalized size = 0.87 \begin {gather*} -\frac {4 \, \pi ^{3} b^{5} x^{5} \operatorname {C}\left (b x\right )^{2} + 7 \, \pi ^{2} b^{5} x^{5} \operatorname {Si}\left (\pi b^{2} x^{2}\right ) + 10 \, \pi b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )^{2} - 3 \, \pi b^{3} x^{3} + 8 \, \pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) \operatorname {C}\left (b x\right ) + 2 \, {\left (3 \, b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 4 \, {\left (\pi ^{2} b^{4} x^{4} - 3\right )} \operatorname {C}\left (b x\right )\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{120 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(4*pi^3*b^5*x^5*fresnel_cos(b*x)^2 + 7*pi^2*b^5*x^5*sin_integral(pi*b^2*x^2) + 10*pi*b^3*x^3*cos(1/2*pi
*b^2*x^2)^2 - 3*pi*b^3*x^3 + 8*pi*b^2*x^2*cos(1/2*pi*b^2*x^2)*fresnel_cos(b*x) + 2*(3*b*x*cos(1/2*pi*b^2*x^2)
- 4*(pi^2*b^4*x^4 - 3)*fresnel_cos(b*x))*sin(1/2*pi*b^2*x^2))/x^5

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} C\left (b x\right )}{x^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnelc(b*x)*sin(1/2*b**2*pi*x**2)/x**6,x)

[Out]

Integral(sin(pi*b**2*x**2/2)*fresnelc(b*x)/x**6, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnel_cos(b*x)*sin(1/2*pi*b^2*x^2)/x^6, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {FresnelC}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right )}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((FresnelC(b*x)*sin((Pi*b^2*x^2)/2))/x^6,x)

[Out]

int((FresnelC(b*x)*sin((Pi*b^2*x^2)/2))/x^6, x)

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