3.84 \(\int \frac {S(b x) \sin (\frac {1}{2} b^2 \pi x^2)}{x^5} \, dx\)

Optimal. Leaf size=153 \[ -\frac {1}{8} \pi ^2 b^4 \text {Int}\left (\frac {S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{x},x\right )+\frac {7 \pi ^2 b^4 C\left (\sqrt {2} b x\right )}{24 \sqrt {2}}-\frac {\pi b^2 S(b x) \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{8 x^2}-\frac {S(b x) \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{4 x^4}+\frac {b \cos \left (\pi b^2 x^2\right )}{24 x^3}-\frac {7 \pi b^3 \sin \left (\pi b^2 x^2\right )}{48 x}-\frac {b}{24 x^3} \]

[Out]

-1/24*b/x^3+1/24*b*cos(b^2*Pi*x^2)/x^3-1/8*b^2*Pi*cos(1/2*b^2*Pi*x^2)*FresnelS(b*x)/x^2-1/4*FresnelS(b*x)*sin(
1/2*b^2*Pi*x^2)/x^4-7/48*b^3*Pi*sin(b^2*Pi*x^2)/x+7/48*b^4*Pi^2*FresnelC(b*x*2^(1/2))*2^(1/2)-1/8*b^4*Pi^2*Uni
ntegrable(FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/x,x)

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Rubi [A]  time = 0.12, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/x^5,x]

[Out]

-b/(24*x^3) + (b*Cos[b^2*Pi*x^2])/(24*x^3) + (7*b^4*Pi^2*FresnelC[Sqrt[2]*b*x])/(24*Sqrt[2]) - (b^2*Pi*Cos[(b^
2*Pi*x^2)/2]*FresnelS[b*x])/(8*x^2) - (FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/(4*x^4) - (7*b^3*Pi*Sin[b^2*Pi*x^2])
/(48*x) - (b^4*Pi^2*Defer[Int][(FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/x, x])/8

Rubi steps

\begin {align*} \int \frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx &=-\frac {b}{24 x^3}-\frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 x^4}-\frac {1}{8} b \int \frac {\cos \left (b^2 \pi x^2\right )}{x^4} \, dx+\frac {1}{4} \left (b^2 \pi \right ) \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{x^3} \, dx\\ &=-\frac {b}{24 x^3}+\frac {b \cos \left (b^2 \pi x^2\right )}{24 x^3}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{8 x^2}-\frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 x^4}+\frac {1}{16} \left (b^3 \pi \right ) \int \frac {\sin \left (b^2 \pi x^2\right )}{x^2} \, dx+\frac {1}{12} \left (b^3 \pi \right ) \int \frac {\sin \left (b^2 \pi x^2\right )}{x^2} \, dx-\frac {1}{8} \left (b^4 \pi ^2\right ) \int \frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x} \, dx\\ &=-\frac {b}{24 x^3}+\frac {b \cos \left (b^2 \pi x^2\right )}{24 x^3}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{8 x^2}-\frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 x^4}-\frac {7 b^3 \pi \sin \left (b^2 \pi x^2\right )}{48 x}-\frac {1}{8} \left (b^4 \pi ^2\right ) \int \frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x} \, dx+\frac {1}{8} \left (b^5 \pi ^2\right ) \int \cos \left (b^2 \pi x^2\right ) \, dx+\frac {1}{6} \left (b^5 \pi ^2\right ) \int \cos \left (b^2 \pi x^2\right ) \, dx\\ &=-\frac {b}{24 x^3}+\frac {b \cos \left (b^2 \pi x^2\right )}{24 x^3}+\frac {7 b^4 \pi ^2 C\left (\sqrt {2} b x\right )}{24 \sqrt {2}}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) S(b x)}{8 x^2}-\frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 x^4}-\frac {7 b^3 \pi \sin \left (b^2 \pi x^2\right )}{48 x}-\frac {1}{8} \left (b^4 \pi ^2\right ) \int \frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x} \, dx\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {S(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/x^5,x]

[Out]

Integrate[(FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/x^5, x]

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fricas [A]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm fresnels}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)*sin(1/2*b^2*pi*x^2)/x^5,x, algorithm="fricas")

[Out]

integral(fresnels(b*x)*sin(1/2*pi*b^2*x^2)/x^5, x)

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giac [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnels}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)*sin(1/2*b^2*pi*x^2)/x^5,x, algorithm="giac")

[Out]

integrate(fresnels(b*x)*sin(1/2*pi*b^2*x^2)/x^5, x)

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maple [A]  time = 0.02, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {S}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/x^5,x)

[Out]

int(FresnelS(b*x)*sin(1/2*b^2*Pi*x^2)/x^5,x)

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maxima [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnels}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)*sin(1/2*b^2*pi*x^2)/x^5,x, algorithm="maxima")

[Out]

integrate(fresnels(b*x)*sin(1/2*pi*b^2*x^2)/x^5, x)

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mupad [A]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {FresnelS}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((FresnelS(b*x)*sin((Pi*b^2*x^2)/2))/x^5,x)

[Out]

int((FresnelS(b*x)*sin((Pi*b^2*x^2)/2))/x^5, x)

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sympy [A]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(fresnels(b*x)*sin(1/2*b**2*pi*x**2)/x**5,x)

[Out]

Integral(sin(pi*b**2*x**2/2)*fresnels(b*x)/x**5, x)

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