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

Optimal. Leaf size=152 \[ -\frac{1}{8} \pi ^2 b^4 \text{Unintegrable}\left (\frac{S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{x},x\right )+\frac{7 \pi ^2 b^4 \text{FresnelC}\left (\sqrt{2} b x\right )}{24 \sqrt{2}}-\frac{S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{4 x^4}-\frac{\pi b^2 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{8 x^2}-\frac{7 \pi b^3 \sin \left (\pi b^2 x^2\right )}{48 x}+\frac{b \cos \left (\pi b^2 x^2\right )}{24 x^3}-\frac{b}{24 x^3} \]

[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*Unintegrable[(FresnelS[b*x]*Sin[(b^2*Pi*x^2)/2])/x, x])/8

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Rubi [A]  time = 0.121331, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., 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*}

Mathematica [A]  time = 0.0303067, size = 0, normalized size = 0. \[ \int \frac{S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{x^5} \, dx \]

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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Maple [A]  time = 0.062, size = 0, normalized size = 0. \begin{align*} \int{\frac{{\it FresnelS} \left ( bx \right ) }{{x}^{5}}\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(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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm fresnels}\left (b x\right ) \sin \left (\frac{1}{2} \, \pi b^{2} x^{2}\right )}{x^{5}}\,{d x} \end{align*}

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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Fricas [A]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin{\left (\frac{\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{x^{5}}\, dx \end{align*}

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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Giac [A]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm fresnels}\left (b x\right ) \sin \left (\frac{1}{2} \, \pi b^{2} x^{2}\right )}{x^{5}}\,{d x} \end{align*}

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)