Optimal. Leaf size=170 \[ -\frac{1}{20} \pi ^2 b^5 \text{Unintegrable}\left (\frac{S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{x},x\right )+\frac{7 \pi ^2 b^5 \text{FresnelC}\left (\sqrt{2} b x\right )}{60 \sqrt{2}}-\frac{b S(b x) \sin \left (\frac{1}{2} \pi b^2 x^2\right )}{10 x^4}-\frac{\pi b^3 S(b x) \cos \left (\frac{1}{2} \pi b^2 x^2\right )}{20 x^2}-\frac{b^2}{60 x^3}-\frac{7 \pi b^4 \sin \left (\pi b^2 x^2\right )}{120 x}+\frac{b^2 \cos \left (\pi b^2 x^2\right )}{60 x^3}-\frac{S(b x)^2}{5 x^5} \]
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Rubi [A] time = 0.153846, 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 = {} \[ \int \frac{S(b x)^2}{x^6} \, dx \]
Verification is Not applicable to the result.
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Rubi steps
\begin{align*} \int \frac{S(b x)^2}{x^6} \, dx &=-\frac{S(b x)^2}{5 x^5}+\frac{1}{5} (2 b) \int \frac{S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{x^5} \, dx\\ &=-\frac{b^2}{60 x^3}-\frac{S(b x)^2}{5 x^5}-\frac{b S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{10 x^4}-\frac{1}{20} b^2 \int \frac{\cos \left (b^2 \pi x^2\right )}{x^4} \, dx+\frac{1}{10} \left (b^3 \pi \right ) \int \frac{\cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{x^3} \, dx\\ &=-\frac{b^2}{60 x^3}+\frac{b^2 \cos \left (b^2 \pi x^2\right )}{60 x^3}-\frac{b^3 \pi \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{20 x^2}-\frac{S(b x)^2}{5 x^5}-\frac{b S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{10 x^4}+\frac{1}{40} \left (b^4 \pi \right ) \int \frac{\sin \left (b^2 \pi x^2\right )}{x^2} \, dx+\frac{1}{30} \left (b^4 \pi \right ) \int \frac{\sin \left (b^2 \pi x^2\right )}{x^2} \, dx-\frac{1}{20} \left (b^5 \pi ^2\right ) \int \frac{S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{x} \, dx\\ &=-\frac{b^2}{60 x^3}+\frac{b^2 \cos \left (b^2 \pi x^2\right )}{60 x^3}-\frac{b^3 \pi \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{20 x^2}-\frac{S(b x)^2}{5 x^5}-\frac{b S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{10 x^4}-\frac{7 b^4 \pi \sin \left (b^2 \pi x^2\right )}{120 x}-\frac{1}{20} \left (b^5 \pi ^2\right ) \int \frac{S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{x} \, dx+\frac{1}{20} \left (b^6 \pi ^2\right ) \int \cos \left (b^2 \pi x^2\right ) \, dx+\frac{1}{15} \left (b^6 \pi ^2\right ) \int \cos \left (b^2 \pi x^2\right ) \, dx\\ &=-\frac{b^2}{60 x^3}+\frac{b^2 \cos \left (b^2 \pi x^2\right )}{60 x^3}+\frac{7 b^5 \pi ^2 C\left (\sqrt{2} b x\right )}{60 \sqrt{2}}-\frac{b^3 \pi \cos \left (\frac{1}{2} b^2 \pi x^2\right ) S(b x)}{20 x^2}-\frac{S(b x)^2}{5 x^5}-\frac{b S(b x) \sin \left (\frac{1}{2} b^2 \pi x^2\right )}{10 x^4}-\frac{7 b^4 \pi \sin \left (b^2 \pi x^2\right )}{120 x}-\frac{1}{20} \left (b^5 \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 = 0.0267984, size = 0, normalized size = 0. \[ \int \frac{S(b x)^2}{x^6} \, dx \]
Verification is Not applicable to the result.
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Maple [A] time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({\it FresnelS} \left ( bx \right ) \right ) ^{2}}{{x}^{6}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm fresnels}\left (b x\right )^{2}}{x^{6}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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 )^{2}}{x^{6}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{S^{2}\left (b x\right )}{x^{6}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm fresnels}\left (b x\right )^{2}}{x^{6}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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