Integrand size = 20, antiderivative size = 20 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=-\frac {b}{40 x^4}+\frac {b \cos \left (b^2 \pi x^2\right )}{40 x^4}+\frac {1}{24} b^5 \pi ^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{15 x^3}-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 x^5}-\frac {b^3 \pi \sin \left (b^2 \pi x^2\right )}{24 x^2}-\frac {1}{15} b^4 \pi ^2 \text {Int}\left (\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2},x\right ) \]
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Not integrable
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 0, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = -\frac {b}{40 x^4}-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 x^5}-\frac {1}{10} b \int \frac {\cos \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 ) \operatorname {FresnelS}(b x)}{x^4} \, dx \\ & = -\frac {b}{40 x^4}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{15 x^3}-\frac {\operatorname {FresnelS}(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 {\cos \left (b^2 \pi x\right )}{x^3} \, dx,x,x^2\right )+\frac {1}{30} \left (b^3 \pi \right ) \int \frac {\sin \left (b^2 \pi x^2\right )}{x^3} \, dx-\frac {1}{15} \left (b^4 \pi ^2\right ) \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2} \, dx \\ & = -\frac {b}{40 x^4}+\frac {b \cos \left (b^2 \pi x^2\right )}{40 x^4}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{15 x^3}-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 x^5}+\frac {1}{60} \left (b^3 \pi \right ) \text {Subst}\left (\int \frac {\sin \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 {\sin \left (b^2 \pi x\right )}{x^2} \, dx,x,x^2\right )-\frac {1}{15} \left (b^4 \pi ^2\right ) \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2} \, dx \\ & = -\frac {b}{40 x^4}+\frac {b \cos \left (b^2 \pi x^2\right )}{40 x^4}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{15 x^3}-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 x^5}-\frac {b^3 \pi \sin \left (b^2 \pi x^2\right )}{24 x^2}-\frac {1}{15} \left (b^4 \pi ^2\right ) \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2} \, dx+\frac {1}{60} \left (b^5 \pi ^2\right ) \text {Subst}\left (\int \frac {\cos \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 {\cos \left (b^2 \pi x\right )}{x} \, dx,x,x^2\right ) \\ & = -\frac {b}{40 x^4}+\frac {b \cos \left (b^2 \pi x^2\right )}{40 x^4}+\frac {1}{24} b^5 \pi ^2 \operatorname {CosIntegral}\left (b^2 \pi x^2\right )-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{15 x^3}-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 x^5}-\frac {b^3 \pi \sin \left (b^2 \pi x^2\right )}{24 x^2}-\frac {1}{15} \left (b^4 \pi ^2\right ) \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2} \, dx \\ \end{align*}
Not integrable
Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx \]
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Not integrable
Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90
\[\int \frac {\operatorname {FresnelS}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{x^{6}}d x\]
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Not integrable
Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int { \frac {\operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{6}} \,d x } \]
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Not integrable
Time = 5.69 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int \frac {\sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{x^{6}}\, dx \]
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Not integrable
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int { \frac {\operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{6}} \,d x } \]
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Not integrable
Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int { \frac {\operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{6}} \,d x } \]
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Not integrable
Time = 4.83 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right )}{x^6} \,d x \]
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