Integrand size = 20, antiderivative size = 20 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=-\frac {b}{24 x^3}+\frac {b \cos \left (b^2 \pi x^2\right )}{24 x^3}+\frac {7 b^4 \pi ^2 \operatorname {FresnelC}\left (\sqrt {2} b x\right )}{24 \sqrt {2}}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{8 x^2}-\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{4 x^4}-\frac {7 b^3 \pi \sin \left (b^2 \pi x^2\right )}{48 x}-\frac {1}{8} b^4 \pi ^2 \text {Int}\left (\frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x},x\right ) \]
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Not integrable
Time = 0.09 (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 = {} \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = -\frac {b}{24 x^3}-\frac {\operatorname {FresnelS}(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 ) \operatorname {FresnelS}(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 ) \operatorname {FresnelS}(b x)}{8 x^2}-\frac {\operatorname {FresnelS}(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 {\operatorname {FresnelS}(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 ) \operatorname {FresnelS}(b x)}{8 x^2}-\frac {\operatorname {FresnelS}(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 {\operatorname {FresnelS}(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 \operatorname {FresnelC}\left (\sqrt {2} b x\right )}{24 \sqrt {2}}-\frac {b^2 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \operatorname {FresnelS}(b x)}{8 x^2}-\frac {\operatorname {FresnelS}(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 {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x} \, dx \\ \end{align*}
Not integrable
Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx \]
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Not integrable
Time = 0.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90
\[\int \frac {\operatorname {FresnelS}\left (b x \right ) \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{x^{5}}d x\]
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Not integrable
Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int { \frac {\operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{5}} \,d x } \]
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Not integrable
Time = 3.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int \frac {\sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} S\left (b x\right )}{x^{5}}\, dx \]
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Not integrable
Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int { \frac {\operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{5}} \,d x } \]
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Not integrable
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int { \frac {\operatorname {S}\left (b x\right ) \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{x^{5}} \,d x } \]
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Not integrable
Time = 4.82 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )\,\sin \left (\frac {\Pi \,b^2\,x^2}{2}\right )}{x^5} \,d x \]
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