Integrand size = 8, antiderivative size = 77 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{40 x^2}-\frac {\operatorname {FresnelS}(b x)}{5 x^5}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{20 x^4}-\frac {1}{80} b^5 \pi ^2 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right ) \]
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Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6561, 3460, 3378, 3380} \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=-\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{20 x^4}-\frac {1}{80} \pi ^2 b^5 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\pi b^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{40 x^2}-\frac {\operatorname {FresnelS}(b x)}{5 x^5} \]
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Rule 3378
Rule 3380
Rule 3460
Rule 6561
Rubi steps \begin{align*} \text {integral}& = -\frac {\operatorname {FresnelS}(b x)}{5 x^5}+\frac {1}{5} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5} \, dx \\ & = -\frac {\operatorname {FresnelS}(b x)}{5 x^5}+\frac {1}{10} b \text {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x^3} \, dx,x,x^2\right ) \\ & = -\frac {\operatorname {FresnelS}(b x)}{5 x^5}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{20 x^4}+\frac {1}{40} \left (b^3 \pi \right ) \text {Subst}\left (\int \frac {\cos \left (\frac {1}{2} b^2 \pi x\right )}{x^2} \, dx,x,x^2\right ) \\ & = -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{40 x^2}-\frac {\operatorname {FresnelS}(b x)}{5 x^5}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{20 x^4}-\frac {1}{80} \left (b^5 \pi ^2\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x} \, dx,x,x^2\right ) \\ & = -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{40 x^2}-\frac {\operatorname {FresnelS}(b x)}{5 x^5}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{20 x^4}-\frac {1}{80} b^5 \pi ^2 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right ) \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{40 x^2}-\frac {\operatorname {FresnelS}(b x)}{5 x^5}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{20 x^4}-\frac {1}{80} b^5 \pi ^2 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right ) \]
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Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.47 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.38
method | result | size |
meijerg | \(-\frac {\pi \,b^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {1}{2}, \frac {3}{2}, \frac {7}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{12 x^{2}}\) | \(29\) |
parts | \(-\frac {\operatorname {FresnelS}\left (b x \right )}{5 x^{5}}+\frac {b \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 x^{4}}+\frac {b^{2} \pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 x^{2}}-\frac {b^{2} \pi \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{5}\) | \(68\) |
derivativedivides | \(b^{5} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{5 b^{5} x^{5}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{20 b^{4} x^{4}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}-\frac {\pi \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{20}\right )\) | \(71\) |
default | \(b^{5} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{5 b^{5} x^{5}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{20 b^{4} x^{4}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}-\frac {\pi \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{20}\right )\) | \(71\) |
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none
Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=-\frac {\pi ^{2} b^{5} x^{5} \operatorname {Si}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 2 \, \pi b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 4 \, b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 16 \, \operatorname {S}\left (b x\right )}{80 \, x^{5}} \]
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Time = 0.68 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.60 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=- \frac {\pi b^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {1}{2}, \frac {3}{2}, \frac {7}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 x^{2} \Gamma \left (\frac {7}{4}\right )} \]
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Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.62 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=-\frac {1}{80} \, {\left (-i \, \pi ^{2} \Gamma \left (-2, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + i \, \pi ^{2} \Gamma \left (-2, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{5} - \frac {\operatorname {S}\left (b x\right )}{5 \, x^{5}} \]
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\[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x^{6}} \,d x } \]
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Timed out. \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^6} \,d x \]
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