Integrand size = 8, antiderivative size = 119 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^9} \, dx=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{280 x^5}+\frac {b^7 \pi ^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x}+\frac {1}{840} b^8 \pi ^4 \operatorname {FresnelS}(b x)-\frac {\operatorname {FresnelS}(b x)}{8 x^8}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^3} \]
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Time = 0.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6561, 3468, 3469, 3432} \[ \int \frac {\operatorname {FresnelS}(b x)}{x^9} \, dx=\frac {1}{840} \pi ^4 b^8 \operatorname {FresnelS}(b x)-\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{56 x^7}+\frac {\pi ^3 b^7 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{840 x}+\frac {\pi ^2 b^5 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{840 x^3}-\frac {\pi b^3 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{280 x^5}-\frac {\operatorname {FresnelS}(b x)}{8 x^8} \]
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Rule 3432
Rule 3468
Rule 3469
Rule 6561
Rubi steps \begin{align*} \text {integral}& = -\frac {\operatorname {FresnelS}(b x)}{8 x^8}+\frac {1}{8} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^8} \, dx \\ & = -\frac {\operatorname {FresnelS}(b x)}{8 x^8}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {1}{56} \left (b^3 \pi \right ) \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6} \, dx \\ & = -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{280 x^5}-\frac {\operatorname {FresnelS}(b x)}{8 x^8}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}-\frac {1}{280} \left (b^5 \pi ^2\right ) \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4} \, dx \\ & = -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{280 x^5}-\frac {\operatorname {FresnelS}(b x)}{8 x^8}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^3}-\frac {1}{840} \left (b^7 \pi ^3\right ) \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2} \, dx \\ & = -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{280 x^5}+\frac {b^7 \pi ^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x}-\frac {\operatorname {FresnelS}(b x)}{8 x^8}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^3}+\frac {1}{840} \left (b^9 \pi ^4\right ) \int \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \\ & = -\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{280 x^5}+\frac {b^7 \pi ^3 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x}+\frac {1}{840} b^8 \pi ^4 \operatorname {FresnelS}(b x)-\frac {\operatorname {FresnelS}(b x)}{8 x^8}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{56 x^7}+\frac {b^5 \pi ^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^3} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.71 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^9} \, dx=\frac {b^3 \pi x^3 \left (-3+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )+\left (-105+b^8 \pi ^4 x^8\right ) \operatorname {FresnelS}(b x)+b x \left (-15+b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{840 x^8} \]
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Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.51 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.24
method | result | size |
meijerg | \(-\frac {\pi \,b^{3} \operatorname {hypergeom}\left (\left [-\frac {5}{4}, \frac {3}{4}\right ], \left [-\frac {1}{4}, \frac {3}{2}, \frac {7}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{30 x^{5}}\) | \(29\) |
derivativedivides | \(b^{8} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{8 b^{8} x^{8}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{56 b^{7} x^{7}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 b^{5} x^{5}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 b^{3} x^{3}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}-\pi \,\operatorname {FresnelS}\left (b x \right )\right )}{3}\right )}{5}\right )}{56}\right )\) | \(109\) |
default | \(b^{8} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{8 b^{8} x^{8}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{56 b^{7} x^{7}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 b^{5} x^{5}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 b^{3} x^{3}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}-\pi \,\operatorname {FresnelS}\left (b x \right )\right )}{3}\right )}{5}\right )}{56}\right )\) | \(109\) |
parts | \(-\frac {\operatorname {FresnelS}\left (b x \right )}{8 x^{8}}+\frac {b \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{7 x^{7}}+\frac {b^{2} \pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 x^{5}}-\frac {b^{2} \pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 x^{3}}+\frac {b^{2} \pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{x}-\frac {b^{2} \pi ^{\frac {3}{2}} \operatorname {FresnelS}\left (\frac {\sqrt {\pi }\, b^{2} x}{\sqrt {b^{2} \pi }}\right )}{\sqrt {b^{2} \pi }}\right )}{3}\right )}{5}\right )}{7}\right )}{8}\) | \(127\) |
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none
Time = 0.26 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.67 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^9} \, dx=\frac {{\left (\pi ^{3} b^{7} x^{7} - 3 \, \pi b^{3} x^{3}\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{4} b^{8} x^{8} - 105\right )} \operatorname {S}\left (b x\right ) + {\left (\pi ^{2} b^{5} x^{5} - 15 \, b x\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{840 \, x^{8}} \]
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Time = 1.72 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.55 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^9} \, dx=\frac {\pi ^{4} b^{8} S\left (b x\right ) \Gamma \left (- \frac {5}{4}\right )}{3584 \Gamma \left (\frac {7}{4}\right )} + \frac {\pi ^{3} b^{7} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {5}{4}\right )}{3584 x \Gamma \left (\frac {7}{4}\right )} + \frac {\pi ^{2} b^{5} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {5}{4}\right )}{3584 x^{3} \Gamma \left (\frac {7}{4}\right )} - \frac {3 \pi b^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {5}{4}\right )}{3584 x^{5} \Gamma \left (\frac {7}{4}\right )} - \frac {15 b \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {5}{4}\right )}{3584 x^{7} \Gamma \left (\frac {7}{4}\right )} - \frac {15 S\left (b x\right ) \Gamma \left (- \frac {5}{4}\right )}{512 x^{8} \Gamma \left (\frac {7}{4}\right )} \]
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Result contains complex when optimal does not.
Time = 0.34 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.51 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^9} \, dx=-\frac {\sqrt {\frac {1}{2}} \left (\pi x^{2}\right )^{\frac {7}{2}} {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {7}{2}, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {7}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{8}}{512 \, x^{7}} - \frac {\operatorname {S}\left (b x\right )}{8 \, x^{8}} \]
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\[ \int \frac {\operatorname {FresnelS}(b x)}{x^9} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x^{9}} \,d x } \]
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Timed out. \[ \int \frac {\operatorname {FresnelS}(b x)}{x^9} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^9} \,d x \]
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