Integrand size = 4, antiderivative size = 26 \[ \int \operatorname {FresnelS}(b x) \, dx=\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{b \pi }+x \operatorname {FresnelS}(b x) \]
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Time = 0.00 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6553} \[ \int \operatorname {FresnelS}(b x) \, dx=\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b}+x \operatorname {FresnelS}(b x) \]
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Rule 6553
Rubi steps \begin{align*} \text {integral}& = \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{b \pi }+x \operatorname {FresnelS}(b x) \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \operatorname {FresnelS}(b x) \, dx=\frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{b \pi }+x \operatorname {FresnelS}(b x) \]
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Time = 0.52 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
parts | \(\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b \pi }+x \,\operatorname {FresnelS}\left (b x \right )\) | \(25\) |
derivativedivides | \(\frac {\operatorname {FresnelS}\left (b x \right ) b x +\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }}{b}\) | \(27\) |
default | \(\frac {\operatorname {FresnelS}\left (b x \right ) b x +\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }}{b}\) | \(27\) |
meijerg | \(\frac {b^{3} \pi \,x^{4} \operatorname {hypergeom}\left (\left [\frac {3}{4}, 1\right ], \left [\frac {3}{2}, \frac {7}{4}, 2\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{24}\) | \(29\) |
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Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \operatorname {FresnelS}(b x) \, dx=\frac {\pi b x \operatorname {S}\left (b x\right ) + \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{\pi b} \]
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Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
Time = 0.57 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \operatorname {FresnelS}(b x) \, dx=\frac {3 x S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {3 \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{4 \pi b \Gamma \left (\frac {7}{4}\right )} \]
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Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \operatorname {FresnelS}(b x) \, dx=\frac {b x \operatorname {S}\left (b x\right ) + \frac {\cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{\pi }}{b} \]
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\[ \int \operatorname {FresnelS}(b x) \, dx=\int { \operatorname {S}\left (b x\right ) \,d x } \]
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Timed out. \[ \int \operatorname {FresnelS}(b x) \, dx=\int \mathrm {FresnelS}\left (b\,x\right ) \,d x \]
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