Integrand size = 8, antiderivative size = 73 \[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\frac {1}{2} i b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};-\frac {1}{2} i b^2 \pi x^2\right )-\frac {1}{2} i b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};\frac {1}{2} i b^2 \pi x^2\right ) \]
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Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6559, 6493, 6495} \[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\frac {1}{2} i b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};-\frac {1}{2} i b^2 \pi x^2\right )-\frac {1}{2} i b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};\frac {1}{2} i b^2 \pi x^2\right ) \]
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Rule 6493
Rule 6495
Rule 6559
Rubi steps \begin{align*} \text {integral}& = \left (-\frac {1}{4}-\frac {i}{4}\right ) \int \frac {\text {erfi}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {\pi } x\right )}{x} \, dx+\left (\frac {1}{4}+\frac {i}{4}\right ) \int \frac {\text {erf}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) b \sqrt {\pi } x\right )}{x} \, dx \\ & = \frac {1}{2} i b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};-\frac {1}{2} i b^2 \pi x^2\right )-\frac {1}{2} i b x \, _2F_2\left (\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};\frac {1}{2} i b^2 \pi x^2\right ) \\ \end{align*}
\[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\int \frac {\operatorname {FresnelS}(b x)}{x} \, dx \]
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Time = 0.55 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.40
method | result | size |
meijerg | \(\frac {\pi \,x^{3} b^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {3}{2}, \frac {7}{4}, \frac {7}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{18}\) | \(29\) |
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\[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x} \,d x } \]
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Time = 0.37 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.63 \[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\frac {\pi b^{3} x^{3} \Gamma ^{2}\left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {3}{2}, \frac {7}{4}, \frac {7}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{32 \Gamma ^{2}\left (\frac {7}{4}\right )} \]
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\[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x} \,d x } \]
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\[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x} \,d x } \]
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Timed out. \[ \int \frac {\operatorname {FresnelS}(b x)}{x} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x} \,d x \]
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