Integrand size = 8, antiderivative size = 52 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\frac {1}{12} b^3 \pi \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\operatorname {FresnelS}(b x)}{3 x^3}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^2} \]
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Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6561, 3460, 3378, 3383} \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=-\frac {b \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{6 x^2}+\frac {1}{12} \pi b^3 \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\operatorname {FresnelS}(b x)}{3 x^3} \]
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Rule 3378
Rule 3383
Rule 3460
Rule 6561
Rubi steps \begin{align*} \text {integral}& = -\frac {\operatorname {FresnelS}(b x)}{3 x^3}+\frac {1}{3} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^3} \, dx \\ & = -\frac {\operatorname {FresnelS}(b x)}{3 x^3}+\frac {1}{6} b \text {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x^2} \, dx,x,x^2\right ) \\ & = -\frac {\operatorname {FresnelS}(b x)}{3 x^3}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^2}+\frac {1}{12} \left (b^3 \pi \right ) \text {Subst}\left (\int \frac {\cos \left (\frac {1}{2} b^2 \pi x\right )}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{12} b^3 \pi \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\operatorname {FresnelS}(b x)}{3 x^3}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^2} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\frac {1}{12} b^3 \pi \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\operatorname {FresnelS}(b x)}{3 x^3}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^2} \]
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Time = 0.61 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.88
method | result | size |
parts | \(-\frac {\operatorname {FresnelS}\left (b x \right )}{3 x^{3}}+\frac {b \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 x^{2}}+\frac {b^{2} \pi \,\operatorname {Ci}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{3}\) | \(46\) |
derivativedivides | \(b^{3} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{3 b^{3} x^{3}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{2} x^{2}}+\frac {\pi \,\operatorname {Ci}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12}\right )\) | \(49\) |
default | \(b^{3} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{3 b^{3} x^{3}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{2} x^{2}}+\frac {\pi \,\operatorname {Ci}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12}\right )\) | \(49\) |
meijerg | \(\frac {\pi ^{\frac {3}{2}} b^{3} \left (-\frac {\pi ^{\frac {3}{2}} x^{4} b^{4} \operatorname {hypergeom}\left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2, \frac {5}{2}, \frac {11}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{21}+\frac {\frac {16 \gamma }{3}-\frac {16 \ln \left (2\right )}{3}-\frac {80}{9}+\frac {32 \ln \left (x \right )}{3}+\frac {16 \ln \left (\pi \right )}{3}+\frac {32 \ln \left (b \right )}{3}}{\sqrt {\pi }}\right )}{64}\) | \(68\) |
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none
Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\frac {\pi b^{3} x^{3} \operatorname {Ci}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 4 \, \operatorname {S}\left (b x\right )}{12 \, x^{3}} \]
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Time = 0.78 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.08 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=- \frac {\pi ^{3} b^{7} x^{4} \Gamma \left (\frac {7}{4}\right ) {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {7}{4} \\ 2, 2, \frac {5}{2}, \frac {11}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{768 \Gamma \left (\frac {11}{4}\right )} + \frac {\pi b^{3} \log {\left (b^{4} x^{4} \right )}}{24} \]
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Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\frac {1}{24} \, {\left (\pi \Gamma \left (-1, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + \pi \Gamma \left (-1, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{3} - \frac {\operatorname {S}\left (b x\right )}{3 \, x^{3}} \]
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\[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x^{4}} \,d x } \]
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Timed out. \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^4} \,d x \]
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