Integrand size = 8, antiderivative size = 84 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=-\frac {4 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac {1}{5} x^5 \operatorname {FresnelC}(b x)+\frac {8 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}-\frac {x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi } \]
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Time = 0.05 (sec) , antiderivative size = 84, 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 = {6562, 3461, 3377, 2717} \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=-\frac {x^4 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi b}+\frac {8 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}-\frac {4 x^2 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}+\frac {1}{5} x^5 \operatorname {FresnelC}(b x) \]
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Rule 2717
Rule 3377
Rule 3461
Rule 6562
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {1}{5} b \int x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \\ & = \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {1}{10} b \text {Subst}\left (\int x^2 \cos \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right ) \\ & = \frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {2 \text {Subst}\left (\int x \sin \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{5 b \pi } \\ & = -\frac {4 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {4 \text {Subst}\left (\int \cos \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{5 b^3 \pi ^2} \\ & = -\frac {4 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac {1}{5} x^5 \operatorname {FresnelC}(b x)+\frac {8 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}-\frac {x^4 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi } \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=-\frac {4 x^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac {1}{5} x^5 \operatorname {FresnelC}(b x)-\frac {\left (-8+b^4 \pi ^2 x^4\right ) \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3} \]
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Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.40 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.31
method | result | size |
meijerg | \(\frac {b \,x^{6} \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {3}{2}\right ], \left [\frac {1}{2}, \frac {5}{4}, \frac {5}{2}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{6}\) | \(26\) |
derivativedivides | \(\frac {\frac {\operatorname {FresnelC}\left (b x \right ) b^{5} x^{5}}{5}-\frac {b^{4} x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }+\frac {-\frac {4 b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }+\frac {8 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi ^{2}}}{\pi }}{b^{5}}\) | \(81\) |
default | \(\frac {\frac {\operatorname {FresnelC}\left (b x \right ) b^{5} x^{5}}{5}-\frac {b^{4} x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }+\frac {-\frac {4 b^{2} x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }+\frac {8 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi ^{2}}}{\pi }}{b^{5}}\) | \(81\) |
parts | \(\frac {x^{5} \operatorname {FresnelC}\left (b x \right )}{5}-\frac {b \left (\frac {x^{4} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }-\frac {4 \left (-\frac {x^{2} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {2 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{4} \pi ^{2}}\right )}{b^{2} \pi }\right )}{5}\) | \(83\) |
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Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.79 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\frac {\pi ^{3} b^{5} x^{5} \operatorname {C}\left (b x\right ) - 4 \, \pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{5 \, \pi ^{3} b^{5}} \]
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Time = 0.63 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.38 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\frac {x^{5} C\left (b x\right ) \Gamma \left (\frac {1}{4}\right )}{20 \Gamma \left (\frac {5}{4}\right )} - \frac {x^{4} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{20 \pi b \Gamma \left (\frac {5}{4}\right )} - \frac {x^{2} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{5 \pi ^{2} b^{3} \Gamma \left (\frac {5}{4}\right )} + \frac {2 \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {1}{4}\right )}{5 \pi ^{3} b^{5} \Gamma \left (\frac {5}{4}\right )} \]
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Time = 0.20 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.73 \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\frac {1}{5} \, x^{5} \operatorname {C}\left (b x\right ) - \frac {4 \, \pi b^{2} x^{2} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{5 \, \pi ^{3} b^{5}} \]
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\[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\int { x^{4} \operatorname {C}\left (b x\right ) \,d x } \]
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Timed out. \[ \int x^4 \operatorname {FresnelC}(b x) \, dx=\int x^4\,\mathrm {FresnelC}\left (b\,x\right ) \,d x \]
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