Integrand size = 8, antiderivative size = 84 \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=-\frac {8 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}+\frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {4 x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2} \]
[Out]
Time = 0.06 (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 = {6561, 3460, 3377, 2718} \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\frac {x^4 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi b}-\frac {8 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^3 b^5}-\frac {4 x^2 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{5 \pi ^2 b^3}+\frac {1}{5} x^5 \operatorname {FresnelS}(b x) \]
[In]
[Out]
Rule 2718
Rule 3377
Rule 3460
Rule 6561
Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {1}{5} b \int x^5 \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \\ & = \frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {1}{10} b \text {Subst}\left (\int x^2 \sin \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right ) \\ & = \frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {2 \text {Subst}\left (\int x \cos \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{5 b \pi } \\ & = \frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {4 x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2}+\frac {4 \text {Subst}\left (\int \sin \left (\frac {1}{2} b^2 \pi x\right ) \, dx,x,x^2\right )}{5 b^3 \pi ^2} \\ & = -\frac {8 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}+\frac {x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b \pi }+\frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {4 x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.85 \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\frac {\left (-8+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^5 \pi ^3}+\frac {1}{5} x^5 \operatorname {FresnelS}(b x)-\frac {4 x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{5 b^3 \pi ^2} \]
[In]
[Out]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.46 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.35
method | result | size |
meijerg | \(\frac {\pi \,b^{3} x^{8} \operatorname {hypergeom}\left (\left [\frac {3}{4}, 2\right ], \left [\frac {3}{2}, \frac {7}{4}, 3\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{48}\) | \(29\) |
derivativedivides | \(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{5} x^{5}}{5}+\frac {b^{4} x^{4} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }-\frac {4 \left (\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{5 \pi }}{b^{5}}\) | \(80\) |
default | \(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{5} x^{5}}{5}+\frac {b^{4} x^{4} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{5 \pi }-\frac {4 \left (\frac {b^{2} x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {2 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi ^{2}}\right )}{5 \pi }}{b^{5}}\) | \(80\) |
parts | \(\frac {x^{5} \operatorname {FresnelS}\left (b x \right )}{5}-\frac {b \left (-\frac {x^{4} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {\frac {4 x^{2} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {8 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{4} \pi ^{2}}}{b^{2} \pi }\right )}{5}\) | \(83\) |
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.77 \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\frac {\pi ^{3} b^{5} x^{5} \operatorname {S}\left (b x\right ) - 4 \, \pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{5 \, \pi ^{3} b^{5}} \]
[In]
[Out]
Time = 1.07 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.44 \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\frac {3 x^{5} S\left (b x\right ) \Gamma \left (\frac {3}{4}\right )}{20 \Gamma \left (\frac {7}{4}\right )} + \frac {3 x^{4} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{20 \pi b \Gamma \left (\frac {7}{4}\right )} - \frac {3 x^{2} \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{5 \pi ^{2} b^{3} \Gamma \left (\frac {7}{4}\right )} - \frac {6 \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (\frac {3}{4}\right )}{5 \pi ^{3} b^{5} \Gamma \left (\frac {7}{4}\right )} \]
[In]
[Out]
none
Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.74 \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\frac {1}{5} \, x^{5} \operatorname {S}\left (b x\right ) - \frac {4 \, \pi b^{2} x^{2} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - {\left (\pi ^{2} b^{4} x^{4} - 8\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )}{5 \, \pi ^{3} b^{5}} \]
[In]
[Out]
\[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\int { x^{4} \operatorname {S}\left (b x\right ) \,d x } \]
[In]
[Out]
Timed out. \[ \int x^4 \operatorname {FresnelS}(b x) \, dx=\int x^4\,\mathrm {FresnelS}\left (b\,x\right ) \,d x \]
[In]
[Out]