Integrand size = 8, antiderivative size = 99 \[ \int x^5 \operatorname {FresnelS}(b x) \, dx=-\frac {5 x \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{2 b^5 \pi ^3}+\frac {x^5 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{6 b \pi }+\frac {5 \operatorname {FresnelC}(b x)}{2 b^6 \pi ^3}+\frac {1}{6} x^6 \operatorname {FresnelS}(b x)-\frac {5 x^3 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 b^3 \pi ^2} \] Output:
-5/2*x*cos(1/2*b^2*Pi*x^2)/b^5/Pi^3+1/6*x^5*cos(1/2*b^2*Pi*x^2)/b/Pi+5/2*F resnelC(b*x)/b^6/Pi^3+1/6*x^6*FresnelS(b*x)-5/6*x^3*sin(1/2*b^2*Pi*x^2)/b^ 3/Pi^2
Time = 0.05 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.80 \[ \int x^5 \operatorname {FresnelS}(b x) \, dx=\frac {b x \left (-15+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )+15 \operatorname {FresnelC}(b x)+b^6 \pi ^3 x^6 \operatorname {FresnelS}(b x)-5 b^3 \pi x^3 \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 b^6 \pi ^3} \] Input:
Integrate[x^5*FresnelS[b*x],x]
Output:
(b*x*(-15 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2] + 15*FresnelC[b*x] + b^6*Pi^ 3*x^6*FresnelS[b*x] - 5*b^3*Pi*x^3*Sin[(b^2*Pi*x^2)/2])/(6*b^6*Pi^3)
Time = 0.42 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6980, 3866, 3867, 3866, 3833}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^5 \operatorname {FresnelS}(b x) \, dx\) |
\(\Big \downarrow \) 6980 |
\(\displaystyle \frac {1}{6} x^6 \operatorname {FresnelS}(b x)-\frac {1}{6} b \int x^6 \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx\) |
\(\Big \downarrow \) 3866 |
\(\displaystyle \frac {1}{6} x^6 \operatorname {FresnelS}(b x)-\frac {1}{6} b \left (\frac {5 \int x^4 \cos \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}-\frac {x^5 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\) |
\(\Big \downarrow \) 3867 |
\(\displaystyle \frac {1}{6} x^6 \operatorname {FresnelS}(b x)-\frac {1}{6} b \left (\frac {5 \left (\frac {x^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {3 \int x^2 \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}\right )}{\pi b^2}-\frac {x^5 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\) |
\(\Big \downarrow \) 3866 |
\(\displaystyle \frac {1}{6} x^6 \operatorname {FresnelS}(b x)-\frac {1}{6} b \left (\frac {5 \left (\frac {x^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {3 \left (\frac {\int \cos \left (\frac {1}{2} b^2 \pi x^2\right )dx}{\pi b^2}-\frac {x \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}\right )}{\pi b^2}-\frac {x^5 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\) |
\(\Big \downarrow \) 3833 |
\(\displaystyle \frac {1}{6} x^6 \operatorname {FresnelS}(b x)-\frac {1}{6} b \left (\frac {5 \left (\frac {x^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}-\frac {3 \left (\frac {\operatorname {FresnelC}(b x)}{\pi b^3}-\frac {x \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )}{\pi b^2}\right )}{\pi b^2}-\frac {x^5 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{\pi b^2}\right )\) |
Input:
Int[x^5*FresnelS[b*x],x]
Output:
(x^6*FresnelS[b*x])/6 - (b*(-((x^5*Cos[(b^2*Pi*x^2)/2])/(b^2*Pi)) + (5*((- 3*(-((x*Cos[(b^2*Pi*x^2)/2])/(b^2*Pi)) + FresnelC[b*x]/(b^3*Pi)))/(b^2*Pi) + (x^3*Sin[(b^2*Pi*x^2)/2])/(b^2*Pi)))/(b^2*Pi)))/6
Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)], x] /; FreeQ[{d, e, f}, x]
Int[((e_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(-e^ (n - 1))*(e*x)^(m - n + 1)*(Cos[c + d*x^n]/(d*n)), x] + Simp[e^n*((m - n + 1)/(d*n)) Int[(e*x)^(m - n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0] && LtQ[n, m + 1]
Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(Sin[c + d*x^n]/(d*n)), x] - Simp[e^n*((m - n + 1)/ (d*n)) Int[(e*x)^(m - n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0] && LtQ[n, m + 1]
Int[FresnelS[(b_.)*(x_)]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1 )*(FresnelS[b*x]/(d*(m + 1))), x] - Simp[b/(d*(m + 1)) Int[(d*x)^(m + 1)* Sin[(Pi/2)*b^2*x^2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 0.50 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.29
method | result | size |
meijerg | \(\frac {\pi \,b^{3} x^{9} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {9}{4}\right ], \left [\frac {3}{2}, \frac {7}{4}, \frac {13}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{54}\) | \(29\) |
derivativedivides | \(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{6} x^{6}}{6}+\frac {b^{5} x^{5} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 \pi }-\frac {5 \left (\frac {b^{3} x^{3} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {3 \left (-\frac {b x \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {\operatorname {FresnelC}\left (b x \right )}{\pi }\right )}{\pi }\right )}{6 \pi }}{b^{6}}\) | \(96\) |
default | \(\frac {\frac {\operatorname {FresnelS}\left (b x \right ) b^{6} x^{6}}{6}+\frac {b^{5} x^{5} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 \pi }-\frac {5 \left (\frac {b^{3} x^{3} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }-\frac {3 \left (-\frac {b x \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{\pi }+\frac {\operatorname {FresnelC}\left (b x \right )}{\pi }\right )}{\pi }\right )}{6 \pi }}{b^{6}}\) | \(96\) |
parts | \(\frac {x^{6} \operatorname {FresnelS}\left (b x \right )}{6}-\frac {b \left (-\frac {x^{5} \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {\frac {5 x^{3} \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }-\frac {15 \left (-\frac {x \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b^{2} \pi }+\frac {\operatorname {FresnelC}\left (\frac {\sqrt {\pi }\, b^{2} x}{\sqrt {b^{2} \pi }}\right )}{b^{2} \sqrt {\pi }\, \sqrt {b^{2} \pi }}\right )}{b^{2} \pi }}{b^{2} \pi }\right )}{6}\) | \(123\) |
Input:
int(x^5*FresnelS(b*x),x,method=_RETURNVERBOSE)
Output:
1/54*Pi*b^3*x^9*hypergeom([3/4,9/4],[3/2,7/4,13/4],-1/16*x^4*Pi^2*b^4)
Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.86 \[ \int x^5 \operatorname {FresnelS}(b x) \, dx=\frac {\pi ^{3} b^{7} x^{6} \operatorname {S}\left (b x\right ) - 5 \, \pi b^{4} x^{3} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{2} b^{6} x^{5} - 15 \, b^{2} x\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 15 \, \sqrt {b^{2}} \operatorname {C}\left (\sqrt {b^{2}} x\right )}{6 \, \pi ^{3} b^{7}} \] Input:
integrate(x^5*fresnel_sin(b*x),x, algorithm="fricas")
Output:
1/6*(pi^3*b^7*x^6*fresnel_sin(b*x) - 5*pi*b^4*x^3*sin(1/2*pi*b^2*x^2) + (p i^2*b^6*x^5 - 15*b^2*x)*cos(1/2*pi*b^2*x^2) + 15*sqrt(b^2)*fresnel_cos(sqr t(b^2)*x))/(pi^3*b^7)
Time = 0.52 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.54 \[ \int x^5 \operatorname {FresnelS}(b x) \, dx=\frac {\pi b^{3} x^{9} \Gamma \left (\frac {3}{4}\right ) \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {3}{4}, \frac {9}{4} \\ \frac {3}{2}, \frac {7}{4}, \frac {13}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{32 \Gamma \left (\frac {7}{4}\right ) \Gamma \left (\frac {13}{4}\right )} \] Input:
integrate(x**5*fresnels(b*x),x)
Output:
pi*b**3*x**9*gamma(3/4)*gamma(9/4)*hyper((3/4, 9/4), (3/2, 7/4, 13/4), -pi **2*b**4*x**4/16)/(32*gamma(7/4)*gamma(13/4))
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.11 \[ \int x^5 \operatorname {FresnelS}(b x) \, dx=\frac {1}{6} \, x^{6} \operatorname {S}\left (b x\right ) - \frac {\sqrt {\frac {1}{2}} {\left (20 \, \sqrt {\frac {1}{2}} \pi ^{2} b^{3} x^{3} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \left (15 i - 15\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {\frac {1}{2} i \, \pi } b x\right ) - \left (15 i + 15\right ) \, \left (\frac {1}{4}\right )^{\frac {1}{4}} \pi \operatorname {erf}\left (\sqrt {-\frac {1}{2} i \, \pi } b x\right ) - 4 \, {\left (\sqrt {\frac {1}{2}} \pi ^{3} b^{5} x^{5} - 15 \, \sqrt {\frac {1}{2}} \pi b x\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right )\right )}}{12 \, \pi ^{4} b^{6}} \] Input:
integrate(x^5*fresnel_sin(b*x),x, algorithm="maxima")
Output:
1/6*x^6*fresnel_sin(b*x) - 1/12*sqrt(1/2)*(20*sqrt(1/2)*pi^2*b^3*x^3*sin(1 /2*pi*b^2*x^2) + (15*I - 15)*(1/4)^(1/4)*pi*erf(sqrt(1/2*I*pi)*b*x) - (15* I + 15)*(1/4)^(1/4)*pi*erf(sqrt(-1/2*I*pi)*b*x) - 4*(sqrt(1/2)*pi^3*b^5*x^ 5 - 15*sqrt(1/2)*pi*b*x)*cos(1/2*pi*b^2*x^2))/(pi^4*b^6)
\[ \int x^5 \operatorname {FresnelS}(b x) \, dx=\int { x^{5} \operatorname {S}\left (b x\right ) \,d x } \] Input:
integrate(x^5*fresnel_sin(b*x),x, algorithm="giac")
Output:
integrate(x^5*fresnel_sin(b*x), x)
Timed out. \[ \int x^5 \operatorname {FresnelS}(b x) \, dx=\int x^5\,\mathrm {FresnelS}\left (b\,x\right ) \,d x \] Input:
int(x^5*FresnelS(b*x),x)
Output:
int(x^5*FresnelS(b*x), x)
\[ \int x^5 \operatorname {FresnelS}(b x) \, dx=\int x^{5} \mathrm {FresnelS}\left (b x \right )d x \] Input:
int(x^5*FresnelS(b*x),x)
Output:
int(x^5*FresnelS(b*x),x)