Integrand size = 8, antiderivative size = 77 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{40 x^2}-\frac {\operatorname {FresnelS}(b x)}{5 x^5}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{20 x^4}-\frac {1}{80} b^5 \pi ^2 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right ) \] Output:
-1/40*b^3*Pi*cos(1/2*b^2*Pi*x^2)/x^2-1/5*FresnelS(b*x)/x^5-1/20*b*sin(1/2* b^2*Pi*x^2)/x^4-1/80*b^5*Pi^2*Si(1/2*b^2*Pi*x^2)
Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{40 x^2}-\frac {\operatorname {FresnelS}(b x)}{5 x^5}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{20 x^4}-\frac {1}{80} b^5 \pi ^2 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right ) \] Input:
Integrate[FresnelS[b*x]/x^6,x]
Output:
-1/40*(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/x^2 - FresnelS[b*x]/(5*x^5) - (b*Sin[(b ^2*Pi*x^2)/2])/(20*x^4) - (b^5*Pi^2*SinIntegral[(b^2*Pi*x^2)/2])/80
Time = 0.47 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.125, Rules used = {6980, 3860, 3042, 3778, 3042, 3778, 25, 3042, 3780}
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.
\(\displaystyle \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx\) |
\(\Big \downarrow \) 6980 |
\(\displaystyle \frac {1}{5} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^5}dx-\frac {\operatorname {FresnelS}(b x)}{5 x^5}\) |
\(\Big \downarrow \) 3860 |
\(\displaystyle \frac {1}{10} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6}dx^2-\frac {\operatorname {FresnelS}(b x)}{5 x^5}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6}dx^2-\frac {\operatorname {FresnelS}(b x)}{5 x^5}\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle \frac {1}{10} b \left (\frac {1}{4} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4}dx^2-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x)}{5 x^5}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} b \left (\frac {1}{4} \pi b^2 \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2+\frac {\pi }{2}\right )}{x^4}dx^2-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x)}{5 x^5}\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle \frac {1}{10} b \left (\frac {1}{4} \pi b^2 \left (\frac {1}{2} \pi b^2 \int -\frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx^2-\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{x^2}\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x)}{5 x^5}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{10} b \left (\frac {1}{4} \pi b^2 \left (-\frac {1}{2} \pi b^2 \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx^2-\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{x^2}\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x)}{5 x^5}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{10} b \left (\frac {1}{4} \pi b^2 \left (-\frac {1}{2} \pi b^2 \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx^2-\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{x^2}\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x)}{5 x^5}\) |
\(\Big \downarrow \) 3780 |
\(\displaystyle \frac {1}{10} b \left (\frac {1}{4} \pi b^2 \left (-\frac {1}{2} \pi b^2 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{x^2}\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x^4}\right )-\frac {\operatorname {FresnelS}(b x)}{5 x^5}\) |
Input:
Int[FresnelS[b*x]/x^6,x]
Output:
-1/5*FresnelS[b*x]/x^5 + (b*(-1/2*Sin[(b^2*Pi*x^2)/2]/x^4 + (b^2*Pi*(-(Cos [(b^2*Pi*x^2)/2]/x^2) - (b^2*Pi*SinIntegral[(b^2*Pi*x^2)/2])/2))/4))/10
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m + 1))), x] - Simp[f/(d*(m + 1)) Int[( c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[m, - 1]
Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sin[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol ] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sin[c + d*x])^ p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simplify[ (m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[ (m + 1)/n], 0]))
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 = 0.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.38
method | result | size |
meijerg | \(-\frac {\pi \,b^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {1}{2}, \frac {3}{2}, \frac {7}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{12 x^{2}}\) | \(29\) |
parts | \(-\frac {\operatorname {FresnelS}\left (b x \right )}{5 x^{5}}+\frac {b \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 x^{4}}+\frac {b^{2} \pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 x^{2}}-\frac {b^{2} \pi \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{5}\) | \(68\) |
derivativedivides | \(b^{5} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{5 b^{5} x^{5}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{20 b^{4} x^{4}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}-\frac {\pi \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{20}\right )\) | \(71\) |
default | \(b^{5} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{5 b^{5} x^{5}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{20 b^{4} x^{4}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}-\frac {\pi \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{20}\right )\) | \(71\) |
Input:
int(FresnelS(b*x)/x^6,x,method=_RETURNVERBOSE)
Output:
-1/12*Pi*b^3/x^2*hypergeom([-1/2,3/4],[1/2,3/2,7/4],-1/16*x^4*Pi^2*b^4)
Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=-\frac {\pi ^{2} b^{5} x^{5} \operatorname {Si}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 2 \, \pi b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 4 \, b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 16 \, \operatorname {S}\left (b x\right )}{80 \, x^{5}} \] Input:
integrate(fresnel_sin(b*x)/x^6,x, algorithm="fricas")
Output:
-1/80*(pi^2*b^5*x^5*sin_integral(1/2*pi*b^2*x^2) + 2*pi*b^3*x^3*cos(1/2*pi *b^2*x^2) + 4*b*x*sin(1/2*pi*b^2*x^2) + 16*fresnel_sin(b*x))/x^5
Time = 0.56 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.60 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=- \frac {\pi b^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {1}{2}, \frac {3}{2}, \frac {7}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 x^{2} \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate(fresnels(b*x)/x**6,x)
Output:
-pi*b**3*gamma(3/4)*hyper((-1/2, 3/4), (1/2, 3/2, 7/4), -pi**2*b**4*x**4/1 6)/(16*x**2*gamma(7/4))
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.62 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=-\frac {1}{80} \, {\left (-i \, \pi ^{2} \Gamma \left (-2, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + i \, \pi ^{2} \Gamma \left (-2, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{5} - \frac {\operatorname {S}\left (b x\right )}{5 \, x^{5}} \] Input:
integrate(fresnel_sin(b*x)/x^6,x, algorithm="maxima")
Output:
-1/80*(-I*pi^2*gamma(-2, 1/2*I*pi*b^2*x^2) + I*pi^2*gamma(-2, -1/2*I*pi*b^ 2*x^2))*b^5 - 1/5*fresnel_sin(b*x)/x^5
\[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x^{6}} \,d x } \] Input:
integrate(fresnel_sin(b*x)/x^6,x, algorithm="giac")
Output:
integrate(fresnel_sin(b*x)/x^6, x)
Timed out. \[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^6} \,d x \] Input:
int(FresnelS(b*x)/x^6,x)
Output:
int(FresnelS(b*x)/x^6, x)
\[ \int \frac {\operatorname {FresnelS}(b x)}{x^6} \, dx=\int \frac {\mathrm {FresnelS}\left (b x \right )}{x^{6}}d x \] Input:
int(FresnelS(b*x)/x^6,x)
Output:
int(FresnelS(b*x)/x^6,x)