Integrand size = 8, antiderivative size = 69 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x}-\frac {1}{12} b^4 \pi ^2 \operatorname {FresnelS}(b x)-\frac {\operatorname {FresnelS}(b x)}{4 x^4}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x^3} \]
-1/12*b^3*Pi*cos(1/2*b^2*Pi*x^2)/x-1/12*b^4*Pi^2*FresnelS(b*x)-1/4*Fresnel S(b*x)/x^4-1/12*b*sin(1/2*b^2*Pi*x^2)/x^3
Time = 0.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=-\frac {b^3 \pi \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x}-\frac {1}{12} b^4 \pi ^2 \operatorname {FresnelS}(b x)-\frac {\operatorname {FresnelS}(b x)}{4 x^4}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{12 x^3} \]
-1/12*(b^3*Pi*Cos[(b^2*Pi*x^2)/2])/x - (b^4*Pi^2*FresnelS[b*x])/12 - Fresn elS[b*x]/(4*x^4) - (b*Sin[(b^2*Pi*x^2)/2])/(12*x^3)
Time = 0.32 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6980, 3868, 3869, 3832}
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.
\(\displaystyle \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx\) |
\(\Big \downarrow \) 6980 |
\(\displaystyle \frac {1}{4} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4}dx-\frac {\operatorname {FresnelS}(b x)}{4 x^4}\) |
\(\Big \downarrow \) 3868 |
\(\displaystyle \frac {1}{4} b \left (\frac {1}{3} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^3}\right )-\frac {\operatorname {FresnelS}(b x)}{4 x^4}\) |
\(\Big \downarrow \) 3869 |
\(\displaystyle \frac {1}{4} b \left (\frac {1}{3} \pi b^2 \left (-\pi b^2 \int \sin \left (\frac {1}{2} b^2 \pi x^2\right )dx-\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{x}\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^3}\right )-\frac {\operatorname {FresnelS}(b x)}{4 x^4}\) |
\(\Big \downarrow \) 3832 |
\(\displaystyle \frac {1}{4} b \left (\frac {1}{3} \pi b^2 \left (-\frac {\cos \left (\frac {1}{2} \pi b^2 x^2\right )}{x}-\pi b \operatorname {FresnelS}(b x)\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{3 x^3}\right )-\frac {\operatorname {FresnelS}(b x)}{4 x^4}\) |
-1/4*FresnelS[b*x]/x^4 + (b*((b^2*Pi*(-(Cos[(b^2*Pi*x^2)/2]/x) - b*Pi*Fres nelS[b*x]))/3 - Sin[(b^2*Pi*x^2)/2]/(3*x^3)))/4
3.1.13.3.1 Defintions of rubi rules used
Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]/(f*Rt[ d, 2]))*FresnelS[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*x) ^(m + 1)*(Sin[c + d*x^n]/(e*(m + 1))), x] - Simp[d*(n/(e^n*(m + 1))) Int[ (e*x)^(m + n)*Cos[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0] & & LtQ[m, -1]
Int[Cos[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_), x_Symbol] :> Simp[(e*x) ^(m + 1)*(Cos[c + d*x^n]/(e*(m + 1))), x] + Simp[d*(n/(e^n*(m + 1))) Int[ (e*x)^(m + n)*Sin[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0] & & LtQ[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]
Time = 0.40 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(b^{4} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{4 b^{4} x^{4}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12 b^{3} x^{3}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}-\pi \,\operatorname {FresnelS}\left (b x \right )\right )}{12}\right )\) | \(65\) |
default | \(b^{4} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{4 b^{4} x^{4}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12 b^{3} x^{3}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{b x}-\pi \,\operatorname {FresnelS}\left (b x \right )\right )}{12}\right )\) | \(65\) |
meijerg | \(\frac {\pi ^{2} b^{4} \left (-\frac {32 \cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi x b}-\frac {32 \sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 \pi ^{2} x^{3} b^{3}}-\frac {32 \left (x^{4} \pi ^{2} b^{4}+3\right ) \operatorname {FresnelS}\left (b x \right )}{3 \pi ^{2} x^{4} b^{4}}\right )}{128}\) | \(79\) |
parts | \(-\frac {\operatorname {FresnelS}\left (b x \right )}{4 x^{4}}+\frac {b \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{3 x^{3}}+\frac {b^{2} \pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{x}-\frac {b^{2} \pi ^{\frac {3}{2}} \operatorname {FresnelS}\left (\frac {\sqrt {\pi }\, b^{2} x}{\sqrt {b^{2} \pi }}\right )}{\sqrt {b^{2} \pi }}\right )}{3}\right )}{4}\) | \(83\) |
b^4*(-1/4*FresnelS(b*x)/b^4/x^4-1/12/b^3/x^3*sin(1/2*b^2*Pi*x^2)+1/12*Pi*( -1/b/x*cos(1/2*b^2*Pi*x^2)-Pi*FresnelS(b*x)))
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.78 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=-\frac {\pi b^{3} x^{3} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + {\left (\pi ^{2} b^{4} x^{4} + 3\right )} \operatorname {S}\left (b x\right )}{12 \, x^{4}} \]
-1/12*(pi*b^3*x^3*cos(1/2*pi*b^2*x^2) + b*x*sin(1/2*pi*b^2*x^2) + (pi^2*b^ 4*x^4 + 3)*fresnel_sin(b*x))/x^4
Time = 0.67 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.59 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=\frac {\pi ^{2} b^{4} S\left (b x\right ) \Gamma \left (- \frac {1}{4}\right )}{64 \Gamma \left (\frac {7}{4}\right )} + \frac {\pi b^{3} \cos {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {1}{4}\right )}{64 x \Gamma \left (\frac {7}{4}\right )} + \frac {b \sin {\left (\frac {\pi b^{2} x^{2}}{2} \right )} \Gamma \left (- \frac {1}{4}\right )}{64 x^{3} \Gamma \left (\frac {7}{4}\right )} + \frac {3 S\left (b x\right ) \Gamma \left (- \frac {1}{4}\right )}{64 x^{4} \Gamma \left (\frac {7}{4}\right )} \]
pi**2*b**4*fresnels(b*x)*gamma(-1/4)/(64*gamma(7/4)) + pi*b**3*cos(pi*b**2 *x**2/2)*gamma(-1/4)/(64*x*gamma(7/4)) + b*sin(pi*b**2*x**2/2)*gamma(-1/4) /(64*x**3*gamma(7/4)) + 3*fresnels(b*x)*gamma(-1/4)/(64*x**4*gamma(7/4))
Result contains complex when optimal does not.
Time = 0.36 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.88 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=-\frac {\sqrt {\frac {1}{2}} \left (\pi x^{2}\right )^{\frac {3}{2}} {\left (-\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {3}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{4}}{64 \, x^{3}} - \frac {\operatorname {S}\left (b x\right )}{4 \, x^{4}} \]
-1/64*sqrt(1/2)*(pi*x^2)^(3/2)*(-(I + 1)*sqrt(2)*gamma(-3/2, 1/2*I*pi*b^2* x^2) + (I - 1)*sqrt(2)*gamma(-3/2, -1/2*I*pi*b^2*x^2))*b^4/x^3 - 1/4*fresn el_sin(b*x)/x^4
\[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x^{5}} \,d x } \]
Timed out. \[ \int \frac {\operatorname {FresnelS}(b x)}{x^5} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^5} \,d x \]