Integrand size = 8, antiderivative size = 52 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\frac {1}{12} b^3 \pi \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\operatorname {FresnelS}(b x)}{3 x^3}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^2} \] Output:
1/12*b^3*Pi*Ci(1/2*b^2*Pi*x^2)-1/3*FresnelS(b*x)/x^3-1/6*b*sin(1/2*b^2*Pi* x^2)/x^2
Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\frac {1}{12} b^3 \pi \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\operatorname {FresnelS}(b x)}{3 x^3}-\frac {b \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{6 x^2} \] Input:
Integrate[FresnelS[b*x]/x^4,x]
Output:
(b^3*Pi*CosIntegral[(b^2*Pi*x^2)/2])/12 - FresnelS[b*x]/(3*x^3) - (b*Sin[( b^2*Pi*x^2)/2])/(6*x^2)
Time = 0.37 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6980, 3860, 3042, 3778, 3042, 3783}
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.
\(\displaystyle \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx\) |
\(\Big \downarrow \) 6980 |
\(\displaystyle \frac {1}{3} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^3}dx-\frac {\operatorname {FresnelS}(b x)}{3 x^3}\) |
\(\Big \downarrow \) 3860 |
\(\displaystyle \frac {1}{6} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4}dx^2-\frac {\operatorname {FresnelS}(b x)}{3 x^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4}dx^2-\frac {\operatorname {FresnelS}(b x)}{3 x^3}\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle \frac {1}{6} b \left (\frac {1}{2} \pi b^2 \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2}dx^2-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{x^2}\right )-\frac {\operatorname {FresnelS}(b x)}{3 x^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{6} b \left (\frac {1}{2} \pi b^2 \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2+\frac {\pi }{2}\right )}{x^2}dx^2-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{x^2}\right )-\frac {\operatorname {FresnelS}(b x)}{3 x^3}\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle \frac {1}{6} b \left (\frac {1}{2} \pi b^2 \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\sin \left (\frac {1}{2} \pi b^2 x^2\right )}{x^2}\right )-\frac {\operatorname {FresnelS}(b x)}{3 x^3}\) |
Input:
Int[FresnelS[b*x]/x^4,x]
Output:
-1/3*FresnelS[b*x]/x^3 + (b*((b^2*Pi*CosIntegral[(b^2*Pi*x^2)/2])/2 - Sin[ (b^2*Pi*x^2)/2]/x^2))/6
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[CosInte gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 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]
Time = 0.56 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.88
method | result | size |
parts | \(-\frac {\operatorname {FresnelS}\left (b x \right )}{3 x^{3}}+\frac {b \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 x^{2}}+\frac {b^{2} \pi \,\operatorname {Ci}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{3}\) | \(46\) |
derivativedivides | \(b^{3} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{3 b^{3} x^{3}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{2} x^{2}}+\frac {\pi \,\operatorname {Ci}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12}\right )\) | \(49\) |
default | \(b^{3} \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{3 b^{3} x^{3}}-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 b^{2} x^{2}}+\frac {\pi \,\operatorname {Ci}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{12}\right )\) | \(49\) |
meijerg | \(\frac {\pi ^{\frac {3}{2}} b^{3} \left (\frac {\frac {16 \gamma }{3}-\frac {16 \ln \left (2\right )}{3}-\frac {80}{9}+\frac {32 \ln \left (x \right )}{3}+\frac {16 \ln \left (\pi \right )}{3}+\frac {32 \ln \left (b \right )}{3}}{\sqrt {\pi }}-\frac {\pi ^{\frac {3}{2}} x^{4} b^{4} \operatorname {hypergeom}\left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2, \frac {5}{2}, \frac {11}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{21}\right )}{64}\) | \(68\) |
Input:
int(FresnelS(b*x)/x^4,x,method=_RETURNVERBOSE)
Output:
-1/3*FresnelS(b*x)/x^3+1/3*b*(-1/2*sin(1/2*b^2*Pi*x^2)/x^2+1/4*b^2*Pi*Ci(1 /2*b^2*Pi*x^2))
Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.85 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\frac {\pi b^{3} x^{3} \operatorname {Ci}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, b x \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 4 \, \operatorname {S}\left (b x\right )}{12 \, x^{3}} \] Input:
integrate(fresnel_sin(b*x)/x^4,x, algorithm="fricas")
Output:
1/12*(pi*b^3*x^3*cos_integral(1/2*pi*b^2*x^2) - 2*b*x*sin(1/2*pi*b^2*x^2) - 4*fresnel_sin(b*x))/x^3
Time = 0.77 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.08 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=- \frac {\pi ^{3} b^{7} x^{4} \Gamma \left (\frac {7}{4}\right ) {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {7}{4} \\ 2, 2, \frac {5}{2}, \frac {11}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{768 \Gamma \left (\frac {11}{4}\right )} + \frac {\pi b^{3} \log {\left (b^{4} x^{4} \right )}}{24} \] Input:
integrate(fresnels(b*x)/x**4,x)
Output:
-pi**3*b**7*x**4*gamma(7/4)*hyper((1, 1, 7/4), (2, 2, 5/2, 11/4), -pi**2*b **4*x**4/16)/(768*gamma(11/4)) + pi*b**3*log(b**4*x**4)/24
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.81 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\frac {1}{24} \, {\left (\pi \Gamma \left (-1, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + \pi \Gamma \left (-1, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{3} - \frac {\operatorname {S}\left (b x\right )}{3 \, x^{3}} \] Input:
integrate(fresnel_sin(b*x)/x^4,x, algorithm="maxima")
Output:
1/24*(pi*gamma(-1, 1/2*I*pi*b^2*x^2) + pi*gamma(-1, -1/2*I*pi*b^2*x^2))*b^ 3 - 1/3*fresnel_sin(b*x)/x^3
\[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\int { \frac {\operatorname {S}\left (b x\right )}{x^{4}} \,d x } \] Input:
integrate(fresnel_sin(b*x)/x^4,x, algorithm="giac")
Output:
integrate(fresnel_sin(b*x)/x^4, x)
Timed out. \[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^4} \,d x \] Input:
int(FresnelS(b*x)/x^4,x)
Output:
int(FresnelS(b*x)/x^4, x)
\[ \int \frac {\operatorname {FresnelS}(b x)}{x^4} \, dx=\int \frac {\mathrm {FresnelS}\left (b x \right )}{x^{4}}d x \] Input:
int(FresnelS(b*x)/x^4,x)
Output:
int(FresnelS(b*x)/x^4,x)