\(\int \frac {\operatorname {FresnelC}(b x)}{x^8} \, dx\) [125]

Optimal result

Integrand size = 8, antiderivative size = 102 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^8} \, dx=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{42 x^6}+\frac {b^5 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{336 x^2}-\frac {\operatorname {FresnelC}(b x)}{7 x^7}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{168 x^4}+\frac {1}{672} b^7 \pi ^3 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right ) \]

[Out]

-1/42*b*cos(1/2*b^2*Pi*x^2)/x^6+1/336*b^5*Pi^2*cos(1/2*b^2*Pi*x^2)/x^2-1/7*FresnelC(b*x)/x^7+1/672*b^7*Pi^3*Si
(1/2*b^2*Pi*x^2)+1/168*b^3*Pi*sin(1/2*b^2*Pi*x^2)/x^4

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6562, 3461, 3378, 3380} \[ \int \frac {\operatorname {FresnelC}(b x)}{x^8} \, dx=-\frac {b \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{42 x^6}+\frac {1}{672} \pi ^3 b^7 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )+\frac {\pi ^2 b^5 \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{336 x^2}+\frac {\pi b^3 \sin \left (\frac {1}{2} \pi b^2 x^2\right )}{168 x^4}-\frac {\operatorname {FresnelC}(b x)}{7 x^7} \]

[In]

Int[FresnelC[b*x]/x^8,x]

[Out]

-1/42*(b*Cos[(b^2*Pi*x^2)/2])/x^6 + (b^5*Pi^2*Cos[(b^2*Pi*x^2)/2])/(336*x^2) - FresnelC[b*x]/(7*x^7) + (b^3*Pi
*Sin[(b^2*Pi*x^2)/2])/(168*x^4) + (b^7*Pi^3*SinIntegral[(b^2*Pi*x^2)/2])/672

Rule 3378

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] - Dist[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]

Rule 3380

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3461

Int[((a_.) + Cos[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Cos[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Simpl
ify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 6562

Int[FresnelC[(b_.)*(x_)]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*(FresnelC[b*x]/(d*(m + 1))), x] -
Dist[b/(d*(m + 1)), Int[(d*x)^(m + 1)*Cos[(Pi/2)*b^2*x^2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\operatorname {FresnelC}(b x)}{7 x^7}+\frac {1}{7} b \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^7} \, dx \\ & = -\frac {\operatorname {FresnelC}(b x)}{7 x^7}+\frac {1}{14} b \text {Subst}\left (\int \frac {\cos \left (\frac {1}{2} b^2 \pi x\right )}{x^4} \, dx,x,x^2\right ) \\ & = -\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{42 x^6}-\frac {\operatorname {FresnelC}(b x)}{7 x^7}-\frac {1}{84} \left (b^3 \pi \right ) \text {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x^3} \, dx,x,x^2\right ) \\ & = -\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{42 x^6}-\frac {\operatorname {FresnelC}(b x)}{7 x^7}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{168 x^4}-\frac {1}{336} \left (b^5 \pi ^2\right ) \text {Subst}\left (\int \frac {\cos \left (\frac {1}{2} b^2 \pi x\right )}{x^2} \, dx,x,x^2\right ) \\ & = -\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{42 x^6}+\frac {b^5 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{336 x^2}-\frac {\operatorname {FresnelC}(b x)}{7 x^7}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{168 x^4}+\frac {1}{672} \left (b^7 \pi ^3\right ) \text {Subst}\left (\int \frac {\sin \left (\frac {1}{2} b^2 \pi x\right )}{x} \, dx,x,x^2\right ) \\ & = -\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{42 x^6}+\frac {b^5 \pi ^2 \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{336 x^2}-\frac {\operatorname {FresnelC}(b x)}{7 x^7}+\frac {b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{168 x^4}+\frac {1}{672} b^7 \pi ^3 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^8} \, dx=\frac {1}{672} \left (\frac {2 b \left (-8+b^4 \pi ^2 x^4\right ) \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^6}-\frac {96 \operatorname {FresnelC}(b x)}{x^7}+\frac {4 b^3 \pi \sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x^4}+b^7 \pi ^3 \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )\right ) \]

[In]

Integrate[FresnelC[b*x]/x^8,x]

[Out]

((2*b*(-8 + b^4*Pi^2*x^4)*Cos[(b^2*Pi*x^2)/2])/x^6 - (96*FresnelC[b*x])/x^7 + (4*b^3*Pi*Sin[(b^2*Pi*x^2)/2])/x
^4 + b^7*Pi^3*SinIntegral[(b^2*Pi*x^2)/2])/672

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.42 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.25

method	result	size
meijerg	\(-\frac {b \operatorname {hypergeom}\left (\left [-\frac {3}{2}, \frac {1}{4}\right ], \left [-\frac {1}{2}, \frac {1}{2}, \frac {5}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{6 x^{6}}\)	\(26\)
parts	\(-\frac {\operatorname {FresnelC}\left (b x \right )}{7 x^{7}}+\frac {b \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{6 x^{6}}-\frac {b^{2} \pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 x^{4}}+\frac {b^{2} \pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 x^{2}}-\frac {b^{2} \pi \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{6}\right )}{7}\)	\(90\)
derivativedivides	\(b^{7} \left (-\frac {\operatorname {FresnelC}\left (b x \right )}{7 b^{7} x^{7}}-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{42 b^{6} x^{6}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 b^{4} x^{4}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}-\frac {\pi \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{42}\right )\)	\(93\)
default	\(b^{7} \left (-\frac {\operatorname {FresnelC}\left (b x \right )}{7 b^{7} x^{7}}-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{42 b^{6} x^{6}}-\frac {\pi \left (-\frac {\sin \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4 b^{4} x^{4}}+\frac {\pi \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b^{2} x^{2}}-\frac {\pi \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{4}\right )}{4}\right )}{42}\right )\)	\(93\)

[In]

int(FresnelC(b*x)/x^8,x,method=_RETURNVERBOSE)

[Out]

-1/6*b/x^6*hypergeom([-3/2,1/4],[-1/2,1/2,5/4],-1/16*x^4*Pi^2*b^4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.76 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^8} \, dx=\frac {\pi ^{3} b^{7} x^{7} \operatorname {Si}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 4 \, \pi b^{3} x^{3} \sin \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + 2 \, {\left (\pi ^{2} b^{5} x^{5} - 8 \, b x\right )} \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 96 \, \operatorname {C}\left (b x\right )}{672 \, x^{7}} \]

[In]

integrate(fresnel_cos(b*x)/x^8,x, algorithm="fricas")

[Out]

1/672*(pi^3*b^7*x^7*sin_integral(1/2*pi*b^2*x^2) + 4*pi*b^3*x^3*sin(1/2*pi*b^2*x^2) + 2*(pi^2*b^5*x^5 - 8*b*x)
*cos(1/2*pi*b^2*x^2) - 96*fresnel_cos(b*x))/x^7

Sympy [A] (verification not implemented)

Time = 0.89 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.43 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^8} \, dx=- \frac {b \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} - \frac {3}{2}, \frac {1}{4} \\ - \frac {1}{2}, \frac {1}{2}, \frac {5}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{24 x^{6} \Gamma \left (\frac {5}{4}\right )} \]

[In]

integrate(fresnelc(b*x)/x**8,x)

[Out]

-b*gamma(1/4)*hyper((-3/2, 1/4), (-1/2, 1/2, 5/4), -pi**2*b**4*x**4/16)/(24*x**6*gamma(5/4))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.47 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^8} \, dx=-\frac {1}{224} \, {\left (-i \, \pi ^{3} \Gamma \left (-3, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) + i \, \pi ^{3} \Gamma \left (-3, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{7} - \frac {\operatorname {C}\left (b x\right )}{7 \, x^{7}} \]

[In]

integrate(fresnel_cos(b*x)/x^8,x, algorithm="maxima")

[Out]

-1/224*(-I*pi^3*gamma(-3, 1/2*I*pi*b^2*x^2) + I*pi^3*gamma(-3, -1/2*I*pi*b^2*x^2))*b^7 - 1/7*fresnel_cos(b*x)/
x^7

Giac [F]

\[ \int \frac {\operatorname {FresnelC}(b x)}{x^8} \, dx=\int { \frac {\operatorname {C}\left (b x\right )}{x^{8}} \,d x } \]

[In]

integrate(fresnel_cos(b*x)/x^8,x, algorithm="giac")

[Out]

integrate(fresnel_cos(b*x)/x^8, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\operatorname {FresnelC}(b x)}{x^8} \, dx=\int \frac {\mathrm {FresnelC}\left (b\,x\right )}{x^8} \,d x \]

[In]

int(FresnelC(b*x)/x^8,x)

[Out]

int(FresnelC(b*x)/x^8, x)