\(\int \frac {\operatorname {FresnelC}(b x)}{x^3} \, dx\) [120]

Optimal result

Integrand size = 8, antiderivative size = 44 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^3} \, dx=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{2 x}-\frac {\operatorname {FresnelC}(b x)}{2 x^2}-\frac {1}{2} b^2 \pi \operatorname {FresnelS}(b x) \]

[Out]

-1/2*b*cos(1/2*b^2*Pi*x^2)/x-1/2*FresnelC(b*x)/x^2-1/2*b^2*Pi*FresnelS(b*x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6562, 3469, 3432} \[ \int \frac {\operatorname {FresnelC}(b x)}{x^3} \, dx=-\frac {1}{2} \pi b^2 \operatorname {FresnelS}(b x)-\frac {b \cos \left (\frac {1}{2} \pi b^2 x^2\right )}{2 x}-\frac {\operatorname {FresnelC}(b x)}{2 x^2} \]

[In]

Int[FresnelC[b*x]/x^3,x]

[Out]

-1/2*(b*Cos[(b^2*Pi*x^2)/2])/x - FresnelC[b*x]/(2*x^2) - (b^2*Pi*FresnelS[b*x])/2

Rule 3432

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]

Rule 3469

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] + Dist[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]

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)}{2 x^2}+\frac {1}{2} b \int \frac {\cos \left (\frac {1}{2} b^2 \pi x^2\right )}{x^2} \, dx \\ & = -\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{2 x}-\frac {\operatorname {FresnelC}(b x)}{2 x^2}-\frac {1}{2} \left (b^3 \pi \right ) \int \sin \left (\frac {1}{2} b^2 \pi x^2\right ) \, dx \\ & = -\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{2 x}-\frac {\operatorname {FresnelC}(b x)}{2 x^2}-\frac {1}{2} b^2 \pi \operatorname {FresnelS}(b x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^3} \, dx=-\frac {b \cos \left (\frac {1}{2} b^2 \pi x^2\right )}{2 x}-\frac {\operatorname {FresnelC}(b x)}{2 x^2}-\frac {1}{2} b^2 \pi \operatorname {FresnelS}(b x) \]

[In]

Integrate[FresnelC[b*x]/x^3,x]

[Out]

-1/2*(b*Cos[(b^2*Pi*x^2)/2])/x - FresnelC[b*x]/(2*x^2) - (b^2*Pi*FresnelS[b*x])/2

Maple [C] (verified)

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

Time = 0.44 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.59

method	result	size
meijerg	\(-\frac {b \operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {1}{4}\right ], \left [\frac {1}{2}, \frac {3}{4}, \frac {5}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{x}\)	\(26\)
derivativedivides	\(b^{2} \left (-\frac {\operatorname {FresnelC}\left (b x \right )}{2 b^{2} x^{2}}-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b x}-\frac {\pi \,\operatorname {FresnelS}\left (b x \right )}{2}\right )\)	\(43\)
default	\(b^{2} \left (-\frac {\operatorname {FresnelC}\left (b x \right )}{2 b^{2} x^{2}}-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2 b x}-\frac {\pi \,\operatorname {FresnelS}\left (b x \right )}{2}\right )\)	\(43\)
parts	\(-\frac {\operatorname {FresnelC}\left (b x \right )}{2 x^{2}}+\frac {b \left (-\frac {\cos \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{x}-\frac {b^{2} \pi ^{\frac {3}{2}} \operatorname {FresnelS}\left (\frac {\sqrt {\pi }\, b^{2} x}{\sqrt {b^{2} \pi }}\right )}{\sqrt {b^{2} \pi }}\right )}{2}\)	\(61\)

[In]

int(FresnelC(b*x)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^3} \, dx=-\frac {\pi \sqrt {b^{2}} b x^{2} \operatorname {S}\left (\sqrt {b^{2}} x\right ) + b x \cos \left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) + \operatorname {C}\left (b x\right )}{2 \, x^{2}} \]

[In]

integrate(fresnel_cos(b*x)/x^3,x, algorithm="fricas")

[Out]

-1/2*(pi*sqrt(b^2)*b*x^2*fresnel_sin(sqrt(b^2)*x) + b*x*cos(1/2*pi*b^2*x^2) + fresnel_cos(b*x))/x^2

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.16 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^3} \, dx=\frac {b \Gamma \left (- \frac {1}{4}\right ) \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} \\ \frac {1}{2}, \frac {3}{4}, \frac {5}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 x \Gamma \left (\frac {3}{4}\right ) \Gamma \left (\frac {5}{4}\right )} \]

[In]

integrate(fresnelc(b*x)/x**3,x)

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.39 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^3} \, dx=-\frac {\sqrt {\frac {1}{2}} \sqrt {\pi x^{2}} {\left (\left (i + 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, \frac {1}{2} i \, \pi b^{2} x^{2}\right ) - \left (i - 1\right ) \, \sqrt {2} \Gamma \left (-\frac {1}{2}, -\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} b^{2}}{16 \, x} - \frac {\operatorname {C}\left (b x\right )}{2 \, x^{2}} \]

[In]

integrate(fresnel_cos(b*x)/x^3,x, algorithm="maxima")

[Out]

-1/16*sqrt(1/2)*sqrt(pi*x^2)*((I + 1)*sqrt(2)*gamma(-1/2, 1/2*I*pi*b^2*x^2) - (I - 1)*sqrt(2)*gamma(-1/2, -1/2
*I*pi*b^2*x^2))*b^2/x - 1/2*fresnel_cos(b*x)/x^2

Giac [F]

\[ \int \frac {\operatorname {FresnelC}(b x)}{x^3} \, dx=\int { \frac {\operatorname {C}\left (b x\right )}{x^{3}} \,d x } \]

[In]

integrate(fresnel_cos(b*x)/x^3,x, algorithm="giac")

[Out]

integrate(fresnel_cos(b*x)/x^3, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\operatorname {FresnelC}(b x)}{x^3} \, dx=\int \frac {\mathrm {FresnelC}\left (b\,x\right )}{x^3} \,d x \]

[In]

int(FresnelC(b*x)/x^3,x)

[Out]

int(FresnelC(b*x)/x^3, x)