\(\int \frac {\operatorname {FresnelC}(b x)}{x^2} \, dx\) [119]

Optimal result

Integrand size = 8, antiderivative size = 27 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^2} \, dx=\frac {1}{2} b \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\operatorname {FresnelC}(b x)}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6562, 3457} \[ \int \frac {\operatorname {FresnelC}(b x)}{x^2} \, dx=\frac {1}{2} b \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\operatorname {FresnelC}(b x)}{x} \]

[In]

Int[FresnelC[b*x]/x^2,x]

[Out]

(b*CosIntegral[(b^2*Pi*x^2)/2])/2 - FresnelC[b*x]/x

Rule 3457

Int[Cos[(d_.)*(x_)^(n_)]/(x_), x_Symbol] :> Simp[CosIntegral[d*x^n]/n, x] /; FreeQ[{d, n}, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^2} \, dx=\frac {1}{2} b \operatorname {CosIntegral}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\operatorname {FresnelC}(b x)}{x} \]

[In]

Integrate[FresnelC[b*x]/x^2,x]

[Out]

(b*CosIntegral[(b^2*Pi*x^2)/2])/2 - FresnelC[b*x]/x

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
parts	\(\frac {b \,\operatorname {Ci}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2}-\frac {\operatorname {FresnelC}\left (b x \right )}{x}\)	\(24\)
derivativedivides	\(b \left (-\frac {\operatorname {FresnelC}\left (b x \right )}{b x}+\frac {\operatorname {Ci}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2}\right )\)	\(28\)
default	\(b \left (-\frac {\operatorname {FresnelC}\left (b x \right )}{b x}+\frac {\operatorname {Ci}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2}\right )\)	\(28\)
meijerg	\(\frac {b \sqrt {\pi }\, \left (-\frac {\pi ^{\frac {3}{2}} x^{4} b^{4} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{4}\right ], \left [\frac {3}{2}, 2, 2, \frac {9}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{10}+\frac {8 \gamma -8 \ln \left (2\right )-16+16 \ln \left (x \right )+8 \ln \left (\pi \right )+16 \ln \left (b \right )}{\sqrt {\pi }}\right )}{16}\)	\(66\)

[In]

int(FresnelC(b*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^2} \, dx=\frac {b x \operatorname {Ci}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, \operatorname {C}\left (b x\right )}{2 \, x} \]

[In]

integrate(fresnel_cos(b*x)/x^2,x, algorithm="fricas")

[Out]

1/2*(b*x*cos_integral(1/2*pi*b^2*x^2) - 2*fresnel_cos(b*x))/x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (20) = 40\).

Time = 0.55 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^2} \, dx=- \frac {\pi ^{2} b^{5} x^{4} \Gamma \left (\frac {5}{4}\right ) {{}_{3}F_{4}\left (\begin {matrix} 1, 1, \frac {5}{4} \\ \frac {3}{2}, 2, 2, \frac {9}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{128 \Gamma \left (\frac {9}{4}\right )} + \frac {b \log {\left (b^{4} x^{4} \right )}}{4} \]

[In]

integrate(fresnelc(b*x)/x**2,x)

[Out]

-pi**2*b**5*x**4*gamma(5/4)*hyper((1, 1, 5/4), (3/2, 2, 2, 9/4), -pi**2*b**4*x**4/16)/(128*gamma(9/4)) + b*log
(b**4*x**4)/4

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\operatorname {FresnelC}(b x)}{x^2} \, dx=\frac {1}{4} \, b {\left ({\rm Ei}\left (\frac {1}{2} i \, \pi b^{2} x^{2}\right ) + {\rm Ei}\left (-\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} - \frac {\operatorname {C}\left (b x\right )}{x} \]

[In]

integrate(fresnel_cos(b*x)/x^2,x, algorithm="maxima")

[Out]

1/4*b*(Ei(1/2*I*pi*b^2*x^2) + Ei(-1/2*I*pi*b^2*x^2)) - fresnel_cos(b*x)/x

Giac [F]

\[ \int \frac {\operatorname {FresnelC}(b x)}{x^2} \, dx=\int { \frac {\operatorname {C}\left (b x\right )}{x^{2}} \,d x } \]

[In]

integrate(fresnel_cos(b*x)/x^2,x, algorithm="giac")

[Out]

integrate(fresnel_cos(b*x)/x^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\operatorname {FresnelC}(b x)}{x^2} \, dx=\int \frac {\mathrm {FresnelC}\left (b\,x\right )}{x^2} \,d x \]

[In]

int(FresnelC(b*x)/x^2,x)

[Out]

int(FresnelC(b*x)/x^2, x)