\(\int \frac {\operatorname {FresnelS}(b x)}{x^2} \, dx\) [10]

Optimal result

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

[Out]

-FresnelS(b*x)/x+1/2*b*Si(1/2*b^2*Pi*x^2)

Rubi [A] (verified)

Time = 0.01 (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 = {6561, 3456} \[ \int \frac {\operatorname {FresnelS}(b x)}{x^2} \, dx=\frac {1}{2} b \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {\operatorname {FresnelS}(b x)}{x} \]

[In]

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

[Out]

-(FresnelS[b*x]/x) + (b*SinIntegral[(b^2*Pi*x^2)/2])/2

Rule 3456

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

Rule 6561

Int[FresnelS[(b_.)*(x_)]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*(FresnelS[b*x]/(d*(m + 1))), x] -
Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\operatorname {FresnelS}(b x)}{x}+b \int \frac {\sin \left (\frac {1}{2} b^2 \pi x^2\right )}{x} \, dx \\ & = -\frac {\operatorname {FresnelS}(b x)}{x}+\frac {1}{2} b \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

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

[In]

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

[Out]

-(FresnelS[b*x]/x) + (b*SinIntegral[(b^2*Pi*x^2)/2])/2

Maple [A] (verified)

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

method	result	size
parts	\(-\frac {\operatorname {FresnelS}\left (b x \right )}{x}+\frac {b \,\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2}\)	\(24\)
derivativedivides	\(b \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{b x}+\frac {\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2}\right )\)	\(28\)
default	\(b \left (-\frac {\operatorname {FresnelS}\left (b x \right )}{b x}+\frac {\operatorname {Si}\left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2}\right )\)	\(28\)
meijerg	\(\frac {b^{3} \pi \,x^{2} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {3}{2}, \frac {3}{2}, \frac {7}{4}\right ], -\frac {x^{4} \pi ^{2} b^{4}}{16}\right )}{12}\)	\(29\)

[In]

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

[Out]

-FresnelS(b*x)/x+1/2*b*Si(1/2*b^2*Pi*x^2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^2} \, dx=\frac {b x \operatorname {Si}\left (\frac {1}{2} \, \pi b^{2} x^{2}\right ) - 2 \, \operatorname {S}\left (b x\right )}{2 \, x} \]

[In]

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

[Out]

1/2*(b*x*sin_integral(1/2*pi*b^2*x^2) - 2*fresnel_sin(b*x))/x

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (20) = 40\).

Time = 0.36 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^2} \, dx=\frac {\pi b^{3} x^{2} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2}, \frac {3}{2}, \frac {7}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 \Gamma \left (\frac {7}{4}\right )} \]

[In]

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

[Out]

pi*b**3*x**2*gamma(3/4)*hyper((1/2, 3/4), (3/2, 3/2, 7/4), -pi**2*b**4*x**4/16)/(16*gamma(7/4))

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.41 \[ \int \frac {\operatorname {FresnelS}(b x)}{x^2} \, dx=-\frac {1}{4} \, b {\left (i \, {\rm Ei}\left (\frac {1}{2} i \, \pi b^{2} x^{2}\right ) - i \, {\rm Ei}\left (-\frac {1}{2} i \, \pi b^{2} x^{2}\right )\right )} - \frac {\operatorname {S}\left (b x\right )}{x} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(fresnel_sin(b*x)/x^2, x)

Mupad [F(-1)]

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

[In]

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

[Out]

int(FresnelS(b*x)/x^2, x)