3.10 \(\int \frac {S(b x)}{x^2} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{2} b \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {S(b x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6426, 3375} \[ \frac {1}{2} b \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {S(b x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3375

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

Rule 6426

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*b^2*x^2)/2], x], x] /; FreeQ[{b, d, m}, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 27, normalized size = 1.00 \[ \frac {1}{2} b \text {Si}\left (\frac {1}{2} b^2 \pi x^2\right )-\frac {S(b x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm fresnels}\left (b x\right )}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(fresnels(b*x)/x^2, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnels}\left (b x\right )}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnels(b*x)/x^2, x)

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maple [A]  time = 0.02, size = 28, normalized size = 1.04 \[ b \left (-\frac {\mathrm {S}\left (b x \right )}{b x}+\frac {\Si \left (\frac {b^{2} \pi \,x^{2}}{2}\right )}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm fresnels}\left (b x\right )}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(fresnels(b*x)/x^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {\mathrm {FresnelS}\left (b\,x\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(FresnelS(b*x)/x^2, x)

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sympy [B]  time = 0.64, size = 42, normalized size = 1.56 \[ \frac {\pi b^{3} x^{2} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{3}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2}, \frac {3}{2}, \frac {7}{4} \end {matrix}\middle | {- \frac {\pi ^{2} b^{4} x^{4}}{16}} \right )}}{16 \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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