3.7 \(\int \frac{\text{Si}(b x)}{x^2} \, dx\)

Optimal. Leaf size=25 \[ b \text{CosIntegral}(b x)-\frac{\text{Si}(b x)}{x}-\frac{\sin (b x)}{x} \]

[Out]

b*CosIntegral[b*x] - Sin[b*x]/x - SinIntegral[b*x]/x

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Rubi [A]  time = 0.0452787, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6503, 12, 3297, 3302} \[ b \text{CosIntegral}(b x)-\frac{\text{Si}(b x)}{x}-\frac{\sin (b x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

b*CosIntegral[b*x] - Sin[b*x]/x - SinIntegral[b*x]/x

Rule 6503

Int[((c_.) + (d_.)*(x_))^(m_.)*SinIntegral[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*SinIntegr
al[a + b*x])/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[((c + d*x)^(m + 1)*Sin[a + b*x])/(a + b*x), x], x] /; F
reeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3297

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 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rubi steps

\begin{align*} \int \frac{\text{Si}(b x)}{x^2} \, dx &=-\frac{\text{Si}(b x)}{x}+b \int \frac{\sin (b x)}{b x^2} \, dx\\ &=-\frac{\text{Si}(b x)}{x}+\int \frac{\sin (b x)}{x^2} \, dx\\ &=-\frac{\sin (b x)}{x}-\frac{\text{Si}(b x)}{x}+b \int \frac{\cos (b x)}{x} \, dx\\ &=b \text{Ci}(b x)-\frac{\sin (b x)}{x}-\frac{\text{Si}(b x)}{x}\\ \end{align*}

Mathematica [A]  time = 0.009746, size = 25, normalized size = 1. \[ b \text{CosIntegral}(b x)-\frac{\text{Si}(b x)}{x}-\frac{\sin (b x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

b*CosIntegral[b*x] - Sin[b*x]/x - SinIntegral[b*x]/x

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Maple [A]  time = 0.054, size = 32, normalized size = 1.3 \begin{align*} b \left ( -{\frac{{\it Si} \left ( bx \right ) }{bx}}-{\frac{\sin \left ( bx \right ) }{bx}}+{\it Ci} \left ( bx \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(-Si(b*x)/b/x-sin(b*x)/b/x+Ci(b*x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\rm Si}\left (b x\right )}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Si(b*x)/x^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{Si}\left (b x\right )}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sin_integral(b*x)/x^2, x)

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Sympy [A]  time = 1.16882, size = 36, normalized size = 1.44 \begin{align*} - \frac{b^{3} x^{2}{{}_{3}F_{4}\left (\begin{matrix} 1, 1, \frac{3}{2} \\ 2, 2, \frac{5}{2}, \frac{5}{2} \end{matrix}\middle |{- \frac{b^{2} x^{2}}{4}} \right )}}{36} + \frac{b \log{\left (b^{2} x^{2} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**3*x**2*hyper((1, 1, 3/2), (2, 2, 5/2, 5/2), -b**2*x**2/4)/36 + b*log(b**2*x**2)/2

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Giac [A]  time = 1.20313, size = 50, normalized size = 2. \begin{align*} \frac{b x \operatorname{Ci}\left (b x\right ) + b x \operatorname{Ci}\left (-b x\right ) - 2 \, \sin \left (b x\right )}{2 \, x} - \frac{\operatorname{Si}\left (b x\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x*cos_integral(b*x) + b*x*cos_integral(-b*x) - 2*sin(b*x))/x - sin_integral(b*x)/x