3.8 \(\int \frac{\text{Si}(b x)}{x^3} \, dx\)

Optimal. Leaf size=46 \[ -\frac{1}{4} b^2 \text{Si}(b x)-\frac{\text{Si}(b x)}{2 x^2}-\frac{\sin (b x)}{4 x^2}-\frac{b \cos (b x)}{4 x} \]

[Out]

-(b*Cos[b*x])/(4*x) - Sin[b*x]/(4*x^2) - (b^2*SinIntegral[b*x])/4 - SinIntegral[b*x]/(2*x^2)

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Rubi [A]  time = 0.0646036, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6503, 12, 3297, 3299} \[ -\frac{1}{4} b^2 \text{Si}(b x)-\frac{\text{Si}(b x)}{2 x^2}-\frac{\sin (b x)}{4 x^2}-\frac{b \cos (b x)}{4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[b*x])/(4*x) - Sin[b*x]/(4*x^2) - (b^2*SinIntegral[b*x])/4 - SinIntegral[b*x]/(2*x^2)

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 3299

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

Rubi steps

\begin{align*} \int \frac{\text{Si}(b x)}{x^3} \, dx &=-\frac{\text{Si}(b x)}{2 x^2}+\frac{1}{2} b \int \frac{\sin (b x)}{b x^3} \, dx\\ &=-\frac{\text{Si}(b x)}{2 x^2}+\frac{1}{2} \int \frac{\sin (b x)}{x^3} \, dx\\ &=-\frac{\sin (b x)}{4 x^2}-\frac{\text{Si}(b x)}{2 x^2}+\frac{1}{4} b \int \frac{\cos (b x)}{x^2} \, dx\\ &=-\frac{b \cos (b x)}{4 x}-\frac{\sin (b x)}{4 x^2}-\frac{\text{Si}(b x)}{2 x^2}-\frac{1}{4} b^2 \int \frac{\sin (b x)}{x} \, dx\\ &=-\frac{b \cos (b x)}{4 x}-\frac{\sin (b x)}{4 x^2}-\frac{1}{4} b^2 \text{Si}(b x)-\frac{\text{Si}(b x)}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0130357, size = 46, normalized size = 1. \[ -\frac{1}{4} b^2 \text{Si}(b x)-\frac{\text{Si}(b x)}{2 x^2}-\frac{\sin (b x)}{4 x^2}-\frac{b \cos (b x)}{4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Cos[b*x])/(4*x) - Sin[b*x]/(4*x^2) - (b^2*SinIntegral[b*x])/4 - SinIntegral[b*x]/(2*x^2)

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Maple [A]  time = 0.052, size = 48, normalized size = 1. \begin{align*}{b}^{2} \left ( -{\frac{{\it Si} \left ( bx \right ) }{2\,{b}^{2}{x}^{2}}}-{\frac{\sin \left ( bx \right ) }{4\,{b}^{2}{x}^{2}}}-{\frac{\cos \left ( bx \right ) }{4\,bx}}-{\frac{{\it Si} \left ( bx \right ) }{4}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*(-1/2*Si(b*x)/b^2/x^2-1/4*sin(b*x)/b^2/x^2-1/4*cos(b*x)/b/x-1/4*Si(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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Si(b*x)/x^3, 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^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sin_integral(b*x)/x^3, x)

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Sympy [A]  time = 0.891122, size = 41, normalized size = 0.89 \begin{align*} - \frac{b^{2} \operatorname{Si}{\left (b x \right )}}{4} - \frac{b \cos{\left (b x \right )}}{4 x} - \frac{\sin{\left (b x \right )}}{4 x^{2}} - \frac{\operatorname{Si}{\left (b x \right )}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**2*Si(b*x)/4 - b*cos(b*x)/(4*x) - sin(b*x)/(4*x**2) - Si(b*x)/(2*x**2)

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Giac [C]  time = 1.35264, size = 201, normalized size = 4.37 \begin{align*} -\frac{b^{2} x^{2} \Im \left ( \operatorname{Ci}\left (b x\right ) \right ) \tan \left (\frac{1}{2} \, b x\right )^{2} - b^{2} x^{2} \Im \left ( \operatorname{Ci}\left (-b x\right ) \right ) \tan \left (\frac{1}{2} \, b x\right )^{2} + 2 \, b^{2} x^{2} \operatorname{Si}\left (b x\right ) \tan \left (\frac{1}{2} \, b x\right )^{2} + b^{2} x^{2} \Im \left ( \operatorname{Ci}\left (b x\right ) \right ) - b^{2} x^{2} \Im \left ( \operatorname{Ci}\left (-b x\right ) \right ) + 2 \, b^{2} x^{2} \operatorname{Si}\left (b x\right ) - 2 \, b x \tan \left (\frac{1}{2} \, b x\right )^{2} + 2 \, b x + 4 \, \tan \left (\frac{1}{2} \, b x\right )}{8 \,{\left (x^{2} \tan \left (\frac{1}{2} \, b x\right )^{2} + x^{2}\right )}} - \frac{\operatorname{Si}\left (b x\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b^2*x^2*imag_part(cos_integral(b*x))*tan(1/2*b*x)^2 - b^2*x^2*imag_part(cos_integral(-b*x))*tan(1/2*b*x)
^2 + 2*b^2*x^2*sin_integral(b*x)*tan(1/2*b*x)^2 + b^2*x^2*imag_part(cos_integral(b*x)) - b^2*x^2*imag_part(cos
_integral(-b*x)) + 2*b^2*x^2*sin_integral(b*x) - 2*b*x*tan(1/2*b*x)^2 + 2*b*x + 4*tan(1/2*b*x))/(x^2*tan(1/2*b
*x)^2 + x^2) - 1/2*sin_integral(b*x)/x^2