3.23 \(\int \frac{\text{Si}(a+b x)}{x^2} \, dx\)

Optimal. Leaf size=46 \[ \frac{b \sin (a) \text{CosIntegral}(b x)}{a}-\frac{b \text{Si}(a+b x)}{a}-\frac{\text{Si}(a+b x)}{x}+\frac{b \cos (a) \text{Si}(b x)}{a} \]

[Out]

(b*CosIntegral[b*x]*Sin[a])/a + (b*Cos[a]*SinIntegral[b*x])/a - (b*SinIntegral[a + b*x])/a - SinIntegral[a + b
*x]/x

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Rubi [A]  time = 0.232405, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6503, 6742, 3303, 3299, 3302} \[ \frac{b \sin (a) \text{CosIntegral}(b x)}{a}-\frac{b \text{Si}(a+b x)}{a}-\frac{\text{Si}(a+b x)}{x}+\frac{b \cos (a) \text{Si}(b x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[SinIntegral[a + b*x]/x^2,x]

[Out]

(b*CosIntegral[b*x]*Sin[a])/a + (b*Cos[a]*SinIntegral[b*x])/a - (b*SinIntegral[a + b*x])/a - SinIntegral[a + 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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 3303

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

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]

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

Mathematica [A]  time = 0.0732792, size = 39, normalized size = 0.85 \[ \frac{b x \sin (a) \text{CosIntegral}(b x)-(a+b x) \text{Si}(a+b x)+b x \cos (a) \text{Si}(b x)}{a x} \]

Antiderivative was successfully verified.

[In]

Integrate[SinIntegral[a + b*x]/x^2,x]

[Out]

(b*x*CosIntegral[b*x]*Sin[a] + b*x*Cos[a]*SinIntegral[b*x] - (a + b*x)*SinIntegral[a + b*x])/(a*x)

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Maple [A]  time = 0.057, size = 48, normalized size = 1. \begin{align*} b \left ( -{\frac{{\it Si} \left ( bx+a \right ) }{bx}}-{\frac{{\it Si} \left ( bx+a \right ) }{a}}+{\frac{{\it Si} \left ( bx \right ) \cos \left ( a \right ) +{\it Ci} \left ( bx \right ) \sin \left ( a \right ) }{a}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(-Si(b*x+a)/b/x-1/a*Si(b*x+a)+1/a*(Si(b*x)*cos(a)+Ci(b*x)*sin(a)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Si(b*x + a)/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 + a\right )}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sin_integral(b*x + a)/x^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{Si}{\left (a + b x \right )}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(Si(a + b*x)/x**2, x)

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Giac [C]  time = 1.31766, size = 244, normalized size = 5.3 \begin{align*} -\frac{{\left (\Im \left ( \operatorname{Ci}\left (b x + a\right ) \right ) \tan \left (\frac{1}{2} \, a\right )^{2} + \Im \left ( \operatorname{Ci}\left (b x\right ) \right ) \tan \left (\frac{1}{2} \, a\right )^{2} - \Im \left ( \operatorname{Ci}\left (-b x - a\right ) \right ) \tan \left (\frac{1}{2} \, a\right )^{2} - \Im \left ( \operatorname{Ci}\left (-b x\right ) \right ) \tan \left (\frac{1}{2} \, a\right )^{2} + 2 \, \operatorname{Si}\left (b x + a\right ) \tan \left (\frac{1}{2} \, a\right )^{2} + 2 \, \operatorname{Si}\left (b x\right ) \tan \left (\frac{1}{2} \, a\right )^{2} - 2 \, \Re \left ( \operatorname{Ci}\left (b x\right ) \right ) \tan \left (\frac{1}{2} \, a\right ) - 2 \, \Re \left ( \operatorname{Ci}\left (-b x\right ) \right ) \tan \left (\frac{1}{2} \, a\right ) + \Im \left ( \operatorname{Ci}\left (b x + a\right ) \right ) - \Im \left ( \operatorname{Ci}\left (b x\right ) \right ) - \Im \left ( \operatorname{Ci}\left (-b x - a\right ) \right ) + \Im \left ( \operatorname{Ci}\left (-b x\right ) \right ) + 2 \, \operatorname{Si}\left (b x + a\right ) - 2 \, \operatorname{Si}\left (b x\right )\right )} b}{2 \,{\left (a \tan \left (\frac{1}{2} \, a\right )^{2} + a\right )}} - \frac{\operatorname{Si}\left (b x + a\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(imag_part(cos_integral(b*x + a))*tan(1/2*a)^2 + imag_part(cos_integral(b*x))*tan(1/2*a)^2 - imag_part(co
s_integral(-b*x - a))*tan(1/2*a)^2 - imag_part(cos_integral(-b*x))*tan(1/2*a)^2 + 2*sin_integral(b*x + a)*tan(
1/2*a)^2 + 2*sin_integral(b*x)*tan(1/2*a)^2 - 2*real_part(cos_integral(b*x))*tan(1/2*a) - 2*real_part(cos_inte
gral(-b*x))*tan(1/2*a) + imag_part(cos_integral(b*x + a)) - imag_part(cos_integral(b*x)) - imag_part(cos_integ
ral(-b*x - a)) + imag_part(cos_integral(-b*x)) + 2*sin_integral(b*x + a) - 2*sin_integral(b*x))*b/(a*tan(1/2*a
)^2 + a) - sin_integral(b*x + a)/x