3.24 \(\int \frac{\text{Si}(a+b x)}{x^3} \, dx\)

Optimal. Leaf size=111 \[ -\frac{b^2 \sin (a) \text{CosIntegral}(b x)}{2 a^2}+\frac{b^2 \text{Si}(a+b x)}{2 a^2}-\frac{b^2 \cos (a) \text{Si}(b x)}{2 a^2}+\frac{b^2 \cos (a) \text{CosIntegral}(b x)}{2 a}-\frac{b^2 \sin (a) \text{Si}(b x)}{2 a}-\frac{\text{Si}(a+b x)}{2 x^2}-\frac{b \sin (a+b x)}{2 a x} \]

[Out]

(b^2*Cos[a]*CosIntegral[b*x])/(2*a) - (b^2*CosIntegral[b*x]*Sin[a])/(2*a^2) - (b*Sin[a + b*x])/(2*a*x) - (b^2*
Cos[a]*SinIntegral[b*x])/(2*a^2) - (b^2*Sin[a]*SinIntegral[b*x])/(2*a) + (b^2*SinIntegral[a + b*x])/(2*a^2) -
SinIntegral[a + b*x]/(2*x^2)

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Rubi [A]  time = 0.325046, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6503, 6742, 3297, 3303, 3299, 3302} \[ -\frac{b^2 \sin (a) \text{CosIntegral}(b x)}{2 a^2}+\frac{b^2 \text{Si}(a+b x)}{2 a^2}-\frac{b^2 \cos (a) \text{Si}(b x)}{2 a^2}+\frac{b^2 \cos (a) \text{CosIntegral}(b x)}{2 a}-\frac{b^2 \sin (a) \text{Si}(b x)}{2 a}-\frac{\text{Si}(a+b x)}{2 x^2}-\frac{b \sin (a+b x)}{2 a x} \]

Antiderivative was successfully verified.

[In]

Int[SinIntegral[a + b*x]/x^3,x]

[Out]

(b^2*Cos[a]*CosIntegral[b*x])/(2*a) - (b^2*CosIntegral[b*x]*Sin[a])/(2*a^2) - (b*Sin[a + b*x])/(2*a*x) - (b^2*
Cos[a]*SinIntegral[b*x])/(2*a^2) - (b^2*Sin[a]*SinIntegral[b*x])/(2*a) + (b^2*SinIntegral[a + b*x])/(2*a^2) -
SinIntegral[a + 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 6742

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

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

Mathematica [A]  time = 0.331378, size = 84, normalized size = 0.76 \[ -\frac{a^2 \text{Si}(a+b x)-b^2 x^2 (a \cos (a)-\sin (a)) \text{CosIntegral}(b x)-b^2 x^2 \text{Si}(a+b x)+b^2 x^2 (a \sin (a)+\cos (a)) \text{Si}(b x)+a b x \sin (a+b x)}{2 a^2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[SinIntegral[a + b*x]/x^3,x]

[Out]

-(-(b^2*x^2*CosIntegral[b*x]*(a*Cos[a] - Sin[a])) + a*b*x*Sin[a + b*x] + b^2*x^2*(Cos[a] + a*Sin[a])*SinIntegr
al[b*x] + a^2*SinIntegral[a + b*x] - b^2*x^2*SinIntegral[a + b*x])/(2*a^2*x^2)

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Maple [A]  time = 0.06, size = 86, normalized size = 0.8 \begin{align*}{b}^{2} \left ( -{\frac{{\it Si} \left ( bx+a \right ) }{2\,{b}^{2}{x}^{2}}}+{\frac{{\it Si} \left ( bx+a \right ) }{2\,{a}^{2}}}+{\frac{1}{2\,a} \left ( -{\frac{\sin \left ( bx+a \right ) }{bx}}-{\it Si} \left ( bx \right ) \sin \left ( a \right ) +{\it Ci} \left ( bx \right ) \cos \left ( a \right ) \right ) }-{\frac{{\it Si} \left ( bx \right ) \cos \left ( a \right ) +{\it Ci} \left ( bx \right ) \sin \left ( a \right ) }{2\,{a}^{2}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*(-1/2*Si(b*x+a)/b^2/x^2+1/2/a^2*Si(b*x+a)+1/2/a*(-sin(b*x+a)/b/x-Si(b*x)*sin(a)+Ci(b*x)*cos(a))-1/2/a^2*(S
i(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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sin_integral(b*x + a)/x^3, 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^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(Si(a + b*x)/x**3, x)

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Giac [C]  time = 1.33659, size = 1092, normalized size = 9.84 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a*b*x*real_part(cos_integral(b*x))*tan(1/2*b*x)^2*tan(1/2*a)^2 + a*b*x*real_part(cos_integral(-b*x))*tan
(1/2*b*x)^2*tan(1/2*a)^2 + 2*a*b*x*imag_part(cos_integral(b*x))*tan(1/2*b*x)^2*tan(1/2*a) - 2*a*b*x*imag_part(
cos_integral(-b*x))*tan(1/2*b*x)^2*tan(1/2*a) + 4*a*b*x*sin_integral(b*x)*tan(1/2*b*x)^2*tan(1/2*a) - b*x*imag
_part(cos_integral(b*x + a))*tan(1/2*b*x)^2*tan(1/2*a)^2 - b*x*imag_part(cos_integral(b*x))*tan(1/2*b*x)^2*tan
(1/2*a)^2 + b*x*imag_part(cos_integral(-b*x - a))*tan(1/2*b*x)^2*tan(1/2*a)^2 + b*x*imag_part(cos_integral(-b*
x))*tan(1/2*b*x)^2*tan(1/2*a)^2 - 2*b*x*sin_integral(b*x + a)*tan(1/2*b*x)^2*tan(1/2*a)^2 - 2*b*x*sin_integral
(b*x)*tan(1/2*b*x)^2*tan(1/2*a)^2 - a*b*x*real_part(cos_integral(b*x))*tan(1/2*b*x)^2 - a*b*x*real_part(cos_in
tegral(-b*x))*tan(1/2*b*x)^2 + 2*b*x*real_part(cos_integral(b*x))*tan(1/2*b*x)^2*tan(1/2*a) + 2*b*x*real_part(
cos_integral(-b*x))*tan(1/2*b*x)^2*tan(1/2*a) + a*b*x*real_part(cos_integral(b*x))*tan(1/2*a)^2 + a*b*x*real_p
art(cos_integral(-b*x))*tan(1/2*a)^2 - b*x*imag_part(cos_integral(b*x + a))*tan(1/2*b*x)^2 + b*x*imag_part(cos
_integral(b*x))*tan(1/2*b*x)^2 + b*x*imag_part(cos_integral(-b*x - a))*tan(1/2*b*x)^2 - b*x*imag_part(cos_inte
gral(-b*x))*tan(1/2*b*x)^2 - 2*b*x*sin_integral(b*x + a)*tan(1/2*b*x)^2 + 2*b*x*sin_integral(b*x)*tan(1/2*b*x)
^2 + 2*a*b*x*imag_part(cos_integral(b*x))*tan(1/2*a) - 2*a*b*x*imag_part(cos_integral(-b*x))*tan(1/2*a) + 4*a*
b*x*sin_integral(b*x)*tan(1/2*a) - b*x*imag_part(cos_integral(b*x + a))*tan(1/2*a)^2 - b*x*imag_part(cos_integ
ral(b*x))*tan(1/2*a)^2 + b*x*imag_part(cos_integral(-b*x - a))*tan(1/2*a)^2 + b*x*imag_part(cos_integral(-b*x)
)*tan(1/2*a)^2 - 2*b*x*sin_integral(b*x + a)*tan(1/2*a)^2 - 2*b*x*sin_integral(b*x)*tan(1/2*a)^2 - a*b*x*real_
part(cos_integral(b*x)) - a*b*x*real_part(cos_integral(-b*x)) + 2*b*x*real_part(cos_integral(b*x))*tan(1/2*a)
+ 2*b*x*real_part(cos_integral(-b*x))*tan(1/2*a) - 4*a*tan(1/2*b*x)^2*tan(1/2*a) - 4*a*tan(1/2*b*x)*tan(1/2*a)
^2 - b*x*imag_part(cos_integral(b*x + a)) + b*x*imag_part(cos_integral(b*x)) + b*x*imag_part(cos_integral(-b*x
- a)) - b*x*imag_part(cos_integral(-b*x)) - 2*b*x*sin_integral(b*x + a) + 2*b*x*sin_integral(b*x) + 4*a*tan(1
/2*b*x) + 4*a*tan(1/2*a))*b/(a^2*x*tan(1/2*b*x)^2*tan(1/2*a)^2 + a^2*x*tan(1/2*b*x)^2 + a^2*x*tan(1/2*a)^2 + a
^2*x) - 1/2*sin_integral(b*x + a)/x^2