3.75 \(\int \frac{\text{CosIntegral}(b x)}{x^2} \, dx\)

Optimal. Leaf size=26 \[ -\frac{\text{CosIntegral}(b x)}{x}-b \text{Si}(b x)-\frac{\cos (b x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.0478876, antiderivative size = 26, 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 = {6504, 12, 3297, 3299} \[ -\frac{\text{CosIntegral}(b x)}{x}-b \text{Si}(b x)-\frac{\cos (b x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6504

Int[CosIntegral[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*CosIntegr
al[a + b*x])/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[((c + d*x)^(m + 1)*Cos[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{Ci}(b x)}{x^2} \, dx &=-\frac{\text{Ci}(b x)}{x}+b \int \frac{\cos (b x)}{b x^2} \, dx\\ &=-\frac{\text{Ci}(b x)}{x}+\int \frac{\cos (b x)}{x^2} \, dx\\ &=-\frac{\cos (b x)}{x}-\frac{\text{Ci}(b x)}{x}-b \int \frac{\sin (b x)}{x} \, dx\\ &=-\frac{\cos (b x)}{x}-\frac{\text{Ci}(b x)}{x}-b \text{Si}(b x)\\ \end{align*}

Mathematica [A]  time = 0.0108275, size = 26, normalized size = 1. \[ -\frac{\text{CosIntegral}(b x)}{x}-b \text{Si}(b x)-\frac{\cos (b x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.047, size = 34, normalized size = 1.3 \begin{align*} b \left ( -{\frac{{\it Ci} \left ( bx \right ) }{bx}}-{\frac{\cos \left ( bx \right ) }{bx}}-{\it Si} \left ( bx \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(-Ci(b*x)/b/x-cos(b*x)/b/x-Si(b*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Ci(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{Ci}\left (b x\right )}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(cos_integral(b*x)/x^2, x)

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Sympy [B]  time = 1.38407, size = 42, normalized size = 1.62 \begin{align*} - \frac{b^{2} x{{}_{3}F_{4}\left (\begin{matrix} \frac{1}{2}, 1, 1 \\ \frac{3}{2}, \frac{3}{2}, 2, 2 \end{matrix}\middle |{- \frac{b^{2} x^{2}}{4}} \right )}}{4} - \frac{\log{\left (b^{2} x^{2} \right )}}{2 x} - \frac{1}{x} - \frac{\gamma }{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**2*x*hyper((1/2, 1, 1), (3/2, 3/2, 2, 2), -b**2*x**2/4)/4 - log(b**2*x**2)/(2*x) - 1/x - EulerGamma/x

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Giac [C]  time = 1.20492, size = 147, normalized size = 5.65 \begin{align*} -\frac{b x \Im \left ( \operatorname{Ci}\left (b x\right ) \right ) \tan \left (\frac{1}{2} \, b x\right )^{2} - b x \Im \left ( \operatorname{Ci}\left (-b x\right ) \right ) \tan \left (\frac{1}{2} \, b x\right )^{2} + 2 \, b x \operatorname{Si}\left (b x\right ) \tan \left (\frac{1}{2} \, b x\right )^{2} + b x \Im \left ( \operatorname{Ci}\left (b x\right ) \right ) - b x \Im \left ( \operatorname{Ci}\left (-b x\right ) \right ) + 2 \, b x \operatorname{Si}\left (b x\right ) - 2 \, \tan \left (\frac{1}{2} \, b x\right )^{2} + 2}{2 \,{\left (x \tan \left (\frac{1}{2} \, b x\right )^{2} + x\right )}} - \frac{\operatorname{Ci}\left (b x\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*x*imag_part(cos_integral(b*x))*tan(1/2*b*x)^2 - b*x*imag_part(cos_integral(-b*x))*tan(1/2*b*x)^2 + 2*b
*x*sin_integral(b*x)*tan(1/2*b*x)^2 + b*x*imag_part(cos_integral(b*x)) - b*x*imag_part(cos_integral(-b*x)) + 2
*b*x*sin_integral(b*x) - 2*tan(1/2*b*x)^2 + 2)/(x*tan(1/2*b*x)^2 + x) - cos_integral(b*x)/x