3.117 \(\int \cos (b x) \text{CosIntegral}(b x) \, dx\)

Optimal. Leaf size=25 \[ \frac{\text{CosIntegral}(b x) \sin (b x)}{b}-\frac{\text{Si}(2 b x)}{2 b} \]

[Out]

(CosIntegral[b*x]*Sin[b*x])/b - SinIntegral[2*b*x]/(2*b)

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Rubi [A]  time = 0.043309, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {6512, 12, 4406, 3299} \[ \frac{\text{CosIntegral}(b x) \sin (b x)}{b}-\frac{\text{Si}(2 b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[b*x]*CosIntegral[b*x],x]

[Out]

(CosIntegral[b*x]*Sin[b*x])/b - SinIntegral[2*b*x]/(2*b)

Rule 6512

Int[Cos[(a_.) + (b_.)*(x_)]*CosIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[(Sin[a + b*x]*CosIntegral[c + d
*x])/b, x] - Dist[d/b, Int[(Sin[a + b*x]*Cos[c + d*x])/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x]

Rule 12

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

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 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]

Rubi steps

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

Mathematica [A]  time = 0.0115221, size = 25, normalized size = 1. \[ \frac{\text{CosIntegral}(b x) \sin (b x)}{b}-\frac{\text{Si}(2 b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[b*x]*CosIntegral[b*x],x]

[Out]

(CosIntegral[b*x]*Sin[b*x])/b - SinIntegral[2*b*x]/(2*b)

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Maple [A]  time = 0.049, size = 22, normalized size = 0.9 \begin{align*}{\frac{1}{b} \left ({\it Ci} \left ( bx \right ) \sin \left ( bx \right ) -{\frac{{\it Si} \left ( 2\,bx \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(Ci(b*x)*cos(b*x),x)

[Out]

1/b*(Ci(b*x)*sin(b*x)-1/2*Si(2*b*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(Ci(b*x)*cos(b*x),x, algorithm="maxima")

[Out]

integrate(Ci(b*x)*cos(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(Ci(b*x)*cos(b*x),x, algorithm="fricas")

[Out]

integral(cos(b*x)*cos_integral(b*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos{\left (b x \right )} \operatorname{Ci}{\left (b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(Ci(b*x)*cos(b*x),x)

[Out]

Integral(cos(b*x)*Ci(b*x), x)

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Giac [C]  time = 1.25457, size = 54, normalized size = 2.16 \begin{align*} \frac{\operatorname{Ci}\left (b x\right ) \sin \left (b x\right )}{b} - \frac{\Im \left ( \operatorname{Ci}\left (2 \, b x\right ) \right ) - \Im \left ( \operatorname{Ci}\left (-2 \, b x\right ) \right ) + 2 \, \operatorname{Si}\left (2 \, b x\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(Ci(b*x)*cos(b*x),x, algorithm="giac")

[Out]

cos_integral(b*x)*sin(b*x)/b - 1/4*(imag_part(cos_integral(2*b*x)) - imag_part(cos_integral(-2*b*x)) + 2*sin_i
ntegral(2*b*x))/b