3.129 \(\int \cos (a+b x) \text {Ci}(a+b x) \, dx\)

Optimal. Leaf size=33 \[ \frac {\text {Ci}(a+b x) \sin (a+b x)}{b}-\frac {\text {Si}(2 a+2 b x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {6512, 4406, 12, 3299} \[ \frac {\text {CosIntegral}(a+b x) \sin (a+b x)}{b}-\frac {\text {Si}(2 a+2 b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 32, normalized size = 0.97 \[ \frac {\text {Ci}(a+b x) \sin (a+b x)}{b}-\frac {\text {Si}(2 (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\cos \left (b x + a\right ) \operatorname {Ci}\left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.15, size = 56, normalized size = 1.70 \[ \frac {\operatorname {Ci}\left (b x + a\right ) \sin \left (b x + a\right )}{b} - \frac {\Im \left (\operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \right ) - \Im \left (\operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \right ) + 2 \, \operatorname {Si}\left (2 \, b x + 2 \, a\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos_integral(b*x + a)*sin(b*x + a)/b - 1/4*(imag_part(cos_integral(2*b*x + 2*a)) - imag_part(cos_integral(-2*b
*x - 2*a)) + 2*sin_integral(2*b*x + 2*a))/b

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maple [A]  time = 0.01, size = 30, normalized size = 0.91 \[ \frac {\Ci \left (b x +a \right ) \sin \left (b x +a \right )-\frac {\Si \left (2 b x +2 a \right )}{2}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\rm Ci}\left (b x + a\right ) \cos \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \mathrm {cosint}\left (a+b\,x\right )\,\cos \left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosint(a + b*x)*cos(a + b*x),x)

[Out]

int(cosint(a + b*x)*cos(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos {\left (a + b x \right )} \operatorname {Ci}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(a + b*x)*Ci(a + b*x), x)

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