3.125 \(\int \text {Ci}(a+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=47 \[ \frac {\text {Ci}(2 a+2 b x)}{2 b}-\frac {\text {Ci}(a+b x) \cos (a+b x)}{b}+\frac {\log (a+b x)}{2 b} \]

[Out]

1/2*Ci(2*b*x+2*a)/b-Ci(b*x+a)*cos(b*x+a)/b+1/2*ln(b*x+a)/b

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {6518, 3312, 3302} \[ \frac {\text {CosIntegral}(2 a+2 b x)}{2 b}-\frac {\text {CosIntegral}(a+b x) \cos (a+b x)}{b}+\frac {\log (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[a + b*x]*CosIntegral[a + b*x])/b) + CosIntegral[2*a + 2*b*x]/(2*b) + Log[a + b*x]/(2*b)

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]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rule 6518

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 46, normalized size = 0.98 \[ \frac {\text {Ci}(2 (a+b x))}{2 b}-\frac {\text {Ci}(a+b x) \cos (a+b x)}{b}+\frac {\log (a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[a + b*x]*CosIntegral[a + b*x])/b) + CosIntegral[2*(a + b*x)]/(2*b) + Log[a + b*x]/(2*b)

________________________________________________________________________________________

fricas [F]  time = 1.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\operatorname {Ci}\left (b x + a\right ) \sin \left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.17, size = 96, normalized size = 2.04 \[ -\frac {\cos \left (b x + a\right ) \operatorname {Ci}\left (b x + a\right )}{b} + \frac {\cos \left (2 \, a\right )^{2} \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) + \cos \left (2 \, a\right )^{2} \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) + \operatorname {Ci}\left (2 \, b x + 2 \, a\right ) \sin \left (2 \, a\right )^{2} + \operatorname {Ci}\left (-2 \, b x - 2 \, a\right ) \sin \left (2 \, a\right )^{2} + 2 \, \log \left (b x + a\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(b*x + a)*cos_integral(b*x + a)/b + 1/4*(cos(2*a)^2*cos_integral(2*b*x + 2*a) + cos(2*a)^2*cos_integral(-2
*b*x - 2*a) + cos_integral(2*b*x + 2*a)*sin(2*a)^2 + cos_integral(-2*b*x - 2*a)*sin(2*a)^2 + 2*log(b*x + a))/b

________________________________________________________________________________________

maple [A]  time = 0.00, size = 44, normalized size = 0.94 \[ \frac {\Ci \left (2 b x +2 a \right )}{2 b}-\frac {\Ci \left (b x +a \right ) \cos \left (b x +a \right )}{b}+\frac {\ln \left (b x +a \right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*Ci(2*b*x+2*a)/b-Ci(b*x+a)*cos(b*x+a)/b+1/2*ln(b*x+a)/b

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \frac {\ln \left (a+b\,x\right )}{2\,b}+\frac {\mathrm {cosint}\left (2\,a+2\,b\,x\right )}{2\,b}-\frac {\mathrm {cosint}\left (a+b\,x\right )\,\cos \left (a+b\,x\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosint(a + b*x)*sin(a + b*x),x)

[Out]

log(a + b*x)/(2*b) + cosint(2*a + 2*b*x)/(2*b) - (cosint(a + b*x)*cos(a + b*x))/b

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b*x)*Ci(a + b*x), x)

________________________________________________________________________________________