\(\int \cos (b x) \text {Si}(b x) \, dx\) [49]

Optimal result

Integrand size = 9, antiderivative size = 34 \[ \int \cos (b x) \text {Si}(b x) \, dx=\frac {\operatorname {CosIntegral}(2 b x)}{2 b}-\frac {\log (x)}{2 b}+\frac {\sin (b x) \text {Si}(b x)}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6652, 12, 3393, 3383} \[ \int \cos (b x) \text {Si}(b x) \, dx=\frac {\operatorname {CosIntegral}(2 b x)}{2 b}+\frac {\text {Si}(b x) \sin (b x)}{b}-\frac {\log (x)}{2 b} \]

[In]

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

[Out]

CosIntegral[2*b*x]/(2*b) - Log[x]/(2*b) + (Sin[b*x]*SinIntegral[b*x])/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 3383

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 3393

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 6652

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

Rubi steps \begin{align*} \text {integral}& = \frac {\sin (b x) \text {Si}(b x)}{b}-\int \frac {\sin ^2(b x)}{b x} \, dx \\ & = \frac {\sin (b x) \text {Si}(b x)}{b}-\frac {\int \frac {\sin ^2(b x)}{x} \, dx}{b} \\ & = \frac {\sin (b x) \text {Si}(b x)}{b}-\frac {\int \left (\frac {1}{2 x}-\frac {\cos (2 b x)}{2 x}\right ) \, dx}{b} \\ & = -\frac {\log (x)}{2 b}+\frac {\sin (b x) \text {Si}(b x)}{b}+\frac {\int \frac {\cos (2 b x)}{x} \, dx}{2 b} \\ & = \frac {\operatorname {CosIntegral}(2 b x)}{2 b}-\frac {\log (x)}{2 b}+\frac {\sin (b x) \text {Si}(b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \cos (b x) \text {Si}(b x) \, dx=\frac {\operatorname {CosIntegral}(2 b x)}{2 b}-\frac {\log (b x)}{2 b}+\frac {\sin (b x) \text {Si}(b x)}{b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.60 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {\operatorname {Si}\left (b x \right ) \sin \left (b x \right )-\frac {\ln \left (b x \right )}{2}+\frac {\operatorname {Ci}\left (2 b x \right )}{2}}{b}\)	\(28\)
default	\(\frac {\operatorname {Si}\left (b x \right ) \sin \left (b x \right )-\frac {\ln \left (b x \right )}{2}+\frac {\operatorname {Ci}\left (2 b x \right )}{2}}{b}\)	\(28\)

[In]

int(cos(b*x)*Si(b*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74 \[ \int \cos (b x) \text {Si}(b x) \, dx=\frac {2 \, \sin \left (b x\right ) \operatorname {Si}\left (b x\right ) + \operatorname {Ci}\left (2 \, b x\right ) - \log \left (x\right )}{2 \, b} \]

[In]

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

[Out]

1/2*(2*sin(b*x)*sin_integral(b*x) + cos_integral(2*b*x) - log(x))/b

Sympy [F]

\[ \int \cos (b x) \text {Si}(b x) \, dx=\int \cos {\left (b x \right )} \operatorname {Si}{\left (b x \right )}\, dx \]

[In]

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

[Out]

Integral(cos(b*x)*Si(b*x), x)

Maxima [F]

\[ \int \cos (b x) \text {Si}(b x) \, dx=\int { \cos \left (b x\right ) \operatorname {Si}\left (b x\right ) \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \cos (b x) \text {Si}(b x) \, dx=\frac {\sin \left (b x\right ) \operatorname {Si}\left (b x\right )}{b} + \frac {\operatorname {Ci}\left (2 \, b x\right ) + \operatorname {Ci}\left (-2 \, b x\right ) - 2 \, \log \left (x\right )}{4 \, b} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \cos (b x) \text {Si}(b x) \, dx=\frac {\mathrm {cosint}\left (2\,b\,x\right )-\ln \left (x\right )+2\,\mathrm {sinint}\left (b\,x\right )\,\sin \left (b\,x\right )}{2\,b} \]

[In]

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

[Out]

(cosint(2*b*x) - log(x) + 2*sinint(b*x)*sin(b*x))/(2*b)