\(\int \frac {\cos (b x) \text {Si}(b x)}{x^2} \, dx\) [47]

Optimal result

Integrand size = 12, antiderivative size = 44 \[ \int \frac {\cos (b x) \text {Si}(b x)}{x^2} \, dx=b \operatorname {CosIntegral}(2 b x)-\frac {\sin (2 b x)}{2 x}-\frac {\cos (b x) \text {Si}(b x)}{x}-\frac {1}{2} b \text {Si}(b x)^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6656, 6818, 12, 4491, 3378, 3383} \[ \int \frac {\cos (b x) \text {Si}(b x)}{x^2} \, dx=b \operatorname {CosIntegral}(2 b x)-\frac {1}{2} b \text {Si}(b x)^2-\frac {\text {Si}(b x) \cos (b x)}{x}-\frac {\sin (2 b x)}{2 x} \]

[In]

Int[(Cos[b*x]*SinIntegral[b*x])/x^2,x]

[Out]

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

Rule 12

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

Rule 3378

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 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 4491

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 6656

Int[Cos[(a_.) + (b_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*SinIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[(e +
f*x)^(m + 1)*Cos[a + b*x]*(SinIntegral[c + d*x]/(f*(m + 1))), x] + (Dist[b/(f*(m + 1)), Int[(e + f*x)^(m + 1)
*Sin[a + b*x]*SinIntegral[c + d*x], x], x] - Dist[d/(f*(m + 1)), Int[(e + f*x)^(m + 1)*Cos[a + b*x]*(Sin[c + d
*x]/(c + d*x)), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[m, -1]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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

Mathematica [F]

\[ \int \frac {\cos (b x) \text {Si}(b x)}{x^2} \, dx=\int \frac {\cos (b x) \text {Si}(b x)}{x^2} \, dx \]

[In]

Integrate[(Cos[b*x]*SinIntegral[b*x])/x^2,x]

[Out]

Integrate[(Cos[b*x]*SinIntegral[b*x])/x^2, x]

Maple [F]

\[\int \frac {\cos \left (b x \right ) \operatorname {Si}\left (b x \right )}{x^{2}}d x\]

[In]

int(cos(b*x)*Si(b*x)/x^2,x)

[Out]

int(cos(b*x)*Si(b*x)/x^2,x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {\cos (b x) \text {Si}(b x)}{x^2} \, dx=-\frac {b x \operatorname {Si}\left (b x\right )^{2} - 2 \, b x \operatorname {Ci}\left (2 \, b x\right ) + 2 \, \cos \left (b x\right ) \sin \left (b x\right ) + 2 \, \cos \left (b x\right ) \operatorname {Si}\left (b x\right )}{2 \, x} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\cos (b x) \text {Si}(b x)}{x^2} \, dx=\int \frac {\cos {\left (b x \right )} \operatorname {Si}{\left (b x \right )}}{x^{2}}\, dx \]

[In]

integrate(cos(b*x)*Si(b*x)/x**2,x)

[Out]

Integral(cos(b*x)*Si(b*x)/x**2, x)

Maxima [F]

\[ \int \frac {\cos (b x) \text {Si}(b x)}{x^2} \, dx=\int { \frac {\cos \left (b x\right ) \operatorname {Si}\left (b x\right )}{x^{2}} \,d x } \]

[In]

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

[Out]

integrate(cos(b*x)*sin_integral(b*x)/x^2, x)

Giac [F]

\[ \int \frac {\cos (b x) \text {Si}(b x)}{x^2} \, dx=\int { \frac {\cos \left (b x\right ) \operatorname {Si}\left (b x\right )}{x^{2}} \,d x } \]

[In]

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

[Out]

integrate(cos(b*x)*sin_integral(b*x)/x^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (b x) \text {Si}(b x)}{x^2} \, dx=\int \frac {\mathrm {sinint}\left (b\,x\right )\,\cos \left (b\,x\right )}{x^2} \,d x \]

[In]

int((sinint(b*x)*cos(b*x))/x^2,x)

[Out]

int((sinint(b*x)*cos(b*x))/x^2, x)