\(\int x \text {Si}(b x) \, dx\) [4]

Optimal result

Integrand size = 6, antiderivative size = 35 \[ \int x \text {Si}(b x) \, dx=\frac {x \cos (b x)}{2 b}-\frac {\sin (b x)}{2 b^2}+\frac {1}{2} x^2 \text {Si}(b x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {6638, 12, 3377, 2717} \[ \int x \text {Si}(b x) \, dx=-\frac {\sin (b x)}{2 b^2}+\frac {1}{2} x^2 \text {Si}(b x)+\frac {x \cos (b x)}{2 b} \]

[In]

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

[Out]

(x*Cos[b*x])/(2*b) - Sin[b*x]/(2*b^2) + (x^2*SinIntegral[b*x])/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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 6638

Int[((c_.) + (d_.)*(x_))^(m_.)*SinIntegral[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(c + d*x)^(m + 1)*(SinIntegr
al[a + b*x]/(d*(m + 1))), x] - Dist[b/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(Sin[a + b*x]/(a + b*x)), x], x] /; F
reeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int x \text {Si}(b x) \, dx=\frac {x \cos (b x)}{2 b}-\frac {\sin (b x)}{2 b^2}+\frac {1}{2} x^2 \text {Si}(b x) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83

method	result	size
parts	\(\frac {x^{2} \operatorname {Si}\left (b x \right )}{2}-\frac {\sin \left (b x \right )-b x \cos \left (b x \right )}{2 b^{2}}\)	\(29\)
derivativedivides	\(\frac {\frac {b^{2} x^{2} \operatorname {Si}\left (b x \right )}{2}-\frac {\sin \left (b x \right )}{2}+\frac {b x \cos \left (b x \right )}{2}}{b^{2}}\)	\(32\)
default	\(\frac {\frac {b^{2} x^{2} \operatorname {Si}\left (b x \right )}{2}-\frac {\sin \left (b x \right )}{2}+\frac {b x \cos \left (b x \right )}{2}}{b^{2}}\)	\(32\)
meijerg	\(\frac {\sqrt {\pi }\, \left (\frac {b x \cos \left (b x \right )}{2 \sqrt {\pi }}-\frac {\sin \left (b x \right )}{2 \sqrt {\pi }}+\frac {b^{2} x^{2} \operatorname {Si}\left (b x \right )}{2 \sqrt {\pi }}\right )}{b^{2}}\)	\(44\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.86 \[ \int x \text {Si}(b x) \, dx=\frac {b^{2} x^{2} \operatorname {Si}\left (b x\right ) + b x \cos \left (b x\right ) - \sin \left (b x\right )}{2 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int x \text {Si}(b x) \, dx=\frac {x^{2} \operatorname {Si}{\left (b x \right )}}{2} + \frac {x \cos {\left (b x \right )}}{2 b} - \frac {\sin {\left (b x \right )}}{2 b^{2}} \]

[In]

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

[Out]

x**2*Si(b*x)/2 + x*cos(b*x)/(2*b) - sin(b*x)/(2*b**2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int x \text {Si}(b x) \, dx=\frac {1}{2} \, x^{2} \operatorname {Si}\left (b x\right ) + \frac {b x \cos \left (b x\right ) - \sin \left (b x\right )}{2 \, b^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int x \text {Si}(b x) \, dx=\frac {1}{2} \, x^{2} \operatorname {Si}\left (b x\right ) + \frac {x \cos \left (b x\right )}{2 \, b} - \frac {\sin \left (b x\right )}{2 \, b^{2}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

(x^2*sinint(b*x))/2 - (sin(b*x) - b*x*cos(b*x))/(2*b^2)