\(\int x^2 \text {Si}(b x) \, dx\) [3]

Optimal result

Integrand size = 8, antiderivative size = 49 \[ \int x^2 \text {Si}(b x) \, dx=-\frac {2 \cos (b x)}{3 b^3}+\frac {x^2 \cos (b x)}{3 b}-\frac {2 x \sin (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(b x) \]

[Out]

-2/3*cos(b*x)/b^3+1/3*x^2*cos(b*x)/b+1/3*x^3*Si(b*x)-2/3*x*sin(b*x)/b^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6638, 12, 3377, 2718} \[ \int x^2 \text {Si}(b x) \, dx=-\frac {2 \cos (b x)}{3 b^3}-\frac {2 x \sin (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(b x)+\frac {x^2 \cos (b x)}{3 b} \]

[In]

Int[x^2*SinIntegral[b*x],x]

[Out]

(-2*Cos[b*x])/(3*b^3) + (x^2*Cos[b*x])/(3*b) - (2*x*Sin[b*x])/(3*b^2) + (x^3*SinIntegral[b*x])/3

Rule 12

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

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[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}{3} x^3 \text {Si}(b x)-\frac {1}{3} b \int \frac {x^2 \sin (b x)}{b} \, dx \\ & = \frac {1}{3} x^3 \text {Si}(b x)-\frac {1}{3} \int x^2 \sin (b x) \, dx \\ & = \frac {x^2 \cos (b x)}{3 b}+\frac {1}{3} x^3 \text {Si}(b x)-\frac {2 \int x \cos (b x) \, dx}{3 b} \\ & = \frac {x^2 \cos (b x)}{3 b}-\frac {2 x \sin (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(b x)+\frac {2 \int \sin (b x) \, dx}{3 b^2} \\ & = -\frac {2 \cos (b x)}{3 b^3}+\frac {x^2 \cos (b x)}{3 b}-\frac {2 x \sin (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(b x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.84 \[ \int x^2 \text {Si}(b x) \, dx=\frac {\left (-2+b^2 x^2\right ) \cos (b x)-2 b x \sin (b x)+b^3 x^3 \text {Si}(b x)}{3 b^3} \]

[In]

Integrate[x^2*SinIntegral[b*x],x]

[Out]

((-2 + b^2*x^2)*Cos[b*x] - 2*b*x*Sin[b*x] + b^3*x^3*SinIntegral[b*x])/(3*b^3)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.88

method	result	size
parts	\(\frac {x^{3} \operatorname {Si}\left (b x \right )}{3}-\frac {-b^{2} x^{2} \cos \left (b x \right )+2 \cos \left (b x \right )+2 b x \sin \left (b x \right )}{3 b^{3}}\)	\(43\)
derivativedivides	\(\frac {\frac {b^{3} x^{3} \operatorname {Si}\left (b x \right )}{3}+\frac {b^{2} x^{2} \cos \left (b x \right )}{3}-\frac {2 \cos \left (b x \right )}{3}-\frac {2 b x \sin \left (b x \right )}{3}}{b^{3}}\)	\(44\)
default	\(\frac {\frac {b^{3} x^{3} \operatorname {Si}\left (b x \right )}{3}+\frac {b^{2} x^{2} \cos \left (b x \right )}{3}-\frac {2 \cos \left (b x \right )}{3}-\frac {2 b x \sin \left (b x \right )}{3}}{b^{3}}\)	\(44\)
meijerg	\(\frac {2 \sqrt {\pi }\, \left (\frac {1}{3 \sqrt {\pi }}-\frac {\left (-\frac {b^{2} x^{2}}{2}+1\right ) \cos \left (b x \right )}{3 \sqrt {\pi }}-\frac {b x \sin \left (b x \right )}{3 \sqrt {\pi }}+\frac {b^{3} x^{3} \operatorname {Si}\left (b x \right )}{6 \sqrt {\pi }}\right )}{b^{3}}\)	\(60\)

[In]

int(x^2*Si(b*x),x,method=_RETURNVERBOSE)

[Out]

1/3*x^3*Si(b*x)-1/3/b^3*(-b^2*x^2*cos(b*x)+2*cos(b*x)+2*b*x*sin(b*x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.80 \[ \int x^2 \text {Si}(b x) \, dx=\frac {b^{3} x^{3} \operatorname {Si}\left (b x\right ) - 2 \, b x \sin \left (b x\right ) + {\left (b^{2} x^{2} - 2\right )} \cos \left (b x\right )}{3 \, b^{3}} \]

[In]

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

[Out]

1/3*(b^3*x^3*sin_integral(b*x) - 2*b*x*sin(b*x) + (b^2*x^2 - 2)*cos(b*x))/b^3

Sympy [A] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int x^2 \text {Si}(b x) \, dx=\frac {x^{3} \operatorname {Si}{\left (b x \right )}}{3} + \frac {x^{2} \cos {\left (b x \right )}}{3 b} - \frac {2 x \sin {\left (b x \right )}}{3 b^{2}} - \frac {2 \cos {\left (b x \right )}}{3 b^{3}} \]

[In]

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

[Out]

x**3*Si(b*x)/3 + x**2*cos(b*x)/(3*b) - 2*x*sin(b*x)/(3*b**2) - 2*cos(b*x)/(3*b**3)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.80 \[ \int x^2 \text {Si}(b x) \, dx=\frac {1}{3} \, x^{3} \operatorname {Si}\left (b x\right ) - \frac {2 \, b x \sin \left (b x\right ) - {\left (b^{2} x^{2} - 2\right )} \cos \left (b x\right )}{3 \, b^{3}} \]

[In]

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

[Out]

1/3*x^3*sin_integral(b*x) - 1/3*(2*b*x*sin(b*x) - (b^2*x^2 - 2)*cos(b*x))/b^3

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.78 \[ \int x^2 \text {Si}(b x) \, dx=\frac {1}{3} \, x^{3} \operatorname {Si}\left (b x\right ) - \frac {2 \, x \sin \left (b x\right )}{3 \, b^{2}} + \frac {{\left (b^{2} x^{2} - 2\right )} \cos \left (b x\right )}{3 \, b^{3}} \]

[In]

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

[Out]

1/3*x^3*sin_integral(b*x) - 2/3*x*sin(b*x)/b^2 + 1/3*(b^2*x^2 - 2)*cos(b*x)/b^3

Mupad [F(-1)]

Timed out. \[ \int x^2 \text {Si}(b x) \, dx=\frac {x^3\,\mathrm {sinint}\left (b\,x\right )}{3}-\frac {\cos \left (b\,x\right )\,\left (\frac {2}{b^3}-\frac {x^2}{b}\right )}{3}-\frac {2\,x\,\sin \left (b\,x\right )}{3\,b^2} \]

[In]

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

[Out]

(x^3*sinint(b*x))/3 - (cos(b*x)*(2/b^3 - x^2/b))/3 - (2*x*sin(b*x))/(3*b^2)