3.3 \(\int x^2 \text {Si}(b x) \, dx\)

Optimal. Leaf size=49 \[ -\frac {2 \cos (b x)}{3 b^3}-\frac {2 x \sin (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Si}(b x)+\frac {x^2 \cos (b x)}{3 b} \]

[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

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Rubi [A]  time = 0.05, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6503, 12, 3296, 2638} \[ -\frac {2 x \sin (b x)}{3 b^2}-\frac {2 \cos (b x)}{3 b^3}+\frac {1}{3} x^3 \text {Si}(b x)+\frac {x^2 \cos (b x)}{3 b} \]

Antiderivative was successfully verified.

[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 2638

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

Rule 3296

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 6503

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*} \int x^2 \text {Si}(b x) \, dx &=\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*}

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Mathematica [A]  time = 0.03, size = 41, normalized size = 0.84 \[ \frac {b^3 x^3 \text {Si}(b x)+\left (b^2 x^2-2\right ) \cos (b x)-2 b x \sin (b x)}{3 b^3} \]

Antiderivative was successfully verified.

[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)

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \operatorname {Si}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*sin_integral(b*x), x)

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giac [A]  time = 0.60, size = 38, normalized size = 0.78 \[ \frac {1}{3} \, x^{3} \operatorname {Si}\left (b x\right ) - \frac {2 \, x \sin \left (b x\right )}{3 \, b^{2}} + \frac {{\left (b^{2} x^{2} - 2\right )} \cos \left (b x\right )}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*Si(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

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maple [A]  time = 0.02, size = 44, normalized size = 0.90 \[ \frac {\frac {b^{3} x^{3} \Si \left (b x \right )}{3}+\frac {b^{2} x^{2} \cos \left (b x \right )}{3}-\frac {2 \cos \left (b x \right )}{3}-\frac {2 b x \sin \left (b x \right )}{3}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm Si}\left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*Si(b*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \frac {x^3\,\mathrm {sinint}\left (b\,x\right )}{3}-\frac {\cos \left (b\,x\right )\,\left (\frac {2}{b^3}-\frac {x^2}{b}\right )}{3}-\frac {2\,x\,\sin \left (b\,x\right )}{3\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 1.21, size = 46, normalized size = 0.94 \[ \frac {x^{3} \operatorname {Si}{\left (b x \right )}}{3} + \frac {x^{2} \cos {\left (b x \right )}}{3 b} - \frac {2 x \sin {\left (b x \right )}}{3 b^{2}} - \frac {2 \cos {\left (b x \right )}}{3 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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