3.1.12 \(\int x \text {Si}(b x)^2 \, dx\) [12]

Optimal. Leaf size=74 \[ -\frac {\text {CosIntegral}(2 b x)}{2 b^2}+\frac {\log (x)}{2 b^2}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \cos (b x) \text {Si}(b x)}{b}-\frac {\sin (b x) \text {Si}(b x)}{b^2}+\frac {1}{2} x^2 \text {Si}(b x)^2 \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6642, 6648, 12, 2644, 30, 6652, 3393, 3383} \begin {gather*} -\frac {\text {CosIntegral}(2 b x)}{2 b^2}-\frac {\text {Si}(b x) \sin (b x)}{b^2}+\frac {\log (x)}{2 b^2}-\frac {\sin ^2(b x)}{2 b^2}+\frac {1}{2} x^2 \text {Si}(b x)^2+\frac {x \text {Si}(b x) \cos (b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*CosIntegral[2*b*x]/b^2 + Log[x]/(2*b^2) - Sin[b*x]^2/(2*b^2) + (x*Cos[b*x]*SinIntegral[b*x])/b - (Sin[b*x
]*SinIntegral[b*x])/b^2 + (x^2*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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2644

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

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 6642

Int[(x_)^(m_.)*SinIntegral[(b_.)*(x_)]^2, x_Symbol] :> Simp[x^(m + 1)*(SinIntegral[b*x]^2/(m + 1)), x] - Dist[
2/(m + 1), Int[x^m*Sin[b*x]*SinIntegral[b*x], x], x] /; FreeQ[b, x] && IGtQ[m, 0]

Rule 6648

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

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*} \int x \text {Si}(b x)^2 \, dx &=\frac {1}{2} x^2 \text {Si}(b x)^2-\int x \sin (b x) \text {Si}(b x) \, dx\\ &=\frac {x \cos (b x) \text {Si}(b x)}{b}+\frac {1}{2} x^2 \text {Si}(b x)^2-\frac {\int \cos (b x) \text {Si}(b x) \, dx}{b}-\int \frac {\cos (b x) \sin (b x)}{b} \, dx\\ &=\frac {x \cos (b x) \text {Si}(b x)}{b}-\frac {\sin (b x) \text {Si}(b x)}{b^2}+\frac {1}{2} x^2 \text {Si}(b x)^2-\frac {\int \cos (b x) \sin (b x) \, dx}{b}+\frac {\int \frac {\sin ^2(b x)}{b x} \, dx}{b}\\ &=\frac {x \cos (b x) \text {Si}(b x)}{b}-\frac {\sin (b x) \text {Si}(b x)}{b^2}+\frac {1}{2} x^2 \text {Si}(b x)^2+\frac {\int \frac {\sin ^2(b x)}{x} \, dx}{b^2}-\frac {\text {Subst}(\int x \, dx,x,\sin (b x))}{b^2}\\ &=-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \cos (b x) \text {Si}(b x)}{b}-\frac {\sin (b x) \text {Si}(b x)}{b^2}+\frac {1}{2} x^2 \text {Si}(b x)^2+\frac {\int \left (\frac {1}{2 x}-\frac {\cos (2 b x)}{2 x}\right ) \, dx}{b^2}\\ &=\frac {\log (x)}{2 b^2}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \cos (b x) \text {Si}(b x)}{b}-\frac {\sin (b x) \text {Si}(b x)}{b^2}+\frac {1}{2} x^2 \text {Si}(b x)^2-\frac {\int \frac {\cos (2 b x)}{x} \, dx}{2 b^2}\\ &=-\frac {\text {Ci}(2 b x)}{2 b^2}+\frac {\log (x)}{2 b^2}-\frac {\sin ^2(b x)}{2 b^2}+\frac {x \cos (b x) \text {Si}(b x)}{b}-\frac {\sin (b x) \text {Si}(b x)}{b^2}+\frac {1}{2} x^2 \text {Si}(b x)^2\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 58, normalized size = 0.78 \begin {gather*} \frac {\cos (2 b x)-2 \text {CosIntegral}(2 b x)+2 \log (x)+4 (b x \cos (b x)-\sin (b x)) \text {Si}(b x)+2 b^2 x^2 \text {Si}(b x)^2}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[2*b*x] - 2*CosIntegral[2*b*x] + 2*Log[x] + 4*(b*x*Cos[b*x] - Sin[b*x])*SinIntegral[b*x] + 2*b^2*x^2*SinIn
tegral[b*x]^2)/(4*b^2)

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Maple [A]
time = 0.38, size = 62, normalized size = 0.84

method	result	size
derivativedivides	\(\frac {\frac {b^{2} x^{2} \sinIntegral \left (b x \right )^{2}}{2}-2 \sinIntegral \left (b x \right ) \left (\frac {\sin \left (b x \right )}{2}-\frac {b x \cos \left (b x \right )}{2}\right )+\frac {\left (\cos ^{2}\left (b x \right )\right )}{2}+\frac {\ln \left (b x \right )}{2}-\frac {\cosineIntegral \left (2 b x \right )}{2}}{b^{2}}\)	\(62\)
default	\(\frac {\frac {b^{2} x^{2} \sinIntegral \left (b x \right )^{2}}{2}-2 \sinIntegral \left (b x \right ) \left (\frac {\sin \left (b x \right )}{2}-\frac {b x \cos \left (b x \right )}{2}\right )+\frac {\left (\cos ^{2}\left (b x \right )\right )}{2}+\frac {\ln \left (b x \right )}{2}-\frac {\cosineIntegral \left (2 b x \right )}{2}}{b^{2}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*sin_integral(b*x)^2, x)

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Fricas [A]
time = 0.37, size = 68, normalized size = 0.92 \begin {gather*} \frac {2 \, b^{2} x^{2} \operatorname {Si}\left (b x\right )^{2} + 4 \, b x \cos \left (b x\right ) \operatorname {Si}\left (b x\right ) + 2 \, \cos \left (b x\right )^{2} - 4 \, \sin \left (b x\right ) \operatorname {Si}\left (b x\right ) - \operatorname {Ci}\left (2 \, b x\right ) - \operatorname {Ci}\left (-2 \, b x\right ) + 2 \, \log \left (x\right )}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \operatorname {Si}^{2}{\left (b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*Si(b*x)**2, x)

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Giac [A]
time = 0.41, size = 65, normalized size = 0.88 \begin {gather*} \frac {1}{2} \, x^{2} \operatorname {Si}\left (b x\right )^{2} + {\left (\frac {x \cos \left (b x\right )}{b} - \frac {\sin \left (b x\right )}{b^{2}}\right )} \operatorname {Si}\left (b x\right ) + \frac {\cos \left (2 \, b x\right ) - \operatorname {Ci}\left (2 \, b x\right ) - \operatorname {Ci}\left (-2 \, b x\right ) + 2 \, \log \left (x\right )}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,{\mathrm {sinint}\left (b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*sinint(b*x)^2, x)

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