3.45 \(\int x^3 \sin (b x) \text {Si}(b x) \, dx\)

Optimal. Leaf size=126 \[ -\frac {3 \text {Ci}(2 b x)}{b^4}-\frac {6 \text {Si}(b x) \sin (b x)}{b^4}+\frac {3 \log (x)}{b^4}-\frac {4 \sin ^2(b x)}{b^4}+\frac {6 x \text {Si}(b x) \cos (b x)}{b^3}+\frac {2 x \sin (b x) \cos (b x)}{b^3}+\frac {3 x^2 \text {Si}(b x) \sin (b x)}{b^2}-\frac {x^2}{b^2}+\frac {x^2 \sin ^2(b x)}{2 b^2}-\frac {x^3 \text {Si}(b x) \cos (b x)}{b} \]

[Out]

-x^2/b^2-3*Ci(2*b*x)/b^4+3*ln(x)/b^4+6*x*cos(b*x)*Si(b*x)/b^3-x^3*cos(b*x)*Si(b*x)/b+2*x*cos(b*x)*sin(b*x)/b^3
-6*Si(b*x)*sin(b*x)/b^4+3*x^2*Si(b*x)*sin(b*x)/b^2-4*sin(b*x)^2/b^4+1/2*x^2*sin(b*x)^2/b^2

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Rubi [A]  time = 0.19, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {6513, 12, 3443, 3310, 30, 6519, 2564, 6517, 3312, 3302} \[ -\frac {3 \text {CosIntegral}(2 b x)}{b^4}+\frac {3 x^2 \text {Si}(b x) \sin (b x)}{b^2}-\frac {6 \text {Si}(b x) \sin (b x)}{b^4}+\frac {6 x \text {Si}(b x) \cos (b x)}{b^3}-\frac {x^2}{b^2}+\frac {x^2 \sin ^2(b x)}{2 b^2}+\frac {3 \log (x)}{b^4}-\frac {4 \sin ^2(b x)}{b^4}+\frac {2 x \sin (b x) \cos (b x)}{b^3}-\frac {x^3 \text {Si}(b x) \cos (b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sin[b*x]*SinIntegral[b*x],x]

[Out]

-(x^2/b^2) - (3*CosIntegral[2*b*x])/b^4 + (3*Log[x])/b^4 + (2*x*Cos[b*x]*Sin[b*x])/b^3 - (4*Sin[b*x]^2)/b^4 +
(x^2*Sin[b*x]^2)/(2*b^2) + (6*x*Cos[b*x]*SinIntegral[b*x])/b^3 - (x^3*Cos[b*x]*SinIntegral[b*x])/b - (6*Sin[b*
x]*SinIntegral[b*x])/b^4 + (3*x^2*Sin[b*x]*SinIntegral[b*x])/b^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 2564

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 3302

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 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 3312

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 3443

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m - n
+ 1)*Sin[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sin[a + b*x^n]^
(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 6513

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 6517

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]

Rule 6519

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

Rubi steps

\begin {align*} \int x^3 \sin (b x) \text {Si}(b x) \, dx &=-\frac {x^3 \cos (b x) \text {Si}(b x)}{b}+\frac {3 \int x^2 \cos (b x) \text {Si}(b x) \, dx}{b}+\int \frac {x^2 \cos (b x) \sin (b x)}{b} \, dx\\ &=-\frac {x^3 \cos (b x) \text {Si}(b x)}{b}+\frac {3 x^2 \sin (b x) \text {Si}(b x)}{b^2}-\frac {6 \int x \sin (b x) \text {Si}(b x) \, dx}{b^2}+\frac {\int x^2 \cos (b x) \sin (b x) \, dx}{b}-\frac {3 \int \frac {x \sin ^2(b x)}{b} \, dx}{b}\\ &=\frac {x^2 \sin ^2(b x)}{2 b^2}+\frac {6 x \cos (b x) \text {Si}(b x)}{b^3}-\frac {x^3 \cos (b x) \text {Si}(b x)}{b}+\frac {3 x^2 \sin (b x) \text {Si}(b x)}{b^2}-\frac {6 \int \cos (b x) \text {Si}(b x) \, dx}{b^3}-\frac {\int x \sin ^2(b x) \, dx}{b^2}-\frac {3 \int x \sin ^2(b x) \, dx}{b^2}-\frac {6 \int \frac {\cos (b x) \sin (b x)}{b} \, dx}{b^2}\\ &=\frac {2 x \cos (b x) \sin (b x)}{b^3}-\frac {\sin ^2(b x)}{b^4}+\frac {x^2 \sin ^2(b x)}{2 b^2}+\frac {6 x \cos (b x) \text {Si}(b x)}{b^3}-\frac {x^3 \cos (b x) \text {Si}(b x)}{b}-\frac {6 \sin (b x) \text {Si}(b x)}{b^4}+\frac {3 x^2 \sin (b x) \text {Si}(b x)}{b^2}-\frac {6 \int \cos (b x) \sin (b x) \, dx}{b^3}+\frac {6 \int \frac {\sin ^2(b x)}{b x} \, dx}{b^3}-\frac {\int x \, dx}{2 b^2}-\frac {3 \int x \, dx}{2 b^2}\\ &=-\frac {x^2}{b^2}+\frac {2 x \cos (b x) \sin (b x)}{b^3}-\frac {\sin ^2(b x)}{b^4}+\frac {x^2 \sin ^2(b x)}{2 b^2}+\frac {6 x \cos (b x) \text {Si}(b x)}{b^3}-\frac {x^3 \cos (b x) \text {Si}(b x)}{b}-\frac {6 \sin (b x) \text {Si}(b x)}{b^4}+\frac {3 x^2 \sin (b x) \text {Si}(b x)}{b^2}+\frac {6 \int \frac {\sin ^2(b x)}{x} \, dx}{b^4}-\frac {6 \operatorname {Subst}(\int x \, dx,x,\sin (b x))}{b^4}\\ &=-\frac {x^2}{b^2}+\frac {2 x \cos (b x) \sin (b x)}{b^3}-\frac {4 \sin ^2(b x)}{b^4}+\frac {x^2 \sin ^2(b x)}{2 b^2}+\frac {6 x \cos (b x) \text {Si}(b x)}{b^3}-\frac {x^3 \cos (b x) \text {Si}(b x)}{b}-\frac {6 \sin (b x) \text {Si}(b x)}{b^4}+\frac {3 x^2 \sin (b x) \text {Si}(b x)}{b^2}+\frac {6 \int \left (\frac {1}{2 x}-\frac {\cos (2 b x)}{2 x}\right ) \, dx}{b^4}\\ &=-\frac {x^2}{b^2}+\frac {3 \log (x)}{b^4}+\frac {2 x \cos (b x) \sin (b x)}{b^3}-\frac {4 \sin ^2(b x)}{b^4}+\frac {x^2 \sin ^2(b x)}{2 b^2}+\frac {6 x \cos (b x) \text {Si}(b x)}{b^3}-\frac {x^3 \cos (b x) \text {Si}(b x)}{b}-\frac {6 \sin (b x) \text {Si}(b x)}{b^4}+\frac {3 x^2 \sin (b x) \text {Si}(b x)}{b^2}-\frac {3 \int \frac {\cos (2 b x)}{x} \, dx}{b^4}\\ &=-\frac {x^2}{b^2}-\frac {3 \text {Ci}(2 b x)}{b^4}+\frac {3 \log (x)}{b^4}+\frac {2 x \cos (b x) \sin (b x)}{b^3}-\frac {4 \sin ^2(b x)}{b^4}+\frac {x^2 \sin ^2(b x)}{2 b^2}+\frac {6 x \cos (b x) \text {Si}(b x)}{b^3}-\frac {x^3 \cos (b x) \text {Si}(b x)}{b}-\frac {6 \sin (b x) \text {Si}(b x)}{b^4}+\frac {3 x^2 \sin (b x) \text {Si}(b x)}{b^2}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 93, normalized size = 0.74 \[ -\frac {4 \text {Si}(b x) \left (b x \left (b^2 x^2-6\right ) \cos (b x)-3 \left (b^2 x^2-2\right ) \sin (b x)\right )+3 b^2 x^2+b^2 x^2 \cos (2 b x)+12 \text {Ci}(2 b x)-4 b x \sin (2 b x)-8 \cos (2 b x)-12 \log (x)}{4 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sin[b*x]*SinIntegral[b*x],x]

[Out]

-1/4*(3*b^2*x^2 - 8*Cos[2*b*x] + b^2*x^2*Cos[2*b*x] + 12*CosIntegral[2*b*x] - 12*Log[x] - 4*b*x*Sin[2*b*x] + 4
*(b*x*(-6 + b^2*x^2)*Cos[b*x] - 3*(-2 + b^2*x^2)*Sin[b*x])*SinIntegral[b*x])/b^4

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fricas [F]  time = 2.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} \sin \left (b x\right ) \operatorname {Si}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*Si(b*x)*sin(b*x),x, algorithm="fricas")

[Out]

integral(x^3*sin(b*x)*sin_integral(b*x), x)

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giac [A]  time = 0.18, size = 106, normalized size = 0.84 \[ -{\left (\frac {{\left (b^{3} x^{3} - 6 \, b x\right )} \cos \left (b x\right )}{b^{4}} - \frac {3 \, {\left (b^{2} x^{2} - 2\right )} \sin \left (b x\right )}{b^{4}}\right )} \operatorname {Si}\left (b x\right ) - \frac {b^{2} x^{2} \cos \left (2 \, b x\right ) + 3 \, b^{2} x^{2} - 4 \, b x \sin \left (2 \, b x\right ) - 8 \, \cos \left (2 \, b x\right ) + 6 \, \operatorname {Ci}\left (2 \, b x\right ) + 6 \, \operatorname {Ci}\left (-2 \, b x\right ) - 12 \, \log \relax (x)}{4 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*Si(b*x)*sin(b*x),x, algorithm="giac")

[Out]

-((b^3*x^3 - 6*b*x)*cos(b*x)/b^4 - 3*(b^2*x^2 - 2)*sin(b*x)/b^4)*sin_integral(b*x) - 1/4*(b^2*x^2*cos(2*b*x) +
3*b^2*x^2 - 4*b*x*sin(2*b*x) - 8*cos(2*b*x) + 6*cos_integral(2*b*x) + 6*cos_integral(-2*b*x) - 12*log(x))/b^4

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maple [A]  time = 0.02, size = 138, normalized size = 1.10 \[ -\frac {x^{3} \cos \left (b x \right ) \Si \left (b x \right )}{b}+\frac {3 x^{2} \Si \left (b x \right ) \sin \left (b x \right )}{b^{2}}+\frac {6 x \cos \left (b x \right ) \Si \left (b x \right )}{b^{3}}-\frac {6 \Si \left (b x \right ) \sin \left (b x \right )}{b^{4}}-\frac {x^{2} \left (\cos ^{2}\left (b x \right )\right )}{2 b^{2}}+\frac {2 x \cos \left (b x \right ) \sin \left (b x \right )}{b^{3}}-\frac {x^{2}}{2 b^{2}}-\frac {\sin ^{2}\left (b x \right )}{b^{4}}+\frac {3 \ln \left (b x \right )}{b^{4}}-\frac {3 \Ci \left (2 b x \right )}{b^{4}}+\frac {3 \left (\cos ^{2}\left (b x \right )\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*Si(b*x)*sin(b*x),x)

[Out]

-x^3*cos(b*x)*Si(b*x)/b+3*x^2*Si(b*x)*sin(b*x)/b^2+6*x*cos(b*x)*Si(b*x)/b^3-6*Si(b*x)*sin(b*x)/b^4-1/2/b^2*x^2
*cos(b*x)^2+2*x*cos(b*x)*sin(b*x)/b^3-1/2*x^2/b^2-sin(b*x)^2/b^4+3/b^4*ln(b*x)-3*Ci(2*b*x)/b^4+3*cos(b*x)^2/b^
4

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} {\rm Si}\left (b x\right ) \sin \left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*Si(b*x)*sin(b*x),x, algorithm="maxima")

[Out]

integrate(x^3*Si(b*x)*sin(b*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\mathrm {sinint}\left (b\,x\right )\,\sin \left (b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sinint(b*x)*sin(b*x),x)

[Out]

int(x^3*sinint(b*x)*sin(b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \sin {\left (b x \right )} \operatorname {Si}{\left (b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*Si(b*x)*sin(b*x),x)

[Out]

Integral(x**3*sin(b*x)*Si(b*x), x)

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