3.52 \(\int x^3 \cos (b x) \text {Si}(b x) \, dx\)

Optimal. Leaf size=128 \[ \frac {3 \text {Si}(2 b x)}{b^4}-\frac {6 \text {Si}(b x) \cos (b x)}{b^4}-\frac {4 \sin (b x) \cos (b x)}{b^4}-\frac {6 x \text {Si}(b x) \sin (b x)}{b^3}+\frac {4 x}{b^3}-\frac {2 x \sin ^2(b x)}{b^3}+\frac {3 x^2 \text {Si}(b x) \cos (b x)}{b^2}+\frac {x^2 \sin (b x) \cos (b x)}{2 b^2}+\frac {x^3 \text {Si}(b x) \sin (b x)}{b}-\frac {x^3}{6 b} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 11, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {6519, 12, 3311, 30, 2635, 8, 6513, 3443, 6511, 4406, 3299} \[ \frac {3 x^2 \text {Si}(b x) \cos (b x)}{b^2}+\frac {3 \text {Si}(2 b x)}{b^4}-\frac {6 x \text {Si}(b x) \sin (b x)}{b^3}-\frac {6 \text {Si}(b x) \cos (b x)}{b^4}+\frac {x^2 \sin (b x) \cos (b x)}{2 b^2}+\frac {4 x}{b^3}-\frac {2 x \sin ^2(b x)}{b^3}-\frac {4 \sin (b x) \cos (b x)}{b^4}+\frac {x^3 \text {Si}(b x) \sin (b x)}{b}-\frac {x^3}{6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2635

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

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[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 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 6511

Int[Sin[(a_.) + (b_.)*(x_)]*SinIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[(Cos[a + b*x]*SinIntegral[c +
d*x])/b, x] + Dist[d/b, Int[(Cos[a + b*x]*Sin[c + d*x])/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x]

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

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Mathematica [A]  time = 0.13, size = 94, normalized size = 0.73 \[ \frac {-2 b^3 x^3+12 \text {Si}(b x) \left (b x \left (b^2 x^2-6\right ) \sin (b x)+3 \left (b^2 x^2-2\right ) \cos (b x)\right )+3 b^2 x^2 \sin (2 b x)+36 \text {Si}(2 b x)+36 b x-24 \sin (2 b x)+12 b x \cos (2 b x)}{12 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*b*x - 2*b^3*x^3 + 12*b*x*Cos[2*b*x] - 24*Sin[2*b*x] + 3*b^2*x^2*Sin[2*b*x] + 12*(3*(-2 + b^2*x^2)*Cos[b*x]
+ b*x*(-6 + b^2*x^2)*Sin[b*x])*SinIntegral[b*x] + 36*SinIntegral[2*b*x])/(12*b^4)

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} \cos \left (b x\right ) \operatorname {Si}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.33, size = 180, normalized size = 1.41 \[ {\left (\frac {3 \, {\left (b^{2} x^{2} - 2\right )} \cos \left (b x\right )}{b^{4}} + \frac {{\left (b^{3} x^{3} - 6 \, b x\right )} \sin \left (b x\right )}{b^{4}}\right )} \operatorname {Si}\left (b x\right ) - \frac {b^{3} x^{3} \tan \left (b x\right )^{2} + b^{3} x^{3} - 3 \, b^{2} x^{2} \tan \left (b x\right ) - 12 \, b x \tan \left (b x\right )^{2} - 9 \, \Im \left (\operatorname {Ci}\left (2 \, b x\right ) \right ) \tan \left (b x\right )^{2} + 9 \, \Im \left (\operatorname {Ci}\left (-2 \, b x\right ) \right ) \tan \left (b x\right )^{2} - 18 \, \operatorname {Si}\left (2 \, b x\right ) \tan \left (b x\right )^{2} - 24 \, b x - 9 \, \Im \left (\operatorname {Ci}\left (2 \, b x\right ) \right ) + 9 \, \Im \left (\operatorname {Ci}\left (-2 \, b x\right ) \right ) - 18 \, \operatorname {Si}\left (2 \, b x\right ) + 24 \, \tan \left (b x\right )}{6 \, {\left (b^{4} \tan \left (b x\right )^{2} + b^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*(b^2*x^2 - 2)*cos(b*x)/b^4 + (b^3*x^3 - 6*b*x)*sin(b*x)/b^4)*sin_integral(b*x) - 1/6*(b^3*x^3*tan(b*x)^2 +
b^3*x^3 - 3*b^2*x^2*tan(b*x) - 12*b*x*tan(b*x)^2 - 9*imag_part(cos_integral(2*b*x))*tan(b*x)^2 + 9*imag_part(c
os_integral(-2*b*x))*tan(b*x)^2 - 18*sin_integral(2*b*x)*tan(b*x)^2 - 24*b*x - 9*imag_part(cos_integral(2*b*x)
) + 9*imag_part(cos_integral(-2*b*x)) - 18*sin_integral(2*b*x) + 24*tan(b*x))/(b^4*tan(b*x)^2 + b^4)

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maple [A]  time = 0.06, size = 111, normalized size = 0.87 \[ \frac {\Si \left (b x \right ) \left (\sin \left (b x \right ) b^{3} x^{3}+3 b^{2} x^{2} \cos \left (b x \right )-6 \cos \left (b x \right )-6 b x \sin \left (b x \right )\right )-b^{2} x^{2} \left (-\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )+2 b x \left (\cos ^{2}\left (b x \right )\right )-4 \sin \left (b x \right ) \cos \left (b x \right )+2 b x +\frac {b^{3} x^{3}}{3}+3 \Si \left (2 b x \right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*Si(b*x)*cos(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^3*sinint(b*x)*cos(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*cos(b*x)*Si(b*x), x)

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