3.1.52 \(\int x^3 \cos (b x) \text {Si}(b x) \, dx\) [52]

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

[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 = 0.14, 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 = {6654, 12, 3392, 30, 2715, 8, 6648, 3524, 6646, 4491, 3380} \begin {gather*} \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} \end {gather*}

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 2715

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] && Integ
erQ[2*n]

Rule 3380

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 3392

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 3524

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 4491

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 6646

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 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 6654

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 = 0.08, size = 94, normalized size = 0.73 \begin {gather*} \frac {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 \left (3 \left (-2+b^2 x^2\right ) \cos (b x)+b x \left (-6+b^2 x^2\right ) \sin (b x)\right ) \text {Si}(b x)+36 \text {Si}(2 b x)}{12 b^4} \end {gather*}

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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Maple [A]
time = 0.53, size = 111, normalized size = 0.87

method	result	size
derivativedivides	\(\frac {\sinIntegral \left (b x \right ) \left (b^{3} x^{3} \sin \left (b x \right )+3 b^{2} x^{2} \cos \left (b x \right )-6 \cos \left (b x \right )-6 b x \sin \left (b x \right )\right )-b^{2} x^{2} \left (-\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )+2 b x \left (\cos ^{2}\left (b x \right )\right )-4 \sin \left (b x \right ) \cos \left (b x \right )+2 b x +\frac {b^{3} x^{3}}{3}+3 \sinIntegral \left (2 b x \right )}{b^{4}}\)	\(111\)
default	\(\frac {\sinIntegral \left (b x \right ) \left (b^{3} x^{3} \sin \left (b x \right )+3 b^{2} x^{2} \cos \left (b x \right )-6 \cos \left (b x \right )-6 b x \sin \left (b x \right )\right )-b^{2} x^{2} \left (-\frac {\sin \left (b x \right ) \cos \left (b x \right )}{2}+\frac {b x}{2}\right )+2 b x \left (\cos ^{2}\left (b x \right )\right )-4 \sin \left (b x \right ) \cos \left (b x \right )+2 b x +\frac {b^{3} x^{3}}{3}+3 \sinIntegral \left (2 b x \right )}{b^{4}}\)	\(111\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*cos(b*x)*Si(b*x),x,method=_RETURNVERBOSE)

[Out]

1/b^4*(Si(b*x)*(b^3*x^3*sin(b*x)+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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 92, normalized size = 0.72 \begin {gather*} -\frac {b^{3} x^{3} - 12 \, b x \cos \left (b x\right )^{2} - 18 \, {\left (b^{2} x^{2} - 2\right )} \cos \left (b x\right ) \operatorname {Si}\left (b x\right ) - 12 \, b x - 3 \, {\left ({\left (b^{2} x^{2} - 8\right )} \cos \left (b x\right ) + 2 \, {\left (b^{3} x^{3} - 6 \, b x\right )} \operatorname {Si}\left (b x\right )\right )} \sin \left (b x\right ) - 18 \, \operatorname {Si}\left (2 \, b x\right )}{6 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(b^3*x^3 - 12*b*x*cos(b*x)^2 - 18*(b^2*x^2 - 2)*cos(b*x)*sin_integral(b*x) - 12*b*x - 3*((b^2*x^2 - 8)*co
s(b*x) + 2*(b^3*x^3 - 6*b*x)*sin_integral(b*x))*sin(b*x) - 18*sin_integral(2*b*x))/b^4

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

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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Giac [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.42, size = 180, normalized size = 1.41 \begin {gather*} {\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cos(b*x)*sin_integral(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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,\mathrm {sinint}\left (b\,x\right )\,\cos \left (b\,x\right ) \,d x \end {gather*}

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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