3.113 \(\int x^3 \text{CosIntegral}(b x) \sin (b x) \, dx\)

Optimal. Leaf size=147 \[ \frac{3 x^2 \text{CosIntegral}(b x) \sin (b x)}{b^2}-\frac{6 \text{CosIntegral}(b x) \sin (b x)}{b^4}+\frac{6 x \text{CosIntegral}(b x) \cos (b x)}{b^3}+\frac{3 \text{Si}(2 b x)}{b^4}+\frac{x^2 \sin (b x) \cos (b x)}{2 b^2}-\frac{5 x}{2 b^3}-\frac{3 x \sin ^2(b x)}{2 b^3}+\frac{x \cos ^2(b x)}{2 b^3}-\frac{4 \sin (b x) \cos (b x)}{b^4}-\frac{x^3 \text{CosIntegral}(b x) \cos (b x)}{b}+\frac{x^3}{6 b} \]

[Out]

(-5*x)/(2*b^3) + x^3/(6*b) + (x*Cos[b*x]^2)/(2*b^3) + (6*x*Cos[b*x]*CosIntegral[b*x])/b^3 - (x^3*Cos[b*x]*CosI
ntegral[b*x])/b - (4*Cos[b*x]*Sin[b*x])/b^4 + (x^2*Cos[b*x]*Sin[b*x])/(2*b^2) - (6*CosIntegral[b*x]*Sin[b*x])/
b^4 + (3*x^2*CosIntegral[b*x]*Sin[b*x])/b^2 - (3*x*Sin[b*x]^2)/(2*b^3) + (3*SinIntegral[2*b*x])/b^4

________________________________________________________________________________________

Rubi [A]  time = 0.182771, antiderivative size = 147, normalized size of antiderivative = 1., 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 = {6520, 12, 3311, 30, 2635, 8, 6514, 3443, 6512, 4406, 3299} \[ \frac{3 x^2 \text{CosIntegral}(b x) \sin (b x)}{b^2}-\frac{6 \text{CosIntegral}(b x) \sin (b x)}{b^4}+\frac{6 x \text{CosIntegral}(b x) \cos (b x)}{b^3}+\frac{3 \text{Si}(2 b x)}{b^4}+\frac{x^2 \sin (b x) \cos (b x)}{2 b^2}-\frac{5 x}{2 b^3}-\frac{3 x \sin ^2(b x)}{2 b^3}+\frac{x \cos ^2(b x)}{2 b^3}-\frac{4 \sin (b x) \cos (b x)}{b^4}-\frac{x^3 \text{CosIntegral}(b x) \cos (b x)}{b}+\frac{x^3}{6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*x)/(2*b^3) + x^3/(6*b) + (x*Cos[b*x]^2)/(2*b^3) + (6*x*Cos[b*x]*CosIntegral[b*x])/b^3 - (x^3*Cos[b*x]*CosI
ntegral[b*x])/b - (4*Cos[b*x]*Sin[b*x])/b^4 + (x^2*Cos[b*x]*Sin[b*x])/(2*b^2) - (6*CosIntegral[b*x]*Sin[b*x])/
b^4 + (3*x^2*CosIntegral[b*x]*Sin[b*x])/b^2 - (3*x*Sin[b*x]^2)/(2*b^3) + (3*SinIntegral[2*b*x])/b^4

Rule 6520

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 8

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

Rule 6514

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

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 6512

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

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 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]

Rubi steps

\begin{align*} \int x^3 \text{Ci}(b x) \sin (b x) \, dx &=-\frac{x^3 \cos (b x) \text{Ci}(b x)}{b}+\frac{3 \int x^2 \cos (b x) \text{Ci}(b x) \, dx}{b}+\int \frac{x^2 \cos ^2(b x)}{b} \, dx\\ &=-\frac{x^3 \cos (b x) \text{Ci}(b x)}{b}+\frac{3 x^2 \text{Ci}(b x) \sin (b x)}{b^2}-\frac{6 \int x \text{Ci}(b x) \sin (b x) \, dx}{b^2}+\frac{\int x^2 \cos ^2(b x) \, dx}{b}-\frac{3 \int \frac{x \cos (b x) \sin (b x)}{b} \, dx}{b}\\ &=\frac{x \cos ^2(b x)}{2 b^3}+\frac{6 x \cos (b x) \text{Ci}(b x)}{b^3}-\frac{x^3 \cos (b x) \text{Ci}(b x)}{b}+\frac{x^2 \cos (b x) \sin (b x)}{2 b^2}+\frac{3 x^2 \text{Ci}(b x) \sin (b x)}{b^2}-\frac{\int \cos ^2(b x) \, dx}{2 b^3}-\frac{6 \int \cos (b x) \text{Ci}(b x) \, dx}{b^3}-\frac{3 \int x \cos (b x) \sin (b x) \, dx}{b^2}-\frac{6 \int \frac{\cos ^2(b x)}{b} \, dx}{b^2}+\frac{\int x^2 \, dx}{2 b}\\ &=\frac{x^3}{6 b}+\frac{x \cos ^2(b x)}{2 b^3}+\frac{6 x \cos (b x) \text{Ci}(b x)}{b^3}-\frac{x^3 \cos (b x) \text{Ci}(b x)}{b}-\frac{\cos (b x) \sin (b x)}{4 b^4}+\frac{x^2 \cos (b x) \sin (b x)}{2 b^2}-\frac{6 \text{Ci}(b x) \sin (b x)}{b^4}+\frac{3 x^2 \text{Ci}(b x) \sin (b x)}{b^2}-\frac{3 x \sin ^2(b x)}{2 b^3}-\frac{\int 1 \, dx}{4 b^3}+\frac{3 \int \sin ^2(b x) \, dx}{2 b^3}-\frac{6 \int \cos ^2(b x) \, dx}{b^3}+\frac{6 \int \frac{\cos (b x) \sin (b x)}{b x} \, dx}{b^3}\\ &=-\frac{x}{4 b^3}+\frac{x^3}{6 b}+\frac{x \cos ^2(b x)}{2 b^3}+\frac{6 x \cos (b x) \text{Ci}(b x)}{b^3}-\frac{x^3 \cos (b x) \text{Ci}(b x)}{b}-\frac{4 \cos (b x) \sin (b x)}{b^4}+\frac{x^2 \cos (b x) \sin (b x)}{2 b^2}-\frac{6 \text{Ci}(b x) \sin (b x)}{b^4}+\frac{3 x^2 \text{Ci}(b x) \sin (b x)}{b^2}-\frac{3 x \sin ^2(b x)}{2 b^3}+\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{5 x}{2 b^3}+\frac{x^3}{6 b}+\frac{x \cos ^2(b x)}{2 b^3}+\frac{6 x \cos (b x) \text{Ci}(b x)}{b^3}-\frac{x^3 \cos (b x) \text{Ci}(b x)}{b}-\frac{4 \cos (b x) \sin (b x)}{b^4}+\frac{x^2 \cos (b x) \sin (b x)}{2 b^2}-\frac{6 \text{Ci}(b x) \sin (b x)}{b^4}+\frac{3 x^2 \text{Ci}(b x) \sin (b x)}{b^2}-\frac{3 x \sin ^2(b x)}{2 b^3}+\frac{6 \int \frac{\sin (2 b x)}{2 x} \, dx}{b^4}\\ &=-\frac{5 x}{2 b^3}+\frac{x^3}{6 b}+\frac{x \cos ^2(b x)}{2 b^3}+\frac{6 x \cos (b x) \text{Ci}(b x)}{b^3}-\frac{x^3 \cos (b x) \text{Ci}(b x)}{b}-\frac{4 \cos (b x) \sin (b x)}{b^4}+\frac{x^2 \cos (b x) \sin (b x)}{2 b^2}-\frac{6 \text{Ci}(b x) \sin (b x)}{b^4}+\frac{3 x^2 \text{Ci}(b x) \sin (b x)}{b^2}-\frac{3 x \sin ^2(b x)}{2 b^3}+\frac{3 \int \frac{\sin (2 b x)}{x} \, dx}{b^4}\\ &=-\frac{5 x}{2 b^3}+\frac{x^3}{6 b}+\frac{x \cos ^2(b x)}{2 b^3}+\frac{6 x \cos (b x) \text{Ci}(b x)}{b^3}-\frac{x^3 \cos (b x) \text{Ci}(b x)}{b}-\frac{4 \cos (b x) \sin (b x)}{b^4}+\frac{x^2 \cos (b x) \sin (b x)}{2 b^2}-\frac{6 \text{Ci}(b x) \sin (b x)}{b^4}+\frac{3 x^2 \text{Ci}(b x) \sin (b x)}{b^2}-\frac{3 x \sin ^2(b x)}{2 b^3}+\frac{3 \text{Si}(2 b x)}{b^4}\\ \end{align*}

Mathematica [A]  time = 0.107106, size = 94, normalized size = 0.64 \[ \frac{-12 \text{CosIntegral}(b x) \left (b x \left (b^2 x^2-6\right ) \cos (b x)-3 \left (b^2 x^2-2\right ) \sin (b x)\right )+2 b^3 x^3+3 b^2 x^2 \sin (2 b x)+36 \text{Si}(2 b x)-36 b x-24 \sin (2 b x)+12 b x \cos (2 b x)}{12 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.076, size = 111, normalized size = 0.8 \begin{align*}{\frac{1}{{b}^{4}} \left ({\it Ci} \left ( bx \right ) \left ( -{b}^{3}{x}^{3}\cos \left ( bx \right ) +3\,{b}^{2}{x}^{2}\sin \left ( bx \right ) -6\,\sin \left ( bx \right ) +6\,bx\cos \left ( bx \right ) \right ) +{b}^{2}{x}^{2} \left ({\frac{\sin \left ( bx \right ) \cos \left ( bx \right ) }{2}}+{\frac{bx}{2}} \right ) +2\,bx \left ( \cos \left ( bx \right ) \right ) ^{2}-4\,\sin \left ( bx \right ) \cos \left ( bx \right ) -4\,bx-{\frac{{x}^{3}{b}^{3}}{3}}+3\,{\it Si} \left ( 2\,bx \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3}{\rm Ci}\left (b x\right ) \sin \left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*Ci(b*x)*sin(b*x), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{3} \operatorname{Ci}\left (b x\right ) \sin \left (b x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \sin{\left (b x \right )} \operatorname{Ci}{\left (b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*sin(b*x)*Ci(b*x), x)

________________________________________________________________________________________

Giac [C]  time = 1.23702, size = 116, normalized size = 0.79 \begin{align*} -{\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{Ci}\left (b x\right ) + \frac{b^{3} x^{3} - 18 \, 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 )}{6 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*Ci(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)*cos_integral(b*x) + 1/6*(b^3*x^3 - 18*b*x + 9
*imag_part(cos_integral(2*b*x)) - 9*imag_part(cos_integral(-2*b*x)) + 18*sin_integral(2*b*x))/b^4