3.79 \(\int x^2 \text{CosIntegral}(b x)^2 \, dx\)

Optimal. Leaf size=112 \[ \frac{4 \text{CosIntegral}(b x) \sin (b x)}{3 b^3}-\frac{4 x \text{CosIntegral}(b x) \cos (b x)}{3 b^2}-\frac{2 \text{Si}(2 b x)}{3 b^3}+\frac{x}{2 b^2}+\frac{x \sin ^2(b x)}{3 b^2}+\frac{5 \sin (b x) \cos (b x)}{6 b^3}+\frac{1}{3} x^3 \text{CosIntegral}(b x)^2-\frac{2 x^2 \text{CosIntegral}(b x) \sin (b x)}{3 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.138129, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {6508, 6514, 12, 3443, 2635, 8, 6520, 6512, 4406, 3299} \[ \frac{4 \text{CosIntegral}(b x) \sin (b x)}{3 b^3}-\frac{4 x \text{CosIntegral}(b x) \cos (b x)}{3 b^2}-\frac{2 \text{Si}(2 b x)}{3 b^3}+\frac{x}{2 b^2}+\frac{x \sin ^2(b x)}{3 b^2}+\frac{5 \sin (b x) \cos (b x)}{6 b^3}+\frac{1}{3} x^3 \text{CosIntegral}(b x)^2-\frac{2 x^2 \text{CosIntegral}(b x) \sin (b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6508

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

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 12

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

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

Mathematica [A]  time = 0.0805696, size = 78, normalized size = 0.7 \[ \frac{4 b^3 x^3 \text{CosIntegral}(b x)^2-8 \text{CosIntegral}(b x) \left (\left (b^2 x^2-2\right ) \sin (b x)+2 b x \cos (b x)\right )-8 \text{Si}(2 b x)+8 b x+5 \sin (2 b x)-2 b x \cos (2 b x)}{12 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*x - 2*b*x*Cos[2*b*x] + 4*b^3*x^3*CosIntegral[b*x]^2 - 8*CosIntegral[b*x]*(2*b*x*Cos[b*x] + (-2 + b^2*x^2)
*Sin[b*x]) + 5*Sin[2*b*x] - 8*SinIntegral[2*b*x])/(12*b^3)

________________________________________________________________________________________

Maple [A]  time = 0.073, size = 84, normalized size = 0.8 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{{b}^{3}{x}^{3} \left ({\it Ci} \left ( bx \right ) \right ) ^{2}}{3}}-2\,{\it Ci} \left ( bx \right ) \left ( 1/3\,{b}^{2}{x}^{2}\sin \left ( bx \right ) -2/3\,\sin \left ( bx \right ) +2/3\,bx\cos \left ( bx \right ) \right ) -{\frac{bx \left ( \cos \left ( bx \right ) \right ) ^{2}}{3}}+{\frac{5\,\sin \left ( bx \right ) \cos \left ( bx \right ) }{6}}+{\frac{5\,bx}{6}}-{\frac{2\,{\it Si} \left ( 2\,bx \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^3*(1/3*b^3*x^3*Ci(b*x)^2-2*Ci(b*x)*(1/3*b^2*x^2*sin(b*x)-2/3*sin(b*x)+2/3*b*x*cos(b*x))-1/3*b*x*cos(b*x)^2
+5/6*sin(b*x)*cos(b*x)+5/6*b*x-2/3*Si(2*b*x))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\rm Ci}\left (b x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*Ci(b*x)^2, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} \operatorname{Ci}\left (b x\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*cos_integral(b*x)^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*Ci(b*x)**2, x)

________________________________________________________________________________________

Giac [C]  time = 1.22111, size = 104, normalized size = 0.93 \begin{align*} \frac{1}{3} \, x^{3} \operatorname{Ci}\left (b x\right )^{2} - \frac{2}{3} \,{\left (\frac{2 \, x \cos \left (b x\right )}{b^{2}} + \frac{{\left (b^{2} x^{2} - 2\right )} \sin \left (b x\right )}{b^{3}}\right )} \operatorname{Ci}\left (b x\right ) + \frac{2 \, b x - \Im \left ( \operatorname{Ci}\left (2 \, b x\right ) \right ) + \Im \left ( \operatorname{Ci}\left (-2 \, b x\right ) \right ) - 2 \, \operatorname{Si}\left (2 \, b x\right )}{3 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*cos_integral(b*x)^2 - 2/3*(2*x*cos(b*x)/b^2 + (b^2*x^2 - 2)*sin(b*x)/b^3)*cos_integral(b*x) + 1/3*(2*b
*x - imag_part(cos_integral(2*b*x)) + imag_part(cos_integral(-2*b*x)) - 2*sin_integral(2*b*x))/b^3