3.80 \(\int x \text {Ci}(b x)^2 \, dx\)

Optimal. Leaf size=75 \[ \frac {\text {Ci}(2 b x)}{2 b^2}-\frac {\text {Ci}(b x) \cos (b x)}{b^2}+\frac {\log (x)}{2 b^2}+\frac {\sin ^2(b x)}{2 b^2}+\frac {1}{2} x^2 \text {Ci}(b x)^2-\frac {x \text {Ci}(b x) \sin (b x)}{b} \]

[Out]

1/2*x^2*Ci(b*x)^2+1/2*Ci(2*b*x)/b^2-Ci(b*x)*cos(b*x)/b^2+1/2*ln(x)/b^2-x*Ci(b*x)*sin(b*x)/b+1/2*sin(b*x)^2/b^2

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Rubi [A]  time = 0.09, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6508, 6514, 12, 2564, 30, 6518, 3312, 3302} \[ \frac {\text {CosIntegral}(2 b x)}{2 b^2}-\frac {\text {CosIntegral}(b x) \cos (b x)}{b^2}+\frac {\log (x)}{2 b^2}+\frac {\sin ^2(b x)}{2 b^2}+\frac {1}{2} x^2 \text {CosIntegral}(b x)^2-\frac {x \text {CosIntegral}(b x) \sin (b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cos[b*x]*CosIntegral[b*x])/b^2) + (x^2*CosIntegral[b*x]^2)/2 + CosIntegral[2*b*x]/(2*b^2) + Log[x]/(2*b^2)
- (x*CosIntegral[b*x]*Sin[b*x])/b + Sin[b*x]^2/(2*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 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 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 6518

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

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 58, normalized size = 0.77 \[ \frac {2 b^2 x^2 \text {Ci}(b x)^2+2 \text {Ci}(2 b x)-4 \text {Ci}(b x) (b x \sin (b x)+\cos (b x))-\cos (2 b x)+2 \log (x)}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x \operatorname {Ci}\left (b x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*cos_integral(b*x)^2, x)

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giac [A]  time = 0.22, size = 65, normalized size = 0.87 \[ \frac {1}{2} \, x^{2} \operatorname {Ci}\left (b x\right )^{2} - {\left (\frac {x \sin \left (b x\right )}{b} + \frac {\cos \left (b x\right )}{b^{2}}\right )} \operatorname {Ci}\left (b x\right ) - \frac {\cos \left (2 \, b x\right ) - \operatorname {Ci}\left (2 \, b x\right ) - \operatorname {Ci}\left (-2 \, b x\right ) - 2 \, \log \relax (x)}{4 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*cos_integral(b*x)^2 - (x*sin(b*x)/b + cos(b*x)/b^2)*cos_integral(b*x) - 1/4*(cos(2*b*x) - cos_integral
(2*b*x) - cos_integral(-2*b*x) - 2*log(x))/b^2

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maple [A]  time = 0.03, size = 70, normalized size = 0.93 \[ \frac {x^{2} \Ci \left (b x \right )^{2}}{2}-\frac {\Ci \left (b x \right ) \cos \left (b x \right )}{b^{2}}-\frac {x \Ci \left (b x \right ) \sin \left (b x \right )}{b}+\frac {\ln \left (b x \right )}{2 b^{2}}+\frac {\Ci \left (2 b x \right )}{2 b^{2}}-\frac {\cos ^{2}\left (b x \right )}{2 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*Ci(b*x)^2-Ci(b*x)*cos(b*x)/b^2-x*Ci(b*x)*sin(b*x)/b+1/2/b^2*ln(b*x)+1/2*Ci(2*b*x)/b^2-1/2/b^2*cos(b*x)
^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x {\rm Ci}\left (b x\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*Ci(b*x)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\mathrm {cosint}\left (b\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*cosint(b*x)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \operatorname {Ci}^{2}{\left (b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*Ci(b*x)**2, x)

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