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

Optimal. Leaf size=74 \[ -\frac{\text{Chi}(2 b x)}{2 b^2}+\frac{\text{Chi}(b x) \cosh (b x)}{b^2}-\frac{\log (x)}{2 b^2}+\frac{\sinh ^2(b x)}{2 b^2}+\frac{1}{2} x^2 \text{Chi}(b x)^2-\frac{x \text{Chi}(b x) \sinh (b x)}{b} \]

[Out]

(Cosh[b*x]*CoshIntegral[b*x])/b^2 + (x^2*CoshIntegral[b*x]^2)/2 - CoshIntegral[2*b*x]/(2*b^2) - Log[x]/(2*b^2)
- (x*CoshIntegral[b*x]*Sinh[b*x])/b + Sinh[b*x]^2/(2*b^2)

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Rubi [A]  time = 0.0917978, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {6537, 6543, 12, 2564, 30, 6547, 3312, 3301} \[ -\frac{\text{Chi}(2 b x)}{2 b^2}+\frac{\text{Chi}(b x) \cosh (b x)}{b^2}-\frac{\log (x)}{2 b^2}+\frac{\sinh ^2(b x)}{2 b^2}+\frac{1}{2} x^2 \text{Chi}(b x)^2-\frac{x \text{Chi}(b x) \sinh (b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[b*x]*CoshIntegral[b*x])/b^2 + (x^2*CoshIntegral[b*x]^2)/2 - CoshIntegral[2*b*x]/(2*b^2) - Log[x]/(2*b^2)
- (x*CoshIntegral[b*x]*Sinh[b*x])/b + Sinh[b*x]^2/(2*b^2)

Rule 6537

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

Rule 6543

Int[Cosh[(a_.) + (b_.)*(x_)]*CoshIntegral[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((
e + f*x)^m*Sinh[a + b*x]*CoshIntegral[c + d*x])/b, x] + (-Dist[d/b, Int[((e + f*x)^m*Sinh[a + b*x]*Cosh[c + d*
x])/(c + d*x), x], x] - Dist[(f*m)/b, Int[(e + f*x)^(m - 1)*Sinh[a + b*x]*CoshIntegral[c + d*x], x], x]) /; Fr
eeQ[{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 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6547

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

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 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

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

Mathematica [A]  time = 0.0427212, size = 57, normalized size = 0.77 \[ \frac{2 b^2 x^2 \text{Chi}(b x)^2-2 \text{Chi}(2 b x)+4 \text{Chi}(b x) (\cosh (b x)-b x \sinh (b x))+\cosh (2 b x)-2 \log (x)}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cosh[2*b*x] + 2*b^2*x^2*CoshIntegral[b*x]^2 - 2*CoshIntegral[2*b*x] - 2*Log[x] + 4*CoshIntegral[b*x]*(Cosh[b*
x] - b*x*Sinh[b*x]))/(4*b^2)

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Maple [A]  time = 0.059, size = 69, normalized size = 0.9 \begin{align*}{\frac{{x}^{2} \left ({\it Chi} \left ( bx \right ) \right ) ^{2}}{2}}-{\frac{x{\it Chi} \left ( bx \right ) \sinh \left ( bx \right ) }{b}}+{\frac{{\it Chi} \left ( bx \right ) \cosh \left ( bx \right ) }{{b}^{2}}}+{\frac{ \left ( \cosh \left ( bx \right ) \right ) ^{2}}{2\,{b}^{2}}}-{\frac{\ln \left ( bx \right ) }{2\,{b}^{2}}}-{\frac{{\it Chi} \left ( 2\,bx \right ) }{2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x^2*Chi(b*x)^2-x*Chi(b*x)*sinh(b*x)/b+Chi(b*x)*cosh(b*x)/b^2+1/2/b^2*cosh(b*x)^2-1/2/b^2*ln(b*x)-1/2*Chi(2
*b*x)/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*Chi(b*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*cosh_integral(b*x)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*Chi(b*x)**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*Chi(b*x)^2, x)