3.72 \(\int x \text{Chi}(b x) \, dx\)

Optimal. Leaf size=35 \[ \frac{\cosh (b x)}{2 b^2}+\frac{1}{2} x^2 \text{Chi}(b x)-\frac{x \sinh (b x)}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0274293, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {6533, 12, 3296, 2638} \[ \frac{\cosh (b x)}{2 b^2}+\frac{1}{2} x^2 \text{Chi}(b x)-\frac{x \sinh (b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6533

Int[CoshIntegral[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*CoshInte
gral[a + b*x])/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[((c + d*x)^(m + 1)*Cosh[a + b*x])/(a + b*x), x], x] /
; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rule 12

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

Rule 3296

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0071446, size = 35, normalized size = 1. \[ \frac{\cosh (b x)}{2 b^2}+\frac{1}{2} x^2 \text{Chi}(b x)-\frac{x \sinh (b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.045, size = 32, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{2}} \left ({\frac{{b}^{2}{x}^{2}{\it Chi} \left ( bx \right ) }{2}}-{\frac{bx\sinh \left ( bx \right ) }{2}}+{\frac{\cosh \left ( bx \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*Chi(b*x), 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 ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x*cosh_integral(b*x), x)

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Sympy [A]  time = 1.79195, size = 53, normalized size = 1.51 \begin{align*} - \frac{x^{2} \log{\left (b x \right )}}{2} + \frac{x^{2} \log{\left (b^{2} x^{2} \right )}}{4} + \frac{x^{2} \operatorname{Chi}\left (b x\right )}{2} - \frac{x \sinh{\left (b x \right )}}{2 b} + \frac{\cosh{\left (b x \right )}}{2 b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2*log(b*x)/2 + x**2*log(b**2*x**2)/4 + x**2*Chi(b*x)/2 - x*sinh(b*x)/(2*b) + cosh(b*x)/(2*b**2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*Chi(b*x), x)