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

Optimal. Leaf size=25 \[ -\frac{\text{Chi}(b x)}{x}+b \text{Shi}(b x)-\frac{\cosh (b x)}{x} \]

[Out]

-(Cosh[b*x]/x) - CoshIntegral[b*x]/x + b*SinhIntegral[b*x]

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Rubi [A]  time = 0.0513678, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6533, 12, 3297, 3298} \[ -\frac{\text{Chi}(b x)}{x}+b \text{Shi}(b x)-\frac{\cosh (b x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cosh[b*x]/x) - CoshIntegral[b*x]/x + b*SinhIntegral[b*x]

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 3297

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

Rule 3298

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

Rubi steps

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

Mathematica [A]  time = 0.0117421, size = 25, normalized size = 1. \[ -\frac{\text{Chi}(b x)}{x}+b \text{Shi}(b x)-\frac{\cosh (b x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cosh[b*x]/x) - CoshIntegral[b*x]/x + b*SinhIntegral[b*x]

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Maple [A]  time = 0.046, size = 32, normalized size = 1.3 \begin{align*} b \left ( -{\frac{{\it Chi} \left ( bx \right ) }{bx}}-{\frac{\cosh \left ( bx \right ) }{bx}}+{\it Shi} \left ( bx \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(-Chi(b*x)/b/x-1/b/x*cosh(b*x)+Shi(b*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Chi(b*x)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(cosh_integral(b*x)/x^2, x)

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Sympy [B]  time = 1.42461, size = 39, normalized size = 1.56 \begin{align*} \frac{b^{2} x{{}_{3}F_{4}\left (\begin{matrix} \frac{1}{2}, 1, 1 \\ \frac{3}{2}, \frac{3}{2}, 2, 2 \end{matrix}\middle |{\frac{b^{2} x^{2}}{4}} \right )}}{4} - \frac{\log{\left (b^{2} x^{2} \right )}}{2 x} - \frac{1}{x} - \frac{\gamma }{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**2*x*hyper((1/2, 1, 1), (3/2, 3/2, 2, 2), b**2*x**2/4)/4 - log(b**2*x**2)/(2*x) - 1/x - EulerGamma/x

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Chi(b*x)/x^2, x)