3.91 \(\int \frac{\text{Chi}(a+b x)}{x^2} \, dx\)

Optimal. Leaf size=46 \[ -\frac{b \text{Chi}(a+b x)}{a}-\frac{\text{Chi}(a+b x)}{x}+\frac{b \cosh (a) \text{Chi}(b x)}{a}+\frac{b \sinh (a) \text{Shi}(b x)}{a} \]

[Out]

(b*Cosh[a]*CoshIntegral[b*x])/a - (b*CoshIntegral[a + b*x])/a - CoshIntegral[a + b*x]/x + (b*Sinh[a]*SinhInteg
ral[b*x])/a

________________________________________________________________________________________

Rubi [A]  time = 0.215945, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6533, 6742, 3303, 3298, 3301} \[ -\frac{b \text{Chi}(a+b x)}{a}-\frac{\text{Chi}(a+b x)}{x}+\frac{b \cosh (a) \text{Chi}(b x)}{a}+\frac{b \sinh (a) \text{Shi}(b x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[CoshIntegral[a + b*x]/x^2,x]

[Out]

(b*Cosh[a]*CoshIntegral[b*x])/a - (b*CoshIntegral[a + b*x])/a - CoshIntegral[a + b*x]/x + (b*Sinh[a]*SinhInteg
ral[b*x])/a

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 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 3303

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

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]

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

Mathematica [A]  time = 0.102557, size = 39, normalized size = 0.85 \[ \frac{-(a+b x) \text{Chi}(a+b x)+b x \cosh (a) \text{Chi}(b x)+b x \sinh (a) \text{Shi}(b x)}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.057, size = 0, normalized size = 0. \begin{align*} \int{\frac{{\it Chi} \left ( bx+a \right ) }{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Chi(b*x + a)/x^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(cosh_integral(b*x + a)/x^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(Chi(a + b*x)/x**2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(Chi(b*x + a)/x^2, x)