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

Optimal. Leaf size=44 \[ \frac{1}{2} b \text{Chi}(b x)^2+b \text{Chi}(2 b x)-\frac{\text{Chi}(b x) \sinh (b x)}{x}-\frac{\sinh (2 b x)}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.0990895, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6551, 6686, 12, 5448, 3297, 3301} \[ \frac{1}{2} b \text{Chi}(b x)^2+b \text{Chi}(2 b x)-\frac{\text{Chi}(b x) \sinh (b x)}{x}-\frac{\sinh (2 b x)}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[(CoshIntegral[b*x]*Sinh[b*x])/x^2,x]

[Out]

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

Rule 6551

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[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 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

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

Mathematica [A]  time = 0.0065755, size = 44, normalized size = 1. \[ \frac{1}{2} b \text{Chi}(b x)^2+b \text{Chi}(2 b x)-\frac{\text{Chi}(b x) \sinh (b x)}{x}-\frac{\sinh (2 b x)}{2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[(CoshIntegral[b*x]*Sinh[b*x])/x^2,x]

[Out]

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

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{{\it Chi} \left ( bx \right ) \sinh \left ( bx \right ) }{{x}^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

integral(cosh_integral(b*x)*sinh(b*x)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(b*x)*Chi(b*x)/x**2, x)

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

[Out]

integrate(Chi(b*x)*sinh(b*x)/x^2, x)