3.50 \(\int x \cosh (b x) \text{Shi}(b x) \, dx\)

Optimal. Leaf size=62 \[ \frac{\text{Shi}(2 b x)}{2 b^2}-\frac{\text{Shi}(b x) \cosh (b x)}{b^2}-\frac{\sinh (b x) \cosh (b x)}{2 b^2}+\frac{x \text{Shi}(b x) \sinh (b x)}{b}+\frac{x}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0720294, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {6548, 12, 2635, 8, 6540, 5448, 3298} \[ \frac{\text{Shi}(2 b x)}{2 b^2}-\frac{\text{Shi}(b x) \cosh (b x)}{b^2}-\frac{\sinh (b x) \cosh (b x)}{2 b^2}+\frac{x \text{Shi}(b x) \sinh (b x)}{b}+\frac{x}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[x*Cosh[b*x]*SinhIntegral[b*x],x]

[Out]

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

Rule 6548

Int[Cosh[(a_.) + (b_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*SinhIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[((
e + f*x)^m*Sinh[a + b*x]*SinhIntegral[c + d*x])/b, x] + (-Dist[d/b, Int[((e + f*x)^m*Sinh[a + b*x]*Sinh[c + d*
x])/(c + d*x), x], x] - Dist[(f*m)/b, Int[(e + f*x)^(m - 1)*Sinh[a + b*x]*SinhIntegral[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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 6540

Int[Sinh[(a_.) + (b_.)*(x_)]*SinhIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[(Cosh[a + b*x]*SinhIntegral[c
+ d*x])/b, x] - Dist[d/b, Int[(Cosh[a + b*x]*Sinh[c + d*x])/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, 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 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 x \cosh (b x) \text{Shi}(b x) \, dx &=\frac{x \sinh (b x) \text{Shi}(b x)}{b}-\frac{\int \sinh (b x) \text{Shi}(b x) \, dx}{b}-\int \frac{\sinh ^2(b x)}{b} \, dx\\ &=-\frac{\cosh (b x) \text{Shi}(b x)}{b^2}+\frac{x \sinh (b x) \text{Shi}(b x)}{b}+\frac{\int \frac{\cosh (b x) \sinh (b x)}{b x} \, dx}{b}-\frac{\int \sinh ^2(b x) \, dx}{b}\\ &=-\frac{\cosh (b x) \sinh (b x)}{2 b^2}-\frac{\cosh (b x) \text{Shi}(b x)}{b^2}+\frac{x \sinh (b x) \text{Shi}(b x)}{b}+\frac{\int \frac{\cosh (b x) \sinh (b x)}{x} \, dx}{b^2}+\frac{\int 1 \, dx}{2 b}\\ &=\frac{x}{2 b}-\frac{\cosh (b x) \sinh (b x)}{2 b^2}-\frac{\cosh (b x) \text{Shi}(b x)}{b^2}+\frac{x \sinh (b x) \text{Shi}(b x)}{b}+\frac{\int \frac{\sinh (2 b x)}{2 x} \, dx}{b^2}\\ &=\frac{x}{2 b}-\frac{\cosh (b x) \sinh (b x)}{2 b^2}-\frac{\cosh (b x) \text{Shi}(b x)}{b^2}+\frac{x \sinh (b x) \text{Shi}(b x)}{b}+\frac{\int \frac{\sinh (2 b x)}{x} \, dx}{2 b^2}\\ &=\frac{x}{2 b}-\frac{\cosh (b x) \sinh (b x)}{2 b^2}-\frac{\cosh (b x) \text{Shi}(b x)}{b^2}+\frac{x \sinh (b x) \text{Shi}(b x)}{b}+\frac{\text{Shi}(2 b x)}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0507832, size = 46, normalized size = 0.74 \[ \frac{2 \text{Shi}(2 b x)+4 \text{Shi}(b x) (b x \sinh (b x)-\cosh (b x))+2 b x-\sinh (2 b x)}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Cosh[b*x]*SinhIntegral[b*x],x]

[Out]

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

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Maple [A]  time = 0.056, size = 46, normalized size = 0.7 \begin{align*}{\frac{1}{{b}^{2}} \left ({\it Shi} \left ( bx \right ) \left ( bx\sinh \left ( bx \right ) -\cosh \left ( bx \right ) \right ) -{\frac{\cosh \left ( bx \right ) \sinh \left ( bx \right ) }{2}}+{\frac{bx}{2}}+{\frac{{\it Shi} \left ( 2\,bx \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*cosh(b*x)*Shi(b*x),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x)*Shi(b*x),x, algorithm="maxima")

[Out]

integrate(x*Shi(b*x)*cosh(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x)*Shi(b*x),x, algorithm="fricas")

[Out]

integral(x*cosh(b*x)*sinh_integral(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x)*Shi(b*x),x)

[Out]

Integral(x*cosh(b*x)*Shi(b*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*cosh(b*x)*Shi(b*x),x, algorithm="giac")

[Out]

integrate(x*Shi(b*x)*cosh(b*x), x)