3.43 \(\int x \sinh (b x) \text{Shi}(b x) \, dx\)

Optimal. Leaf size=61 \[ \frac{\text{Chi}(2 b x)}{2 b^2}-\frac{\text{Shi}(b x) \sinh (b x)}{b^2}-\frac{\log (x)}{2 b^2}-\frac{\sinh ^2(b x)}{2 b^2}+\frac{x \text{Shi}(b x) \cosh (b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.0900407, antiderivative size = 61, 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 = {6542, 12, 2564, 30, 6546, 3312, 3301} \[ \frac{\text{Chi}(2 b x)}{2 b^2}-\frac{\text{Shi}(b x) \sinh (b x)}{b^2}-\frac{\log (x)}{2 b^2}-\frac{\sinh ^2(b x)}{2 b^2}+\frac{x \text{Shi}(b x) \cosh (b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6542

Int[((e_.) + (f_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]*SinhIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[((
e + f*x)^m*Cosh[a + b*x]*SinhIntegral[c + d*x])/b, x] + (-Dist[d/b, Int[((e + f*x)^m*Cosh[a + b*x]*Sinh[c + d*
x])/(c + d*x), x], x] - Dist[(f*m)/b, Int[(e + f*x)^(m - 1)*Cosh[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 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
 !(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6546

Int[Cosh[(a_.) + (b_.)*(x_)]*SinhIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[(Sinh[a + b*x]*SinhIntegral[c
 + d*x])/b, x] - Dist[d/b, Int[(Sinh[a + b*x]*Sinh[c + d*x])/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x]

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && 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 x \sinh (b x) \text{Shi}(b x) \, dx &=\frac{x \cosh (b x) \text{Shi}(b x)}{b}-\frac{\int \cosh (b x) \text{Shi}(b x) \, dx}{b}-\int \frac{\cosh (b x) \sinh (b x)}{b} \, dx\\ &=\frac{x \cosh (b x) \text{Shi}(b x)}{b}-\frac{\sinh (b x) \text{Shi}(b x)}{b^2}-\frac{\int \cosh (b x) \sinh (b x) \, dx}{b}+\frac{\int \frac{\sinh ^2(b x)}{b x} \, dx}{b}\\ &=\frac{x \cosh (b x) \text{Shi}(b x)}{b}-\frac{\sinh (b x) \text{Shi}(b x)}{b^2}+\frac{\int \frac{\sinh ^2(b x)}{x} \, dx}{b^2}+\frac{\operatorname{Subst}(\int x \, dx,x,i \sinh (b x))}{b^2}\\ &=-\frac{\sinh ^2(b x)}{2 b^2}+\frac{x \cosh (b x) \text{Shi}(b x)}{b}-\frac{\sinh (b x) \text{Shi}(b x)}{b^2}-\frac{\int \left (\frac{1}{2 x}-\frac{\cosh (2 b x)}{2 x}\right ) \, dx}{b^2}\\ &=-\frac{\log (x)}{2 b^2}-\frac{\sinh ^2(b x)}{2 b^2}+\frac{x \cosh (b x) \text{Shi}(b x)}{b}-\frac{\sinh (b x) \text{Shi}(b x)}{b^2}+\frac{\int \frac{\cosh (2 b x)}{x} \, dx}{2 b^2}\\ &=\frac{\text{Chi}(2 b x)}{2 b^2}-\frac{\log (x)}{2 b^2}-\frac{\sinh ^2(b x)}{2 b^2}+\frac{x \cosh (b x) \text{Shi}(b x)}{b}-\frac{\sinh (b x) \text{Shi}(b x)}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.0831388, size = 44, normalized size = 0.72 \[ -\frac{-2 \text{Chi}(2 b x)+\text{Shi}(b x) (4 \sinh (b x)-4 b x \cosh (b x))+\cosh (2 b x)+2 \log (x)}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 58, normalized size = 1. \begin{align*}{\frac{x\cosh \left ( bx \right ){\it Shi} \left ( bx \right ) }{b}}-{\frac{{\it Shi} \left ( bx \right ) \sinh \left ( bx \right ) }{{b}^{2}}}-{\frac{ \left ( \cosh \left ( bx \right ) \right ) ^{2}}{2\,{b}^{2}}}-{\frac{\ln \left ( bx \right ) }{2\,{b}^{2}}}+{\frac{{\it Chi} \left ( 2\,bx \right ) }{2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*Shi(b*x)*sinh(b*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x \sinh \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*Shi(b*x)*sinh(b*x),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \sinh{\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*Shi(b*x)*sinh(b*x),x)

[Out]

Integral(x*sinh(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 ) \sinh \left (b x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*Shi(b*x)*sinh(b*x), x)