3.56 \(\int x \sinh (a+b x) \text{Shi}(a+b x) \, dx\)

Optimal. Leaf size=97 \[ \frac{\text{Chi}(2 a+2 b x)}{2 b^2}+\frac{a \text{Shi}(2 a+2 b x)}{2 b^2}-\frac{\text{Shi}(a+b x) \sinh (a+b x)}{b^2}-\frac{\log (a+b x)}{2 b^2}-\frac{\cosh (2 a+2 b x)}{4 b^2}+\frac{x \text{Shi}(a+b x) \cosh (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.251543, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.643, Rules used = {6542, 5617, 6741, 6742, 2638, 3298, 6546, 3312, 3301} \[ \frac{\text{Chi}(2 a+2 b x)}{2 b^2}+\frac{a \text{Shi}(2 a+2 b x)}{2 b^2}-\frac{\text{Shi}(a+b x) \sinh (a+b x)}{b^2}-\frac{\log (a+b x)}{2 b^2}-\frac{\cosh (2 a+2 b x)}{4 b^2}+\frac{x \text{Shi}(a+b x) \cosh (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cosh[2*a + 2*b*x]/(4*b^2) + CoshIntegral[2*a + 2*b*x]/(2*b^2) - Log[a + b*x]/(2*b^2) + (x*Cosh[a + b*x]*SinhI
ntegral[a + b*x])/b - (Sinh[a + b*x]*SinhIntegral[a + b*x])/b^2 + (a*SinhIntegral[2*a + 2*b*x])/(2*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 5617

Int[Cosh[w_]^(p_.)*(u_.)*Sinh[v_]^(p_.), x_Symbol] :> Dist[1/2^p, Int[u*Sinh[2*v]^p, x], x] /; EqQ[w, v] && In
tegerQ[p]

Rule 6741

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

Rule 6742

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 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 (a+b x) \text{Shi}(a+b x) \, dx &=\frac{x \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{\int \cosh (a+b x) \text{Shi}(a+b x) \, dx}{b}-\int \frac{x \cosh (a+b x) \sinh (a+b x)}{a+b x} \, dx\\ &=\frac{x \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{\sinh (a+b x) \text{Shi}(a+b x)}{b^2}-\frac{1}{2} \int \frac{x \sinh (2 (a+b x))}{a+b x} \, dx+\frac{\int \frac{\sinh ^2(a+b x)}{a+b x} \, dx}{b}\\ &=\frac{x \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{\sinh (a+b x) \text{Shi}(a+b x)}{b^2}-\frac{1}{2} \int \frac{x \sinh (2 a+2 b x)}{a+b x} \, dx-\frac{\int \left (\frac{1}{2 (a+b x)}-\frac{\cosh (2 a+2 b x)}{2 (a+b x)}\right ) \, dx}{b}\\ &=-\frac{\log (a+b x)}{2 b^2}+\frac{x \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{\sinh (a+b x) \text{Shi}(a+b x)}{b^2}-\frac{1}{2} \int \left (\frac{\sinh (2 a+2 b x)}{b}+\frac{a \sinh (2 a+2 b x)}{b (-a-b x)}\right ) \, dx+\frac{\int \frac{\cosh (2 a+2 b x)}{a+b x} \, dx}{2 b}\\ &=\frac{\text{Chi}(2 a+2 b x)}{2 b^2}-\frac{\log (a+b x)}{2 b^2}+\frac{x \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{\sinh (a+b x) \text{Shi}(a+b x)}{b^2}-\frac{\int \sinh (2 a+2 b x) \, dx}{2 b}-\frac{a \int \frac{\sinh (2 a+2 b x)}{-a-b x} \, dx}{2 b}\\ &=-\frac{\cosh (2 a+2 b x)}{4 b^2}+\frac{\text{Chi}(2 a+2 b x)}{2 b^2}-\frac{\log (a+b x)}{2 b^2}+\frac{x \cosh (a+b x) \text{Shi}(a+b x)}{b}-\frac{\sinh (a+b x) \text{Shi}(a+b x)}{b^2}+\frac{a \text{Shi}(2 a+2 b x)}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.173726, size = 73, normalized size = 0.75 \[ \frac{2 \text{Chi}(2 (a+b x))+2 a \text{Shi}(2 (a+b x))+4 \text{Shi}(a+b x) (b x \cosh (a+b x)-\sinh (a+b x))-2 \log (a+b x)-\cosh (2 (a+b x))}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.056, size = 89, normalized size = 0.9 \begin{align*}{\frac{x\cosh \left ( bx+a \right ){\it Shi} \left ( bx+a \right ) }{b}}-{\frac{{\it Shi} \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{{b}^{2}}}-{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{2\,{b}^{2}}}-{\frac{\ln \left ( bx+a \right ) }{2\,{b}^{2}}}+{\frac{{\it Chi} \left ( 2\,bx+2\,a \right ) }{2\,{b}^{2}}}+{\frac{a{\it Shi} \left ( 2\,bx+2\,a \right ) }{2\,{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*Shi(b*x + a)*sinh(b*x + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*Shi(b*x+a)*sinh(b*x+a),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*Shi(b*x+a)*sinh(b*x+a),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*Shi(b*x + a)*sinh(b*x + a), x)