3.45 \(\int x^3 \sinh (b x) \text{Shi}(b x) \, dx\)

Optimal. Leaf size=125 \[ \frac{3 \text{Chi}(2 b x)}{b^4}-\frac{3 x^2 \text{Shi}(b x) \sinh (b x)}{b^2}-\frac{6 \text{Shi}(b x) \sinh (b x)}{b^4}+\frac{6 x \text{Shi}(b x) \cosh (b x)}{b^3}-\frac{x^2}{b^2}-\frac{x^2 \sinh ^2(b x)}{2 b^2}-\frac{3 \log (x)}{b^4}-\frac{4 \sinh ^2(b x)}{b^4}+\frac{2 x \sinh (b x) \cosh (b x)}{b^3}+\frac{x^3 \text{Shi}(b x) \cosh (b x)}{b} \]

[Out]

-(x^2/b^2) + (3*CoshIntegral[2*b*x])/b^4 - (3*Log[x])/b^4 + (2*x*Cosh[b*x]*Sinh[b*x])/b^3 - (4*Sinh[b*x]^2)/b^
4 - (x^2*Sinh[b*x]^2)/(2*b^2) + (6*x*Cosh[b*x]*SinhIntegral[b*x])/b^3 + (x^3*Cosh[b*x]*SinhIntegral[b*x])/b -
(6*Sinh[b*x]*SinhIntegral[b*x])/b^4 - (3*x^2*Sinh[b*x]*SinhIntegral[b*x])/b^2

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Rubi [A]  time = 0.204213, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 10, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.833, Rules used = {6542, 12, 5372, 3310, 30, 6548, 2564, 6546, 3312, 3301} \[ \frac{3 \text{Chi}(2 b x)}{b^4}-\frac{3 x^2 \text{Shi}(b x) \sinh (b x)}{b^2}-\frac{6 \text{Shi}(b x) \sinh (b x)}{b^4}+\frac{6 x \text{Shi}(b x) \cosh (b x)}{b^3}-\frac{x^2}{b^2}-\frac{x^2 \sinh ^2(b x)}{2 b^2}-\frac{3 \log (x)}{b^4}-\frac{4 \sinh ^2(b x)}{b^4}+\frac{2 x \sinh (b x) \cosh (b x)}{b^3}+\frac{x^3 \text{Shi}(b x) \cosh (b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*Sinh[b*x]*SinhIntegral[b*x],x]

[Out]

-(x^2/b^2) + (3*CoshIntegral[2*b*x])/b^4 - (3*Log[x])/b^4 + (2*x*Cosh[b*x]*Sinh[b*x])/b^3 - (4*Sinh[b*x]^2)/b^
4 - (x^2*Sinh[b*x]^2)/(2*b^2) + (6*x*Cosh[b*x]*SinhIntegral[b*x])/b^3 + (x^3*Cosh[b*x]*SinhIntegral[b*x])/b -
(6*Sinh[b*x]*SinhIntegral[b*x])/b^4 - (3*x^2*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 5372

Int[Cosh[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sinh[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[(x^(m -
n + 1)*Sinh[a + b*x^n]^(p + 1))/(b*n*(p + 1)), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sinh[a + b*x
^n]^(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 3310

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

Rule 30

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

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 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 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^3 \sinh (b x) \text{Shi}(b x) \, dx &=\frac{x^3 \cosh (b x) \text{Shi}(b x)}{b}-\frac{3 \int x^2 \cosh (b x) \text{Shi}(b x) \, dx}{b}-\int \frac{x^2 \cosh (b x) \sinh (b x)}{b} \, dx\\ &=\frac{x^3 \cosh (b x) \text{Shi}(b x)}{b}-\frac{3 x^2 \sinh (b x) \text{Shi}(b x)}{b^2}+\frac{6 \int x \sinh (b x) \text{Shi}(b x) \, dx}{b^2}-\frac{\int x^2 \cosh (b x) \sinh (b x) \, dx}{b}+\frac{3 \int \frac{x \sinh ^2(b x)}{b} \, dx}{b}\\ &=-\frac{x^2 \sinh ^2(b x)}{2 b^2}+\frac{6 x \cosh (b x) \text{Shi}(b x)}{b^3}+\frac{x^3 \cosh (b x) \text{Shi}(b x)}{b}-\frac{3 x^2 \sinh (b x) \text{Shi}(b x)}{b^2}-\frac{6 \int \cosh (b x) \text{Shi}(b x) \, dx}{b^3}+\frac{\int x \sinh ^2(b x) \, dx}{b^2}+\frac{3 \int x \sinh ^2(b x) \, dx}{b^2}-\frac{6 \int \frac{\cosh (b x) \sinh (b x)}{b} \, dx}{b^2}\\ &=\frac{2 x \cosh (b x) \sinh (b x)}{b^3}-\frac{\sinh ^2(b x)}{b^4}-\frac{x^2 \sinh ^2(b x)}{2 b^2}+\frac{6 x \cosh (b x) \text{Shi}(b x)}{b^3}+\frac{x^3 \cosh (b x) \text{Shi}(b x)}{b}-\frac{6 \sinh (b x) \text{Shi}(b x)}{b^4}-\frac{3 x^2 \sinh (b x) \text{Shi}(b x)}{b^2}-\frac{6 \int \cosh (b x) \sinh (b x) \, dx}{b^3}+\frac{6 \int \frac{\sinh ^2(b x)}{b x} \, dx}{b^3}-\frac{\int x \, dx}{2 b^2}-\frac{3 \int x \, dx}{2 b^2}\\ &=-\frac{x^2}{b^2}+\frac{2 x \cosh (b x) \sinh (b x)}{b^3}-\frac{\sinh ^2(b x)}{b^4}-\frac{x^2 \sinh ^2(b x)}{2 b^2}+\frac{6 x \cosh (b x) \text{Shi}(b x)}{b^3}+\frac{x^3 \cosh (b x) \text{Shi}(b x)}{b}-\frac{6 \sinh (b x) \text{Shi}(b x)}{b^4}-\frac{3 x^2 \sinh (b x) \text{Shi}(b x)}{b^2}+\frac{6 \int \frac{\sinh ^2(b x)}{x} \, dx}{b^4}+\frac{6 \operatorname{Subst}(\int x \, dx,x,i \sinh (b x))}{b^4}\\ &=-\frac{x^2}{b^2}+\frac{2 x \cosh (b x) \sinh (b x)}{b^3}-\frac{4 \sinh ^2(b x)}{b^4}-\frac{x^2 \sinh ^2(b x)}{2 b^2}+\frac{6 x \cosh (b x) \text{Shi}(b x)}{b^3}+\frac{x^3 \cosh (b x) \text{Shi}(b x)}{b}-\frac{6 \sinh (b x) \text{Shi}(b x)}{b^4}-\frac{3 x^2 \sinh (b x) \text{Shi}(b x)}{b^2}-\frac{6 \int \left (\frac{1}{2 x}-\frac{\cosh (2 b x)}{2 x}\right ) \, dx}{b^4}\\ &=-\frac{x^2}{b^2}-\frac{3 \log (x)}{b^4}+\frac{2 x \cosh (b x) \sinh (b x)}{b^3}-\frac{4 \sinh ^2(b x)}{b^4}-\frac{x^2 \sinh ^2(b x)}{2 b^2}+\frac{6 x \cosh (b x) \text{Shi}(b x)}{b^3}+\frac{x^3 \cosh (b x) \text{Shi}(b x)}{b}-\frac{6 \sinh (b x) \text{Shi}(b x)}{b^4}-\frac{3 x^2 \sinh (b x) \text{Shi}(b x)}{b^2}+\frac{3 \int \frac{\cosh (2 b x)}{x} \, dx}{b^4}\\ &=-\frac{x^2}{b^2}+\frac{3 \text{Chi}(2 b x)}{b^4}-\frac{3 \log (x)}{b^4}+\frac{2 x \cosh (b x) \sinh (b x)}{b^3}-\frac{4 \sinh ^2(b x)}{b^4}-\frac{x^2 \sinh ^2(b x)}{2 b^2}+\frac{6 x \cosh (b x) \text{Shi}(b x)}{b^3}+\frac{x^3 \cosh (b x) \text{Shi}(b x)}{b}-\frac{6 \sinh (b x) \text{Shi}(b x)}{b^4}-\frac{3 x^2 \sinh (b x) \text{Shi}(b x)}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.162853, size = 93, normalized size = 0.74 \[ -\frac{-4 \text{Shi}(b x) \left (b x \left (b^2 x^2+6\right ) \cosh (b x)-3 \left (b^2 x^2+2\right ) \sinh (b x)\right )+3 b^2 x^2+b^2 x^2 \cosh (2 b x)-12 \text{Chi}(2 b x)-4 b x \sinh (2 b x)+8 \cosh (2 b x)+12 \log (x)}{4 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sinh[b*x]*SinhIntegral[b*x],x]

[Out]

-(3*b^2*x^2 + 8*Cosh[2*b*x] + b^2*x^2*Cosh[2*b*x] - 12*CoshIntegral[2*b*x] + 12*Log[x] - 4*b*x*Sinh[2*b*x] - 4
*(b*x*(6 + b^2*x^2)*Cosh[b*x] - 3*(2 + b^2*x^2)*Sinh[b*x])*SinhIntegral[b*x])/(4*b^4)

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Maple [A]  time = 0.056, size = 126, normalized size = 1. \begin{align*}{\frac{{x}^{3}\cosh \left ( bx \right ){\it Shi} \left ( bx \right ) }{b}}-3\,{\frac{{x}^{2}{\it Shi} \left ( bx \right ) \sinh \left ( bx \right ) }{{b}^{2}}}+6\,{\frac{x\cosh \left ( bx \right ){\it Shi} \left ( bx \right ) }{{b}^{3}}}-6\,{\frac{{\it Shi} \left ( bx \right ) \sinh \left ( bx \right ) }{{b}^{4}}}-{\frac{{x}^{2} \left ( \cosh \left ( bx \right ) \right ) ^{2}}{2\,{b}^{2}}}+2\,{\frac{x\cosh \left ( bx \right ) \sinh \left ( bx \right ) }{{b}^{3}}}-{\frac{{x}^{2}}{2\,{b}^{2}}}-4\,{\frac{ \left ( \cosh \left ( bx \right ) \right ) ^{2}}{{b}^{4}}}-3\,{\frac{\ln \left ( bx \right ) }{{b}^{4}}}+3\,{\frac{{\it Chi} \left ( 2\,bx \right ) }{{b}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*Shi(b*x)*sinh(b*x),x)

[Out]

x^3*cosh(b*x)*Shi(b*x)/b-3*x^2*Shi(b*x)*sinh(b*x)/b^2+6*x*cosh(b*x)*Shi(b*x)/b^3-6*Shi(b*x)*sinh(b*x)/b^4-1/2/
b^2*x^2*cosh(b*x)^2+2*x*cosh(b*x)*sinh(b*x)/b^3-1/2*x^2/b^2-4*cosh(b*x)^2/b^4-3/b^4*ln(b*x)+3*Chi(2*b*x)/b^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3}{\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^3*Shi(b*x)*sinh(b*x),x, algorithm="maxima")

[Out]

integrate(x^3*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^{3} \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^3*Shi(b*x)*sinh(b*x),x, algorithm="fricas")

[Out]

integral(x^3*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^{3} \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**3*Shi(b*x)*sinh(b*x),x)

[Out]

Integral(x**3*sinh(b*x)*Shi(b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3}{\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^3*Shi(b*x)*sinh(b*x),x, algorithm="giac")

[Out]

integrate(x^3*Shi(b*x)*sinh(b*x), x)