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

Optimal. Leaf size=63 \[ \frac{3 x^2 \sinh (b x)}{4 b^2}+\frac{3 \sinh (b x)}{2 b^4}-\frac{3 x \cosh (b x)}{2 b^3}+\frac{1}{4} x^4 \text{Shi}(b x)-\frac{x^3 \cosh (b x)}{4 b} \]

[Out]

(-3*x*Cosh[b*x])/(2*b^3) - (x^3*Cosh[b*x])/(4*b) + (3*Sinh[b*x])/(2*b^4) + (3*x^2*Sinh[b*x])/(4*b^2) + (x^4*Si
nhIntegral[b*x])/4

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Rubi [A]  time = 0.0839202, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6532, 12, 3296, 2637} \[ \frac{3 x^2 \sinh (b x)}{4 b^2}+\frac{3 \sinh (b x)}{2 b^4}-\frac{3 x \cosh (b x)}{2 b^3}+\frac{1}{4} x^4 \text{Shi}(b x)-\frac{x^3 \cosh (b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x*Cosh[b*x])/(2*b^3) - (x^3*Cosh[b*x])/(4*b) + (3*Sinh[b*x])/(2*b^4) + (3*x^2*Sinh[b*x])/(4*b^2) + (x^4*Si
nhIntegral[b*x])/4

Rule 6532

Int[((c_.) + (d_.)*(x_))^(m_.)*SinhIntegral[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[((c + d*x)^(m + 1)*SinhInte
gral[a + b*x])/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[((c + d*x)^(m + 1)*Sinh[a + b*x])/(a + b*x), x], x] /
; FreeQ[{a, b, c, d, 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 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x^3 \text{Shi}(b x) \, dx &=\frac{1}{4} x^4 \text{Shi}(b x)-\frac{1}{4} b \int \frac{x^3 \sinh (b x)}{b} \, dx\\ &=\frac{1}{4} x^4 \text{Shi}(b x)-\frac{1}{4} \int x^3 \sinh (b x) \, dx\\ &=-\frac{x^3 \cosh (b x)}{4 b}+\frac{1}{4} x^4 \text{Shi}(b x)+\frac{3 \int x^2 \cosh (b x) \, dx}{4 b}\\ &=-\frac{x^3 \cosh (b x)}{4 b}+\frac{3 x^2 \sinh (b x)}{4 b^2}+\frac{1}{4} x^4 \text{Shi}(b x)-\frac{3 \int x \sinh (b x) \, dx}{2 b^2}\\ &=-\frac{3 x \cosh (b x)}{2 b^3}-\frac{x^3 \cosh (b x)}{4 b}+\frac{3 x^2 \sinh (b x)}{4 b^2}+\frac{1}{4} x^4 \text{Shi}(b x)+\frac{3 \int \cosh (b x) \, dx}{2 b^3}\\ &=-\frac{3 x \cosh (b x)}{2 b^3}-\frac{x^3 \cosh (b x)}{4 b}+\frac{3 \sinh (b x)}{2 b^4}+\frac{3 x^2 \sinh (b x)}{4 b^2}+\frac{1}{4} x^4 \text{Shi}(b x)\\ \end{align*}

Mathematica [A]  time = 0.0361031, size = 53, normalized size = 0.84 \[ \frac{3 \left (b^2 x^2+2\right ) \sinh (b x)}{4 b^4}-\frac{x \left (b^2 x^2+6\right ) \cosh (b x)}{4 b^3}+\frac{1}{4} x^4 \text{Shi}(b x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 56, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{4}} \left ({\frac{{b}^{4}{x}^{4}{\it Shi} \left ( bx \right ) }{4}}-{\frac{{b}^{3}{x}^{3}\cosh \left ( bx \right ) }{4}}+{\frac{3\,{b}^{2}{x}^{2}\sinh \left ( bx \right ) }{4}}-{\frac{3\,bx\cosh \left ( bx \right ) }{2}}+{\frac{3\,\sinh \left ( bx \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate(x^3*Shi(b*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{3} \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),x, algorithm="fricas")

[Out]

integral(x^3*sinh_integral(b*x), x)

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Sympy [A]  time = 1.26648, size = 61, normalized size = 0.97 \begin{align*} \frac{x^{4} \operatorname{Shi}{\left (b x \right )}}{4} - \frac{x^{3} \cosh{\left (b x \right )}}{4 b} + \frac{3 x^{2} \sinh{\left (b x \right )}}{4 b^{2}} - \frac{3 x \cosh{\left (b x \right )}}{2 b^{3}} + \frac{3 \sinh{\left (b x \right )}}{2 b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*Shi(b*x),x)

[Out]

x**4*Shi(b*x)/4 - x**3*cosh(b*x)/(4*b) + 3*x**2*sinh(b*x)/(4*b**2) - 3*x*cosh(b*x)/(2*b**3) + 3*sinh(b*x)/(2*b
**4)

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

[Out]

integrate(x^3*Shi(b*x), x)