3.11 \(\int x^2 \text{Shi}(b x)^2 \, dx\)

Optimal. Leaf size=112 \[ \frac{2 \text{Shi}(2 b x)}{3 b^3}+\frac{4 x \text{Shi}(b x) \sinh (b x)}{3 b^2}-\frac{4 \text{Shi}(b x) \cosh (b x)}{3 b^3}+\frac{5 x}{6 b^2}+\frac{x \sinh ^2(b x)}{3 b^2}-\frac{5 \sinh (b x) \cosh (b x)}{6 b^3}+\frac{1}{3} x^3 \text{Shi}(b x)^2-\frac{2 x^2 \text{Shi}(b x) \cosh (b x)}{3 b} \]

[Out]

(5*x)/(6*b^2) - (5*Cosh[b*x]*Sinh[b*x])/(6*b^3) + (x*Sinh[b*x]^2)/(3*b^2) - (4*Cosh[b*x]*SinhIntegral[b*x])/(3
*b^3) - (2*x^2*Cosh[b*x]*SinhIntegral[b*x])/(3*b) + (4*x*Sinh[b*x]*SinhIntegral[b*x])/(3*b^2) + (x^3*SinhInteg
ral[b*x]^2)/3 + (2*SinhIntegral[2*b*x])/(3*b^3)

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Rubi [A]  time = 0.146728, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {6536, 6542, 12, 5372, 2635, 8, 6548, 6540, 5448, 3298} \[ \frac{2 \text{Shi}(2 b x)}{3 b^3}+\frac{4 x \text{Shi}(b x) \sinh (b x)}{3 b^2}-\frac{4 \text{Shi}(b x) \cosh (b x)}{3 b^3}+\frac{5 x}{6 b^2}+\frac{x \sinh ^2(b x)}{3 b^2}-\frac{5 \sinh (b x) \cosh (b x)}{6 b^3}+\frac{1}{3} x^3 \text{Shi}(b x)^2-\frac{2 x^2 \text{Shi}(b x) \cosh (b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[x^2*SinhIntegral[b*x]^2,x]

[Out]

(5*x)/(6*b^2) - (5*Cosh[b*x]*Sinh[b*x])/(6*b^3) + (x*Sinh[b*x]^2)/(3*b^2) - (4*Cosh[b*x]*SinhIntegral[b*x])/(3
*b^3) - (2*x^2*Cosh[b*x]*SinhIntegral[b*x])/(3*b) + (4*x*Sinh[b*x]*SinhIntegral[b*x])/(3*b^2) + (x^3*SinhInteg
ral[b*x]^2)/3 + (2*SinhIntegral[2*b*x])/(3*b^3)

Rule 6536

Int[(x_)^(m_.)*SinhIntegral[(b_.)*(x_)]^2, x_Symbol] :> Simp[(x^(m + 1)*SinhIntegral[b*x]^2)/(m + 1), x] - Dis
t[2/(m + 1), Int[x^m*Sinh[b*x]*SinhIntegral[b*x], x], x] /; FreeQ[b, x] && IGtQ[m, 0]

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 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 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 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^2 \text{Shi}(b x)^2 \, dx &=\frac{1}{3} x^3 \text{Shi}(b x)^2-\frac{2}{3} \int x^2 \sinh (b x) \text{Shi}(b x) \, dx\\ &=-\frac{2 x^2 \cosh (b x) \text{Shi}(b x)}{3 b}+\frac{1}{3} x^3 \text{Shi}(b x)^2+\frac{2}{3} \int \frac{x \cosh (b x) \sinh (b x)}{b} \, dx+\frac{4 \int x \cosh (b x) \text{Shi}(b x) \, dx}{3 b}\\ &=-\frac{2 x^2 \cosh (b x) \text{Shi}(b x)}{3 b}+\frac{4 x \sinh (b x) \text{Shi}(b x)}{3 b^2}+\frac{1}{3} x^3 \text{Shi}(b x)^2-\frac{4 \int \sinh (b x) \text{Shi}(b x) \, dx}{3 b^2}+\frac{2 \int x \cosh (b x) \sinh (b x) \, dx}{3 b}-\frac{4 \int \frac{\sinh ^2(b x)}{b} \, dx}{3 b}\\ &=\frac{x \sinh ^2(b x)}{3 b^2}-\frac{4 \cosh (b x) \text{Shi}(b x)}{3 b^3}-\frac{2 x^2 \cosh (b x) \text{Shi}(b x)}{3 b}+\frac{4 x \sinh (b x) \text{Shi}(b x)}{3 b^2}+\frac{1}{3} x^3 \text{Shi}(b x)^2-\frac{\int \sinh ^2(b x) \, dx}{3 b^2}+\frac{4 \int \frac{\cosh (b x) \sinh (b x)}{b x} \, dx}{3 b^2}-\frac{4 \int \sinh ^2(b x) \, dx}{3 b^2}\\ &=-\frac{5 \cosh (b x) \sinh (b x)}{6 b^3}+\frac{x \sinh ^2(b x)}{3 b^2}-\frac{4 \cosh (b x) \text{Shi}(b x)}{3 b^3}-\frac{2 x^2 \cosh (b x) \text{Shi}(b x)}{3 b}+\frac{4 x \sinh (b x) \text{Shi}(b x)}{3 b^2}+\frac{1}{3} x^3 \text{Shi}(b x)^2+\frac{4 \int \frac{\cosh (b x) \sinh (b x)}{x} \, dx}{3 b^3}+\frac{\int 1 \, dx}{6 b^2}+\frac{2 \int 1 \, dx}{3 b^2}\\ &=\frac{5 x}{6 b^2}-\frac{5 \cosh (b x) \sinh (b x)}{6 b^3}+\frac{x \sinh ^2(b x)}{3 b^2}-\frac{4 \cosh (b x) \text{Shi}(b x)}{3 b^3}-\frac{2 x^2 \cosh (b x) \text{Shi}(b x)}{3 b}+\frac{4 x \sinh (b x) \text{Shi}(b x)}{3 b^2}+\frac{1}{3} x^3 \text{Shi}(b x)^2+\frac{4 \int \frac{\sinh (2 b x)}{2 x} \, dx}{3 b^3}\\ &=\frac{5 x}{6 b^2}-\frac{5 \cosh (b x) \sinh (b x)}{6 b^3}+\frac{x \sinh ^2(b x)}{3 b^2}-\frac{4 \cosh (b x) \text{Shi}(b x)}{3 b^3}-\frac{2 x^2 \cosh (b x) \text{Shi}(b x)}{3 b}+\frac{4 x \sinh (b x) \text{Shi}(b x)}{3 b^2}+\frac{1}{3} x^3 \text{Shi}(b x)^2+\frac{2 \int \frac{\sinh (2 b x)}{x} \, dx}{3 b^3}\\ &=\frac{5 x}{6 b^2}-\frac{5 \cosh (b x) \sinh (b x)}{6 b^3}+\frac{x \sinh ^2(b x)}{3 b^2}-\frac{4 \cosh (b x) \text{Shi}(b x)}{3 b^3}-\frac{2 x^2 \cosh (b x) \text{Shi}(b x)}{3 b}+\frac{4 x \sinh (b x) \text{Shi}(b x)}{3 b^2}+\frac{1}{3} x^3 \text{Shi}(b x)^2+\frac{2 \text{Shi}(2 b x)}{3 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0763325, size = 78, normalized size = 0.7 \[ \frac{4 b^3 x^3 \text{Shi}(b x)^2-8 \text{Shi}(b x) \left (\left (b^2 x^2+2\right ) \cosh (b x)-2 b x \sinh (b x)\right )+8 \text{Shi}(2 b x)+8 b x-5 \sinh (2 b x)+2 b x \cosh (2 b x)}{12 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*SinhIntegral[b*x]^2,x]

[Out]

(8*b*x + 2*b*x*Cosh[2*b*x] - 5*Sinh[2*b*x] - 8*((2 + b^2*x^2)*Cosh[b*x] - 2*b*x*Sinh[b*x])*SinhIntegral[b*x] +
4*b^3*x^3*SinhIntegral[b*x]^2 + 8*SinhIntegral[2*b*x])/(12*b^3)

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Maple [A]  time = 0.057, size = 84, normalized size = 0.8 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{{b}^{3}{x}^{3} \left ({\it Shi} \left ( bx \right ) \right ) ^{2}}{3}}-2\,{\it Shi} \left ( bx \right ) \left ( 1/3\,{b}^{2}{x}^{2}\cosh \left ( bx \right ) -2/3\,bx\sinh \left ( bx \right ) +2/3\,\cosh \left ( bx \right ) \right ) +{\frac{bx \left ( \cosh \left ( bx \right ) \right ) ^{2}}{3}}-{\frac{5\,\cosh \left ( bx \right ) \sinh \left ( bx \right ) }{6}}+{\frac{bx}{2}}+{\frac{2\,{\it Shi} \left ( 2\,bx \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*Shi(b*x)^2,x)

[Out]

1/b^3*(1/3*b^3*x^3*Shi(b*x)^2-2*Shi(b*x)*(1/3*b^2*x^2*cosh(b*x)-2/3*b*x*sinh(b*x)+2/3*cosh(b*x))+1/3*b*x*cosh(
b*x)^2-5/6*cosh(b*x)*sinh(b*x)+1/2*b*x+2/3*Shi(2*b*x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\rm Shi}\left (b x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*Shi(b*x)^2,x, algorithm="maxima")

[Out]

integrate(x^2*Shi(b*x)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{2} \operatorname{Shi}\left (b x\right )^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*Shi(b*x)^2,x, algorithm="fricas")

[Out]

integral(x^2*sinh_integral(b*x)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{Shi}^{2}{\left (b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*Shi(b*x)**2,x)

[Out]

Integral(x**2*Shi(b*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2}{\rm Shi}\left (b x\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*Shi(b*x)^2,x, algorithm="giac")

[Out]

integrate(x^2*Shi(b*x)^2, x)