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

Optimal. Leaf size=128 \[ \frac {3 \text {Shi}(2 b x)}{b^4}-\frac {6 \text {Shi}(b x) \cosh (b x)}{b^4}-\frac {4 \sinh (b x) \cosh (b x)}{b^4}+\frac {6 x \text {Shi}(b x) \sinh (b x)}{b^3}+\frac {4 x}{b^3}+\frac {2 x \sinh ^2(b x)}{b^3}-\frac {3 x^2 \text {Shi}(b x) \cosh (b x)}{b^2}-\frac {x^2 \sinh (b x) \cosh (b x)}{2 b^2}+\frac {x^3 \text {Shi}(b x) \sinh (b x)}{b}+\frac {x^3}{6 b} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 11, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {6548, 12, 3311, 30, 2635, 8, 6542, 5372, 6540, 5448, 3298} \[ -\frac {3 x^2 \text {Shi}(b x) \cosh (b x)}{b^2}+\frac {3 \text {Shi}(2 b x)}{b^4}+\frac {6 x \text {Shi}(b x) \sinh (b x)}{b^3}-\frac {6 \text {Shi}(b x) \cosh (b x)}{b^4}-\frac {x^2 \sinh (b x) \cosh (b x)}{2 b^2}+\frac {4 x}{b^3}+\frac {2 x \sinh ^2(b x)}{b^3}-\frac {4 \sinh (b x) \cosh (b x)}{b^4}+\frac {x^3 \text {Shi}(b x) \sinh (b x)}{b}+\frac {x^3}{6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -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 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 3311

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

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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 94, normalized size = 0.73 \[ \frac {2 b^3 x^3+12 \text {Shi}(b x) \left (b x \left (b^2 x^2+6\right ) \sinh (b x)-3 \left (b^2 x^2+2\right ) \cosh (b x)\right )-3 b^2 x^2 \sinh (2 b x)+36 \text {Shi}(2 b x)+36 b x-24 \sinh (2 b x)+12 b x \cosh (2 b x)}{12 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*b*x + 2*b^3*x^3 + 12*b*x*Cosh[2*b*x] - 24*Sinh[2*b*x] - 3*b^2*x^2*Sinh[2*b*x] + 12*(-3*(2 + b^2*x^2)*Cosh[
b*x] + b*x*(6 + b^2*x^2)*Sinh[b*x])*SinhIntegral[b*x] + 36*SinhIntegral[2*b*x])/(12*b^4)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{3} \cosh \left (b x\right ) \operatorname {Shi}\left (b x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x)*Shi(b*x),x, algorithm="fricas")

[Out]

integral(x^3*cosh(b*x)*sinh_integral(b*x), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} {\rm Shi}\left (b x\right ) \cosh \left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x)*Shi(b*x),x, algorithm="giac")

[Out]

integrate(x^3*Shi(b*x)*cosh(b*x), x)

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maple [A]  time = 0.03, size = 104, normalized size = 0.81 \[ \frac {\Shi \left (b x \right ) \left (\sinh \left (b x \right ) b^{3} x^{3}-3 b^{2} x^{2} \cosh \left (b x \right )+6 b x \sinh \left (b x \right )-6 \cosh \left (b x \right )\right )-\frac {b^{2} x^{2} \cosh \left (b x \right ) \sinh \left (b x \right )}{2}+\frac {b^{3} x^{3}}{6}+2 b x \left (\cosh ^{2}\left (b x \right )\right )-4 \sinh \left (b x \right ) \cosh \left (b x \right )+2 b x +3 \Shi \left (2 b x \right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} {\rm Shi}\left (b x\right ) \cosh \left (b x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*cosh(b*x)*Shi(b*x),x, algorithm="maxima")

[Out]

integrate(x^3*Shi(b*x)*cosh(b*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\mathrm {sinhint}\left (b\,x\right )\,\mathrm {cosh}\left (b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*sinhint(b*x)*cosh(b*x),x)

[Out]

int(x^3*sinhint(b*x)*cosh(b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \cosh {\left (b x \right )} \operatorname {Shi}{\left (b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*cosh(b*x)*Shi(b*x), x)

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