3.18 \(\int x^3 \text{Shi}(a+b x) \, dx\)

Optimal. Leaf size=184 \[ -\frac{a^4 \text{Shi}(a+b x)}{4 b^4}+\frac{a^2 \sinh (a+b x)}{4 b^4}+\frac{a^3 \cosh (a+b x)}{4 b^4}-\frac{a^2 x \cosh (a+b x)}{4 b^3}+\frac{3 x^2 \sinh (a+b x)}{4 b^2}+\frac{a x^2 \cosh (a+b x)}{4 b^2}-\frac{a x \sinh (a+b x)}{2 b^3}+\frac{3 \sinh (a+b x)}{2 b^4}+\frac{a \cosh (a+b x)}{2 b^4}-\frac{3 x \cosh (a+b x)}{2 b^3}+\frac{1}{4} x^4 \text{Shi}(a+b x)-\frac{x^3 \cosh (a+b x)}{4 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.380206, antiderivative size = 184, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6532, 6742, 2638, 3296, 2637, 3298} \[ -\frac{a^4 \text{Shi}(a+b x)}{4 b^4}+\frac{a^2 \sinh (a+b x)}{4 b^4}+\frac{a^3 \cosh (a+b x)}{4 b^4}-\frac{a^2 x \cosh (a+b x)}{4 b^3}+\frac{3 x^2 \sinh (a+b x)}{4 b^2}+\frac{a x^2 \cosh (a+b x)}{4 b^2}-\frac{a x \sinh (a+b x)}{2 b^3}+\frac{3 \sinh (a+b x)}{2 b^4}+\frac{a \cosh (a+b x)}{2 b^4}-\frac{3 x \cosh (a+b x)}{2 b^3}+\frac{1}{4} x^4 \text{Shi}(a+b x)-\frac{x^3 \cosh (a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*SinhIntegral[a + b*x],x]

[Out]

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

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

Mathematica [A]  time = 0.199126, size = 94, normalized size = 0.51 \[ \frac{\left (b^4 x^4-a^4\right ) \text{Shi}(a+b x)+\left (a^2-2 a b x+3 b^2 x^2+6\right ) \sinh (a+b x)+\left (-a^2 b x+a^3+a b^2 x^2+2 a-b^3 x^3-6 b x\right ) \cosh (a+b x)}{4 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.051, size = 156, normalized size = 0.9 \begin{align*}{\frac{1}{{b}^{4}} \left ({\frac{{\it Shi} \left ( bx+a \right ){b}^{4}{x}^{4}}{4}}-{\frac{ \left ( bx+a \right ) ^{3}\cosh \left ( bx+a \right ) }{4}}+{\frac{3\, \left ( bx+a \right ) ^{2}\sinh \left ( bx+a \right ) }{4}}-{\frac{ \left ( 3\,bx+3\,a \right ) \cosh \left ( bx+a \right ) }{2}}+{\frac{3\,\sinh \left ( bx+a \right ) }{2}}+a \left ( \left ( bx+a \right ) ^{2}\cosh \left ( bx+a \right ) -2\, \left ( bx+a \right ) \sinh \left ( bx+a \right ) +2\,\cosh \left ( bx+a \right ) \right ) -{\frac{3\,{a}^{2} \left ( \left ( bx+a \right ) \cosh \left ( bx+a \right ) -\sinh \left ( bx+a \right ) \right ) }{2}}+{a}^{3}\cosh \left ( bx+a \right ) -{\frac{{a}^{4}{\it Shi} \left ( bx+a \right ) }{4}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3}{\rm Shi}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*Shi(b*x + a), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x^{3} \operatorname{Shi}\left (b x + a\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^3*sinh_integral(b*x + a), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{Shi}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*Shi(a + b*x), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3}{\rm Shi}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*Shi(b*x + a), x)