3.55 \(\int x^2 \sinh (a+b x) \text {Shi}(a+b x) \, dx\)

Optimal. Leaf size=186 \[ -\frac {a^2 \text {Shi}(2 a+2 b x)}{2 b^3}-\frac {a \text {Chi}(2 a+2 b x)}{b^3}-\frac {\text {Shi}(2 a+2 b x)}{b^3}+\frac {2 \text {Shi}(a+b x) \cosh (a+b x)}{b^3}+\frac {a \log (a+b x)}{b^3}+\frac {\sinh (2 a+2 b x)}{8 b^3}+\frac {a \cosh (2 a+2 b x)}{4 b^3}+\frac {\sinh (a+b x) \cosh (a+b x)}{b^3}-\frac {2 x \text {Shi}(a+b x) \sinh (a+b x)}{b^2}-\frac {x \cosh (2 a+2 b x)}{4 b^2}+\frac {x^2 \text {Shi}(a+b x) \cosh (a+b x)}{b}-\frac {x}{b^2} \]

[Out]

-x/b^2-a*Chi(2*b*x+2*a)/b^3+1/4*a*cosh(2*b*x+2*a)/b^3-1/4*x*cosh(2*b*x+2*a)/b^2+a*ln(b*x+a)/b^3+2*cosh(b*x+a)*
Shi(b*x+a)/b^3+x^2*cosh(b*x+a)*Shi(b*x+a)/b-Shi(2*b*x+2*a)/b^3-1/2*a^2*Shi(2*b*x+2*a)/b^3+cosh(b*x+a)*sinh(b*x
+a)/b^3-2*x*Shi(b*x+a)*sinh(b*x+a)/b^2+1/8*sinh(2*b*x+2*a)/b^3

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Rubi [A]  time = 0.58, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 16, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6542, 5617, 6741, 6742, 2638, 3296, 2637, 3298, 6548, 2635, 8, 3312, 3301, 6540, 5448, 12} \[ -\frac {a^2 \text {Shi}(2 a+2 b x)}{2 b^3}-\frac {a \text {Chi}(2 a+2 b x)}{b^3}-\frac {\text {Shi}(2 a+2 b x)}{b^3}-\frac {2 x \text {Shi}(a+b x) \sinh (a+b x)}{b^2}+\frac {2 \text {Shi}(a+b x) \cosh (a+b x)}{b^3}+\frac {a \log (a+b x)}{b^3}+\frac {\sinh (2 a+2 b x)}{8 b^3}+\frac {a \cosh (2 a+2 b x)}{4 b^3}-\frac {x \cosh (2 a+2 b x)}{4 b^2}+\frac {\sinh (a+b x) \cosh (a+b x)}{b^3}+\frac {x^2 \text {Shi}(a+b x) \cosh (a+b x)}{b}-\frac {x}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sinh[a + b*x]*SinhIntegral[a + b*x],x]

[Out]

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

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 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 2637

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

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 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 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]

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 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 5617

Int[Cosh[w_]^(p_.)*(u_.)*Sinh[v_]^(p_.), x_Symbol] :> Dist[1/2^p, Int[u*Sinh[2*v]^p, x], x] /; EqQ[w, v] && In
tegerQ[p]

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]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int x^2 \sinh (a+b x) \text {Shi}(a+b x) \, dx &=\frac {x^2 \cosh (a+b x) \text {Shi}(a+b x)}{b}-\frac {2 \int x \cosh (a+b x) \text {Shi}(a+b x) \, dx}{b}-\int \frac {x^2 \cosh (a+b x) \sinh (a+b x)}{a+b x} \, dx\\ &=\frac {x^2 \cosh (a+b x) \text {Shi}(a+b x)}{b}-\frac {2 x \sinh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {1}{2} \int \frac {x^2 \sinh (2 (a+b x))}{a+b x} \, dx+\frac {2 \int \sinh (a+b x) \text {Shi}(a+b x) \, dx}{b^2}+\frac {2 \int \frac {x \sinh ^2(a+b x)}{a+b x} \, dx}{b}\\ &=\frac {2 \cosh (a+b x) \text {Shi}(a+b x)}{b^3}+\frac {x^2 \cosh (a+b x) \text {Shi}(a+b x)}{b}-\frac {2 x \sinh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {1}{2} \int \frac {x^2 \sinh (2 a+2 b x)}{a+b x} \, dx-\frac {2 \int \frac {\cosh (a+b x) \sinh (a+b x)}{a+b x} \, dx}{b^2}+\frac {2 \int \left (\frac {\sinh ^2(a+b x)}{b}-\frac {a \sinh ^2(a+b x)}{b (a+b x)}\right ) \, dx}{b}\\ &=\frac {2 \cosh (a+b x) \text {Shi}(a+b x)}{b^3}+\frac {x^2 \cosh (a+b x) \text {Shi}(a+b x)}{b}-\frac {2 x \sinh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {1}{2} \int \left (-\frac {a \sinh (2 a+2 b x)}{b^2}+\frac {x \sinh (2 a+2 b x)}{b}+\frac {a^2 \sinh (2 a+2 b x)}{b^2 (a+b x)}\right ) \, dx+\frac {2 \int \sinh ^2(a+b x) \, dx}{b^2}-\frac {2 \int \frac {\sinh (2 a+2 b x)}{2 (a+b x)} \, dx}{b^2}-\frac {(2 a) \int \frac {\sinh ^2(a+b x)}{a+b x} \, dx}{b^2}\\ &=\frac {\cosh (a+b x) \sinh (a+b x)}{b^3}+\frac {2 \cosh (a+b x) \text {Shi}(a+b x)}{b^3}+\frac {x^2 \cosh (a+b x) \text {Shi}(a+b x)}{b}-\frac {2 x \sinh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {\int 1 \, dx}{b^2}-\frac {\int \frac {\sinh (2 a+2 b x)}{a+b x} \, dx}{b^2}+\frac {a \int \sinh (2 a+2 b x) \, dx}{2 b^2}+\frac {(2 a) \int \left (\frac {1}{2 (a+b x)}-\frac {\cosh (2 a+2 b x)}{2 (a+b x)}\right ) \, dx}{b^2}-\frac {a^2 \int \frac {\sinh (2 a+2 b x)}{a+b x} \, dx}{2 b^2}-\frac {\int x \sinh (2 a+2 b x) \, dx}{2 b}\\ &=-\frac {x}{b^2}+\frac {a \cosh (2 a+2 b x)}{4 b^3}-\frac {x \cosh (2 a+2 b x)}{4 b^2}+\frac {a \log (a+b x)}{b^3}+\frac {\cosh (a+b x) \sinh (a+b x)}{b^3}+\frac {2 \cosh (a+b x) \text {Shi}(a+b x)}{b^3}+\frac {x^2 \cosh (a+b x) \text {Shi}(a+b x)}{b}-\frac {2 x \sinh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {\text {Shi}(2 a+2 b x)}{b^3}-\frac {a^2 \text {Shi}(2 a+2 b x)}{2 b^3}+\frac {\int \cosh (2 a+2 b x) \, dx}{4 b^2}-\frac {a \int \frac {\cosh (2 a+2 b x)}{a+b x} \, dx}{b^2}\\ &=-\frac {x}{b^2}+\frac {a \cosh (2 a+2 b x)}{4 b^3}-\frac {x \cosh (2 a+2 b x)}{4 b^2}-\frac {a \text {Chi}(2 a+2 b x)}{b^3}+\frac {a \log (a+b x)}{b^3}+\frac {\cosh (a+b x) \sinh (a+b x)}{b^3}+\frac {\sinh (2 a+2 b x)}{8 b^3}+\frac {2 \cosh (a+b x) \text {Shi}(a+b x)}{b^3}+\frac {x^2 \cosh (a+b x) \text {Shi}(a+b x)}{b}-\frac {2 x \sinh (a+b x) \text {Shi}(a+b x)}{b^2}-\frac {\text {Shi}(2 a+2 b x)}{b^3}-\frac {a^2 \text {Shi}(2 a+2 b x)}{2 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 123, normalized size = 0.66 \[ \frac {-4 a^2 \text {Shi}(2 (a+b x))+8 \text {Shi}(a+b x) \left (\left (b^2 x^2+2\right ) \cosh (a+b x)-2 b x \sinh (a+b x)\right )-8 a \text {Chi}(2 (a+b x))-8 \text {Shi}(2 (a+b x))+8 a \log (a+b x)+5 \sinh (2 (a+b x))+2 a \cosh (2 (a+b x))-2 b x \cosh (2 (a+b x))-8 b x}{8 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sinh[a + b*x]*SinhIntegral[a + b*x],x]

[Out]

(-8*b*x + 2*a*Cosh[2*(a + b*x)] - 2*b*x*Cosh[2*(a + b*x)] - 8*a*CoshIntegral[2*(a + b*x)] + 8*a*Log[a + b*x] +
5*Sinh[2*(a + b*x)] + 8*((2 + b^2*x^2)*Cosh[a + b*x] - 2*b*x*Sinh[a + b*x])*SinhIntegral[a + b*x] - 8*SinhInt
egral[2*(a + b*x)] - 4*a^2*SinhIntegral[2*(a + b*x)])/(8*b^3)

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{2} \sinh \left (b x + a\right ) \operatorname {Shi}\left (b x + a\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^2*sinh(b*x + a)*sinh_integral(b*x + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm Shi}\left (b x + a\right ) \sinh \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*Shi(b*x + a)*sinh(b*x + a), x)

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maple [A]  time = 0.03, size = 170, normalized size = 0.91 \[ \frac {x^{2} \cosh \left (b x +a \right ) \Shi \left (b x +a \right )}{b}-\frac {2 x \Shi \left (b x +a \right ) \sinh \left (b x +a \right )}{b^{2}}+\frac {2 \cosh \left (b x +a \right ) \Shi \left (b x +a \right )}{b^{3}}-\frac {\left (\cosh ^{2}\left (b x +a \right )\right ) x}{2 b^{2}}+\frac {a \left (\cosh ^{2}\left (b x +a \right )\right )}{2 b^{3}}+\frac {5 \cosh \left (b x +a \right ) \sinh \left (b x +a \right )}{4 b^{3}}-\frac {3 x}{4 b^{2}}-\frac {3 a}{4 b^{3}}-\frac {\Shi \left (2 b x +2 a \right )}{b^{3}}+\frac {a \ln \left (b x +a \right )}{b^{3}}-\frac {a \Chi \left (2 b x +2 a \right )}{b^{3}}-\frac {a^{2} \Shi \left (2 b x +2 a \right )}{2 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*Shi(b*x+a)*sinh(b*x+a),x)

[Out]

x^2*cosh(b*x+a)*Shi(b*x+a)/b-2*x*Shi(b*x+a)*sinh(b*x+a)/b^2+2*cosh(b*x+a)*Shi(b*x+a)/b^3-1/2/b^2*cosh(b*x+a)^2
*x+1/2/b^3*a*cosh(b*x+a)^2+5/4*cosh(b*x+a)*sinh(b*x+a)/b^3-3/4*x/b^2-3/4/b^3*a-Shi(2*b*x+2*a)/b^3+a*ln(b*x+a)/
b^3-a*Chi(2*b*x+2*a)/b^3-1/2*a^2*Shi(2*b*x+2*a)/b^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} {\rm Shi}\left (b x + a\right ) \sinh \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*Shi(b*x + a)*sinh(b*x + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {sinhint}\left (a+b\,x\right )\,\mathrm {sinh}\left (a+b\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sinhint(a + b*x)*sinh(a + b*x),x)

[Out]

int(x^2*sinhint(a + b*x)*sinh(a + b*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \sinh {\left (a + b x \right )} \operatorname {Shi}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*Shi(b*x+a)*sinh(b*x+a),x)

[Out]

Integral(x**2*sinh(a + b*x)*Shi(a + b*x), x)

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