\(\int x^2 \text {Shi}(b x)^2 \, dx\) [11]

Optimal result

Integrand size = 10, antiderivative size = 112 \[ \int x^2 \text {Shi}(b x)^2 \, dx=\frac {5 x}{6 b^2}-\frac {5 \cosh (b x) \sinh (b x)}{6 b^3}+\frac {x \sinh ^2(b x)}{3 b^2}-\frac {4 \cosh (b x) \text {Shi}(b x)}{3 b^3}-\frac {2 x^2 \cosh (b x) \text {Shi}(b x)}{3 b}+\frac {4 x \sinh (b x) \text {Shi}(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Shi}(b x)^2+\frac {2 \text {Shi}(2 b x)}{3 b^3} \]

[Out]

5/6*x/b^2-4/3*cosh(b*x)*Shi(b*x)/b^3-2/3*x^2*cosh(b*x)*Shi(b*x)/b+1/3*x^3*Shi(b*x)^2+2/3*Shi(2*b*x)/b^3-5/6*co
sh(b*x)*sinh(b*x)/b^3+4/3*x*Shi(b*x)*sinh(b*x)/b^2+1/3*x*sinh(b*x)^2/b^2

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {6671, 6677, 12, 5480, 2715, 8, 6683, 6675, 5556, 3379} \[ \int x^2 \text {Shi}(b x)^2 \, dx=\frac {2 \text {Shi}(2 b x)}{3 b^3}-\frac {4 \text {Shi}(b x) \cosh (b x)}{3 b^3}-\frac {5 \sinh (b x) \cosh (b x)}{6 b^3}+\frac {4 x \text {Shi}(b x) \sinh (b x)}{3 b^2}+\frac {5 x}{6 b^2}+\frac {x \sinh ^2(b x)}{3 b^2}+\frac {1}{3} x^3 \text {Shi}(b x)^2-\frac {2 x^2 \text {Shi}(b x) \cosh (b x)}{3 b} \]

[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 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 2715

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] && Integ
erQ[2*n]

Rule 3379

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 5480

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 5556

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 6671

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 6675

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 6677

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 6683

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*} \text {integral}& = \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] (verified)

Time = 0.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.70 \[ \int x^2 \text {Shi}(b x)^2 \, dx=\frac {8 b x+2 b x \cosh (2 b x)-5 \sinh (2 b x)-8 \left (\left (2+b^2 x^2\right ) \cosh (b x)-2 b x \sinh (b x)\right ) \text {Shi}(b x)+4 b^3 x^3 \text {Shi}(b x)^2+8 \text {Shi}(2 b x)}{12 b^3} \]

[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)

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {\frac {b^{3} x^{3} \operatorname {Shi}\left (b x \right )^{2}}{3}-2 \,\operatorname {Shi}\left (b x \right ) \left (\frac {b^{2} x^{2} \cosh \left (b x \right )}{3}-\frac {2 b x \sinh \left (b x \right )}{3}+\frac {2 \cosh \left (b x \right )}{3}\right )+\frac {b x \cosh \left (b x \right )^{2}}{3}-\frac {5 \cosh \left (b x \right ) \sinh \left (b x \right )}{6}+\frac {b x}{2}+\frac {2 \,\operatorname {Shi}\left (2 b x \right )}{3}}{b^{3}}\)	\(84\)
default	\(\frac {\frac {b^{3} x^{3} \operatorname {Shi}\left (b x \right )^{2}}{3}-2 \,\operatorname {Shi}\left (b x \right ) \left (\frac {b^{2} x^{2} \cosh \left (b x \right )}{3}-\frac {2 b x \sinh \left (b x \right )}{3}+\frac {2 \cosh \left (b x \right )}{3}\right )+\frac {b x \cosh \left (b x \right )^{2}}{3}-\frac {5 \cosh \left (b x \right ) \sinh \left (b x \right )}{6}+\frac {b x}{2}+\frac {2 \,\operatorname {Shi}\left (2 b x \right )}{3}}{b^{3}}\)	\(84\)

[In]

int(x^2*Shi(b*x)^2,x,method=_RETURNVERBOSE)

[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))

Fricas [F]

\[ \int x^2 \text {Shi}(b x)^2 \, dx=\int { x^{2} {\rm Shi}\left (b x\right )^{2} \,d x } \]

[In]

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

[Out]

integral(x^2*sinh_integral(b*x)^2, x)

Sympy [F]

\[ \int x^2 \text {Shi}(b x)^2 \, dx=\int x^{2} \operatorname {Shi}^{2}{\left (b x \right )}\, dx \]

[In]

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

[Out]

Integral(x**2*Shi(b*x)**2, x)

Maxima [F]

\[ \int x^2 \text {Shi}(b x)^2 \, dx=\int { x^{2} {\rm Shi}\left (b x\right )^{2} \,d x } \]

[In]

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

[Out]

integrate(x^2*Shi(b*x)^2, x)

Giac [F]

\[ \int x^2 \text {Shi}(b x)^2 \, dx=\int { x^{2} {\rm Shi}\left (b x\right )^{2} \,d x } \]

[In]

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

[Out]

integrate(x^2*Shi(b*x)^2, x)

Mupad [F(-1)]

Timed out. \[ \int x^2 \text {Shi}(b x)^2 \, dx=\int x^2\,{\mathrm {sinhint}\left (b\,x\right )}^2 \,d x \]

[In]

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

[Out]

int(x^2*sinhint(b*x)^2, x)