\(\int x^2 \text {Shi}(b x) \, dx\) [3]

Optimal result

Integrand size = 8, antiderivative size = 49 \[ \int x^2 \text {Shi}(b x) \, dx=-\frac {2 \cosh (b x)}{3 b^3}-\frac {x^2 \cosh (b x)}{3 b}+\frac {2 x \sinh (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Shi}(b x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6667, 12, 3377, 2718} \[ \int x^2 \text {Shi}(b x) \, dx=-\frac {2 \cosh (b x)}{3 b^3}+\frac {2 x \sinh (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Shi}(b x)-\frac {x^2 \cosh (b x)}{3 b} \]

[In]

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

[Out]

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

Rule 12

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

Rule 2718

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

Rule 3377

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 6667

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]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90 \[ \int x^2 \text {Shi}(b x) \, dx=-\frac {\left (2+b^2 x^2\right ) \cosh (b x)}{3 b^3}+\frac {2 x \sinh (b x)}{3 b^2}+\frac {1}{3} x^3 \text {Shi}(b x) \]

[In]

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

[Out]

-1/3*((2 + b^2*x^2)*Cosh[b*x])/b^3 + (2*x*Sinh[b*x])/(3*b^2) + (x^3*SinhIntegral[b*x])/3

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.86

method	result	size
parts	\(\frac {x^{3} \operatorname {Shi}\left (b x \right )}{3}-\frac {b^{2} x^{2} \cosh \left (b x \right )-2 b x \sinh \left (b x \right )+2 \cosh \left (b x \right )}{3 b^{3}}\)	\(42\)
derivativedivides	\(\frac {\frac {b^{3} x^{3} \operatorname {Shi}\left (b x \right )}{3}-\frac {b^{2} x^{2} \cosh \left (b x \right )}{3}+\frac {2 b x \sinh \left (b x \right )}{3}-\frac {2 \cosh \left (b x \right )}{3}}{b^{3}}\)	\(44\)
default	\(\frac {\frac {b^{3} x^{3} \operatorname {Shi}\left (b x \right )}{3}-\frac {b^{2} x^{2} \cosh \left (b x \right )}{3}+\frac {2 b x \sinh \left (b x \right )}{3}-\frac {2 \cosh \left (b x \right )}{3}}{b^{3}}\)	\(44\)
meijerg	\(\frac {2 \sqrt {\pi }\, \left (\frac {1}{3 \sqrt {\pi }}-\frac {\left (\frac {b^{2} x^{2}}{2}+1\right ) \cosh \left (b x \right )}{3 \sqrt {\pi }}+\frac {b x \sinh \left (b x \right )}{3 \sqrt {\pi }}+\frac {b^{3} x^{3} \operatorname {Shi}\left (b x \right )}{6 \sqrt {\pi }}\right )}{b^{3}}\)	\(60\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral(x^2*sinh_integral(b*x), x)

Sympy [A] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.94 \[ \int x^2 \text {Shi}(b x) \, dx=\frac {x^{3} \operatorname {Shi}{\left (b x \right )}}{3} - \frac {x^{2} \cosh {\left (b x \right )}}{3 b} + \frac {2 x \sinh {\left (b x \right )}}{3 b^{2}} - \frac {2 \cosh {\left (b x \right )}}{3 b^{3}} \]

[In]

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

[Out]

x**3*Shi(b*x)/3 - x**2*cosh(b*x)/(3*b) + 2*x*sinh(b*x)/(3*b**2) - 2*cosh(b*x)/(3*b**3)

Maxima [F]

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

[In]

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

[Out]

integrate(x^2*Shi(b*x), x)

Giac [F]

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

[In]

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

[Out]

integrate(x^2*Shi(b*x), x)

Mupad [F(-1)]

Timed out. \[ \int x^2 \text {Shi}(b x) \, dx=\frac {x^3\,\mathrm {sinhint}\left (b\,x\right )}{3}-\frac {\frac {2\,\mathrm {cosh}\left (b\,x\right )}{3}+\frac {b^2\,x^2\,\mathrm {cosh}\left (b\,x\right )}{3}-\frac {2\,b\,x\,\mathrm {sinh}\left (b\,x\right )}{3}}{b^3} \]

[In]

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

[Out]

(x^3*sinhint(b*x))/3 - ((2*cosh(b*x))/3 + (b^2*x^2*cosh(b*x))/3 - (2*b*x*sinh(b*x))/3)/b^3