\(\int \frac {\sinh (b x) \text {Shi}(b x)}{x^3} \, dx\) [39]

Optimal result

Integrand size = 12, antiderivative size = 96 \[ \int \frac {\sinh (b x) \text {Shi}(b x)}{x^3} \, dx=b^2 \text {Chi}(2 b x)-\frac {b \cosh (b x) \sinh (b x)}{2 x}-\frac {\sinh ^2(b x)}{4 x^2}-\frac {b \sinh (2 b x)}{4 x}-\frac {b \cosh (b x) \text {Shi}(b x)}{2 x}-\frac {\sinh (b x) \text {Shi}(b x)}{2 x^2}+\frac {1}{4} b^2 \text {Shi}(b x)^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {6679, 6685, 6818, 12, 5556, 3378, 3382, 3395, 29, 3393} \[ \int \frac {\sinh (b x) \text {Shi}(b x)}{x^3} \, dx=b^2 \text {Chi}(2 b x)+\frac {1}{4} b^2 \text {Shi}(b x)^2-\frac {\text {Shi}(b x) \sinh (b x)}{2 x^2}-\frac {b \text {Shi}(b x) \cosh (b x)}{2 x}-\frac {\sinh ^2(b x)}{4 x^2}-\frac {b \sinh (2 b x)}{4 x}-\frac {b \sinh (b x) \cosh (b x)}{2 x} \]

[In]

Int[(Sinh[b*x]*SinhIntegral[b*x])/x^3,x]

[Out]

b^2*CoshIntegral[2*b*x] - (b*Cosh[b*x]*Sinh[b*x])/(2*x) - Sinh[b*x]^2/(4*x^2) - (b*Sinh[2*b*x])/(4*x) - (b*Cos
h[b*x]*SinhIntegral[b*x])/(2*x) - (Sinh[b*x]*SinhIntegral[b*x])/(2*x^2) + (b^2*SinhIntegral[b*x]^2)/4

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 3378

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

Rule 3382

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 3393

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 3395

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

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 6679

Int[((e_.) + (f_.)*(x_))^(m_)*Sinh[(a_.) + (b_.)*(x_)]*SinhIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[(e
+ f*x)^(m + 1)*Sinh[a + b*x]*(SinhIntegral[c + d*x]/(f*(m + 1))), x] + (-Dist[b/(f*(m + 1)), Int[(e + f*x)^(m
+ 1)*Cosh[a + b*x]*SinhIntegral[c + d*x], x], x] - Dist[d/(f*(m + 1)), Int[(e + f*x)^(m + 1)*Sinh[a + b*x]*(Si
nh[c + d*x]/(c + d*x)), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[m, -1]

Rule 6685

Int[Cosh[(a_.) + (b_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.)*SinhIntegral[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[(e
+ f*x)^(m + 1)*Cosh[a + b*x]*(SinhIntegral[c + d*x]/(f*(m + 1))), x] + (-Dist[b/(f*(m + 1)), Int[(e + f*x)^(m
+ 1)*Sinh[a + b*x]*SinhIntegral[c + d*x], x], x] - Dist[d/(f*(m + 1)), Int[(e + f*x)^(m + 1)*Cosh[a + b*x]*(S
inh[c + d*x]/(c + d*x)), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[m, -1]

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int \frac {\sinh (b x) \text {Shi}(b x)}{x^3} \, dx=b^2 \text {Chi}(2 b x)-\frac {b \cosh (b x) \sinh (b x)}{2 x}-\frac {\sinh ^2(b x)}{4 x^2}-\frac {b \sinh (2 b x)}{4 x}-\frac {b \cosh (b x) \text {Shi}(b x)}{2 x}-\frac {\sinh (b x) \text {Shi}(b x)}{2 x^2}+\frac {1}{4} b^2 \text {Shi}(b x)^2 \]

[In]

Integrate[(Sinh[b*x]*SinhIntegral[b*x])/x^3,x]

[Out]

b^2*CoshIntegral[2*b*x] - (b*Cosh[b*x]*Sinh[b*x])/(2*x) - Sinh[b*x]^2/(4*x^2) - (b*Sinh[2*b*x])/(4*x) - (b*Cos
h[b*x]*SinhIntegral[b*x])/(2*x) - (Sinh[b*x]*SinhIntegral[b*x])/(2*x^2) + (b^2*SinhIntegral[b*x]^2)/4

Maple [F]

\[\int \frac {\operatorname {Shi}\left (b x \right ) \sinh \left (b x \right )}{x^{3}}d x\]

[In]

int(Shi(b*x)*sinh(b*x)/x^3,x)

[Out]

int(Shi(b*x)*sinh(b*x)/x^3,x)

Fricas [F]

\[ \int \frac {\sinh (b x) \text {Shi}(b x)}{x^3} \, dx=\int { \frac {{\rm Shi}\left (b x\right ) \sinh \left (b x\right )}{x^{3}} \,d x } \]

[In]

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

[Out]

integral(sinh(b*x)*sinh_integral(b*x)/x^3, x)

Sympy [F]

\[ \int \frac {\sinh (b x) \text {Shi}(b x)}{x^3} \, dx=\int \frac {\sinh {\left (b x \right )} \operatorname {Shi}{\left (b x \right )}}{x^{3}}\, dx \]

[In]

integrate(Shi(b*x)*sinh(b*x)/x**3,x)

[Out]

Integral(sinh(b*x)*Shi(b*x)/x**3, x)

Maxima [F]

\[ \int \frac {\sinh (b x) \text {Shi}(b x)}{x^3} \, dx=\int { \frac {{\rm Shi}\left (b x\right ) \sinh \left (b x\right )}{x^{3}} \,d x } \]

[In]

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

[Out]

integrate(Shi(b*x)*sinh(b*x)/x^3, x)

Giac [F]

\[ \int \frac {\sinh (b x) \text {Shi}(b x)}{x^3} \, dx=\int { \frac {{\rm Shi}\left (b x\right ) \sinh \left (b x\right )}{x^{3}} \,d x } \]

[In]

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

[Out]

integrate(Shi(b*x)*sinh(b*x)/x^3, x)

Mupad [F(-1)]

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

[In]

int((sinhint(b*x)*sinh(b*x))/x^3,x)

[Out]

int((sinhint(b*x)*sinh(b*x))/x^3, x)