\(\int \sinh (5 x) \text {Shi}(2 x) \, dx\) [53]

Optimal result

Integrand size = 9, antiderivative size = 29 \[ \int \sinh (5 x) \text {Shi}(2 x) \, dx=\frac {1}{5} \cosh (5 x) \text {Shi}(2 x)+\frac {\text {Shi}(3 x)}{10}-\frac {\text {Shi}(7 x)}{10} \]

[Out]

1/5*cosh(5*x)*Shi(2*x)+1/10*Shi(3*x)-1/10*Shi(7*x)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6675, 12, 5580, 3379} \[ \int \sinh (5 x) \text {Shi}(2 x) \, dx=\frac {\text {Shi}(3 x)}{10}-\frac {\text {Shi}(7 x)}{10}+\frac {1}{5} \text {Shi}(2 x) \cosh (5 x) \]

[In]

Int[Sinh[5*x]*SinhIntegral[2*x],x]

[Out]

(Cosh[5*x]*SinhIntegral[2*x])/5 + SinhIntegral[3*x]/10 - SinhIntegral[7*x]/10

Rule 12

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

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 5580

Int[Cosh[(c_.) + (d_.)*(x_)]^(q_.)*((e_.) + (f_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(p_.), x_Symbol] :> Int
[ExpandTrigReduce[(e + f*x)^m, Sinh[a + b*x]^p*Cosh[c + d*x]^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && I
GtQ[p, 0] && IGtQ[q, 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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} \cosh (5 x) \text {Shi}(2 x)-\frac {2}{5} \int \frac {\cosh (5 x) \sinh (2 x)}{2 x} \, dx \\ & = \frac {1}{5} \cosh (5 x) \text {Shi}(2 x)-\frac {1}{5} \int \frac {\cosh (5 x) \sinh (2 x)}{x} \, dx \\ & = \frac {1}{5} \cosh (5 x) \text {Shi}(2 x)-\frac {1}{5} \int \left (-\frac {\sinh (3 x)}{2 x}+\frac {\sinh (7 x)}{2 x}\right ) \, dx \\ & = \frac {1}{5} \cosh (5 x) \text {Shi}(2 x)+\frac {1}{10} \int \frac {\sinh (3 x)}{x} \, dx-\frac {1}{10} \int \frac {\sinh (7 x)}{x} \, dx \\ & = \frac {1}{5} \cosh (5 x) \text {Shi}(2 x)+\frac {\text {Shi}(3 x)}{10}-\frac {\text {Shi}(7 x)}{10} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \sinh (5 x) \text {Shi}(2 x) \, dx=\frac {1}{10} (2 \cosh (5 x) \text {Shi}(2 x)+\text {Shi}(3 x)-\text {Shi}(7 x)) \]

[In]

Integrate[Sinh[5*x]*SinhIntegral[2*x],x]

[Out]

(2*Cosh[5*x]*SinhIntegral[2*x] + SinhIntegral[3*x] - SinhIntegral[7*x])/10

Maple [A] (verified)

Time = 1.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {\cosh \left (5 x \right ) \operatorname {Shi}\left (2 x \right )}{5}+\frac {\operatorname {Shi}\left (3 x \right )}{10}-\frac {\operatorname {Shi}\left (7 x \right )}{10}\)	\(24\)

[In]

int(Shi(2*x)*sinh(5*x),x,method=_RETURNVERBOSE)

[Out]

1/5*cosh(5*x)*Shi(2*x)+1/10*Shi(3*x)-1/10*Shi(7*x)

Fricas [F]

\[ \int \sinh (5 x) \text {Shi}(2 x) \, dx=\int { {\rm Shi}\left (2 \, x\right ) \sinh \left (5 \, x\right ) \,d x } \]

[In]

integrate(Shi(2*x)*sinh(5*x),x, algorithm="fricas")

[Out]

integral(sinh(5*x)*sinh_integral(2*x), x)

Sympy [F]

\[ \int \sinh (5 x) \text {Shi}(2 x) \, dx=\int \sinh {\left (5 x \right )} \operatorname {Shi}{\left (2 x \right )}\, dx \]

[In]

integrate(Shi(2*x)*sinh(5*x),x)

[Out]

Integral(sinh(5*x)*Shi(2*x), x)

Maxima [F]

\[ \int \sinh (5 x) \text {Shi}(2 x) \, dx=\int { {\rm Shi}\left (2 \, x\right ) \sinh \left (5 \, x\right ) \,d x } \]

[In]

integrate(Shi(2*x)*sinh(5*x),x, algorithm="maxima")

[Out]

integrate(Shi(2*x)*sinh(5*x), x)

Giac [F]

\[ \int \sinh (5 x) \text {Shi}(2 x) \, dx=\int { {\rm Shi}\left (2 \, x\right ) \sinh \left (5 \, x\right ) \,d x } \]

[In]

integrate(Shi(2*x)*sinh(5*x),x, algorithm="giac")

[Out]

integrate(Shi(2*x)*sinh(5*x), x)

Mupad [F(-1)]

Timed out. \[ \int \sinh (5 x) \text {Shi}(2 x) \, dx=\int \mathrm {sinhint}\left (2\,x\right )\,\mathrm {sinh}\left (5\,x\right ) \,d x \]

[In]

int(sinhint(2*x)*sinh(5*x),x)

[Out]

int(sinhint(2*x)*sinh(5*x), x)