### 3.255 $$\int \frac{\cosh (a+b x) \sinh (a+b x)}{x} \, dx$$

Optimal. Leaf size=27 $\frac{1}{2} \sinh (2 a) \text{Chi}(2 b x)+\frac{1}{2} \cosh (2 a) \text{Shi}(2 b x)$

[Out]

(CoshIntegral[2*b*x]*Sinh[2*a])/2 + (Cosh[2*a]*SinhIntegral[2*b*x])/2

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Rubi [A]  time = 0.0745568, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.312, Rules used = {5448, 12, 3303, 3298, 3301} $\frac{1}{2} \sinh (2 a) \text{Chi}(2 b x)+\frac{1}{2} \cosh (2 a) \text{Shi}(2 b x)$

Antiderivative was successfully veriﬁed.

[In]

Int[(Cosh[a + b*x]*Sinh[a + b*x])/x,x]

[Out]

(CoshIntegral[2*b*x]*Sinh[2*a])/2 + (Cosh[2*a]*SinhIntegral[2*b*x])/2

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 12

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

Rule 3303

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

Rubi steps

\begin{align*} \int \frac{\cosh (a+b x) \sinh (a+b x)}{x} \, dx &=\int \frac{\sinh (2 a+2 b x)}{2 x} \, dx\\ &=\frac{1}{2} \int \frac{\sinh (2 a+2 b x)}{x} \, dx\\ &=\frac{1}{2} \cosh (2 a) \int \frac{\sinh (2 b x)}{x} \, dx+\frac{1}{2} \sinh (2 a) \int \frac{\cosh (2 b x)}{x} \, dx\\ &=\frac{1}{2} \text{Chi}(2 b x) \sinh (2 a)+\frac{1}{2} \cosh (2 a) \text{Shi}(2 b x)\\ \end{align*}

Mathematica [A]  time = 0.0254978, size = 25, normalized size = 0.93 $\frac{1}{2} (\sinh (2 a) \text{Chi}(2 b x)+\cosh (2 a) \text{Shi}(2 b x))$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Cosh[a + b*x]*Sinh[a + b*x])/x,x]

[Out]

(CoshIntegral[2*b*x]*Sinh[2*a] + Cosh[2*a]*SinhIntegral[2*b*x])/2

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Maple [A]  time = 0.021, size = 26, normalized size = 1. \begin{align*}{\frac{{{\rm e}^{-2\,a}}{\it Ei} \left ( 1,2\,bx \right ) }{4}}-{\frac{{{\rm e}^{2\,a}}{\it Ei} \left ( 1,-2\,bx \right ) }{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)*sinh(b*x+a)/x,x)

[Out]

1/4*exp(-2*a)*Ei(1,2*b*x)-1/4*exp(2*a)*Ei(1,-2*b*x)

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Maxima [A]  time = 1.32902, size = 31, normalized size = 1.15 \begin{align*} \frac{1}{4} \,{\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} - \frac{1}{4} \,{\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x,x, algorithm="maxima")

[Out]

1/4*Ei(2*b*x)*e^(2*a) - 1/4*Ei(-2*b*x)*e^(-2*a)

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Fricas [A]  time = 2.02355, size = 109, normalized size = 4.04 \begin{align*} \frac{1}{4} \,{\left ({\rm Ei}\left (2 \, b x\right ) -{\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + \frac{1}{4} \,{\left ({\rm Ei}\left (2 \, b x\right ) +{\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x,x, algorithm="fricas")

[Out]

1/4*(Ei(2*b*x) - Ei(-2*b*x))*cosh(2*a) + 1/4*(Ei(2*b*x) + Ei(-2*b*x))*sinh(2*a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{x}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x,x)

[Out]

Integral(sinh(a + b*x)*cosh(a + b*x)/x, x)

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Giac [A]  time = 1.14183, size = 31, normalized size = 1.15 \begin{align*} \frac{1}{4} \,{\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} - \frac{1}{4} \,{\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x,x, algorithm="giac")

[Out]

1/4*Ei(2*b*x)*e^(2*a) - 1/4*Ei(-2*b*x)*e^(-2*a)