### 3.256 $$\int \frac{\cosh (a+b x) \sinh (a+b x)}{x^2} \, dx$$

Optimal. Leaf size=39 $b \cosh (2 a) \text{Chi}(2 b x)+b \sinh (2 a) \text{Shi}(2 b x)-\frac{\sinh (2 a+2 b x)}{2 x}$

[Out]

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

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Rubi [A]  time = 0.0948402, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.375, Rules used = {5448, 12, 3297, 3303, 3298, 3301} $b \cosh (2 a) \text{Chi}(2 b x)+b \sinh (2 a) \text{Shi}(2 b x)-\frac{\sinh (2 a+2 b x)}{2 x}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Cosh[a + b*x]*Sinh[a + b*x])/x^2,x]

[Out]

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

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 3297

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

Mathematica [A]  time = 0.0709192, size = 42, normalized size = 1.08 $\frac{1}{2} \left (2 b \cosh (2 a) \text{Chi}(2 b x)+2 b \sinh (2 a) \text{Shi}(2 b x)-\frac{\sinh (2 (a+b x))}{x}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Cosh[a + b*x]*Sinh[a + b*x])/x^2,x]

[Out]

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

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Maple [A]  time = 0.026, size = 56, normalized size = 1.4 \begin{align*}{\frac{{{\rm e}^{-2\,bx-2\,a}}}{4\,x}}-{\frac{b{{\rm e}^{-2\,a}}{\it Ei} \left ( 1,2\,bx \right ) }{2}}-{\frac{{{\rm e}^{2\,bx+2\,a}}}{4\,x}}-{\frac{b{{\rm e}^{2\,a}}{\it Ei} \left ( 1,-2\,bx \right ) }{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)*sinh(b*x+a)/x^2,x)

[Out]

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

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Maxima [A]  time = 1.34828, size = 36, normalized size = 0.92 \begin{align*} \frac{1}{2} \, b e^{\left (-2 \, a\right )} \Gamma \left (-1, 2 \, b x\right ) + \frac{1}{2} \, b e^{\left (2 \, a\right )} \Gamma \left (-1, -2 \, b x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x^2,x, algorithm="maxima")

[Out]

1/2*b*e^(-2*a)*gamma(-1, 2*b*x) + 1/2*b*e^(2*a)*gamma(-1, -2*b*x)

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Fricas [A]  time = 1.89209, size = 174, normalized size = 4.46 \begin{align*} \frac{{\left (b x{\rm Ei}\left (2 \, b x\right ) + b x{\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) - 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) +{\left (b x{\rm Ei}\left (2 \, b x\right ) - b x{\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{2 \, x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x**2,x)

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.13854, size = 70, normalized size = 1.79 \begin{align*} \frac{2 \, b x{\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + 2 \, b x{\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}}{4 \, x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x^2,x, algorithm="giac")

[Out]

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