### 3.257 $$\int \frac{\cosh (a+b x) \sinh (a+b x)}{x^3} \, dx$$

Optimal. Leaf size=60 $b^2 \sinh (2 a) \text{Chi}(2 b x)+b^2 \cosh (2 a) \text{Shi}(2 b x)-\frac{\sinh (2 a+2 b x)}{4 x^2}-\frac{b \cosh (2 a+2 b x)}{2 x}$

[Out]

-(b*Cosh[2*a + 2*b*x])/(2*x) + b^2*CoshIntegral[2*b*x]*Sinh[2*a] - Sinh[2*a + 2*b*x]/(4*x^2) + b^2*Cosh[2*a]*S
inhIntegral[2*b*x]

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Rubi [A]  time = 0.123143, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 7, 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} $b^2 \sinh (2 a) \text{Chi}(2 b x)+b^2 \cosh (2 a) \text{Shi}(2 b x)-\frac{\sinh (2 a+2 b x)}{4 x^2}-\frac{b \cosh (2 a+2 b x)}{2 x}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Cosh[a + b*x]*Sinh[a + b*x])/x^3,x]

[Out]

-(b*Cosh[2*a + 2*b*x])/(2*x) + b^2*CoshIntegral[2*b*x]*Sinh[2*a] - Sinh[2*a + 2*b*x]/(4*x^2) + b^2*Cosh[2*a]*S
inhIntegral[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^3} \, dx &=\int \frac{\sinh (2 a+2 b x)}{2 x^3} \, dx\\ &=\frac{1}{2} \int \frac{\sinh (2 a+2 b x)}{x^3} \, dx\\ &=-\frac{\sinh (2 a+2 b x)}{4 x^2}+\frac{1}{2} b \int \frac{\cosh (2 a+2 b x)}{x^2} \, dx\\ &=-\frac{b \cosh (2 a+2 b x)}{2 x}-\frac{\sinh (2 a+2 b x)}{4 x^2}+b^2 \int \frac{\sinh (2 a+2 b x)}{x} \, dx\\ &=-\frac{b \cosh (2 a+2 b x)}{2 x}-\frac{\sinh (2 a+2 b x)}{4 x^2}+\left (b^2 \cosh (2 a)\right ) \int \frac{\sinh (2 b x)}{x} \, dx+\left (b^2 \sinh (2 a)\right ) \int \frac{\cosh (2 b x)}{x} \, dx\\ &=-\frac{b \cosh (2 a+2 b x)}{2 x}+b^2 \text{Chi}(2 b x) \sinh (2 a)-\frac{\sinh (2 a+2 b x)}{4 x^2}+b^2 \cosh (2 a) \text{Shi}(2 b x)\\ \end{align*}

Mathematica [A]  time = 0.174611, size = 61, normalized size = 1.02 $\frac{1}{2} \left (2 b^2 \sinh (2 a) \text{Chi}(2 b x)+2 b^2 \cosh (2 a) \text{Shi}(2 b x)-\frac{\sinh (2 (a+b x))+2 b x \cosh (2 (a+b x))}{2 x^2}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Cosh[a + b*x]*Sinh[a + b*x])/x^3,x]

[Out]

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

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Maple [A]  time = 0.032, size = 90, normalized size = 1.5 \begin{align*} -{\frac{b{{\rm e}^{-2\,bx-2\,a}}}{4\,x}}+{\frac{{{\rm e}^{-2\,bx-2\,a}}}{8\,{x}^{2}}}+{\frac{{b}^{2}{{\rm e}^{-2\,a}}{\it Ei} \left ( 1,2\,bx \right ) }{2}}-{\frac{{{\rm e}^{2\,bx+2\,a}}}{8\,{x}^{2}}}-{\frac{b{{\rm e}^{2\,bx+2\,a}}}{4\,x}}-{\frac{{b}^{2}{{\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^3,x)

[Out]

-1/4*b*exp(-2*b*x-2*a)/x+1/8*exp(-2*b*x-2*a)/x^2+1/2*b^2*exp(-2*a)*Ei(1,2*b*x)-1/8*exp(2*b*x+2*a)/x^2-1/4*b*ex
p(2*b*x+2*a)/x-1/2*b^2*exp(2*a)*Ei(1,-2*b*x)

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Maxima [A]  time = 1.23321, size = 41, normalized size = 0.68 \begin{align*} b^{2} e^{\left (-2 \, a\right )} \Gamma \left (-2, 2 \, b x\right ) - b^{2} e^{\left (2 \, a\right )} \Gamma \left (-2, -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^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.94455, size = 257, normalized size = 4.28 \begin{align*} -\frac{b x \cosh \left (b x + a\right )^{2} + b x \sinh \left (b x + a\right )^{2} -{\left (b^{2} x^{2}{\rm Ei}\left (2 \, b x\right ) - b^{2} x^{2}{\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) -{\left (b^{2} x^{2}{\rm Ei}\left (2 \, b x\right ) + b^{2} x^{2}{\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{2 \, x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x^3,x, algorithm="fricas")

[Out]

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

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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**3,x)

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.13612, size = 116, normalized size = 1.93 \begin{align*} \frac{4 \, b^{2} x^{2}{\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} - 4 \, b^{2} x^{2}{\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - 2 \, b x e^{\left (2 \, b x + 2 \, a\right )} - 2 \, b x e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}}{8 \, x^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x^3,x, algorithm="giac")

[Out]

1/8*(4*b^2*x^2*Ei(2*b*x)*e^(2*a) - 4*b^2*x^2*Ei(-2*b*x)*e^(-2*a) - 2*b*x*e^(2*b*x + 2*a) - 2*b*x*e^(-2*b*x - 2
*a) - e^(2*b*x + 2*a) + e^(-2*b*x - 2*a))/x^2