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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 60, normalized size of antiderivative = 1.00, 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 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 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]

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 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]

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 = 0.18, 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

________________________________________________________________________________________

fricas [A]  time = 0.52, size = 104, normalized size = 1.73 \[ -\frac {b x \cosh \left (b x + a\right )^{2} + b x \sinh \left (b x + a\right )^{2} - {\left (b^{2} x^{2} {\rm Ei}\left (2 \, b x\right ) - b^{2} x^{2} {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) + \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) - {\left (b^{2} x^{2} {\rm Ei}\left (2 \, b x\right ) + b^{2} x^{2} {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{2 \, x^{2}} \]

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

________________________________________________________________________________________

giac [A]  time = 0.11, size = 86, normalized size = 1.43 \[ \frac {4 \, b^{2} x^{2} {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} - 4 \, b^{2} x^{2} {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - 2 \, b x e^{\left (2 \, b x + 2 \, a\right )} - 2 \, b x e^{\left (-2 \, b x - 2 \, a\right )} - e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}}{8 \, x^{2}} \]

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

________________________________________________________________________________________

maple [A]  time = 0.12, size = 90, normalized size = 1.50 \[ -\frac {b \,{\mathrm e}^{-2 b x -2 a}}{4 x}+\frac {{\mathrm e}^{-2 b x -2 a}}{8 x^{2}}+\frac {b^{2} {\mathrm e}^{-2 a} \Ei \left (1, 2 b x \right )}{2}-\frac {{\mathrm e}^{2 b x +2 a}}{8 x^{2}}-\frac {b \,{\mathrm e}^{2 b x +2 a}}{4 x}-\frac {b^{2} {\mathrm e}^{2 a} \Ei \left (1, -2 b x \right )}{2} \]

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)

________________________________________________________________________________________

maxima [A]  time = 0.38, size = 30, normalized size = 0.50 \[ b^{2} e^{\left (-2 \, a\right )} \Gamma \left (-2, 2 \, b x\right ) - b^{2} e^{\left (2 \, a\right )} \Gamma \left (-2, -2 \, b x\right ) \]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {cosh}\left (a+b\,x\right )\,\mathrm {sinh}\left (a+b\,x\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(a + b*x)*sinh(a + b*x))/x^3,x)

[Out]

int((cosh(a + b*x)*sinh(a + b*x))/x^3, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)*sinh(b*x+a)/x**3,x)

[Out]

Integral(sinh(a + b*x)*cosh(a + b*x)/x**3, x)

________________________________________________________________________________________