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

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Rubi [A]  time = 0.09, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 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^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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.47, size = 65, normalized size = 1.67 \[ \frac {{\left (b x {\rm Ei}\left (2 \, b x\right ) + b x {\rm Ei}\left (-2 \, b x\right )\right )} \cosh \left (2 \, a\right ) - 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + {\left (b x {\rm Ei}\left (2 \, b x\right ) - b x {\rm Ei}\left (-2 \, b x\right )\right )} \sinh \left (2 \, a\right )}{2 \, x} \]

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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giac [A]  time = 0.14, size = 52, normalized size = 1.33 \[ \frac {2 \, b x {\rm Ei}\left (2 \, b x\right ) e^{\left (2 \, a\right )} + 2 \, b x {\rm Ei}\left (-2 \, b x\right ) e^{\left (-2 \, a\right )} - e^{\left (2 \, b x + 2 \, a\right )} + e^{\left (-2 \, b x - 2 \, a\right )}}{4 \, x} \]

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

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maple [A]  time = 0.13, size = 56, normalized size = 1.44 \[ \frac {{\mathrm e}^{-2 b x -2 a}}{4 x}-\frac {b \,{\mathrm e}^{-2 a} \Ei \left (1, 2 b x \right )}{2}-\frac {{\mathrm e}^{2 b x +2 a}}{4 x}-\frac {b \,{\mathrm e}^{2 a} \Ei \left (1, -2 b x \right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)*sinh(b*x+a)/x^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 = 0.39, size = 27, normalized size = 0.69 \[ \frac {1}{2} \, b e^{\left (-2 \, a\right )} \Gamma \left (-1, 2 \, b x\right ) + \frac {1}{2} \, b e^{\left (2 \, a\right )} \Gamma \left (-1, -2 \, b x\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\mathrm {cosh}\left (a+b\,x\right )\,\mathrm {sinh}\left (a+b\,x\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh {\left (a + b x \right )} \cosh {\left (a + b x \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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