3.279 \(\int \frac {\cosh (x) \sinh (x)}{x^3} \, dx\)

Optimal. Leaf size=27 \[ \text {Shi}(2 x)-\frac {\sinh (2 x)}{4 x^2}-\frac {\cosh (2 x)}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5448, 12, 3297, 3298} \[ \text {Shi}(2 x)-\frac {\sinh (2 x)}{4 x^2}-\frac {\cosh (2 x)}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cosh[2*x]/(2*x) - Sinh[2*x]/(4*x^2) + SinhIntegral[2*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 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 (x) \sinh (x)}{x^3} \, dx &=\int \frac {\sinh (2 x)}{2 x^3} \, dx\\ &=\frac {1}{2} \int \frac {\sinh (2 x)}{x^3} \, dx\\ &=-\frac {\sinh (2 x)}{4 x^2}+\frac {1}{2} \int \frac {\cosh (2 x)}{x^2} \, dx\\ &=-\frac {\cosh (2 x)}{2 x}-\frac {\sinh (2 x)}{4 x^2}+\int \frac {\sinh (2 x)}{x} \, dx\\ &=-\frac {\cosh (2 x)}{2 x}-\frac {\sinh (2 x)}{4 x^2}+\text {Shi}(2 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ \text {Shi}(2 x)-\frac {\sinh (2 x)}{4 x^2}-\frac {\cosh (2 x)}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*Cosh[2*x]/x - Sinh[2*x]/(4*x^2) + SinhIntegral[2*x]

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fricas [A]  time = 0.58, size = 43, normalized size = 1.59 \[ \frac {x^{2} {\rm Ei}\left (2 \, x\right ) - x^{2} {\rm Ei}\left (-2 \, x\right ) - x \cosh \relax (x)^{2} - x \sinh \relax (x)^{2} - \cosh \relax (x) \sinh \relax (x)}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.11, size = 48, normalized size = 1.78 \[ \frac {4 \, x^{2} {\rm Ei}\left (2 \, x\right ) - 4 \, x^{2} {\rm Ei}\left (-2 \, x\right ) - 2 \, x e^{\left (2 \, x\right )} - 2 \, x e^{\left (-2 \, x\right )} - e^{\left (2 \, x\right )} + e^{\left (-2 \, x\right )}}{8 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 24, normalized size = 0.89 \[ -\frac {\cosh \left (2 x \right )}{2 x}+\Shi \left (2 x \right )-\frac {\sinh \left (2 x \right )}{4 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)*sinh(x)/x^3,x)

[Out]

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

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maxima [A]  time = 0.34, size = 13, normalized size = 0.48 \[ \Gamma \left (-2, 2 \, x\right ) - \Gamma \left (-2, -2 \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sinh(x)/x^3,x, algorithm="maxima")

[Out]

gamma(-2, 2*x) - gamma(-2, -2*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {\mathrm {cosh}\relax (x)\,\mathrm {sinh}\relax (x)}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(x)*sinh(x))/x^3,x)

[Out]

int((cosh(x)*sinh(x))/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)*sinh(x)/x**3,x)

[Out]

Integral(sinh(x)*cosh(x)/x**3, x)

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