3.28 \(\int \frac{\cosh (a+\frac{b}{x})}{x^3} \, dx\)

Optimal. Leaf size=29 \[ \frac{\cosh \left (a+\frac{b}{x}\right )}{b^2}-\frac{\sinh \left (a+\frac{b}{x}\right )}{b x} \]

[Out]

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

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Rubi [A]  time = 0.0300225, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5321, 3296, 2638} \[ \frac{\cosh \left (a+\frac{b}{x}\right )}{b^2}-\frac{\sinh \left (a+\frac{b}{x}\right )}{b x} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b/x]/x^3,x]

[Out]

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

Rule 5321

Int[((a_.) + Cosh[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Cosh[c + d*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IntegerQ[Sim
plify[(m + 1)/n]] && (EqQ[p, 1] || EqQ[m, n - 1] || (IntegerQ[p] && GtQ[Simplify[(m + 1)/n], 0]))

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\cosh \left (a+\frac{b}{x}\right )}{x^3} \, dx &=-\operatorname{Subst}\left (\int x \cosh (a+b x) \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\sinh \left (a+\frac{b}{x}\right )}{b x}+\frac{\operatorname{Subst}\left (\int \sinh (a+b x) \, dx,x,\frac{1}{x}\right )}{b}\\ &=\frac{\cosh \left (a+\frac{b}{x}\right )}{b^2}-\frac{\sinh \left (a+\frac{b}{x}\right )}{b x}\\ \end{align*}

Mathematica [A]  time = 0.0232162, size = 29, normalized size = 1. \[ \frac{x \cosh \left (a+\frac{b}{x}\right )-b \sinh \left (a+\frac{b}{x}\right )}{b^2 x} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b/x]/x^3,x]

[Out]

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

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Maple [A]  time = 0.01, size = 44, normalized size = 1.5 \begin{align*} -{\frac{1}{{b}^{2}} \left ( \left ( a+{\frac{b}{x}} \right ) \sinh \left ( a+{\frac{b}{x}} \right ) -\cosh \left ( a+{\frac{b}{x}} \right ) -a\sinh \left ( a+{\frac{b}{x}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.23743, size = 63, normalized size = 2.17 \begin{align*} \frac{1}{4} \, b{\left (\frac{e^{\left (-a\right )} \Gamma \left (3, \frac{b}{x}\right )}{b^{3}} + \frac{e^{a} \Gamma \left (3, -\frac{b}{x}\right )}{b^{3}}\right )} - \frac{\cosh \left (a + \frac{b}{x}\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b/x)/x^3,x, algorithm="maxima")

[Out]

1/4*b*(e^(-a)*gamma(3, b/x)/b^3 + e^a*gamma(3, -b/x)/b^3) - 1/2*cosh(a + b/x)/x^2

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Fricas [A]  time = 1.71844, size = 72, normalized size = 2.48 \begin{align*} \frac{x \cosh \left (\frac{a x + b}{x}\right ) - b \sinh \left (\frac{a x + b}{x}\right )}{b^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.55675, size = 29, normalized size = 1. \begin{align*} \begin{cases} - \frac{\sinh{\left (a + \frac{b}{x} \right )}}{b x} + \frac{\cosh{\left (a + \frac{b}{x} \right )}}{b^{2}} & \text{for}\: b \neq 0 \\- \frac{\cosh{\left (a \right )}}{2 x^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-sinh(a + b/x)/(b*x) + cosh(a + b/x)/b**2, Ne(b, 0)), (-cosh(a)/(2*x**2), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh \left (a + \frac{b}{x}\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cosh(a + b/x)/x^3, x)