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 = 0.03, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 2638

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

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 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]))

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*}

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Mathematica [A]  time = 0.03, size = 29, normalized size = 1.00 \[ \frac {x \cosh \left (a+\frac {b}{x}\right )-b \sinh \left (a+\frac {b}{x}\right )}{b^2 x} \]

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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fricas [A]  time = 0.42, size = 33, normalized size = 1.14 \[ \frac {x \cosh \left (\frac {a x + b}{x}\right ) - b \sinh \left (\frac {a x + b}{x}\right )}{b^{2} x} \]

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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giac [B]  time = 0.14, size = 93, normalized size = 3.21 \[ \frac {a e^{\left (\frac {a x + b}{x}\right )} - a e^{\left (-\frac {a x + b}{x}\right )} - \frac {{\left (a x + b\right )} e^{\left (\frac {a x + b}{x}\right )}}{x} + \frac {{\left (a x + b\right )} e^{\left (-\frac {a x + b}{x}\right )}}{x} + e^{\left (\frac {a x + b}{x}\right )} + e^{\left (-\frac {a x + b}{x}\right )}}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 44, normalized size = 1.52 \[ -\frac {\left (a +\frac {b}{x}\right ) \sinh \left (a +\frac {b}{x}\right )-\cosh \left (a +\frac {b}{x}\right )-a \sinh \left (a +\frac {b}{x}\right )}{b^{2}} \]

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 = 0.38, size = 47, normalized size = 1.62 \[ \frac {1}{4} \, b {\left (\frac {e^{\left (-a\right )} \Gamma \left (3, \frac {b}{x}\right )}{b^{3}} + \frac {e^{a} \Gamma \left (3, -\frac {b}{x}\right )}{b^{3}}\right )} - \frac {\cosh \left (a + \frac {b}{x}\right )}{2 \, x^{2}} \]

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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mupad [B]  time = 0.89, size = 29, normalized size = 1.00 \[ \frac {\mathrm {cosh}\left (a+\frac {b}{x}\right )}{b^2}-\frac {\mathrm {sinh}\left (a+\frac {b}{x}\right )}{b\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b/x)/x^3,x)

[Out]

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

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sympy [A]  time = 1.86, size = 29, normalized size = 1.00 \[ \begin {cases} - \frac {\sinh {\left (a + \frac {b}{x} \right )}}{b x} + \frac {\cosh {\left (a + \frac {b}{x} \right )}}{b^{2}} & \text {for}\: b \neq 0 \\- \frac {\cosh {\relax (a )}}{2 x^{2}} & \text {otherwise} \end {cases} \]

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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