3.29 \(\int \frac{\cosh (a+\frac{b}{x})}{x^4} \, dx\)

Optimal. Leaf size=46 \[ -\frac{2 \sinh \left (a+\frac{b}{x}\right )}{b^3}+\frac{2 \cosh \left (a+\frac{b}{x}\right )}{b^2 x}-\frac{\sinh \left (a+\frac{b}{x}\right )}{b x^2} \]

[Out]

(2*Cosh[a + b/x])/(b^2*x) - (2*Sinh[a + b/x])/b^3 - Sinh[a + b/x]/(b*x^2)

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Rubi [A]  time = 0.0530366, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5321, 3296, 2637} \[ -\frac{2 \sinh \left (a+\frac{b}{x}\right )}{b^3}+\frac{2 \cosh \left (a+\frac{b}{x}\right )}{b^2 x}-\frac{\sinh \left (a+\frac{b}{x}\right )}{b x^2} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b/x]/x^4,x]

[Out]

(2*Cosh[a + b/x])/(b^2*x) - (2*Sinh[a + b/x])/b^3 - Sinh[a + b/x]/(b*x^2)

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\cosh \left (a+\frac{b}{x}\right )}{x^4} \, dx &=-\operatorname{Subst}\left (\int x^2 \cosh (a+b x) \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\sinh \left (a+\frac{b}{x}\right )}{b x^2}+\frac{2 \operatorname{Subst}\left (\int x \sinh (a+b x) \, dx,x,\frac{1}{x}\right )}{b}\\ &=\frac{2 \cosh \left (a+\frac{b}{x}\right )}{b^2 x}-\frac{\sinh \left (a+\frac{b}{x}\right )}{b x^2}-\frac{2 \operatorname{Subst}\left (\int \cosh (a+b x) \, dx,x,\frac{1}{x}\right )}{b^2}\\ &=\frac{2 \cosh \left (a+\frac{b}{x}\right )}{b^2 x}-\frac{2 \sinh \left (a+\frac{b}{x}\right )}{b^3}-\frac{\sinh \left (a+\frac{b}{x}\right )}{b x^2}\\ \end{align*}

Mathematica [A]  time = 0.0417849, size = 39, normalized size = 0.85 \[ \frac{2 b x \cosh \left (a+\frac{b}{x}\right )-\left (b^2+2 x^2\right ) \sinh \left (a+\frac{b}{x}\right )}{b^3 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b/x]/x^4,x]

[Out]

(2*b*x*Cosh[a + b/x] - (b^2 + 2*x^2)*Sinh[a + b/x])/(b^3*x^2)

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Maple [B]  time = 0.013, size = 94, normalized size = 2. \begin{align*} -{\frac{1}{{b}^{3}} \left ( \left ( a+{\frac{b}{x}} \right ) ^{2}\sinh \left ( a+{\frac{b}{x}} \right ) -2\, \left ( a+{\frac{b}{x}} \right ) \cosh \left ( a+{\frac{b}{x}} \right ) +2\,\sinh \left ( a+{\frac{b}{x}} \right ) -2\,a \left ( \left ( a+{\frac{b}{x}} \right ) \sinh \left ( a+{\frac{b}{x}} \right ) -\cosh \left ( a+{\frac{b}{x}} \right ) \right ) +{a}^{2}\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^4,x)

[Out]

-1/b^3*((a+b/x)^2*sinh(a+b/x)-2*(a+b/x)*cosh(a+b/x)+2*sinh(a+b/x)-2*a*((a+b/x)*sinh(a+b/x)-cosh(a+b/x))+a^2*si
nh(a+b/x))

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Maxima [C]  time = 1.19481, size = 65, normalized size = 1.41 \begin{align*} \frac{1}{6} \, b{\left (\frac{e^{\left (-a\right )} \Gamma \left (4, \frac{b}{x}\right )}{b^{4}} - \frac{e^{a} \Gamma \left (4, -\frac{b}{x}\right )}{b^{4}}\right )} - \frac{\cosh \left (a + \frac{b}{x}\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b/x)/x^4,x, algorithm="maxima")

[Out]

1/6*b*(e^(-a)*gamma(4, b/x)/b^4 - e^a*gamma(4, -b/x)/b^4) - 1/3*cosh(a + b/x)/x^3

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Fricas [A]  time = 1.70482, size = 96, normalized size = 2.09 \begin{align*} \frac{2 \, b x \cosh \left (\frac{a x + b}{x}\right ) -{\left (b^{2} + 2 \, x^{2}\right )} \sinh \left (\frac{a x + b}{x}\right )}{b^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b/x)/x^4,x, algorithm="fricas")

[Out]

(2*b*x*cosh((a*x + b)/x) - (b^2 + 2*x^2)*sinh((a*x + b)/x))/(b^3*x^2)

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Sympy [A]  time = 4.4698, size = 46, normalized size = 1. \begin{align*} \begin{cases} - \frac{\sinh{\left (a + \frac{b}{x} \right )}}{b x^{2}} + \frac{2 \cosh{\left (a + \frac{b}{x} \right )}}{b^{2} x} - \frac{2 \sinh{\left (a + \frac{b}{x} \right )}}{b^{3}} & \text{for}\: b \neq 0 \\- \frac{\cosh{\left (a \right )}}{3 x^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b/x)/x**4,x)

[Out]

Piecewise((-sinh(a + b/x)/(b*x**2) + 2*cosh(a + b/x)/(b**2*x) - 2*sinh(a + b/x)/b**3, Ne(b, 0)), (-cosh(a)/(3*
x**3), 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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b/x)/x^4,x, algorithm="giac")

[Out]

integrate(cosh(a + b/x)/x^4, x)