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 = 0.05, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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^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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.49, size = 43, normalized size = 0.93 \[ \frac {2 \, b x \cosh \left (\frac {a x + b}{x}\right ) - {\left (b^{2} + 2 \, x^{2}\right )} \sinh \left (\frac {a x + b}{x}\right )}{b^{3} x^{2}} \]

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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giac [B]  time = 0.14, size = 216, normalized size = 4.70 \[ -\frac {a^{2} e^{\left (\frac {a x + b}{x}\right )} - a^{2} e^{\left (-\frac {a x + b}{x}\right )} + 2 \, a e^{\left (\frac {a x + b}{x}\right )} - \frac {2 \, {\left (a x + b\right )} a e^{\left (\frac {a x + b}{x}\right )}}{x} + 2 \, a e^{\left (-\frac {a x + b}{x}\right )} + \frac {2 \, {\left (a x + b\right )} a e^{\left (-\frac {a x + b}{x}\right )}}{x} + \frac {{\left (a x + b\right )}^{2} e^{\left (\frac {a x + b}{x}\right )}}{x^{2}} - \frac {2 \, {\left (a x + b\right )} e^{\left (\frac {a x + b}{x}\right )}}{x} - \frac {{\left (a x + b\right )}^{2} e^{\left (-\frac {a x + b}{x}\right )}}{x^{2}} - \frac {2 \, {\left (a x + b\right )} e^{\left (-\frac {a x + b}{x}\right )}}{x} + 2 \, e^{\left (\frac {a x + b}{x}\right )} - 2 \, e^{\left (-\frac {a x + b}{x}\right )}}{2 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a^2*e^((a*x + b)/x) - a^2*e^(-(a*x + b)/x) + 2*a*e^((a*x + b)/x) - 2*(a*x + b)*a*e^((a*x + b)/x)/x + 2*a
*e^(-(a*x + b)/x) + 2*(a*x + b)*a*e^(-(a*x + b)/x)/x + (a*x + b)^2*e^((a*x + b)/x)/x^2 - 2*(a*x + b)*e^((a*x +
b)/x)/x - (a*x + b)^2*e^(-(a*x + b)/x)/x^2 - 2*(a*x + b)*e^(-(a*x + b)/x)/x + 2*e^((a*x + b)/x) - 2*e^(-(a*x
+ b)/x))/b^3

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maple [B]  time = 0.08, size = 94, normalized size = 2.04 \[ -\frac {\left (a +\frac {b}{x}\right )^{2} \sinh \left (a +\frac {b}{x}\right )-2 \cosh \left (a +\frac {b}{x}\right ) \left (a +\frac {b}{x}\right )+2 \sinh \left (a +\frac {b}{x}\right )-2 a \left (\left (a +\frac {b}{x}\right ) \sinh \left (a +\frac {b}{x}\right )-\cosh \left (a +\frac {b}{x}\right )\right )+a^{2} \sinh \left (a +\frac {b}{x}\right )}{b^{3}} \]

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*cosh(a+b/x)*(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 = 0.35, size = 48, normalized size = 1.04 \[ \frac {1}{6} \, b {\left (\frac {e^{\left (-a\right )} \Gamma \left (4, \frac {b}{x}\right )}{b^{4}} - \frac {e^{a} \Gamma \left (4, -\frac {b}{x}\right )}{b^{4}}\right )} - \frac {\cosh \left (a + \frac {b}{x}\right )}{3 \, x^{3}} \]

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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mupad [B]  time = 0.93, size = 66, normalized size = 1.43 \[ \frac {{\mathrm {e}}^{-a-\frac {b}{x}}\,\left (\frac {x}{b^2}+\frac {1}{2\,b}+\frac {x^2}{b^3}\right )}{x^2}-\frac {{\mathrm {e}}^{a+\frac {b}{x}}\,\left (\frac {1}{2\,b}-\frac {x}{b^2}+\frac {x^2}{b^3}\right )}{x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b/x)/x^4,x)

[Out]

(exp(- a - b/x)*(x/b^2 + 1/(2*b) + x^2/b^3))/x^2 - (exp(a + b/x)*(1/(2*b) - x/b^2 + x^2/b^3))/x^2

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sympy [A]  time = 3.13, size = 46, normalized size = 1.00 \[ \begin {cases} - \frac {\sinh {\left (a + \frac {b}{x} \right )}}{b x^{2}} + \frac {2 \cosh {\left (a + \frac {b}{x} \right )}}{b^{2} x} - \frac {2 \sinh {\left (a + \frac {b}{x} \right )}}{b^{3}} & \text {for}\: b \neq 0 \\- \frac {\cosh {\relax (a )}}{3 x^{3}} & \text {otherwise} \end {cases} \]

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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