3.48 \(\int \frac {\sinh (a+\frac {b}{x^2})}{x^3} \, dx\)

Optimal. Leaf size=15 \[ -\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5320, 2638} \[ -\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x^2]/x^3,x]

[Out]

-Cosh[a + b/x^2]/(2*b)

Rule 2638

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

Rule 5320

Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simpli
fy[(m + 1)/n] - 1)*(a + b*Sinh[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 {\sinh \left (a+\frac {b}{x^2}\right )}{x^3} \, dx &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \sinh (a+b x) \, dx,x,\frac {1}{x^2}\right )\right )\\ &=-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \[ -\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b/x^2]/x^3,x]

[Out]

-1/2*Cosh[a + b/x^2]/b

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fricas [A]  time = 0.47, size = 17, normalized size = 1.13 \[ -\frac {\cosh \left (\frac {a x^{2} + b}{x^{2}}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/x^3,x, algorithm="fricas")

[Out]

-1/2*cosh((a*x^2 + b)/x^2)/b

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giac [B]  time = 0.16, size = 27, normalized size = 1.80 \[ -\frac {{\left (e^{\left (2 \, a + \frac {b}{x^{2}}\right )} + e^{\left (-\frac {b}{x^{2}}\right )}\right )} e^{\left (-a\right )}}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/x^3,x, algorithm="giac")

[Out]

-1/4*(e^(2*a + b/x^2) + e^(-b/x^2))*e^(-a)/b

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maple [A]  time = 0.00, size = 14, normalized size = 0.93 \[ -\frac {\cosh \left (a +\frac {b}{x^{2}}\right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a+b/x^2)/x^3,x)

[Out]

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

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maxima [A]  time = 0.32, size = 13, normalized size = 0.87 \[ -\frac {\cosh \left (a + \frac {b}{x^{2}}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/x^3,x, algorithm="maxima")

[Out]

-1/2*cosh(a + b/x^2)/b

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mupad [B]  time = 0.37, size = 13, normalized size = 0.87 \[ -\frac {\mathrm {cosh}\left (a+\frac {b}{x^2}\right )}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b/x^2)/x^3,x)

[Out]

-cosh(a + b/x^2)/(2*b)

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sympy [A]  time = 2.74, size = 22, normalized size = 1.47 \[ \begin {cases} - \frac {\cosh {\left (a + \frac {b}{x^{2}} \right )}}{2 b} & \text {for}\: b \neq 0 \\- \frac {\sinh {\relax (a )}}{2 x^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x**2)/x**3,x)

[Out]

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

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