3.50 \(\int \frac{\sinh (a+\frac{b}{x^2})}{x^5} \, dx\)

Optimal. Leaf size=34 \[ \frac{\sinh \left (a+\frac{b}{x^2}\right )}{2 b^2}-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{2 b x^2} \]

[Out]

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

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Rubi [A]  time = 0.0335038, antiderivative size = 34, 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 = {5320, 3296, 2637} \[ \frac{\sinh \left (a+\frac{b}{x^2}\right )}{2 b^2}-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{2 b x^2} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x^2]/x^5,x]

[Out]

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

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]))

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{\sinh \left (a+\frac{b}{x^2}\right )}{x^5} \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int x \sinh (a+b x) \, dx,x,\frac{1}{x^2}\right )\right )\\ &=-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{2 b x^2}+\frac{\operatorname{Subst}\left (\int \cosh (a+b x) \, dx,x,\frac{1}{x^2}\right )}{2 b}\\ &=-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{2 b x^2}+\frac{\sinh \left (a+\frac{b}{x^2}\right )}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0273929, size = 34, normalized size = 1. \[ \frac{x^2 \sinh \left (a+\frac{b}{x^2}\right )-b \cosh \left (a+\frac{b}{x^2}\right )}{2 b^2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b/x^2]/x^5,x]

[Out]

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

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Maple [A]  time = 0.025, size = 55, normalized size = 1.6 \begin{align*} -{\frac{-{x}^{2}+b}{4\,{x}^{2}{b}^{2}}{{\rm e}^{{\frac{a{x}^{2}+b}{{x}^{2}}}}}}-{\frac{{x}^{2}+b}{4\,{x}^{2}{b}^{2}}{{\rm e}^{-{\frac{a{x}^{2}+b}{{x}^{2}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a+b/x^2)/x^5,x)

[Out]

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

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Maxima [C]  time = 1.31073, size = 65, normalized size = 1.91 \begin{align*} -\frac{1}{8} \, b{\left (\frac{e^{\left (-a\right )} \Gamma \left (3, \frac{b}{x^{2}}\right )}{b^{3}} - \frac{e^{a} \Gamma \left (3, -\frac{b}{x^{2}}\right )}{b^{3}}\right )} - \frac{\sinh \left (a + \frac{b}{x^{2}}\right )}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/x^5,x, algorithm="maxima")

[Out]

-1/8*b*(e^(-a)*gamma(3, b/x^2)/b^3 - e^a*gamma(3, -b/x^2)/b^3) - 1/4*sinh(a + b/x^2)/x^4

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Fricas [A]  time = 1.71071, size = 93, normalized size = 2.74 \begin{align*} \frac{x^{2} \sinh \left (\frac{a x^{2} + b}{x^{2}}\right ) - b \cosh \left (\frac{a x^{2} + b}{x^{2}}\right )}{2 \, b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 15.0673, size = 37, normalized size = 1.09 \begin{align*} \begin{cases} - \frac{\cosh{\left (a + \frac{b}{x^{2}} \right )}}{2 b x^{2}} + \frac{\sinh{\left (a + \frac{b}{x^{2}} \right )}}{2 b^{2}} & \text{for}\: b \neq 0 \\- \frac{\sinh{\left (a \right )}}{4 x^{4}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x**2)/x**5,x)

[Out]

Piecewise((-cosh(a + b/x**2)/(2*b*x**2) + sinh(a + b/x**2)/(2*b**2), Ne(b, 0)), (-sinh(a)/(4*x**4), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (a + \frac{b}{x^{2}}\right )}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/x^5,x, algorithm="giac")

[Out]

integrate(sinh(a + b/x^2)/x^5, x)