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

Optimal. Leaf size=62 \[ \frac{3 \sinh \left (a+\frac{b}{x}\right )}{b^2 x^2}+\frac{6 \sinh \left (a+\frac{b}{x}\right )}{b^4}-\frac{6 \cosh \left (a+\frac{b}{x}\right )}{b^3 x}-\frac{\cosh \left (a+\frac{b}{x}\right )}{b x^3} \]

[Out]

-(Cosh[a + b/x]/(b*x^3)) - (6*Cosh[a + b/x])/(b^3*x) + (6*Sinh[a + b/x])/b^4 + (3*Sinh[a + b/x])/(b^2*x^2)

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Rubi [A]  time = 0.0795417, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {5320, 3296, 2637} \[ \frac{3 \sinh \left (a+\frac{b}{x}\right )}{b^2 x^2}+\frac{6 \sinh \left (a+\frac{b}{x}\right )}{b^4}-\frac{6 \cosh \left (a+\frac{b}{x}\right )}{b^3 x}-\frac{\cosh \left (a+\frac{b}{x}\right )}{b x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x]/x^5,x]

[Out]

-(Cosh[a + b/x]/(b*x^3)) - (6*Cosh[a + b/x])/(b^3*x) + (6*Sinh[a + b/x])/b^4 + (3*Sinh[a + b/x])/(b^2*x^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}\right )}{x^5} \, dx &=-\operatorname{Subst}\left (\int x^3 \sinh (a+b x) \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\cosh \left (a+\frac{b}{x}\right )}{b x^3}+\frac{3 \operatorname{Subst}\left (\int x^2 \cosh (a+b x) \, dx,x,\frac{1}{x}\right )}{b}\\ &=-\frac{\cosh \left (a+\frac{b}{x}\right )}{b x^3}+\frac{3 \sinh \left (a+\frac{b}{x}\right )}{b^2 x^2}-\frac{6 \operatorname{Subst}\left (\int x \sinh (a+b x) \, dx,x,\frac{1}{x}\right )}{b^2}\\ &=-\frac{\cosh \left (a+\frac{b}{x}\right )}{b x^3}-\frac{6 \cosh \left (a+\frac{b}{x}\right )}{b^3 x}+\frac{3 \sinh \left (a+\frac{b}{x}\right )}{b^2 x^2}+\frac{6 \operatorname{Subst}\left (\int \cosh (a+b x) \, dx,x,\frac{1}{x}\right )}{b^3}\\ &=-\frac{\cosh \left (a+\frac{b}{x}\right )}{b x^3}-\frac{6 \cosh \left (a+\frac{b}{x}\right )}{b^3 x}+\frac{6 \sinh \left (a+\frac{b}{x}\right )}{b^4}+\frac{3 \sinh \left (a+\frac{b}{x}\right )}{b^2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.058288, size = 48, normalized size = 0.77 \[ \frac{3 x \left (b^2+2 x^2\right ) \sinh \left (a+\frac{b}{x}\right )-b \left (b^2+6 x^2\right ) \cosh \left (a+\frac{b}{x}\right )}{b^4 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b/x]/x^5,x]

[Out]

(-(b*(b^2 + 6*x^2)*Cosh[a + b/x]) + 3*x*(b^2 + 2*x^2)*Sinh[a + b/x])/(b^4*x^3)

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Maple [B]  time = 0.009, size = 165, normalized size = 2.7 \begin{align*} -{\frac{1}{{b}^{4}} \left ( \left ( a+{\frac{b}{x}} \right ) ^{3}\cosh \left ( a+{\frac{b}{x}} \right ) -3\, \left ( a+{\frac{b}{x}} \right ) ^{2}\sinh \left ( a+{\frac{b}{x}} \right ) +6\, \left ( a+{\frac{b}{x}} \right ) \cosh \left ( a+{\frac{b}{x}} \right ) -6\,\sinh \left ( a+{\frac{b}{x}} \right ) -3\,a \left ( \left ( a+{\frac{b}{x}} \right ) ^{2}\cosh \left ( a+{\frac{b}{x}} \right ) -2\, \left ( a+{\frac{b}{x}} \right ) \sinh \left ( a+{\frac{b}{x}} \right ) +2\,\cosh \left ( a+{\frac{b}{x}} \right ) \right ) +3\,{a}^{2} \left ( \left ( a+{\frac{b}{x}} \right ) \cosh \left ( a+{\frac{b}{x}} \right ) -\sinh \left ( a+{\frac{b}{x}} \right ) \right ) -{a}^{3}\cosh \left ( a+{\frac{b}{x}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b^4*((a+b/x)^3*cosh(a+b/x)-3*(a+b/x)^2*sinh(a+b/x)+6*(a+b/x)*cosh(a+b/x)-6*sinh(a+b/x)-3*a*((a+b/x)^2*cosh(
a+b/x)-2*(a+b/x)*sinh(a+b/x)+2*cosh(a+b/x))+3*a^2*((a+b/x)*cosh(a+b/x)-sinh(a+b/x))-a^3*cosh(a+b/x))

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Maxima [C]  time = 1.23791, size = 65, normalized size = 1.05 \begin{align*} -\frac{1}{8} \, b{\left (\frac{e^{\left (-a\right )} \Gamma \left (5, \frac{b}{x}\right )}{b^{5}} - \frac{e^{a} \Gamma \left (5, -\frac{b}{x}\right )}{b^{5}}\right )} - \frac{\sinh \left (a + \frac{b}{x}\right )}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*b*(e^(-a)*gamma(5, b/x)/b^5 - e^a*gamma(5, -b/x)/b^5) - 1/4*sinh(a + b/x)/x^4

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Fricas [A]  time = 1.7854, size = 116, normalized size = 1.87 \begin{align*} -\frac{{\left (b^{3} + 6 \, b x^{2}\right )} \cosh \left (\frac{a x + b}{x}\right ) - 3 \,{\left (b^{2} x + 2 \, x^{3}\right )} \sinh \left (\frac{a x + b}{x}\right )}{b^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((b^3 + 6*b*x^2)*cosh((a*x + b)/x) - 3*(b^2*x + 2*x^3)*sinh((a*x + b)/x))/(b^4*x^3)

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Sympy [A]  time = 9.01725, size = 61, normalized size = 0.98 \begin{align*} \begin{cases} - \frac{\cosh{\left (a + \frac{b}{x} \right )}}{b x^{3}} + \frac{3 \sinh{\left (a + \frac{b}{x} \right )}}{b^{2} x^{2}} - \frac{6 \cosh{\left (a + \frac{b}{x} \right )}}{b^{3} x} + \frac{6 \sinh{\left (a + \frac{b}{x} \right )}}{b^{4}} & \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)/x**5,x)

[Out]

Piecewise((-cosh(a + b/x)/(b*x**3) + 3*sinh(a + b/x)/(b**2*x**2) - 6*cosh(a + b/x)/(b**3*x) + 6*sinh(a + b/x)/
b**4, 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}\right )}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(a + b/x)/x^5, x)