3.52 \(\int \frac{\sinh (a+\frac{b}{x^2})}{x^7} \, dx\)

Optimal. Leaf size=47 \[ \frac{\sinh \left (a+\frac{b}{x^2}\right )}{b^2 x^2}-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{b^3}-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{2 b x^4} \]

[Out]

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

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Rubi [A]  time = 0.0586295, antiderivative size = 47, 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 = {5320, 3296, 2638} \[ \frac{\sinh \left (a+\frac{b}{x^2}\right )}{b^2 x^2}-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{b^3}-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{2 b x^4} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x^2]/x^7,x]

[Out]

-(Cosh[a + b/x^2]/b^3) - Cosh[a + b/x^2]/(2*b*x^4) + Sinh[a + b/x^2]/(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 2638

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

Rubi steps

\begin{align*} \int \frac{\sinh \left (a+\frac{b}{x^2}\right )}{x^7} \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int x^2 \sinh (a+b x) \, dx,x,\frac{1}{x^2}\right )\right )\\ &=-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{2 b x^4}+\frac{\operatorname{Subst}\left (\int x \cosh (a+b x) \, dx,x,\frac{1}{x^2}\right )}{b}\\ &=-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{2 b x^4}+\frac{\sinh \left (a+\frac{b}{x^2}\right )}{b^2 x^2}-\frac{\operatorname{Subst}\left (\int \sinh (a+b x) \, dx,x,\frac{1}{x^2}\right )}{b^2}\\ &=-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{b^3}-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{2 b x^4}+\frac{\sinh \left (a+\frac{b}{x^2}\right )}{b^2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.043287, size = 44, normalized size = 0.94 \[ \frac{2 b x^2 \sinh \left (a+\frac{b}{x^2}\right )-\left (b^2+2 x^4\right ) \cosh \left (a+\frac{b}{x^2}\right )}{2 b^3 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b/x^2]/x^7,x]

[Out]

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

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Maple [A]  time = 0.035, size = 73, normalized size = 1.6 \begin{align*} -{\frac{2\,{x}^{4}-2\,b{x}^{2}+{b}^{2}}{4\,{b}^{3}{x}^{4}}{{\rm e}^{{\frac{a{x}^{2}+b}{{x}^{2}}}}}}-{\frac{2\,{x}^{4}+2\,b{x}^{2}+{b}^{2}}{4\,{b}^{3}{x}^{4}}{{\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^7,x)

[Out]

-1/4*(2*x^4-2*b*x^2+b^2)/b^3/x^4*exp((a*x^2+b)/x^2)-1/4*(2*x^4+2*b*x^2+b^2)/b^3/x^4*exp(-(a*x^2+b)/x^2)

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Maxima [C]  time = 1.1839, size = 63, normalized size = 1.34 \begin{align*} -\frac{1}{12} \, b{\left (\frac{e^{\left (-a\right )} \Gamma \left (4, \frac{b}{x^{2}}\right )}{b^{4}} + \frac{e^{a} \Gamma \left (4, -\frac{b}{x^{2}}\right )}{b^{4}}\right )} - \frac{\sinh \left (a + \frac{b}{x^{2}}\right )}{6 \, x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/x^7,x, algorithm="maxima")

[Out]

-1/12*b*(e^(-a)*gamma(4, b/x^2)/b^4 + e^a*gamma(4, -b/x^2)/b^4) - 1/6*sinh(a + b/x^2)/x^6

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Fricas [A]  time = 1.65089, size = 115, normalized size = 2.45 \begin{align*} \frac{2 \, b x^{2} \sinh \left (\frac{a x^{2} + b}{x^{2}}\right ) -{\left (2 \, x^{4} + b^{2}\right )} \cosh \left (\frac{a x^{2} + b}{x^{2}}\right )}{2 \, b^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/x^7,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 48.2079, size = 51, normalized size = 1.09 \begin{align*} \begin{cases} - \frac{\cosh{\left (a + \frac{b}{x^{2}} \right )}}{2 b x^{4}} + \frac{\sinh{\left (a + \frac{b}{x^{2}} \right )}}{b^{2} x^{2}} - \frac{\cosh{\left (a + \frac{b}{x^{2}} \right )}}{b^{3}} & \text{for}\: b \neq 0 \\- \frac{\sinh{\left (a \right )}}{6 x^{6}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x**2)/x**7,x)

[Out]

Piecewise((-cosh(a + b/x**2)/(2*b*x**4) + sinh(a + b/x**2)/(b**2*x**2) - cosh(a + b/x**2)/b**3, Ne(b, 0)), (-s
inh(a)/(6*x**6), 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^{7}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/x^7,x, algorithm="giac")

[Out]

integrate(sinh(a + b/x^2)/x^7, x)