3.1.50 \(\int \frac {\sinh (a+\frac {b}{x^2})}{x^5} \, dx\) [50]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5428, 3377, 2717} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x^2]/x^5,x]

[Out]

-1/2*Cosh[a + b/x^2]/(b*x^2) + Sinh[a + b/x^2]/(2*b^2)

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

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 5428

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^5} \, dx &=-\left (\frac {1}{2} \text {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 {\text {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*}

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Mathematica [A]
time = 0.02, size = 34, normalized size = 1.00 \begin {gather*} \frac {-b \cosh \left (a+\frac {b}{x^2}\right )+x^2 \sinh \left (a+\frac {b}{x^2}\right )}{2 b^2 x^2} \end {gather*}

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 = 0.31, size = 55, normalized size = 1.62

method	result	size
risch	\(-\frac {\left (-x^{2}+b \right ) {\mathrm e}^{\frac {a \,x^{2}+b}{x^{2}}}}{4 b^{2} x^{2}}-\frac {\left (x^{2}+b \right ) {\mathrm e}^{-\frac {a \,x^{2}+b}{x^{2}}}}{4 b^{2} x^{2}}\)	\(55\)
meijerg	\(-\frac {\cosh \left (a \right ) \left (\frac {\cosh \left (\frac {b}{x^{2}}\right ) b}{x^{2}}-\sinh \left (\frac {b}{x^{2}}\right )\right )}{2 b^{2}}+\frac {\sqrt {\pi }\, \sinh \left (a \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\cosh \left (\frac {b}{x^{2}}\right )}{2 \sqrt {\pi }}-\frac {b \sinh \left (\frac {b}{x^{2}}\right )}{2 \sqrt {\pi }\, x^{2}}\right )}{b^{2}}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.30, size = 48, normalized size = 1.41 \begin {gather*} -\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 {gather*}

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 = 0.34, size = 40, normalized size = 1.18 \begin {gather*} \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 {gather*}

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 = 1.32, size = 37, normalized size = 1.09 \begin {gather*} \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 {gather*}

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 [A]
time = 0.42, size = 43, normalized size = 1.26 \begin {gather*} -\frac {{\left ({\left (\frac {b}{x^{2}} - 1\right )} e^{\left (2 \, a + \frac {b}{x^{2}}\right )} + {\left (\frac {b}{x^{2}} + 1\right )} e^{\left (-\frac {b}{x^{2}}\right )}\right )} e^{\left (-a\right )}}{4 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.41, size = 58, normalized size = 1.71 \begin {gather*} -\frac {{\mathrm {e}}^{a+\frac {b}{x^2}}\,\left (\frac {1}{4\,b}-\frac {x^2}{4\,b^2}\right )}{x^2}-\frac {{\mathrm {e}}^{-a-\frac {b}{x^2}}\,\left (\frac {1}{4\,b}+\frac {x^2}{4\,b^2}\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b/x^2)/x^5,x)

[Out]

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

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