\(\int \frac {\sinh (a+\frac {b}{x^2})}{x^7} \, dx\) [52]

Optimal result

Integrand size = 12, antiderivative size = 47 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{b^3}-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x^4}+\frac {\sinh \left (a+\frac {b}{x^2}\right )}{b^2 x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {5428, 3377, 2718} \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{b^3}+\frac {\sinh \left (a+\frac {b}{x^2}\right )}{b^2 x^2}-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x^4} \]

[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 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[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*} \text {integral}& = -\left (\frac {1}{2} \text {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 {\text {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 {\text {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] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.94 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=\frac {-\left (\left (b^2+2 x^4\right ) \cosh \left (a+\frac {b}{x^2}\right )\right )+2 b x^2 \sinh \left (a+\frac {b}{x^2}\right )}{2 b^3 x^4} \]

[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)

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.55

method	result	size
risch	\(-\frac {\left (2 x^{4}-2 x^{2} b +b^{2}\right ) {\mathrm e}^{\frac {a \,x^{2}+b}{x^{2}}}}{4 b^{3} x^{4}}-\frac {\left (2 x^{4}+2 x^{2} b +b^{2}\right ) {\mathrm e}^{-\frac {a \,x^{2}+b}{x^{2}}}}{4 b^{3} x^{4}}\)	\(73\)
parallelrisch	\(\frac {4 x^{4}-4 \tanh \left (\frac {a \,x^{2}+b}{2 x^{2}}\right ) x^{2} b +\tanh \left (\frac {a \,x^{2}+b}{2 x^{2}}\right )^{2} b^{2}+b^{2}}{2 x^{4} b^{3} \left (\tanh \left (\frac {a \,x^{2}+b}{2 x^{2}}\right )^{2}-1\right )}\)	\(75\)
meijerg	\(-\frac {2 \sqrt {\pi }\, \cosh \left (a \right ) \left (-\frac {1}{2 \sqrt {\pi }}+\frac {\left (\frac {b^{2}}{2 x^{4}}+1\right ) \cosh \left (\frac {b}{x^{2}}\right )}{2 \sqrt {\pi }}-\frac {b \sinh \left (\frac {b}{x^{2}}\right )}{2 \sqrt {\pi }\, x^{2}}\right )}{b^{3}}-\frac {2 i \sqrt {\pi }\, \sinh \left (a \right ) \left (\frac {i b \cosh \left (\frac {b}{x^{2}}\right )}{2 \sqrt {\pi }\, x^{2}}-\frac {i \left (\frac {3 b^{2}}{2 x^{4}}+3\right ) \sinh \left (\frac {b}{x^{2}}\right )}{6 \sqrt {\pi }}\right )}{b^{3}}\)	\(104\)

[In]

int(sinh(a+b/x^2)/x^7,x,method=_RETURNVERBOSE)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.06 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=\frac {2 \, b x^{2} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) - {\left (2 \, x^{4} + b^{2}\right )} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right )}{2 \, b^{3} x^{4}} \]

[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)

Sympy [A] (verification not implemented)

Time = 1.63 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.09 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=\begin {cases} - \frac {\cosh {\left (a + \frac {b}{x^{2}} \right )}}{2 b x^{4}} + \frac {\sinh {\left (a + \frac {b}{x^{2}} \right )}}{b^{2} x^{2}} - \frac {\cosh {\left (a + \frac {b}{x^{2}} \right )}}{b^{3}} & \text {for}\: b \neq 0 \\- \frac {\sinh {\left (a \right )}}{6 x^{6}} & \text {otherwise} \end {cases} \]

[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))

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=-\frac {1}{12} \, b {\left (\frac {e^{\left (-a\right )} \Gamma \left (4, \frac {b}{x^{2}}\right )}{b^{4}} + \frac {e^{a} \Gamma \left (4, -\frac {b}{x^{2}}\right )}{b^{4}}\right )} - \frac {\sinh \left (a + \frac {b}{x^{2}}\right )}{6 \, x^{6}} \]

[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

Giac [F]

\[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=\int { \frac {\sinh \left (a + \frac {b}{x^{2}}\right )}{x^{7}} \,d x } \]

[In]

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

[Out]

integrate(sinh(a + b/x^2)/x^7, x)

Mupad [B] (verification not implemented)

Time = 1.11 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.57 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^7} \, dx=-\frac {{\mathrm {e}}^{a+\frac {b}{x^2}}\,\left (\frac {1}{4\,b}-\frac {x^2}{2\,b^2}+\frac {x^4}{2\,b^3}\right )}{x^4}-\frac {{\mathrm {e}}^{-a-\frac {b}{x^2}}\,\left (\frac {1}{4\,b}+\frac {x^2}{2\,b^2}+\frac {x^4}{2\,b^3}\right )}{x^4} \]

[In]

int(sinh(a + b/x^2)/x^7,x)

[Out]

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