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

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

[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

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 2637

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

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 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]))

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 = 0.06, 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)

________________________________________________________________________________________

fricas [A]  time = 0.38, size = 53, normalized size = 0.85 \[ -\frac {{\left (b^{3} + 6 \, b x^{2}\right )} \cosh \left (\frac {a x + b}{x}\right ) - 3 \, {\left (b^{2} x + 2 \, x^{3}\right )} \sinh \left (\frac {a x + b}{x}\right )}{b^{4} x^{3}} \]

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)

________________________________________________________________________________________

giac [B]  time = 0.19, size = 386, normalized size = 6.23 \[ \frac {a^{3} e^{\left (\frac {a x + b}{x}\right )} + a^{3} e^{\left (-\frac {a x + b}{x}\right )} + 3 \, a^{2} e^{\left (\frac {a x + b}{x}\right )} - \frac {3 \, {\left (a x + b\right )} a^{2} e^{\left (\frac {a x + b}{x}\right )}}{x} - 3 \, a^{2} e^{\left (-\frac {a x + b}{x}\right )} - \frac {3 \, {\left (a x + b\right )} a^{2} e^{\left (-\frac {a x + b}{x}\right )}}{x} + 6 \, a e^{\left (\frac {a x + b}{x}\right )} + \frac {3 \, {\left (a x + b\right )}^{2} a e^{\left (\frac {a x + b}{x}\right )}}{x^{2}} - \frac {6 \, {\left (a x + b\right )} a e^{\left (\frac {a x + b}{x}\right )}}{x} + 6 \, a e^{\left (-\frac {a x + b}{x}\right )} + \frac {3 \, {\left (a x + b\right )}^{2} a e^{\left (-\frac {a x + b}{x}\right )}}{x^{2}} + \frac {6 \, {\left (a x + b\right )} a e^{\left (-\frac {a x + b}{x}\right )}}{x} - \frac {{\left (a x + b\right )}^{3} e^{\left (\frac {a x + b}{x}\right )}}{x^{3}} + \frac {3 \, {\left (a x + b\right )}^{2} e^{\left (\frac {a x + b}{x}\right )}}{x^{2}} - \frac {6 \, {\left (a x + b\right )} e^{\left (\frac {a x + b}{x}\right )}}{x} - \frac {{\left (a x + b\right )}^{3} e^{\left (-\frac {a x + b}{x}\right )}}{x^{3}} - \frac {3 \, {\left (a x + b\right )}^{2} e^{\left (-\frac {a x + b}{x}\right )}}{x^{2}} - \frac {6 \, {\left (a x + b\right )} e^{\left (-\frac {a x + b}{x}\right )}}{x} + 6 \, e^{\left (\frac {a x + b}{x}\right )} - 6 \, e^{\left (-\frac {a x + b}{x}\right )}}{2 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.02, size = 165, normalized size = 2.66 \[ -\frac {\left (a +\frac {b}{x}\right )^{3} \cosh \left (a +\frac {b}{x}\right )-3 \sinh \left (a +\frac {b}{x}\right ) \left (a +\frac {b}{x}\right )^{2}+6 \left (a +\frac {b}{x}\right ) \cosh \left (a +\frac {b}{x}\right )-6 \sinh \left (a +\frac {b}{x}\right )-3 a \left (\left (a +\frac {b}{x}\right )^{2} \cosh \left (a +\frac {b}{x}\right )-2 \sinh \left (a +\frac {b}{x}\right ) \left (a +\frac {b}{x}\right )+2 \cosh \left (a +\frac {b}{x}\right )\right )+3 a^{2} \left (\left (a +\frac {b}{x}\right ) \cosh \left (a +\frac {b}{x}\right )-\sinh \left (a +\frac {b}{x}\right )\right )-a^{3} \cosh \left (a +\frac {b}{x}\right )}{b^{4}} \]

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

________________________________________________________________________________________

maxima [C]  time = 0.39, size = 48, normalized size = 0.77 \[ -\frac {1}{8} \, b {\left (\frac {e^{\left (-a\right )} \Gamma \left (5, \frac {b}{x}\right )}{b^{5}} - \frac {e^{a} \Gamma \left (5, -\frac {b}{x}\right )}{b^{5}}\right )} - \frac {\sinh \left (a + \frac {b}{x}\right )}{4 \, x^{4}} \]

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

________________________________________________________________________________________

mupad [B]  time = 0.44, size = 85, normalized size = 1.37 \[ \frac {{\mathrm {e}}^{a+\frac {b}{x}}\,\left (\frac {3\,x}{2\,b^2}-\frac {1}{2\,b}-\frac {3\,x^2}{b^3}+\frac {3\,x^3}{b^4}\right )}{x^3}-\frac {{\mathrm {e}}^{-a-\frac {b}{x}}\,\left (\frac {3\,x}{2\,b^2}+\frac {1}{2\,b}+\frac {3\,x^2}{b^3}+\frac {3\,x^3}{b^4}\right )}{x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b/x)/x^5,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 4.90, size = 61, normalized size = 0.98 \[ \begin {cases} - \frac {\cosh {\left (a + \frac {b}{x} \right )}}{b x^{3}} + \frac {3 \sinh {\left (a + \frac {b}{x} \right )}}{b^{2} x^{2}} - \frac {6 \cosh {\left (a + \frac {b}{x} \right )}}{b^{3} x} + \frac {6 \sinh {\left (a + \frac {b}{x} \right )}}{b^{4}} & \text {for}\: b \neq 0 \\- \frac {\sinh {\relax (a )}}{4 x^{4}} & \text {otherwise} \end {cases} \]

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))

________________________________________________________________________________________