3.34 \(\int \frac {\cosh (a+\frac {b}{x^2})}{x^4} \, dx\)

Optimal. Leaf size=75 \[ -\frac {\sqrt {\pi } e^{-a} \text {erf}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{3/2}}+\frac {\sqrt {\pi } e^a \text {erfi}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{3/2}}-\frac {\sinh \left (a+\frac {b}{x^2}\right )}{2 b x} \]

[Out]

-1/2*sinh(a+b/x^2)/b/x-1/8*erf(b^(1/2)/x)*Pi^(1/2)/b^(3/2)/exp(a)+1/8*exp(a)*erfi(b^(1/2)/x)*Pi^(1/2)/b^(3/2)

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Rubi [A]  time = 0.05, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5347, 5325, 5298, 2204, 2205} \[ -\frac {\sqrt {\pi } e^{-a} \text {Erf}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{3/2}}+\frac {\sqrt {\pi } e^a \text {Erfi}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{3/2}}-\frac {\sinh \left (a+\frac {b}{x^2}\right )}{2 b x} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b/x^2]/x^4,x]

[Out]

-(Sqrt[Pi]*Erf[Sqrt[b]/x])/(8*b^(3/2)*E^a) + (E^a*Sqrt[Pi]*Erfi[Sqrt[b]/x])/(8*b^(3/2)) - Sinh[a + b/x^2]/(2*b
*x)

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 5298

Int[Sinh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Dist[1/2, Int[E^(c + d*x^n), x], x] - Dist[1/2, Int[E^(-c - d*
x^n), x], x] /; FreeQ[{c, d}, x] && IGtQ[n, 1]

Rule 5325

Int[Cosh[(c_.) + (d_.)*(x_)^(n_)]*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[(e^(n - 1)*(e*x)^(m - n + 1)*Sinh[c +
d*x^n])/(d*n), x] - Dist[(e^n*(m - n + 1))/(d*n), Int[(e*x)^(m - n)*Sinh[c + d*x^n], x], x] /; FreeQ[{c, d, e}
, x] && IGtQ[n, 0] && LtQ[0, n, m + 1]

Rule 5347

Int[((a_.) + Cosh[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> -Subst[Int[(a + b*Cosh[c + d/
x^n])^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, c, d}, x] && IntegerQ[p] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\cosh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx &=-\operatorname {Subst}\left (\int x^2 \cosh \left (a+b x^2\right ) \, dx,x,\frac {1}{x}\right )\\ &=-\frac {\sinh \left (a+\frac {b}{x^2}\right )}{2 b x}+\frac {\operatorname {Subst}\left (\int \sinh \left (a+b x^2\right ) \, dx,x,\frac {1}{x}\right )}{2 b}\\ &=-\frac {\sinh \left (a+\frac {b}{x^2}\right )}{2 b x}-\frac {\operatorname {Subst}\left (\int e^{-a-b x^2} \, dx,x,\frac {1}{x}\right )}{4 b}+\frac {\operatorname {Subst}\left (\int e^{a+b x^2} \, dx,x,\frac {1}{x}\right )}{4 b}\\ &=-\frac {e^{-a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{3/2}}+\frac {e^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{3/2}}-\frac {\sinh \left (a+\frac {b}{x^2}\right )}{2 b x}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 74, normalized size = 0.99 \[ \frac {\sqrt {\pi } x (\sinh (a)-\cosh (a)) \text {erf}\left (\frac {\sqrt {b}}{x}\right )+\sqrt {\pi } x (\sinh (a)+\cosh (a)) \text {erfi}\left (\frac {\sqrt {b}}{x}\right )-4 \sqrt {b} \sinh \left (a+\frac {b}{x^2}\right )}{8 b^{3/2} x} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b/x^2]/x^4,x]

[Out]

(Sqrt[Pi]*x*Erf[Sqrt[b]/x]*(-Cosh[a] + Sinh[a]) + Sqrt[Pi]*x*Erfi[Sqrt[b]/x]*(Cosh[a] + Sinh[a]) - 4*Sqrt[b]*S
inh[a + b/x^2])/(8*b^(3/2)*x)

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fricas [B]  time = 0.46, size = 250, normalized size = 3.33 \[ -\frac {2 \, b \cosh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} + \sqrt {\pi } {\left (x \cosh \relax (a) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + x \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \relax (a) + {\left (x \cosh \relax (a) + x \sinh \relax (a)\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {-b} \operatorname {erf}\left (\frac {\sqrt {-b}}{x}\right ) + \sqrt {\pi } {\left (x \cosh \relax (a) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) - x \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \relax (a) + {\left (x \cosh \relax (a) - x \sinh \relax (a)\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {b} \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right ) + 4 \, b \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) + 2 \, b \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} - 2 \, b}{8 \, {\left (b^{2} x \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + b^{2} x \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b/x^2)/x^4,x, algorithm="fricas")

[Out]

-1/8*(2*b*cosh((a*x^2 + b)/x^2)^2 + sqrt(pi)*(x*cosh(a)*cosh((a*x^2 + b)/x^2) + x*cosh((a*x^2 + b)/x^2)*sinh(a
) + (x*cosh(a) + x*sinh(a))*sinh((a*x^2 + b)/x^2))*sqrt(-b)*erf(sqrt(-b)/x) + sqrt(pi)*(x*cosh(a)*cosh((a*x^2
+ b)/x^2) - x*cosh((a*x^2 + b)/x^2)*sinh(a) + (x*cosh(a) - x*sinh(a))*sinh((a*x^2 + b)/x^2))*sqrt(b)*erf(sqrt(
b)/x) + 4*b*cosh((a*x^2 + b)/x^2)*sinh((a*x^2 + b)/x^2) + 2*b*sinh((a*x^2 + b)/x^2)^2 - 2*b)/(b^2*x*cosh((a*x^
2 + b)/x^2) + b^2*x*sinh((a*x^2 + b)/x^2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh \left (a + \frac {b}{x^{2}}\right )}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b/x^2)/x^4,x, algorithm="giac")

[Out]

integrate(cosh(a + b/x^2)/x^4, x)

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maple [A]  time = 0.13, size = 82, normalized size = 1.09 \[ \frac {{\mathrm e}^{-a} {\mathrm e}^{-\frac {b}{x^{2}}}}{4 b x}-\frac {{\mathrm e}^{-a} \sqrt {\pi }\, \erf \left (\frac {\sqrt {b}}{x}\right )}{8 b^{\frac {3}{2}}}-\frac {{\mathrm e}^{a} {\mathrm e}^{\frac {b}{x^{2}}}}{4 x b}+\frac {{\mathrm e}^{a} \sqrt {\pi }\, \erf \left (\frac {\sqrt {-b}}{x}\right )}{8 b \sqrt {-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a+b/x^2)/x^4,x)

[Out]

1/4*exp(-a)/b/x*exp(-b/x^2)-1/8*exp(-a)/b^(3/2)*Pi^(1/2)*erf(b^(1/2)/x)-1/4*exp(a)*exp(b/x^2)/x/b+1/8*exp(a)/b
*Pi^(1/2)/(-b)^(1/2)*erf((-b)^(1/2)/x)

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maxima [A]  time = 0.35, size = 63, normalized size = 0.84 \[ \frac {1}{6} \, b {\left (\frac {e^{\left (-a\right )} \Gamma \left (\frac {5}{2}, \frac {b}{x^{2}}\right )}{x^{5} \left (\frac {b}{x^{2}}\right )^{\frac {5}{2}}} - \frac {e^{a} \Gamma \left (\frac {5}{2}, -\frac {b}{x^{2}}\right )}{x^{5} \left (-\frac {b}{x^{2}}\right )^{\frac {5}{2}}}\right )} - \frac {\cosh \left (a + \frac {b}{x^{2}}\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b/x^2)/x^4,x, algorithm="maxima")

[Out]

1/6*b*(e^(-a)*gamma(5/2, b/x^2)/(x^5*(b/x^2)^(5/2)) - e^a*gamma(5/2, -b/x^2)/(x^5*(-b/x^2)^(5/2))) - 1/3*cosh(
a + b/x^2)/x^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {cosh}\left (a+\frac {b}{x^2}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b/x^2)/x^4,x)

[Out]

int(cosh(a + b/x^2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh {\left (a + \frac {b}{x^{2}} \right )}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(a+b/x**2)/x**4,x)

[Out]

Integral(cosh(a + b/x**2)/x**4, x)

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