\(\int \frac {\cosh (a+\frac {b}{x^2})}{x^4} \, dx\) [34]

Optimal result

Integrand size = 12, antiderivative size = 75 \[ \int \frac {\cosh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=-\frac {e^{-a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{3/2}}+\frac {e^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{3/2}}-\frac {\sinh \left (a+\frac {b}{x^2}\right )}{2 b x} \]

[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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5455, 5433, 5406, 2235, 2236} \[ \int \frac {\cosh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=-\frac {\sqrt {\pi } e^{-a} \text {erf}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{3/2}}+\frac {\sqrt {\pi } e^a \text {erfi}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{3/2}}-\frac {\sinh \left (a+\frac {b}{x^2}\right )}{2 b x} \]

[In]

Int[Cosh[a + b/x^2]/x^4,x]

[Out]

-1/8*(Sqrt[Pi]*Erf[Sqrt[b]/x])/(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 2235

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 2236

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 5406

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 5433

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 5455

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*} \text {integral}& = -\text {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 {\text {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 {\text {Subst}\left (\int e^{-a-b x^2} \, dx,x,\frac {1}{x}\right )}{4 b}+\frac {\text {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*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int \frac {\cosh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\frac {\sqrt {\pi } x \text {erf}\left (\frac {\sqrt {b}}{x}\right ) (-\cosh (a)+\sinh (a))+\sqrt {\pi } x \text {erfi}\left (\frac {\sqrt {b}}{x}\right ) (\cosh (a)+\sinh (a))-4 \sqrt {b} \sinh \left (a+\frac {b}{x^2}\right )}{8 b^{3/2} x} \]

[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)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09

method	result	size
risch	\(\frac {{\mathrm e}^{-a} {\mathrm e}^{-\frac {b}{x^{2}}}}{4 b x}-\frac {\operatorname {erf}\left (\frac {\sqrt {b}}{x}\right ) \sqrt {\pi }\, {\mathrm e}^{-a}}{8 b^{\frac {3}{2}}}-\frac {{\mathrm e}^{a} {\mathrm e}^{\frac {b}{x^{2}}}}{4 x b}+\frac {{\mathrm e}^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b}}{x}\right )}{8 b \sqrt {-b}}\)	\(82\)
meijerg	\(\frac {\sqrt {\pi }\, \cosh \left (a \right ) \sqrt {2}\, \sqrt {i b}\, \left (\frac {\sqrt {2}\, \left (i b \right )^{\frac {3}{2}} {\mathrm e}^{\frac {b}{x^{2}}}}{4 \sqrt {\pi }\, x b}-\frac {\sqrt {2}\, \left (i b \right )^{\frac {3}{2}} {\mathrm e}^{-\frac {b}{x^{2}}}}{4 \sqrt {\pi }\, x b}+\frac {\left (i b \right )^{\frac {3}{2}} \sqrt {2}\, \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{\frac {3}{2}}}-\frac {\left (i b \right )^{\frac {3}{2}} \sqrt {2}\, \operatorname {erfi}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{\frac {3}{2}}}\right )}{2 b^{2}}-\frac {i \sqrt {\pi }\, \sinh \left (a \right ) \sqrt {2}\, \sqrt {i b}\, \left (\frac {\sqrt {2}\, \left (i b \right )^{\frac {5}{2}} {\mathrm e}^{-\frac {b}{x^{2}}}}{4 \sqrt {\pi }\, x \,b^{2}}+\frac {\sqrt {2}\, \left (i b \right )^{\frac {5}{2}} {\mathrm e}^{\frac {b}{x^{2}}}}{4 \sqrt {\pi }\, x \,b^{2}}-\frac {\left (i b \right )^{\frac {5}{2}} \sqrt {2}\, \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{\frac {5}{2}}}-\frac {\left (i b \right )^{\frac {5}{2}} \sqrt {2}\, \operatorname {erfi}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{\frac {5}{2}}}\right )}{2 b^{2}}\)	\(237\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (55) = 110\).

Time = 0.25 (sec) , antiderivative size = 250, normalized size of antiderivative = 3.33 \[ \int \frac {\cosh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=-\frac {2 \, b \cosh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} + \sqrt {\pi } {\left (x \cosh \left (a\right ) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + x \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (a\right ) + {\left (x \cosh \left (a\right ) + x \sinh \left (a\right )\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {-b} \operatorname {erf}\left (\frac {\sqrt {-b}}{x}\right ) + \sqrt {\pi } {\left (x \cosh \left (a\right ) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) - x \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (a\right ) + {\left (x \cosh \left (a\right ) - x \sinh \left (a\right )\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {b} \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right ) + 4 \, b \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) + 2 \, b \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} - 2 \, b}{8 \, {\left (b^{2} x \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + b^{2} x \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )}} \]

[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))

Sympy [F]

\[ \int \frac {\cosh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\int \frac {\cosh {\left (a + \frac {b}{x^{2}} \right )}}{x^{4}}\, dx \]

[In]

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

[Out]

Integral(cosh(a + b/x**2)/x**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.84 \[ \int \frac {\cosh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\frac {1}{6} \, b {\left (\frac {e^{\left (-a\right )} \Gamma \left (\frac {5}{2}, \frac {b}{x^{2}}\right )}{x^{5} \left (\frac {b}{x^{2}}\right )^{\frac {5}{2}}} - \frac {e^{a} \Gamma \left (\frac {5}{2}, -\frac {b}{x^{2}}\right )}{x^{5} \left (-\frac {b}{x^{2}}\right )^{\frac {5}{2}}}\right )} - \frac {\cosh \left (a + \frac {b}{x^{2}}\right )}{3 \, x^{3}} \]

[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

Giac [F]

\[ \int \frac {\cosh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\int { \frac {\cosh \left (a + \frac {b}{x^{2}}\right )}{x^{4}} \,d x } \]

[In]

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

[Out]

integrate(cosh(a + b/x^2)/x^4, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cosh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\int \frac {\mathrm {cosh}\left (a+\frac {b}{x^2}\right )}{x^4} \,d x \]

[In]

int(cosh(a + b/x^2)/x^4,x)

[Out]

int(cosh(a + b/x^2)/x^4, x)