Integrand size = 12, antiderivative size = 75 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x}+\frac {e^{-a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{3/2}}+\frac {e^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{3/2}} \]
-1/2*cosh(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)
Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\frac {-4 \sqrt {b} \cosh \left (a+\frac {b}{x^2}\right )+\sqrt {\pi } x \text {erf}\left (\frac {\sqrt {b}}{x}\right ) (\cosh (a)-\sinh (a))+\sqrt {\pi } x \text {erfi}\left (\frac {\sqrt {b}}{x}\right ) (\cosh (a)+\sinh (a))}{8 b^{3/2} x} \]
(-4*Sqrt[b]*Cosh[a + b/x^2] + Sqrt[Pi]*x*Erf[Sqrt[b]/x]*(Cosh[a] - Sinh[a] ) + Sqrt[Pi]*x*Erfi[Sqrt[b]/x]*(Cosh[a] + Sinh[a]))/(8*b^(3/2)*x)
Time = 0.34 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {5869, 5847, 5822, 2633, 2634}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 5869 |
\(\displaystyle -\int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^2}d\frac {1}{x}\) |
\(\Big \downarrow \) 5847 |
\(\displaystyle \frac {\int \cosh \left (a+\frac {b}{x^2}\right )d\frac {1}{x}}{2 b}-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x}\) |
\(\Big \downarrow \) 5822 |
\(\displaystyle \frac {\frac {1}{2} \int e^{-a-\frac {b}{x^2}}d\frac {1}{x}+\frac {1}{2} \int e^{a+\frac {b}{x^2}}d\frac {1}{x}}{2 b}-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x}\) |
\(\Big \downarrow \) 2633 |
\(\displaystyle \frac {\frac {1}{2} \int e^{-a-\frac {b}{x^2}}d\frac {1}{x}+\frac {\sqrt {\pi } e^a \text {erfi}\left (\frac {\sqrt {b}}{x}\right )}{4 \sqrt {b}}}{2 b}-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x}\) |
\(\Big \downarrow \) 2634 |
\(\displaystyle \frac {\frac {\sqrt {\pi } e^{-a} \text {erf}\left (\frac {\sqrt {b}}{x}\right )}{4 \sqrt {b}}+\frac {\sqrt {\pi } e^a \text {erfi}\left (\frac {\sqrt {b}}{x}\right )}{4 \sqrt {b}}}{2 b}-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x}\) |
-1/2*Cosh[a + b/x^2]/(b*x) + ((Sqrt[Pi]*Erf[Sqrt[b]/x])/(4*Sqrt[b]*E^a) + (E^a*Sqrt[Pi]*Erfi[Sqrt[b]/x])/(4*Sqrt[b]))/(2*b)
3.1.49.3.1 Defintions of rubi rules used
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]
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] /; Fr eeQ[{F, a, b, c, d}, x] && NegQ[b]
Int[Cosh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[1/2 Int[E^(c + d*x^n ), x], x] + Simp[1/2 Int[E^(-c - d*x^n), x], x] /; FreeQ[{c, d}, x] && IG tQ[n, 1]
Int[((e_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[e^( n - 1)*(e*x)^(m - n + 1)*(Cosh[c + d*x^n]/(d*n)), x] - Simp[e^n*((m - n + 1 )/(d*n)) Int[(e*x)^(m - n)*Cosh[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[n, 0] && LtQ[0, n, m + 1]
Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbo l] :> -Subst[Int[(a + b*Sinh[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]
Time = 0.60 (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 | \(-\frac {i \sqrt {\pi }\, \cosh \left (a \right ) \sqrt {2}\, \sqrt {i b}\, \left (\frac {\sqrt {2}\, \left (i b \right )^{\frac {5}{2}} {\mathrm e}^{-\frac {b}{x^{2}}}}{4 \sqrt {\pi }\, x \,b^{2}}+\frac {\sqrt {2}\, \left (i b \right )^{\frac {5}{2}} {\mathrm e}^{\frac {b}{x^{2}}}}{4 \sqrt {\pi }\, x \,b^{2}}-\frac {\left (i b \right )^{\frac {5}{2}} \sqrt {2}\, \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{\frac {5}{2}}}-\frac {\left (i b \right )^{\frac {5}{2}} \sqrt {2}\, \operatorname {erfi}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{\frac {5}{2}}}\right )}{2 b^{2}}+\frac {\sqrt {\pi }\, \sinh \left (a \right ) \sqrt {2}\, \sqrt {i b}\, \left (\frac {\sqrt {2}\, \left (i b \right )^{\frac {3}{2}} {\mathrm e}^{\frac {b}{x^{2}}}}{4 \sqrt {\pi }\, x b}-\frac {\sqrt {2}\, \left (i b \right )^{\frac {3}{2}} {\mathrm e}^{-\frac {b}{x^{2}}}}{4 \sqrt {\pi }\, x b}+\frac {\left (i b \right )^{\frac {3}{2}} \sqrt {2}\, \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{\frac {3}{2}}}-\frac {\left (i b \right )^{\frac {3}{2}} \sqrt {2}\, \operatorname {erfi}\left (\frac {\sqrt {b}}{x}\right )}{8 b^{\frac {3}{2}}}\right )}{2 b^{2}}\) | \(237\) |
-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*Pi^(1/2)/(-b)^(1/2)*erf((-b)^(1/2)/x)
Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (55) = 110\).
Time = 0.25 (sec) , antiderivative size = 251, normalized size of antiderivative = 3.35 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=-\frac {2 \, b \cosh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} + \sqrt {\pi } {\left (x \cosh \left (a\right ) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + x \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (a\right ) + {\left (x \cosh \left (a\right ) + x \sinh \left (a\right )\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {-b} \operatorname {erf}\left (\frac {\sqrt {-b}}{x}\right ) - \sqrt {\pi } {\left (x \cosh \left (a\right ) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) - x \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (a\right ) + {\left (x \cosh \left (a\right ) - x \sinh \left (a\right )\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {b} \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right ) + 4 \, b \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) + 2 \, b \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} + 2 \, b}{8 \, {\left (b^{2} x \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + b^{2} x \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )}} \]
-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))
\[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\int \frac {\sinh {\left (a + \frac {b}{x^{2}} \right )}}{x^{4}}\, dx \]
Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.83 \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=-\frac {1}{6} \, b {\left (\frac {e^{\left (-a\right )} \Gamma \left (\frac {5}{2}, \frac {b}{x^{2}}\right )}{x^{5} \left (\frac {b}{x^{2}}\right )^{\frac {5}{2}}} + \frac {e^{a} \Gamma \left (\frac {5}{2}, -\frac {b}{x^{2}}\right )}{x^{5} \left (-\frac {b}{x^{2}}\right )^{\frac {5}{2}}}\right )} - \frac {\sinh \left (a + \frac {b}{x^{2}}\right )}{3 \, x^{3}} \]
-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*sinh(a + b/x^2)/x^3
\[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\int { \frac {\sinh \left (a + \frac {b}{x^{2}}\right )}{x^{4}} \,d x } \]
Timed out. \[ \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx=\int \frac {\mathrm {sinh}\left (a+\frac {b}{x^2}\right )}{x^4} \,d x \]