3.1.49 \(\int \frac {\sinh (a+\frac {b}{x^2})}{x^4} \, dx\) [49]

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

[Out]

-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)

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Rubi [A]
time = 0.04, 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 = {5454, 5432, 5407, 2235, 2236} \begin {gather*} \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 {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x^2]/x^4,x]

[Out]

-1/2*Cosh[a + b/x^2]/(b*x) + (Sqrt[Pi]*Erf[Sqrt[b]/x])/(8*b^(3/2)*E^a) + (E^a*Sqrt[Pi]*Erfi[Sqrt[b]/x])/(8*b^(
3/2))

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 5407

Int[Cosh[(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 5432

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] - Dist[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]

Rule 5454

Int[(x_)^(m_.)*((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -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]

Rubi steps

\begin {align*} \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^4} \, dx &=-\text {Subst}\left (\int x^2 \sinh \left (a+b x^2\right ) \, dx,x,\frac {1}{x}\right )\\ &=-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x}+\frac {\text {Subst}\left (\int \cosh \left (a+b x^2\right ) \, dx,x,\frac {1}{x}\right )}{2 b}\\ &=-\frac {\cosh \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 {\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}}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 74, normalized size = 0.99 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b/x^2]/x^4,x]

[Out]

(-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)

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Maple [A]
time = 0.35, size = 82, normalized size = 1.09

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.30, size = 62, normalized size = 0.83 \begin {gather*} -\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(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*sinh
(a + b/x^2)/x^3

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (55) = 110\).
time = 0.40, size = 251, normalized size = 3.35 \begin {gather*} -\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sinh {\left (a + \frac {b}{x^{2}} \right )}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b/x**2)/x**4, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(a + b/x^2)/x^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {sinh}\left (a+\frac {b}{x^2}\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b/x^2)/x^4,x)

[Out]

int(sinh(a + b/x^2)/x^4, x)

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