3.47 \(\int \frac{\sinh (a+\frac{b}{x^2})}{x^2} \, dx\)

Optimal. Leaf size=57 \[ \frac{\sqrt{\pi } e^{-a} \text{Erf}\left (\frac{\sqrt{b}}{x}\right )}{4 \sqrt{b}}-\frac{\sqrt{\pi } e^a \text{Erfi}\left (\frac{\sqrt{b}}{x}\right )}{4 \sqrt{b}} \]

[Out]

(Sqrt[Pi]*Erf[Sqrt[b]/x])/(4*Sqrt[b]*E^a) - (E^a*Sqrt[Pi]*Erfi[Sqrt[b]/x])/(4*Sqrt[b])

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Rubi [A]  time = 0.0311093, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5346, 5298, 2204, 2205} \[ \frac{\sqrt{\pi } e^{-a} \text{Erf}\left (\frac{\sqrt{b}}{x}\right )}{4 \sqrt{b}}-\frac{\sqrt{\pi } e^a \text{Erfi}\left (\frac{\sqrt{b}}{x}\right )}{4 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x^2]/x^2,x]

[Out]

(Sqrt[Pi]*Erf[Sqrt[b]/x])/(4*Sqrt[b]*E^a) - (E^a*Sqrt[Pi]*Erfi[Sqrt[b]/x])/(4*Sqrt[b])

Rule 5346

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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0343331, size = 50, normalized size = 0.88 \[ \frac{\sqrt{\pi } \left ((\cosh (a)-\sinh (a)) \text{Erf}\left (\frac{\sqrt{b}}{x}\right )-(\sinh (a)+\cosh (a)) \text{Erfi}\left (\frac{\sqrt{b}}{x}\right )\right )}{4 \sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b/x^2]/x^2,x]

[Out]

(Sqrt[Pi]*(Erf[Sqrt[b]/x]*(Cosh[a] - Sinh[a]) - Erfi[Sqrt[b]/x]*(Cosh[a] + Sinh[a])))/(4*Sqrt[b])

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Maple [A]  time = 0.027, size = 44, normalized size = 0.8 \begin{align*}{\frac{\sqrt{\pi }{{\rm e}^{-a}}}{4}{\it Erf} \left ({\frac{1}{x}\sqrt{b}} \right ){\frac{1}{\sqrt{b}}}}-{\frac{{{\rm e}^{a}}\sqrt{\pi }}{4}{\it Erf} \left ({\frac{1}{x}\sqrt{-b}} \right ){\frac{1}{\sqrt{-b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a+b/x^2)/x^2,x)

[Out]

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

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Maxima [A]  time = 1.17297, size = 84, normalized size = 1.47 \begin{align*} -\frac{1}{2} \, b{\left (\frac{e^{\left (-a\right )} \Gamma \left (\frac{3}{2}, \frac{b}{x^{2}}\right )}{x^{3} \left (\frac{b}{x^{2}}\right )^{\frac{3}{2}}} + \frac{e^{a} \Gamma \left (\frac{3}{2}, -\frac{b}{x^{2}}\right )}{x^{3} \left (-\frac{b}{x^{2}}\right )^{\frac{3}{2}}}\right )} - \frac{\sinh \left (a + \frac{b}{x^{2}}\right )}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.82332, size = 158, normalized size = 2.77 \begin{align*} \frac{\sqrt{\pi } \sqrt{-b}{\left (\cosh \left (a\right ) + \sinh \left (a\right )\right )} \operatorname{erf}\left (\frac{\sqrt{-b}}{x}\right ) + \sqrt{\pi } \sqrt{b}{\left (\cosh \left (a\right ) - \sinh \left (a\right )\right )} \operatorname{erf}\left (\frac{\sqrt{b}}{x}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b/x**2)/x**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (a + \frac{b}{x^{2}}\right )}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sinh(a + b/x^2)/x^2, x)