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

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

[Out]

-Cosh[a + b/x^2]/(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))

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Rubi [A]  time = 0.0480669, antiderivative size = 75, normalized size of antiderivative = 1., 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 = {5346, 5324, 5299, 2204, 2205} \[ \frac{\sqrt{\pi } e^{-a} \text{Erf}\left (\frac{\sqrt{b}}{x}\right )}{8 b^{3/2}}+\frac{\sqrt{\pi } e^a \text{Erfi}\left (\frac{\sqrt{b}}{x}\right )}{8 b^{3/2}}-\frac{\cosh \left (a+\frac{b}{x^2}\right )}{2 b x} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x^2]/x^4,x]

[Out]

-Cosh[a + b/x^2]/(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 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 5324

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 5299

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 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^4} \, dx &=-\operatorname{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{\operatorname{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{\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{\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*}

Mathematica [A]  time = 0.0710922, size = 74, normalized size = 0.99 \[ \frac{\sqrt{\pi } x (\cosh (a)-\sinh (a)) \text{Erf}\left (\frac{\sqrt{b}}{x}\right )+\sqrt{\pi } x (\sinh (a)+\cosh (a)) \text{Erfi}\left (\frac{\sqrt{b}}{x}\right )-4 \sqrt{b} \cosh \left (a+\frac{b}{x^2}\right )}{8 b^{3/2} x} \]

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 = 0.035, size = 82, normalized size = 1.1 \begin{align*} -{\frac{{{\rm e}^{-a}}}{4\,bx}{{\rm e}^{-{\frac{b}{{x}^{2}}}}}}+{\frac{{{\rm e}^{-a}}\sqrt{\pi }}{8}{\it Erf} \left ({\frac{1}{x}\sqrt{b}} \right ){b}^{-{\frac{3}{2}}}}-{\frac{{{\rm e}^{a}}}{4\,bx}{{\rm e}^{{\frac{b}{{x}^{2}}}}}}+{\frac{{{\rm e}^{a}}\sqrt{\pi }}{8\,b}{\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^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 = 1.26549, size = 84, normalized size = 1.12 \begin{align*} -\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{align*}

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]  time = 1.77908, size = 653, normalized size = 8.71 \begin{align*} -\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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh{\left (a + \frac{b}{x^{2}} \right )}}{x^{4}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\sinh \left (a + \frac{b}{x^{2}}\right )}{x^{4}}\,{d x} \end{align*}

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)