3.51 \(\int \frac {\sinh (a+\frac {b}{x^2})}{x^6} \, dx\)

Optimal. Leaf size=93 \[ \frac {3 \sqrt {\pi } e^{-a} \text {erf}\left (\frac {\sqrt {b}}{x}\right )}{16 b^{5/2}}-\frac {3 \sqrt {\pi } e^a \text {erfi}\left (\frac {\sqrt {b}}{x}\right )}{16 b^{5/2}}+\frac {3 \sinh \left (a+\frac {b}{x^2}\right )}{4 b^2 x}-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x^3} \]

[Out]

-1/2*cosh(a+b/x^2)/b/x^3+3/4*sinh(a+b/x^2)/b^2/x+3/16*erf(b^(1/2)/x)*Pi^(1/2)/b^(5/2)/exp(a)-3/16*exp(a)*erfi(
b^(1/2)/x)*Pi^(1/2)/b^(5/2)

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Rubi [A]  time = 0.07, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5346, 5324, 5325, 5298, 2204, 2205} \[ \frac {3 \sqrt {\pi } e^{-a} \text {Erf}\left (\frac {\sqrt {b}}{x}\right )}{16 b^{5/2}}-\frac {3 \sqrt {\pi } e^a \text {Erfi}\left (\frac {\sqrt {b}}{x}\right )}{16 b^{5/2}}+\frac {3 \sinh \left (a+\frac {b}{x^2}\right )}{4 b^2 x}-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x^3} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x^2]/x^6,x]

[Out]

-Cosh[a + b/x^2]/(2*b*x^3) + (3*Sqrt[Pi]*Erf[Sqrt[b]/x])/(16*b^(5/2)*E^a) - (3*E^a*Sqrt[Pi]*Erfi[Sqrt[b]/x])/(
16*b^(5/2)) + (3*Sinh[a + b/x^2])/(4*b^2*x)

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]

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 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 5325

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 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]

Rubi steps

\begin {align*} \int \frac {\sinh \left (a+\frac {b}{x^2}\right )}{x^6} \, dx &=-\operatorname {Subst}\left (\int x^4 \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^3}+\frac {3 \operatorname {Subst}\left (\int x^2 \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^3}+\frac {3 \sinh \left (a+\frac {b}{x^2}\right )}{4 b^2 x}-\frac {3 \operatorname {Subst}\left (\int \sinh \left (a+b x^2\right ) \, dx,x,\frac {1}{x}\right )}{4 b^2}\\ &=-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x^3}+\frac {3 \sinh \left (a+\frac {b}{x^2}\right )}{4 b^2 x}+\frac {3 \operatorname {Subst}\left (\int e^{-a-b x^2} \, dx,x,\frac {1}{x}\right )}{8 b^2}-\frac {3 \operatorname {Subst}\left (\int e^{a+b x^2} \, dx,x,\frac {1}{x}\right )}{8 b^2}\\ &=-\frac {\cosh \left (a+\frac {b}{x^2}\right )}{2 b x^3}+\frac {3 e^{-a} \sqrt {\pi } \text {erf}\left (\frac {\sqrt {b}}{x}\right )}{16 b^{5/2}}-\frac {3 e^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b}}{x}\right )}{16 b^{5/2}}+\frac {3 \sinh \left (a+\frac {b}{x^2}\right )}{4 b^2 x}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 97, normalized size = 1.04 \[ \frac {3 \sqrt {\pi } x^3 (\cosh (a)-\sinh (a)) \text {erf}\left (\frac {\sqrt {b}}{x}\right )-3 \sqrt {\pi } x^3 (\sinh (a)+\cosh (a)) \text {erfi}\left (\frac {\sqrt {b}}{x}\right )+4 \sqrt {b} \left (3 x^2 \sinh \left (a+\frac {b}{x^2}\right )-2 b \cosh \left (a+\frac {b}{x^2}\right )\right )}{16 b^{5/2} x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b/x^2]/x^6,x]

[Out]

(3*Sqrt[Pi]*x^3*Erf[Sqrt[b]/x]*(Cosh[a] - Sinh[a]) - 3*Sqrt[Pi]*x^3*Erfi[Sqrt[b]/x]*(Cosh[a] + Sinh[a]) + 4*Sq
rt[b]*(-2*b*Cosh[a + b/x^2] + 3*x^2*Sinh[a + b/x^2]))/(16*b^(5/2)*x^3)

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fricas [B]  time = 0.45, size = 313, normalized size = 3.37 \[ -\frac {6 \, b x^{2} - 2 \, {\left (3 \, b x^{2} - 2 \, b^{2}\right )} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} - 3 \, \sqrt {\pi } {\left (x^{3} \cosh \relax (a) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + x^{3} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \relax (a) + {\left (x^{3} \cosh \relax (a) + x^{3} \sinh \relax (a)\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {-b} \operatorname {erf}\left (\frac {\sqrt {-b}}{x}\right ) - 3 \, \sqrt {\pi } {\left (x^{3} \cosh \relax (a) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) - x^{3} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \relax (a) + {\left (x^{3} \cosh \relax (a) - x^{3} \sinh \relax (a)\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )} \sqrt {b} \operatorname {erf}\left (\frac {\sqrt {b}}{x}\right ) - 4 \, {\left (3 \, b x^{2} - 2 \, b^{2}\right )} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) - 2 \, {\left (3 \, b x^{2} - 2 \, b^{2}\right )} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} + 4 \, b^{2}}{16 \, {\left (b^{3} x^{3} \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + b^{3} x^{3} \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/x^6,x, algorithm="fricas")

[Out]

-1/16*(6*b*x^2 - 2*(3*b*x^2 - 2*b^2)*cosh((a*x^2 + b)/x^2)^2 - 3*sqrt(pi)*(x^3*cosh(a)*cosh((a*x^2 + b)/x^2) +
x^3*cosh((a*x^2 + b)/x^2)*sinh(a) + (x^3*cosh(a) + x^3*sinh(a))*sinh((a*x^2 + b)/x^2))*sqrt(-b)*erf(sqrt(-b)/
x) - 3*sqrt(pi)*(x^3*cosh(a)*cosh((a*x^2 + b)/x^2) - x^3*cosh((a*x^2 + b)/x^2)*sinh(a) + (x^3*cosh(a) - x^3*si
nh(a))*sinh((a*x^2 + b)/x^2))*sqrt(b)*erf(sqrt(b)/x) - 4*(3*b*x^2 - 2*b^2)*cosh((a*x^2 + b)/x^2)*sinh((a*x^2 +
b)/x^2) - 2*(3*b*x^2 - 2*b^2)*sinh((a*x^2 + b)/x^2)^2 + 4*b^2)/(b^3*x^3*cosh((a*x^2 + b)/x^2) + b^3*x^3*sinh(
(a*x^2 + b)/x^2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh \left (a + \frac {b}{x^{2}}\right )}{x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/x^6,x, algorithm="giac")

[Out]

integrate(sinh(a + b/x^2)/x^6, x)

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maple [A]  time = 0.06, size = 117, normalized size = 1.26 \[ -\frac {{\mathrm e}^{-a} {\mathrm e}^{-\frac {b}{x^{2}}}}{4 b \,x^{3}}-\frac {3 \,{\mathrm e}^{-a} {\mathrm e}^{-\frac {b}{x^{2}}}}{8 b^{2} x}+\frac {3 \,{\mathrm e}^{-a} \sqrt {\pi }\, \erf \left (\frac {\sqrt {b}}{x}\right )}{16 b^{\frac {5}{2}}}-\frac {{\mathrm e}^{a} {\mathrm e}^{\frac {b}{x^{2}}}}{4 x^{3} b}+\frac {3 \,{\mathrm e}^{a} {\mathrm e}^{\frac {b}{x^{2}}}}{8 b^{2} x}-\frac {3 \,{\mathrm e}^{a} \sqrt {\pi }\, \erf \left (\frac {\sqrt {-b}}{x}\right )}{16 b^{2} \sqrt {-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a+b/x^2)/x^6,x)

[Out]

-1/4*exp(-a)/b/x^3*exp(-b/x^2)-3/8*exp(-a)/b^2/x*exp(-b/x^2)+3/16*exp(-a)/b^(5/2)*Pi^(1/2)*erf(b^(1/2)/x)-1/4*
exp(a)*exp(b/x^2)/x^3/b+3/8*exp(a)/b^2*exp(b/x^2)/x-3/16*exp(a)/b^2*Pi^(1/2)/(-b)^(1/2)*erf((-b)^(1/2)/x)

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maxima [A]  time = 0.77, size = 62, normalized size = 0.67 \[ -\frac {1}{10} \, b {\left (\frac {e^{\left (-a\right )} \Gamma \left (\frac {7}{2}, \frac {b}{x^{2}}\right )}{x^{7} \left (\frac {b}{x^{2}}\right )^{\frac {7}{2}}} + \frac {e^{a} \Gamma \left (\frac {7}{2}, -\frac {b}{x^{2}}\right )}{x^{7} \left (-\frac {b}{x^{2}}\right )^{\frac {7}{2}}}\right )} - \frac {\sinh \left (a + \frac {b}{x^{2}}\right )}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x^2)/x^6,x, algorithm="maxima")

[Out]

-1/10*b*(e^(-a)*gamma(7/2, b/x^2)/(x^7*(b/x^2)^(7/2)) + e^a*gamma(7/2, -b/x^2)/(x^7*(-b/x^2)^(7/2))) - 1/5*sin
h(a + b/x^2)/x^5

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {sinh}\left (a+\frac {b}{x^2}\right )}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b/x^2)/x^6,x)

[Out]

int(sinh(a + b/x^2)/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sinh {\left (a + \frac {b}{x^{2}} \right )}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(a+b/x**2)/x**6,x)

[Out]

Integral(sinh(a + b/x**2)/x**6, x)

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