3.1.45 \(\int \sinh (a+\frac {b}{x^2}) \, dx\) [45]

Optimal. Leaf size=67 \[ -\frac {1}{2} \sqrt {b} e^{-a} \sqrt {\pi } \text {Erf}\left (\frac {\sqrt {b}}{x}\right )-\frac {1}{2} \sqrt {b} e^a \sqrt {\pi } \text {Erfi}\left (\frac {\sqrt {b}}{x}\right )+x \sinh \left (a+\frac {b}{x^2}\right ) \]

[Out]

x*sinh(a+b/x^2)-1/2*erf(b^(1/2)/x)*b^(1/2)*Pi^(1/2)/exp(a)-1/2*exp(a)*erfi(b^(1/2)/x)*b^(1/2)*Pi^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5410, 5434, 5407, 2235, 2236} \begin {gather*} -\frac {1}{2} \sqrt {\pi } e^{-a} \sqrt {b} \text {Erf}\left (\frac {\sqrt {b}}{x}\right )-\frac {1}{2} \sqrt {\pi } e^a \sqrt {b} \text {Erfi}\left (\frac {\sqrt {b}}{x}\right )+x \sinh \left (a+\frac {b}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sinh[a + b/x^2],x]

[Out]

-1/2*(Sqrt[b]*Sqrt[Pi]*Erf[Sqrt[b]/x])/E^a - (Sqrt[b]*E^a*Sqrt[Pi]*Erfi[Sqrt[b]/x])/2 + x*Sinh[a + b/x^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 5410

Int[((a_.) + (b_.)*Sinh[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> -Subst[Int[(a + b*Sinh[c + d/x^n])^p/x^2
, x], x, 1/x] /; FreeQ[{a, b, c, d}, x] && ILtQ[n, 0] && IntegerQ[p]

Rule 5434

Int[((e_.)*(x_))^(m_)*Sinh[(c_.) + (d_.)*(x_)^(n_)], x_Symbol] :> Simp[(e*x)^(m + 1)*(Sinh[c + d*x^n]/(e*(m +
1))), x] - Dist[d*(n/(e^n*(m + 1))), Int[(e*x)^(m + n)*Cosh[c + d*x^n], x], x] /; FreeQ[{c, d, e}, x] && IGtQ[
n, 0] && LtQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 70, normalized size = 1.04 \begin {gather*} x \cosh \left (\frac {b}{x^2}\right ) \sinh (a)-\frac {1}{2} \sqrt {b} \sqrt {\pi } \left (\text {Erf}\left (\frac {\sqrt {b}}{x}\right ) (\cosh (a)-\sinh (a))+\text {Erfi}\left (\frac {\sqrt {b}}{x}\right ) (\cosh (a)+\sinh (a))\right )+x \cosh (a) \sinh \left (\frac {b}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sinh[a + b/x^2],x]

[Out]

x*Cosh[b/x^2]*Sinh[a] - (Sqrt[b]*Sqrt[Pi]*(Erf[Sqrt[b]/x]*(Cosh[a] - Sinh[a]) + Erfi[Sqrt[b]/x]*(Cosh[a] + Sin
h[a])))/2 + x*Cosh[a]*Sinh[b/x^2]

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Maple [A]
time = 0.35, size = 70, normalized size = 1.04

method	result	size
risch	\(-\frac {\sqrt {\pi }\, \erf \left (\frac {\sqrt {b}}{x}\right ) {\mathrm e}^{-a} \sqrt {b}}{2}-\frac {{\mathrm e}^{-a} {\mathrm e}^{-\frac {b}{x^{2}}} x}{2}+\frac {{\mathrm e}^{a} {\mathrm e}^{\frac {b}{x^{2}}} x}{2}-\frac {{\mathrm e}^{a} b \sqrt {\pi }\, \erf \left (\frac {\sqrt {-b}}{x}\right )}{2 \sqrt {-b}}\)	\(70\)
meijerg	\(\frac {i \sqrt {\pi }\, \cosh \left (a \right ) \sqrt {2}\, \sqrt {i b}\, \left (\frac {2 x \sqrt {2}\, \sqrt {i b}\, {\mathrm e}^{-\frac {b}{x^{2}}}}{\sqrt {\pi }\, b}-\frac {2 x \sqrt {2}\, \sqrt {i b}\, {\mathrm e}^{\frac {b}{x^{2}}}}{\sqrt {\pi }\, b}+\frac {2 \sqrt {i b}\, \sqrt {2}\, \erf \left (\frac {\sqrt {b}}{x}\right )}{\sqrt {b}}+\frac {2 \sqrt {i b}\, \sqrt {2}\, \erfi \left (\frac {\sqrt {b}}{x}\right )}{\sqrt {b}}\right )}{8}-\frac {\sqrt {\pi }\, \sinh \left (a \right ) \sqrt {2}\, \sqrt {i b}\, \left (-\frac {2 x \sqrt {2}\, {\mathrm e}^{\frac {b}{x^{2}}}}{\sqrt {\pi }\, \sqrt {i b}}-\frac {2 x \sqrt {2}\, {\mathrm e}^{-\frac {b}{x^{2}}}}{\sqrt {\pi }\, \sqrt {i b}}-\frac {2 \sqrt {2}\, \sqrt {b}\, \erf \left (\frac {\sqrt {b}}{x}\right )}{\sqrt {i b}}+\frac {2 \sqrt {2}\, \sqrt {b}\, \erfi \left (\frac {\sqrt {b}}{x}\right )}{\sqrt {i b}}\right )}{8}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 71, normalized size = 1.06 \begin {gather*} -\frac {1}{2} \, b {\left (\frac {\sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {\frac {b}{x^{2}}}\right ) - 1\right )} e^{\left (-a\right )}}{x \sqrt {\frac {b}{x^{2}}}} + \frac {\sqrt {\pi } {\left (\operatorname {erf}\left (\sqrt {-\frac {b}{x^{2}}}\right ) - 1\right )} e^{a}}{x \sqrt {-\frac {b}{x^{2}}}}\right )} + x \sinh \left (a + \frac {b}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*(sqrt(pi)*(erf(sqrt(b/x^2)) - 1)*e^(-a)/(x*sqrt(b/x^2)) + sqrt(pi)*(erf(sqrt(-b/x^2)) - 1)*e^a/(x*sqrt(
-b/x^2))) + x*sinh(a + b/x^2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (49) = 98\).
time = 0.41, size = 228, normalized size = 3.40 \begin {gather*} \frac {x \cosh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} + \sqrt {\pi } {\left (\cosh \left (a\right ) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (a\right ) + {\left (\cosh \left (a\right ) + \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 (\cosh \left (a\right ) \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) - \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (a\right ) + {\left (\cosh \left (a\right ) - \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 ) + 2 \, x \cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) \sinh \left (\frac {a x^{2} + b}{x^{2}}\right ) + x \sinh \left (\frac {a x^{2} + b}{x^{2}}\right )^{2} - x}{2 \, {\left (\cosh \left (\frac {a x^{2} + b}{x^{2}}\right ) + \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, algorithm="fricas")

[Out]

1/2*(x*cosh((a*x^2 + b)/x^2)^2 + sqrt(pi)*(cosh(a)*cosh((a*x^2 + b)/x^2) + cosh((a*x^2 + b)/x^2)*sinh(a) + (co
sh(a) + sinh(a))*sinh((a*x^2 + b)/x^2))*sqrt(-b)*erf(sqrt(-b)/x) - sqrt(pi)*(cosh(a)*cosh((a*x^2 + b)/x^2) - c
osh((a*x^2 + b)/x^2)*sinh(a) + (cosh(a) - sinh(a))*sinh((a*x^2 + b)/x^2))*sqrt(b)*erf(sqrt(b)/x) + 2*x*cosh((a
*x^2 + b)/x^2)*sinh((a*x^2 + b)/x^2) + x*sinh((a*x^2 + b)/x^2)^2 - x)/(cosh((a*x^2 + b)/x^2) + 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 \sinh {\left (a + \frac {b}{x^{2}} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sinh(a + b/x**2), 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, algorithm="giac")

[Out]

integrate(sinh(a + b/x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(a + b/x^2),x)

[Out]

int(sinh(a + b/x^2), x)

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