3.128 \(\int \frac{1}{\sqrt{-1-\sinh ^2(x)}} \, dx\)

Optimal. Leaf size=16 \[ \frac{\cosh (x) \tan ^{-1}(\sinh (x))}{\sqrt{-\cosh ^2(x)}} \]

[Out]

(ArcTan[Sinh[x]]*Cosh[x])/Sqrt[-Cosh[x]^2]

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Rubi [A]  time = 0.0237084, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3176, 3207, 3770} \[ \frac{\cosh (x) \tan ^{-1}(\sinh (x))}{\sqrt{-\cosh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-1 - Sinh[x]^2],x]

[Out]

(ArcTan[Sinh[x]]*Cosh[x])/Sqrt[-Cosh[x]^2]

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{-1-\sinh ^2(x)}} \, dx &=\int \frac{1}{\sqrt{-\cosh ^2(x)}} \, dx\\ &=\frac{\cosh (x) \int \text{sech}(x) \, dx}{\sqrt{-\cosh ^2(x)}}\\ &=\frac{\tan ^{-1}(\sinh (x)) \cosh (x)}{\sqrt{-\cosh ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0079731, size = 21, normalized size = 1.31 \[ \frac{2 \cosh (x) \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )}{\sqrt{-\cosh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-1 - Sinh[x]^2],x]

[Out]

(2*ArcTan[Tanh[x/2]]*Cosh[x])/Sqrt[-Cosh[x]^2]

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Maple [B]  time = 0.04, size = 34, normalized size = 2.1 \begin{align*} -{\frac{\cosh \left ( x \right ) }{\sinh \left ( x \right ) }\sqrt{- \left ( \sinh \left ( x \right ) \right ) ^{2}}{\it Artanh} \left ({\frac{1}{\sqrt{- \left ( \sinh \left ( x \right ) \right ) ^{2}}}} \right ){\frac{1}{\sqrt{- \left ( \cosh \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-1-sinh(x)^2)^(1/2),x)

[Out]

-cosh(x)*(-sinh(x)^2)^(1/2)*arctanh(1/(-sinh(x)^2)^(1/2))/sinh(x)/(-cosh(x)^2)^(1/2)

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Maxima [C]  time = 1.92463, size = 7, normalized size = 0.44 \begin{align*} -2 i \, \arctan \left (e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1-sinh(x)^2)^(1/2),x, algorithm="maxima")

[Out]

-2*I*arctan(e^x)

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Fricas [C]  time = 1.77536, size = 39, normalized size = 2.44 \begin{align*} \log \left (e^{x} + i\right ) - \log \left (e^{x} - i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1-sinh(x)^2)^(1/2),x, algorithm="fricas")

[Out]

log(e^x + I) - log(e^x - I)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{- \sinh ^{2}{\left (x \right )} - 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1-sinh(x)**2)**(1/2),x)

[Out]

Integral(1/sqrt(-sinh(x)**2 - 1), x)

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Giac [C]  time = 1.15006, size = 7, normalized size = 0.44 \begin{align*} -2 i \, \arctan \left (e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-1-sinh(x)^2)^(1/2),x, algorithm="giac")

[Out]

-2*I*arctan(e^x)