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)/(-cosh(x)^2)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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*}

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Mathematica [A]  time = 0.01, 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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fricas [C]  time = 1.03, size = 13, normalized size = 0.81 \[ \log \left (e^{x} + i\right ) - \log \left (e^{x} - i\right ) \]

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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giac [C]  time = 0.14, size = 5, normalized size = 0.31 \[ -2 i \, \arctan \left (e^{x}\right ) \]

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)

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maple [B]  time = 0.07, size = 34, normalized size = 2.12 \[ -\frac {\cosh \relax (x ) \sqrt {-\left (\sinh ^{2}\relax (x )\right )}\, \arctanh \left (\frac {1}{\sqrt {-\left (\sinh ^{2}\relax (x )\right )}}\right )}{\sinh \relax (x ) \sqrt {-\left (\cosh ^{2}\relax (x )\right )}} \]

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 = 0.59, size = 5, normalized size = 0.31 \[ -2 i \, \arctan \left (e^{x}\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \[ \int \frac {1}{\sqrt {-{\mathrm {sinh}\relax (x)}^2-1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(- sinh(x)^2 - 1)^(1/2),x)

[Out]

int(1/(- sinh(x)^2 - 1)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- \sinh ^{2}{\relax (x )} - 1}}\, dx \]

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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