3.174 \(\int \sqrt{-1+\text{sech}^2(x)} \, dx\)

Optimal. Leaf size=16 \[ \sqrt{-\tanh ^2(x)} \coth (x) \log (\cosh (x)) \]

[Out]

Coth[x]*Log[Cosh[x]]*Sqrt[-Tanh[x]^2]

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Rubi [A]  time = 0.0195109, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {4121, 3658, 3475} \[ \sqrt{-\tanh ^2(x)} \coth (x) \log (\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + Sech[x]^2],x]

[Out]

Coth[x]*Log[Cosh[x]]*Sqrt[-Tanh[x]^2]

Rule 4121

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

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[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 3475

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

Rubi steps

\begin{align*} \int \sqrt{-1+\text{sech}^2(x)} \, dx &=\int \sqrt{-\tanh ^2(x)} \, dx\\ &=\left (\coth (x) \sqrt{-\tanh ^2(x)}\right ) \int \tanh (x) \, dx\\ &=\coth (x) \log (\cosh (x)) \sqrt{-\tanh ^2(x)}\\ \end{align*}

Mathematica [A]  time = 0.0064218, size = 16, normalized size = 1. \[ \sqrt{-\tanh ^2(x)} \coth (x) \log (\cosh (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + Sech[x]^2],x]

[Out]

Coth[x]*Log[Cosh[x]]*Sqrt[-Tanh[x]^2]

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Maple [B]  time = 0.122, size = 81, normalized size = 5.1 \begin{align*} -{\frac{ \left ({{\rm e}^{2\,x}}+1 \right ) x}{{{\rm e}^{2\,x}}-1}\sqrt{-{\frac{ \left ({{\rm e}^{2\,x}}-1 \right ) ^{2}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}}+{\frac{ \left ({{\rm e}^{2\,x}}+1 \right ) \ln \left ({{\rm e}^{2\,x}}+1 \right ) }{{{\rm e}^{2\,x}}-1}\sqrt{-{\frac{ \left ({{\rm e}^{2\,x}}-1 \right ) ^{2}}{ \left ({{\rm e}^{2\,x}}+1 \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1+sech(x)^2)^(1/2),x)

[Out]

-1/(exp(2*x)-1)*(exp(2*x)+1)*(-(exp(2*x)-1)^2/(exp(2*x)+1)^2)^(1/2)*x+1/(exp(2*x)-1)*(exp(2*x)+1)*(-(exp(2*x)-
1)^2/(exp(2*x)+1)^2)^(1/2)*ln(exp(2*x)+1)

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Maxima [C]  time = 1.70793, size = 18, normalized size = 1.12 \begin{align*} -i \, x - i \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*x - I*log(e^(-2*x) + 1)

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Fricas [A]  time = 1.97561, size = 4, normalized size = 0.25 \begin{align*} 0 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+sech(x)**2)**(1/2),x)

[Out]

Integral(sqrt(sech(x)**2 - 1), x)

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Giac [C]  time = 1.12261, size = 42, normalized size = 2.62 \begin{align*} i \, x \mathrm{sgn}\left (-e^{\left (4 \, x\right )} + 1\right ) - i \, \log \left (e^{\left (2 \, x\right )} + 1\right ) \mathrm{sgn}\left (-e^{\left (4 \, x\right )} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*x*sgn(-e^(4*x) + 1) - I*log(e^(2*x) + 1)*sgn(-e^(4*x) + 1)