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

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

[Out]

(Log[Sinh[x]]*Tanh[x])/Sqrt[-Tanh[x]^2]

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Rubi [A]  time = 0.0219229, 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} \[ \frac{\tanh (x) \log (\sinh (x))}{\sqrt{-\tanh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-1 + Sech[x]^2],x]

[Out]

(Log[Sinh[x]]*Tanh[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 \frac{1}{\sqrt{-1+\text{sech}^2(x)}} \, dx &=\int \frac{1}{\sqrt{-\tanh ^2(x)}} \, dx\\ &=\frac{\tanh (x) \int \coth (x) \, dx}{\sqrt{-\tanh ^2(x)}}\\ &=\frac{\log (\sinh (x)) \tanh (x)}{\sqrt{-\tanh ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0086223, size = 16, normalized size = 1. \[ \frac{\tanh (x) \log (\sinh (x))}{\sqrt{-\tanh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-1 + Sech[x]^2],x]

[Out]

(Log[Sinh[x]]*Tanh[x])/Sqrt[-Tanh[x]^2]

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Maple [B]  time = 0.101, size = 81, normalized size = 5.1 \begin{align*} -{\frac{ \left ({{\rm e}^{2\,x}}-1 \right ) x}{{{\rm e}^{2\,x}}+1}{\frac{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}{\frac{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/(-1+sech(x)^2)^(1/2),x)

[Out]

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

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Maxima [C]  time = 1.70369, size = 30, normalized size = 1.88 \begin{align*} i \, x + i \, \log \left (e^{\left (-x\right )} + 1\right ) + i \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

I*x + I*log(e^(-x) + 1) + I*log(e^(-x) - 1)

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Fricas [A]  time = 1.96996, size = 4, normalized size = 0.25 \begin{align*} 0 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-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 \frac{1}{\sqrt{\operatorname{sech}^{2}{\left (x \right )} - 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(sech(x)**2 - 1), x)

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Giac [C]  time = 1.14028, size = 50, normalized size = 3.12 \begin{align*} -\frac{i \, x}{\mathrm{sgn}\left (-e^{\left (4 \, x\right )} + 1\right )} + \frac{i \, \log \left (-i \, e^{\left (2 \, x\right )} + i\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/(-1+sech(x)^2)^(1/2),x, algorithm="giac")

[Out]

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