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]

ln(sinh(x))*tanh(x)/(-tanh(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 = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 3475

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

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 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]

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*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \[ \frac {\tanh (x) \log (\sinh (x))}{\sqrt {-\tanh ^2(x)}} \]

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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fricas [A]  time = 0.49, size = 1, normalized size = 0.06 \[ 0 \]

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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giac [C]  time = 0.14, size = 37, normalized size = 2.31 \[ -\frac {i \, x}{\mathrm {sgn}\left (-e^{\left (4 \, x\right )} + 1\right )} + \frac {i \, \log \left (-i \, e^{\left (2 \, x\right )} + i\right )}{\mathrm {sgn}\left (-e^{\left (4 \, x\right )} + 1\right )} \]

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)

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maple [B]  time = 0.30, size = 81, normalized size = 5.06 \[ -\frac {\left ({\mathrm e}^{2 x}-1\right ) x}{\sqrt {-\frac {\left ({\mathrm e}^{2 x}-1\right )^{2}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )}+\frac {\left ({\mathrm e}^{2 x}-1\right ) \ln \left ({\mathrm e}^{2 x}-1\right )}{\sqrt {-\frac {\left ({\mathrm e}^{2 x}-1\right )^{2}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )} \]

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

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maxima [C]  time = 0.44, size = 22, normalized size = 1.38 \[ i \, x + i \, \log \left (e^{\left (-x\right )} + 1\right ) + i \, \log \left (e^{\left (-x\right )} - 1\right ) \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1/cosh(x)^2 - 1)^(1/2),x)

[Out]

int(1/(1/cosh(x)^2 - 1)^(1/2), x)

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

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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