3.987 \(\int \frac {\text {sech}^2(x)}{-\tanh ^2(x)+\tanh ^3(x)} \, dx\)

Optimal. Leaf size=15 \[ \coth (x)+\log (1-\tanh (x))-\log (\tanh (x)) \]

[Out]

coth(x)+ln(1-tanh(x))-ln(tanh(x))

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Rubi [A]  time = 0.06, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4342, 44} \[ \coth (x)+\log (1-\tanh (x))-\log (\tanh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(-Tanh[x]^2 + Tanh[x]^3),x]

[Out]

Coth[x] + Log[1 - Tanh[x]] - Log[Tanh[x]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 4342

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^2, x_Symbol] :> With[{d = FreeFactors[Tan[c*(a + b*x)], x]}, Dist[d/
(b*c), Subst[Int[SubstFor[1, Tan[c*(a + b*x)]/d, u, x], x], x, Tan[c*(a + b*x)]/d], x] /; FunctionOfQ[Tan[c*(a
+ b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && NonsumQ[u] && (EqQ[F, Sec] || EqQ[F, sec])

Rubi steps

\begin {align*} \int \frac {\text {sech}^2(x)}{-\tanh ^2(x)+\tanh ^3(x)} \, dx &=\operatorname {Subst}\left (\int \frac {1}{(-1+x) x^2} \, dx,x,\tanh (x)\right )\\ &=\operatorname {Subst}\left (\int \left (\frac {1}{-1+x}-\frac {1}{x^2}-\frac {1}{x}\right ) \, dx,x,\tanh (x)\right )\\ &=\coth (x)+\log (1-\tanh (x))-\log (\tanh (x))\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 11, normalized size = 0.73 \[ -x+\coth (x)-\log (\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^2/(-Tanh[x]^2 + Tanh[x]^3),x]

[Out]

-x + Coth[x] - Log[Sinh[x]]

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fricas [B]  time = 0.42, size = 53, normalized size = 3.53 \[ -\frac {{\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} - 1\right )} \log \left (\frac {2 \, \sinh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right ) - 2}{\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(-tanh(x)^2+tanh(x)^3),x, algorithm="fricas")

[Out]

-((cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*log(2*sinh(x)/(cosh(x) - sinh(x))) - 2)/(cosh(x)^2 + 2*cosh(
x)*sinh(x) + sinh(x)^2 - 1)

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giac [A]  time = 0.12, size = 26, normalized size = 1.73 \[ \frac {e^{\left (2 \, x\right )} + 1}{e^{\left (2 \, x\right )} - 1} - \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(-tanh(x)^2+tanh(x)^3),x, algorithm="giac")

[Out]

(e^(2*x) + 1)/(e^(2*x) - 1) - log(abs(e^(2*x) - 1))

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maple [B]  time = 0.24, size = 32, normalized size = 2.13 \[ \frac {\tanh \left (\frac {x}{2}\right )}{2}+2 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+\frac {1}{2 \tanh \left (\frac {x}{2}\right )}-\ln \left (\tanh \left (\frac {x}{2}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2/(-tanh(x)^2+tanh(x)^3),x)

[Out]

1/2*tanh(1/2*x)+2*ln(tanh(1/2*x)-1)+1/2/tanh(1/2*x)-ln(tanh(1/2*x))

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maxima [B]  time = 0.30, size = 32, normalized size = 2.13 \[ -2 \, x - \frac {2}{e^{\left (-2 \, x\right )} - 1} - \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(-tanh(x)^2+tanh(x)^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.66, size = 20, normalized size = 1.33 \[ \frac {2}{{\mathrm {e}}^{2\,x}-1}-\ln \left ({\mathrm {e}}^{2\,x}-1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(cosh(x)^2*(tanh(x)^2 - tanh(x)^3)),x)

[Out]

2/(exp(2*x) - 1) - log(exp(2*x) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2/(-tanh(x)**2+tanh(x)**3),x)

[Out]

Integral(sech(x)**2/((tanh(x) - 1)*tanh(x)**2), x)

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