3.109 \(\int \frac {\text {sech}(x)}{1+\coth (x)} \, dx\)

Optimal. Leaf size=10 \[ -\sinh (x)+\cosh (x)+\tan ^{-1}(\sinh (x)) \]

[Out]

arctan(sinh(x))+cosh(x)-sinh(x)

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Rubi [A]  time = 0.11, antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.778, Rules used = {3518, 3108, 3107, 2638, 2592, 321, 203} \[ -\sinh (x)+\cosh (x)+\tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]/(1 + Coth[x]),x]

[Out]

ArcTan[Sinh[x]] + Cosh[x] - Sinh[x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 2638

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

Rule 3107

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_.), x_Symbol] :> Int[ExpandTrig[cos[c + d*x]^m*sin[c + d*x]^n*(a*cos[c + d*x] + b*sin[c +
d*x])^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IGtQ[p, 0]

Rule 3108

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[a^p*b^p, Int[(Cos[c + d*x]^m*Sin[c + d*x]^n)/(b*Cos[c + d*x] + a*Sin
[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[a^2 + b^2, 0] && ILtQ[p, 0]

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 16, normalized size = 1.60 \[ -\sinh (x)+\cosh (x)+2 \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]/(1 + Coth[x]),x]

[Out]

2*ArcTan[Tanh[x/2]] + Cosh[x] - Sinh[x]

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fricas [B]  time = 0.39, size = 23, normalized size = 2.30 \[ \frac {2 \, {\left (\cosh \relax (x) + \sinh \relax (x)\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) + 1}{\cosh \relax (x) + \sinh \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)/(1+coth(x)),x, algorithm="fricas")

[Out]

(2*(cosh(x) + sinh(x))*arctan(cosh(x) + sinh(x)) + 1)/(cosh(x) + sinh(x))

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giac [A]  time = 0.11, size = 10, normalized size = 1.00 \[ 2 \, \arctan \left (e^{x}\right ) + e^{\left (-x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)/(1+coth(x)),x, algorithm="giac")

[Out]

2*arctan(e^x) + e^(-x)

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maple [A]  time = 0.10, size = 19, normalized size = 1.90 \[ \frac {2}{\tanh \left (\frac {x}{2}\right )+1}+2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)/(1+coth(x)),x)

[Out]

2/(tanh(1/2*x)+1)+2*arctan(tanh(1/2*x))

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maxima [A]  time = 0.50, size = 12, normalized size = 1.20 \[ -2 \, \arctan \left (e^{\left (-x\right )}\right ) + e^{\left (-x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)/(1+coth(x)),x, algorithm="maxima")

[Out]

-2*arctan(e^(-x)) + e^(-x)

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mupad [B]  time = 1.28, size = 10, normalized size = 1.00 \[ {\mathrm {e}}^{-x}+2\,\mathrm {atan}\left ({\mathrm {e}}^x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)*(coth(x) + 1)),x)

[Out]

exp(-x) + 2*atan(exp(x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)/(1+coth(x)),x)

[Out]

Integral(sech(x)/(coth(x) + 1), x)

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