3.160 \(\int \frac {\text {csch}(x)}{a+a \cosh (x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {1}{2 (a \cosh (x)+a)}-\frac {\tanh ^{-1}(\cosh (x))}{2 a} \]

[Out]

-1/2*arctanh(cosh(x))/a+1/2/(a+a*cosh(x))

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Rubi [A]  time = 0.05, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2667, 44, 206} \[ \frac {1}{2 (a \cosh (x)+a)}-\frac {\tanh ^{-1}(\cosh (x))}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]/(a + a*Cosh[x]),x]

[Out]

-ArcTanh[Cosh[x]]/(2*a) + 1/(2*(a + a*Cosh[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 206

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

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 42, normalized size = 1.83 \[ \frac {1-2 \cosh ^2\left (\frac {x}{2}\right ) \left (\log \left (\cosh \left (\frac {x}{2}\right )\right )-\log \left (\sinh \left (\frac {x}{2}\right )\right )\right )}{2 a (\cosh (x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]/(a + a*Cosh[x]),x]

[Out]

(1 - 2*Cosh[x/2]^2*(Log[Cosh[x/2]] - Log[Sinh[x/2]]))/(2*a*(1 + Cosh[x]))

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fricas [B]  time = 1.75, size = 103, normalized size = 4.48 \[ -\frac {{\left (\cosh \relax (x)^{2} + 2 \, {\left (\cosh \relax (x) + 1\right )} \sinh \relax (x) + \sinh \relax (x)^{2} + 2 \, \cosh \relax (x) + 1\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) - {\left (\cosh \relax (x)^{2} + 2 \, {\left (\cosh \relax (x) + 1\right )} \sinh \relax (x) + \sinh \relax (x)^{2} + 2 \, \cosh \relax (x) + 1\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right ) - 2 \, \cosh \relax (x) - 2 \, \sinh \relax (x)}{2 \, {\left (a \cosh \relax (x)^{2} + a \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (a \cosh \relax (x) + a\right )} \sinh \relax (x) + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+a*cosh(x)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.15, size = 52, normalized size = 2.26 \[ -\frac {\log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{4 \, a} + \frac {\log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{4 \, a} + \frac {e^{\left (-x\right )} + e^{x} + 6}{4 \, a {\left (e^{\left (-x\right )} + e^{x} + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+a*cosh(x)),x, algorithm="giac")

[Out]

-1/4*log(e^(-x) + e^x + 2)/a + 1/4*log(e^(-x) + e^x - 2)/a + 1/4*(e^(-x) + e^x + 6)/(a*(e^(-x) + e^x + 2))

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maple [A]  time = 0.07, size = 23, normalized size = 1.00 \[ -\frac {\tanh ^{2}\left (\frac {x}{2}\right )}{4 a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)/(a+a*cosh(x)),x)

[Out]

-1/4/a*tanh(1/2*x)^2+1/2/a*ln(tanh(1/2*x))

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maxima [B]  time = 0.30, size = 47, normalized size = 2.04 \[ \frac {e^{\left (-x\right )}}{2 \, a e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a} - \frac {\log \left (e^{\left (-x\right )} + 1\right )}{2 \, a} + \frac {\log \left (e^{\left (-x\right )} - 1\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+a*cosh(x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.93, size = 51, normalized size = 2.22 \[ \frac {1}{a\,\left ({\mathrm {e}}^x+1\right )}-\frac {1}{a\,\left ({\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1\right )}-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {-a^2}}{a}\right )}{\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)*(a + a*cosh(x))),x)

[Out]

1/(a*(exp(x) + 1)) - 1/(a*(exp(2*x) + 2*exp(x) + 1)) - atan((exp(x)*(-a^2)^(1/2))/a)/(-a^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(a+a*cosh(x)),x)

[Out]

Integral(csch(x)/(cosh(x) + 1), x)/a

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