3.183 \(\int \frac {\coth (x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=54 \[ -\frac {a \log (a+b \cosh (x))}{a^2-b^2}+\frac {\log (1-\cosh (x))}{2 (a+b)}+\frac {\log (\cosh (x)+1)}{2 (a-b)} \]

[Out]

1/2*ln(1-cosh(x))/(a+b)+1/2*ln(1+cosh(x))/(a-b)-a*ln(a+b*cosh(x))/(a^2-b^2)

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Rubi [A]  time = 0.07, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2721, 801} \[ -\frac {a \log (a+b \cosh (x))}{a^2-b^2}+\frac {\log (1-\cosh (x))}{2 (a+b)}+\frac {\log (\cosh (x)+1)}{2 (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]/(a + b*Cosh[x]),x]

[Out]

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 38, normalized size = 0.70 \[ -\frac {a \log (a+b \cosh (x))-a \log (\sinh (x))+b \log \left (\tanh \left (\frac {x}{2}\right )\right )}{a^2-b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]/(a + b*Cosh[x]),x]

[Out]

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

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fricas [A]  time = 0.56, size = 60, normalized size = 1.11 \[ -\frac {a \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) - {\left (a + b\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) - {\left (a - b\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right )}{a^{2} - b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.12, size = 67, normalized size = 1.24 \[ -\frac {a b \log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{a^{2} b - b^{3}} + \frac {\log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{2 \, {\left (a - b\right )}} + \frac {\log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{2 \, {\left (a + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*b*log(abs(b*(e^(-x) + e^x) + 2*a))/(a^2*b - b^3) + 1/2*log(e^(-x) + e^x + 2)/(a - b) + 1/2*log(e^(-x) + e^x
- 2)/(a + b)

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maple [A]  time = 0.09, size = 53, normalized size = 0.98 \[ -\frac {a \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right )}{\left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a +b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)/(a+b*cosh(x)),x)

[Out]

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

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maxima [A]  time = 0.30, size = 59, normalized size = 1.09 \[ -\frac {a \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{a^{2} - b^{2}} + \frac {\log \left (e^{\left (-x\right )} + 1\right )}{a - b} + \frac {\log \left (e^{\left (-x\right )} - 1\right )}{a + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.43, size = 148, normalized size = 2.74 \[ \frac {\ln \left (128\,a\,b-128\,a^2-32\,b^2+128\,a^2\,{\mathrm {e}}^x+32\,b^2\,{\mathrm {e}}^x-128\,a\,b\,{\mathrm {e}}^x\right )}{a+b}+\frac {\ln \left (-128\,a\,b-128\,a^2-32\,b^2-128\,a^2\,{\mathrm {e}}^x-32\,b^2\,{\mathrm {e}}^x-128\,a\,b\,{\mathrm {e}}^x\right )}{a-b}-\frac {a\,\ln \left (16\,a^2\,b-4\,b^3\,{\mathrm {e}}^{2\,x}-4\,b^3+32\,a^3\,{\mathrm {e}}^x-8\,a\,b^2\,{\mathrm {e}}^x+16\,a^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^2-b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)/(a + b*cosh(x)),x)

[Out]

log(128*a*b - 128*a^2 - 32*b^2 + 128*a^2*exp(x) + 32*b^2*exp(x) - 128*a*b*exp(x))/(a + b) + log(- 128*a*b - 12
8*a^2 - 32*b^2 - 128*a^2*exp(x) - 32*b^2*exp(x) - 128*a*b*exp(x))/(a - b) - (a*log(16*a^2*b - 4*b^3*exp(2*x) -
4*b^3 + 32*a^3*exp(x) - 8*a*b^2*exp(x) + 16*a^2*b*exp(2*x)))/(a^2 - b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(a+b*cosh(x)),x)

[Out]

Integral(coth(x)/(a + b*cosh(x)), x)

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