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

Optimal. Leaf size=53 \[ \frac {b \log (a+b \cosh (x))}{a^2-b^2}+\frac {\log (1-\cosh (x))}{2 (a+b)}-\frac {\log (\cosh (x)+1)}{2 (a-b)} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2668, 706, 31, 633} \[ \frac {b \log (a+b \cosh (x))}{a^2-b^2}+\frac {\log (1-\cosh (x))}{2 (a+b)}-\frac {\log (\cosh (x)+1)}{2 (a-b)} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]/(a + b*Cosh[x]),x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 2668

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*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.07, size = 37, normalized size = 0.70 \[ \frac {b \log (a+b \cosh (x))+a \log \left (\tanh \left (\frac {x}{2}\right )\right )-b \log (\sinh (x))}{a^2-b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]/(a + b*Cosh[x]),x]

[Out]

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

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fricas [A]  time = 0.62, size = 58, normalized size = 1.09 \[ \frac {b \log \left (\frac {2 \, {\left (b \cosh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) - {\left (a + b\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) + 1\right ) + {\left (a - b\right )} \log \left (\cosh \relax (x) + \sinh \relax (x) - 1\right )}{a^{2} - b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 67, normalized size = 1.26 \[ \frac {b^{2} \log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{a^{2} b - b^{3}} - \frac {\log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{2 \, {\left (a - b\right )}} + \frac {\log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{2 \, {\left (a + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*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 = 0.07, size = 52, normalized size = 0.98 \[ \frac {b \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-\left (\tanh ^{2}\left (\frac {x}{2}\right )\right ) b -a -b \right )}{\left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a +b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b/(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 = 0.38, size = 59, normalized size = 1.11 \[ \frac {b \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{a^{2} - b^{2}} - \frac {\log \left (e^{\left (-x\right )} + 1\right )}{a - b} + \frac {\log \left (e^{\left (-x\right )} - 1\right )}{a + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*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 = 1.29, size = 160, normalized size = 3.02 \[ \frac {\ln \left (128\,a\,b^2-128\,a^2\,b+32\,a^3-32\,a^3\,{\mathrm {e}}^x-128\,a\,b^2\,{\mathrm {e}}^x+128\,a^2\,b\,{\mathrm {e}}^x\right )}{a+b}-\frac {\ln \left (128\,a\,b^2+128\,a^2\,b+32\,a^3+32\,a^3\,{\mathrm {e}}^x+128\,a\,b^2\,{\mathrm {e}}^x+128\,a^2\,b\,{\mathrm {e}}^x\right )}{a-b}+\frac {b\,\ln \left (16\,b^3\,{\mathrm {e}}^{2\,x}-4\,a^2\,b+16\,b^3-8\,a^3\,{\mathrm {e}}^x+32\,a\,b^2\,{\mathrm {e}}^x-4\,a^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^2-b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)*(a + b*cosh(x))),x)

[Out]

log(128*a*b^2 - 128*a^2*b + 32*a^3 - 32*a^3*exp(x) - 128*a*b^2*exp(x) + 128*a^2*b*exp(x))/(a + b) - log(128*a*
b^2 + 128*a^2*b + 32*a^3 + 32*a^3*exp(x) + 128*a*b^2*exp(x) + 128*a^2*b*exp(x))/(a - b) + (b*log(16*b^3*exp(2*
x) - 4*a^2*b + 16*b^3 - 8*a^3*exp(x) + 32*a*b^2*exp(x) - 4*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 \[ \int \frac {\operatorname {csch}{\relax (x )}}{a + b \cosh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csch(x)/(a + b*cosh(x)), x)

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