3.120 \(\int \frac{\coth (x)}{a+b \text{sech}(x)} \, dx\)

Optimal. Leaf size=66 \[ -\frac{b^2 \log (a+b \text{sech}(x))}{a \left (a^2-b^2\right )}+\frac{\log (1-\text{sech}(x))}{2 (a+b)}+\frac{\log (\text{sech}(x)+1)}{2 (a-b)}+\frac{\log (\cosh (x))}{a} \]

[Out]

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

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Rubi [A]  time = 0.106399, antiderivative size = 66, normalized size of antiderivative = 1., 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 = {3885, 894} \[ -\frac{b^2 \log (a+b \text{sech}(x))}{a \left (a^2-b^2\right )}+\frac{\log (1-\text{sech}(x))}{2 (a+b)}+\frac{\log (\text{sech}(x)+1)}{2 (a-b)}+\frac{\log (\cosh (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]/(a + b*Sech[x]),x]

[Out]

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

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rubi steps

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

Mathematica [A]  time = 0.0853563, size = 44, normalized size = 0.67 \[ -\frac{a^2 (-\log (\sinh (x)))+b^2 \log (a \cosh (x)+b)+a b \log \left (\tanh \left (\frac{x}{2}\right )\right )}{a^3-a b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]/(a + b*Sech[x]),x]

[Out]

-((b^2*Log[b + a*Cosh[x]] - a^2*Log[Sinh[x]] + a*b*Log[Tanh[x/2]])/(a^3 - a*b^2))

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Maple [A]  time = 0.029, size = 78, normalized size = 1.2 \begin{align*} -{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{1}{a+b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-{\frac{1}{a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{{b}^{2}}{a \left ( a+b \right ) \left ( a-b \right ) }\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}- \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b+a+b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.07207, size = 90, normalized size = 1.36 \begin{align*} -\frac{b^{2} \log \left (2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a\right )}{a^{3} - a b^{2}} + \frac{x}{a} + \frac{\log \left (e^{\left (-x\right )} + 1\right )}{a - b} + \frac{\log \left (e^{\left (-x\right )} - 1\right )}{a + b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.33004, size = 220, normalized size = 3.33 \begin{align*} -\frac{b^{2} \log \left (\frac{2 \,{\left (a \cosh \left (x\right ) + b\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) +{\left (a^{2} - b^{2}\right )} x -{\left (a^{2} + a b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) -{\left (a^{2} - a b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{a^{3} - a b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*log(2*(a*cosh(x) + b)/(cosh(x) - sinh(x))) + (a^2 - b^2)*x - (a^2 + a*b)*log(cosh(x) + sinh(x) + 1) - (a
^2 - a*b)*log(cosh(x) + sinh(x) - 1))/(a^3 - a*b^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth{\left (x \right )}}{a + b \operatorname{sech}{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(coth(x)/(a + b*sech(x)), x)

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Giac [A]  time = 1.1175, size = 90, normalized size = 1.36 \begin{align*} -\frac{b^{2} \log \left ({\left | a{\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b \right |}\right )}{a^{3} - a b^{2}} + \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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