3.119 \(\int \frac{\text{sech}^2(x)}{a+b \coth (x)} \, dx\)

Optimal. Leaf size=29 \[ -\frac{b \log (\tanh (x))}{a^2}-\frac{b \log (a+b \coth (x))}{a^2}+\frac{\tanh (x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.0554407, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3516, 44} \[ -\frac{b \log (\tanh (x))}{a^2}-\frac{b \log (a+b \coth (x))}{a^2}+\frac{\tanh (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(a + b*Coth[x]),x]

[Out]

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

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

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])

Rubi steps

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

Mathematica [A]  time = 0.0806487, size = 27, normalized size = 0.93 \[ \frac{-b \log (a \sinh (x)+b \cosh (x))+a \tanh (x)+b \log (\cosh (x))}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^2/(a + b*Coth[x]),x]

[Out]

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

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Maple [A]  time = 0.04, size = 59, normalized size = 2. \begin{align*} 2\,{\frac{\tanh \left ( x/2 \right ) }{a \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}+{\frac{b}{{a}^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }-{\frac{b}{{a}^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}b+2\,a\tanh \left ( x/2 \right ) +b \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2/(a+b*coth(x)),x)

[Out]

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

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Maxima [A]  time = 1.76628, size = 62, normalized size = 2.14 \begin{align*} -\frac{b \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{a^{2}} + \frac{b \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{2}} + \frac{2}{a e^{\left (-2 \, x\right )} + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(a+b*coth(x)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.61689, size = 362, normalized size = 12.48 \begin{align*} -\frac{{\left (b \cosh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) \sinh \left (x\right ) + b \sinh \left (x\right )^{2} + b\right )} \log \left (\frac{2 \,{\left (b \cosh \left (x\right ) + a \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) -{\left (b \cosh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) \sinh \left (x\right ) + b \sinh \left (x\right )^{2} + b\right )} \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 2 \, a}{a^{2} \cosh \left (x\right )^{2} + 2 \, a^{2} \cosh \left (x\right ) \sinh \left (x\right ) + a^{2} \sinh \left (x\right )^{2} + a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(a+b*coth(x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{2}{\left (x \right )}}{a + b \coth{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2/(a+b*coth(x)),x)

[Out]

Integral(sech(x)**2/(a + b*coth(x)), x)

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Giac [B]  time = 1.1468, size = 103, normalized size = 3.55 \begin{align*} -\frac{{\left (a b + b^{2}\right )} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} - a + b \right |}\right )}{a^{3} + a^{2} b} + \frac{b \log \left (e^{\left (2 \, x\right )} + 1\right )}{a^{2}} - \frac{b e^{\left (2 \, x\right )} + 2 \, a + b}{a^{2}{\left (e^{\left (2 \, x\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(a+b*coth(x)),x, algorithm="giac")

[Out]

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