3.143 \(\int \frac{\tanh (x)}{a+b \coth (x)} \, dx\)

Optimal. Leaf size=51 \[ -\frac{b x}{a^2-b^2}+\frac{b^2 \log (a \sinh (x)+b \cosh (x))}{a \left (a^2-b^2\right )}+\frac{\log (\cosh (x))}{a} \]

[Out]

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

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Rubi [A]  time = 0.0824597, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {3571, 3530, 3475} \[ -\frac{b x}{a^2-b^2}+\frac{b^2 \log (a \sinh (x)+b \cosh (x))}{a \left (a^2-b^2\right )}+\frac{\log (\cosh (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]/(a + b*Coth[x]),x]

[Out]

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

Rule 3571

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

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0755132, size = 46, normalized size = 0.9 \[ \frac{\left (a^2-b^2\right ) \log (\cosh (x))+b (b \log (a \sinh (x)+b \cosh (x))-a x)}{a^3-a b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]/(a + b*Coth[x]),x]

[Out]

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

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Maple [A]  time = 0.039, size = 88, normalized size = 1.7 \begin{align*} -8\,{\frac{\ln \left ( \tanh \left ( x/2 \right ) +1 \right ) }{8\,a-8\,b}}-8\,{\frac{\ln \left ( \tanh \left ( x/2 \right ) -1 \right ) }{8\,a+8\,b}}+{\frac{1}{a}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }+{\frac{{b}^{2}}{ \left ( a+b \right ) \left ( a-b \right ) a}\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(tanh(x)/(a+b*coth(x)),x)

[Out]

-8/(8*a-8*b)*ln(tanh(1/2*x)+1)-8/(8*a+8*b)*ln(tanh(1/2*x)-1)+1/a*ln(tanh(1/2*x)^2+1)+b^2/(a+b)/(a-b)/a*ln(tanh
(1/2*x)^2*b+2*a*tanh(1/2*x)+b)

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Maxima [A]  time = 1.74219, size = 68, normalized size = 1.33 \begin{align*} \frac{b^{2} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} + a + b\right )}{a^{3} - a b^{2}} + \frac{x}{a + b} + \frac{\log \left (e^{\left (-2 \, x\right )} + 1\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.81707, size = 185, normalized size = 3.63 \begin{align*} \frac{b^{2} \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 (a^{2} + a b\right )} x +{\left (a^{2} - b^{2}\right )} \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{3} - a b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tanh(x)/(a + b*coth(x)), x)

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Giac [A]  time = 1.19928, size = 77, normalized size = 1.51 \begin{align*} \frac{b^{2} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} - a + b \right |}\right )}{a^{3} - a b^{2}} - \frac{x}{a - b} + \frac{\log \left (e^{\left (2 \, x\right )} + 1\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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