3.81 \(\int \frac{1}{a+b \coth (c+d x)} \, dx\)

Optimal. Leaf size=50 \[ \frac{a x}{a^2-b^2}-\frac{b \log (a \sinh (c+d x)+b \cosh (c+d x))}{d \left (a^2-b^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0539133, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3484, 3530} \[ \frac{a x}{a^2-b^2}-\frac{b \log (a \sinh (c+d x)+b \cosh (c+d x))}{d \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Coth[c + d*x])^(-1),x]

[Out]

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

Rule 3484

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

Rubi steps

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

Mathematica [A]  time = 0.0792178, size = 64, normalized size = 1.28 \[ \frac{(b-a) \log (1-\coth (c+d x))+(a+b) \log (\coth (c+d x)+1)-2 b \log (a+b \coth (c+d x))}{2 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Coth[c + d*x])^(-1),x]

[Out]

((-a + b)*Log[1 - Coth[c + d*x]] + (a + b)*Log[1 + Coth[c + d*x]] - 2*b*Log[a + b*Coth[c + d*x]])/(2*(a - b)*(
a + b)*d)

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Maple [A]  time = 0.019, size = 76, normalized size = 1.5 \begin{align*}{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ) }{d \left ( 2\,a-2\,b \right ) }}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ) }{d \left ( 2\,b+2\,a \right ) }}-{\frac{b\ln \left ( a+b{\rm coth} \left (dx+c\right ) \right ) }{d \left ( a-b \right ) \left ( a+b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*coth(d*x+c)),x)

[Out]

1/d/(2*a-2*b)*ln(coth(d*x+c)+1)-1/d/(2*b+2*a)*ln(coth(d*x+c)-1)-1/d*b/(a-b)/(a+b)*ln(a+b*coth(d*x+c))

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Maxima [A]  time = 1.12232, size = 70, normalized size = 1.4 \begin{align*} -\frac{b \log \left (-{\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a + b\right )}{{\left (a^{2} - b^{2}\right )} d} + \frac{d x + c}{{\left (a + b\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.88622, size = 149, normalized size = 2.98 \begin{align*} \frac{{\left (a + b\right )} d x - b \log \left (\frac{2 \,{\left (b \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right )\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{{\left (a^{2} - b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.14948, size = 236, normalized size = 4.72 \begin{align*} \begin{cases} \frac{\tilde{\infty } x}{\coth{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\- \frac{d x \tanh{\left (c + d x \right )}}{2 b d \tanh{\left (c + d x \right )} - 2 b d} + \frac{d x}{2 b d \tanh{\left (c + d x \right )} - 2 b d} - \frac{1}{2 b d \tanh{\left (c + d x \right )} - 2 b d} & \text{for}\: a = - b \\\frac{d x \tanh{\left (c + d x \right )}}{2 b d \tanh{\left (c + d x \right )} + 2 b d} + \frac{d x}{2 b d \tanh{\left (c + d x \right )} + 2 b d} + \frac{1}{2 b d \tanh{\left (c + d x \right )} + 2 b d} & \text{for}\: a = b \\\frac{x}{a + b \coth{\left (c \right )}} & \text{for}\: d = 0 \\\frac{x - \frac{\log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{d}}{b} & \text{for}\: a = 0 \\\frac{a d x}{a^{2} d - b^{2} d} - \frac{b d x}{a^{2} d - b^{2} d} + \frac{b \log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{a^{2} d - b^{2} d} - \frac{b \log{\left (\tanh{\left (c + d x \right )} + \frac{b}{a} \right )}}{a^{2} d - b^{2} d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c)),x)

[Out]

Piecewise((zoo*x/coth(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (-d*x*tanh(c + d*x)/(2*b*d*tanh(c + d*x) - 2*b*d) +
d*x/(2*b*d*tanh(c + d*x) - 2*b*d) - 1/(2*b*d*tanh(c + d*x) - 2*b*d), Eq(a, -b)), (d*x*tanh(c + d*x)/(2*b*d*ta
nh(c + d*x) + 2*b*d) + d*x/(2*b*d*tanh(c + d*x) + 2*b*d) + 1/(2*b*d*tanh(c + d*x) + 2*b*d), Eq(a, b)), (x/(a +
b*coth(c)), Eq(d, 0)), ((x - log(tanh(c + d*x) + 1)/d)/b, Eq(a, 0)), (a*d*x/(a**2*d - b**2*d) - b*d*x/(a**2*d
- b**2*d) + b*log(tanh(c + d*x) + 1)/(a**2*d - b**2*d) - b*log(tanh(c + d*x) + b/a)/(a**2*d - b**2*d), True))

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Giac [A]  time = 1.14268, size = 85, normalized size = 1.7 \begin{align*} -\frac{b \log \left ({\left | a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} - a + b \right |}\right )}{a^{2} d - b^{2} d} + \frac{d x + c}{a d - b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c)),x, algorithm="giac")

[Out]

-b*log(abs(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) - a + b))/(a^2*d - b^2*d) + (d*x + c)/(a*d - b*d)