3.82 \(\int \frac{1}{(a+b \coth (c+d x))^2} \, dx\)

Optimal. Leaf size=85 \[ \frac{b}{d \left (a^2-b^2\right ) (a+b \coth (c+d x))}-\frac{2 a b \log (a \sinh (c+d x)+b \cosh (c+d x))}{d \left (a^2-b^2\right )^2}+\frac{x \left (a^2+b^2\right )}{\left (a^2-b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0946267, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3483, 3531, 3530} \[ \frac{b}{d \left (a^2-b^2\right ) (a+b \coth (c+d x))}-\frac{2 a b \log (a \sinh (c+d x)+b \cosh (c+d x))}{d \left (a^2-b^2\right )^2}+\frac{x \left (a^2+b^2\right )}{\left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Coth[c + d*x])^(-2),x]

[Out]

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

Rule 3483

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

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*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[a*c + b*d, 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))^2} \, dx &=\frac{b}{\left (a^2-b^2\right ) d (a+b \coth (c+d x))}+\frac{\int \frac{a-b \coth (c+d x)}{a+b \coth (c+d x)} \, dx}{a^2-b^2}\\ &=\frac{\left (a^2+b^2\right ) x}{\left (a^2-b^2\right )^2}+\frac{b}{\left (a^2-b^2\right ) d (a+b \coth (c+d x))}-\frac{(2 i a b) \int \frac{-i b-i a \coth (c+d x)}{a+b \coth (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac{\left (a^2+b^2\right ) x}{\left (a^2-b^2\right )^2}+\frac{b}{\left (a^2-b^2\right ) d (a+b \coth (c+d x))}-\frac{2 a b \log (b \cosh (c+d x)+a \sinh (c+d x))}{\left (a^2-b^2\right )^2 d}\\ \end{align*}

Mathematica [A]  time = 1.4201, size = 100, normalized size = 1.18 \[ \frac{\frac{2 b \left (\frac{b^3-a^2 b}{a \tanh (c+d x)+b}-2 a^2 \log (a \tanh (c+d x)+b)\right )}{a \left (a^2-b^2\right )^2}-\frac{\log (1-\tanh (c+d x))}{(a+b)^2}+\frac{\log (\tanh (c+d x)+1)}{(a-b)^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Coth[c + d*x])^(-2),x]

[Out]

(-(Log[1 - Tanh[c + d*x]]/(a + b)^2) + Log[1 + Tanh[c + d*x]]/(a - b)^2 + (2*b*(-2*a^2*Log[b + a*Tanh[c + d*x]
] + (-(a^2*b) + b^3)/(b + a*Tanh[c + d*x])))/(a*(a^2 - b^2)^2))/(2*d)

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Maple [A]  time = 0.025, size = 101, normalized size = 1.2 \begin{align*}{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ) }{2\,d \left ( a-b \right ) ^{2}}}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ) }{2\,d \left ( a+b \right ) ^{2}}}+{\frac{b}{d \left ( a-b \right ) \left ( a+b \right ) \left ( a+b{\rm coth} \left (dx+c\right ) \right ) }}-2\,{\frac{ab\ln \left ( a+b{\rm coth} \left (dx+c\right ) \right ) }{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*coth(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.16493, size = 167, normalized size = 1.96 \begin{align*} -\frac{2 \, a b \log \left (-{\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a + b\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac{2 \, b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} -{\left (a^{4} - 2 \, a^{3} b + 2 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} + \frac{d x + c}{{\left (a^{2} + 2 \, a b + 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))^2,x, algorithm="maxima")

[Out]

-2*a*b*log(-(a - b)*e^(-2*d*x - 2*c) + a + b)/((a^4 - 2*a^2*b^2 + b^4)*d) - 2*b^2/((a^4 - 2*a^2*b^2 + b^4 - (a
^4 - 2*a^3*b + 2*a*b^3 - b^4)*e^(-2*d*x - 2*c))*d) + (d*x + c)/((a^2 + 2*a*b + b^2)*d)

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Fricas [B]  time = 2.67566, size = 980, normalized size = 11.53 \begin{align*} \frac{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x \cosh \left (d x + c\right )^{2} + 2 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x \sinh \left (d x + c\right )^{2} - 2 \, a b^{2} + 2 \, b^{3} -{\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} d x + 2 \,{\left (a^{2} b - a b^{2} -{\left (a^{2} b + a b^{2}\right )} \cosh \left (d x + c\right )^{2} - 2 \,{\left (a^{2} b + a b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) -{\left (a^{2} b + a b^{2}\right )} \sinh \left (d x + c\right )^{2}\right )} \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^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} d \cosh \left (d x + c\right )^{2} + 2 \,{\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) +{\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} d \sinh \left (d x + c\right )^{2} -{\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c))^2,x, algorithm="fricas")

[Out]

((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)^2 + 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)*sin
h(d*x + c) + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*sinh(d*x + c)^2 - 2*a*b^2 + 2*b^3 - (a^3 + a^2*b - a*b^2 - b^
3)*d*x + 2*(a^2*b - a*b^2 - (a^2*b + a*b^2)*cosh(d*x + c)^2 - 2*(a^2*b + a*b^2)*cosh(d*x + c)*sinh(d*x + c) -
(a^2*b + a*b^2)*sinh(d*x + c)^2)*log(2*(b*cosh(d*x + c) + a*sinh(d*x + c))/(cosh(d*x + c) - sinh(d*x + c))))/(
(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*d*cosh(d*x + c)^2 + 2*(a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3
+ a*b^4 + b^5)*d*cosh(d*x + c)*sinh(d*x + c) + (a^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5)*d*sinh(d*x
+ c)^2 - (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*d)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c))**2,x)

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.1549, size = 185, normalized size = 2.18 \begin{align*} -\frac{2 \, a 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^{4} d - 2 \, a^{2} b^{2} d + b^{4} d} + \frac{d x + c}{a^{2} d - 2 \, a b d + b^{2} d} - \frac{2 \,{\left (a b^{2} - b^{3}\right )}}{{\left (a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} - a + b\right )}{\left (a + b\right )}^{2}{\left (a - b\right )}^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*coth(d*x+c))^2,x, algorithm="giac")

[Out]

-2*a*b*log(abs(a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) - a + b))/(a^4*d - 2*a^2*b^2*d + b^4*d) + (d*x + c)/(a^2*
d - 2*a*b*d + b^2*d) - 2*(a*b^2 - b^3)/((a*e^(2*d*x + 2*c) + b*e^(2*d*x + 2*c) - a + b)*(a + b)^2*(a - b)^2*d)