\(\int \frac {1}{(a+b \coth (c+d x))^2} \, dx\) [82]

Optimal result

Integrand size = 12, antiderivative size = 85 \[ \int \frac {1}{(a+b \coth (c+d x))^2} \, dx=\frac {\left (a^2+b^2\right ) x}{\left (a^2-b^2\right )^2}+\frac {b}{\left (a^2-b^2\right ) d (a+b \coth (c+d x))}-\frac {2 a b \log (b \cosh (c+d x)+a \sinh (c+d x))}{\left (a^2-b^2\right )^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3564, 3612, 3611} \[ \int \frac {1}{(a+b \coth (c+d x))^2} \, dx=\frac {b}{d \left (a^2-b^2\right ) (a+b \coth (c+d x))}-\frac {2 a b \log (a \sinh (c+d x)+b \cosh (c+d x))}{d \left (a^2-b^2\right )^2}+\frac {x \left (a^2+b^2\right )}{\left (a^2-b^2\right )^2} \]

[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 3564

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 3611

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c/(b*f))
*Log[RemoveContent[a*Cos[e + f*x] + b*Sin[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] && EqQ[a*c + b*d, 0]

Rule 3612

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 1.58 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.18 \[ \int \frac {1}{(a+b \coth (c+d x))^2} \, dx=\frac {-\frac {\log (1-\tanh (c+d x))}{(a+b)^2}+\frac {\log (1+\tanh (c+d x))}{(a-b)^2}+\frac {2 b \left (-2 a^2 \log (b+a \tanh (c+d x))+\frac {-a^2 b+b^3}{b+a \tanh (c+d x)}\right )}{a \left (a^2-b^2\right )^2}}{2 d} \]

[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)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {\frac {b}{\left (a -b \right ) \left (a +b \right ) \left (a +b \coth \left (d x +c \right )\right )}-\frac {2 a b \ln \left (a +b \coth \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {\ln \left (\coth \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}+\frac {\ln \left (\coth \left (d x +c \right )+1\right )}{2 \left (a -b \right )^{2}}}{d}\)	\(93\)
default	\(\frac {\frac {b}{\left (a -b \right ) \left (a +b \right ) \left (a +b \coth \left (d x +c \right )\right )}-\frac {2 a b \ln \left (a +b \coth \left (d x +c \right )\right )}{\left (a +b \right )^{2} \left (a -b \right )^{2}}-\frac {\ln \left (\coth \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{2}}+\frac {\ln \left (\coth \left (d x +c \right )+1\right )}{2 \left (a -b \right )^{2}}}{d}\)	\(93\)
parallelrisch	\(\frac {\left (-2 a^{3} b \tanh \left (d x +c \right )-2 a^{2} b^{2}\right ) \ln \left (b +a \tanh \left (d x +c \right )\right )+\left (2 a^{3} b \tanh \left (d x +c \right )+2 a^{2} b^{2}\right ) \ln \left (1-\tanh \left (d x +c \right )\right )+\left (a^{2} d x \left (a +b \right ) \tanh \left (d x +c \right )+\left (a^{2} d x +b \left (d x -1\right ) a +b^{2}\right ) b \right ) \left (a +b \right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \left (b +a \tanh \left (d x +c \right )\right ) d a}\)	\(139\)
risch	\(\frac {x}{a^{2}+2 a b +b^{2}}+\frac {4 a b x}{a^{4}-2 a^{2} b^{2}+b^{4}}+\frac {4 a b c}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 b^{2}}{\left (a -b \right ) d \left (a^{2}+2 a b +b^{2}\right ) \left ({\mathrm e}^{2 d x +2 c} a +b \,{\mathrm e}^{2 d x +2 c}-a +b \right )}-\frac {2 a b \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {a -b}{a +b}\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}\)	\(168\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 426 vs. \(2 (85) = 170\).

Time = 0.26 (sec) , antiderivative size = 426, normalized size of antiderivative = 5.01 \[ \int \frac {1}{(a+b \coth (c+d x))^2} \, dx=\frac {{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x \sinh \left (d x + c\right )^{2} - 2 \, a b^{2} + 2 \, b^{3} - {\left (a^{3} + a^{2} b - a b^{2} - b^{3}\right )} d x + 2 \, {\left (a^{2} b - a b^{2} - {\left (a^{2} b + a b^{2}\right )} \cosh \left (d x + c\right )^{2} - 2 \, {\left (a^{2} b + a b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) - {\left (a^{2} b + a b^{2}\right )} \sinh \left (d x + c\right )^{2}\right )} \log \left (\frac {2 \, {\left (b \cosh \left (d x + c\right ) + a \sinh \left (d x + c\right )\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{{\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{5} + a^{4} b - 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4} + b^{5}\right )} d \sinh \left (d x + c\right )^{2} - {\left (a^{5} - a^{4} b - 2 \, a^{3} b^{2} + 2 \, a^{2} b^{3} + a b^{4} - b^{5}\right )} d} \]

[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)

Sympy [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \coth (c+d x))^2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> Invalid NaN comparison

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.46 \[ \int \frac {1}{(a+b \coth (c+d x))^2} \, dx=-\frac {2 \, a b \log \left (-{\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a + b\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d} - \frac {2 \, b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{3} b + 2 \, a b^{3} - b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} + \frac {d x + c}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.53 \[ \int \frac {1}{(a+b \coth (c+d x))^2} \, dx=-\frac {\frac {2 \, a b \log \left ({\left | a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} - a + b \right |}\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac {d x + c}{a^{2} - 2 \, a b + b^{2}} + \frac {2 \, {\left (a b^{2} - b^{3}\right )}}{{\left (a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} - a + b\right )} {\left (a + b\right )}^{2} {\left (a - b\right )}^{2}}}{d} \]

[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 - 2*a^2*b^2 + b^4) - (d*x + c)/(a^2 - 2*a
*b + b^2) + 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

Mupad [B] (verification not implemented)

Time = 2.00 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.22 \[ \int \frac {1}{(a+b \coth (c+d x))^2} \, dx=\frac {x}{{\left (a-b\right )}^2}-\frac {2\,a\,b\,\ln \left (b-a+a\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )}{d\,a^4-2\,d\,a^2\,b^2+d\,b^4}-\frac {2\,b^2}{d\,{\left (a+b\right )}^2\,\left (a-b\right )\,\left (b-a+{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )\right )} \]

[In]

int(1/(a + b*coth(c + d*x))^2,x)

[Out]

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