\(\int \frac {\coth ^2(x)}{a+b \tanh (x)} \, dx\) [140]

Optimal result

Integrand size = 13, antiderivative size = 60 \[ \int \frac {\coth ^2(x)}{a+b \tanh (x)} \, dx=\frac {a x}{a^2-b^2}-\frac {\coth (x)}{a}-\frac {b \log (\sinh (x))}{a^2}-\frac {b^3 \log (a \cosh (x)+b \sinh (x))}{a^2 \left (a^2-b^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3650, 3732, 3611, 3556} \[ \int \frac {\coth ^2(x)}{a+b \tanh (x)} \, dx=\frac {a x}{a^2-b^2}-\frac {b^3 \log (a \cosh (x)+b \sinh (x))}{a^2 \left (a^2-b^2\right )}-\frac {b \log (\sinh (x))}{a^2}-\frac {\coth (x)}{a} \]

[In]

Int[Coth[x]^2/(a + b*Tanh[x]),x]

[Out]

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

Rule 3556

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

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 3650

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3732

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\coth (x)}{a}-\frac {i \int \frac {\coth (x) \left (-i b+i a \tanh (x)+i b \tanh ^2(x)\right )}{a+b \tanh (x)} \, dx}{a} \\ & = \frac {a x}{a^2-b^2}-\frac {\coth (x)}{a}-\frac {b \int \coth (x) \, dx}{a^2}-\frac {\left (i b^3\right ) \int \frac {-i b-i a \tanh (x)}{a+b \tanh (x)} \, dx}{a^2 \left (a^2-b^2\right )} \\ & = \frac {a x}{a^2-b^2}-\frac {\coth (x)}{a}-\frac {b \log (\sinh (x))}{a^2}-\frac {b^3 \log (a \cosh (x)+b \sinh (x))}{a^2 \left (a^2-b^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10 \[ \int \frac {\coth ^2(x)}{a+b \tanh (x)} \, dx=-\frac {\coth (x)}{a}-\frac {\log (1-\coth (x))}{2 (a+b)}+\frac {\log (1+\coth (x))}{2 (a-b)}-\frac {b^3 \log (b+a \coth (x))}{a^2 \left (a^2-b^2\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(-\frac {b \ln \left (\tanh \left (x \right )\right )}{a^{2}}-\frac {1}{a \tanh \left (x \right )}+\frac {\ln \left (1+\tanh \left (x \right )\right )}{2 a -2 b}-\frac {\ln \left (\tanh \left (x \right )-1\right )}{2 a +2 b}-\frac {b^{3} \ln \left (a +b \tanh \left (x \right )\right )}{a^{2} \left (a -b \right ) \left (a +b \right )}\)	\(78\)
default	\(-\frac {b \ln \left (\tanh \left (x \right )\right )}{a^{2}}-\frac {1}{a \tanh \left (x \right )}+\frac {\ln \left (1+\tanh \left (x \right )\right )}{2 a -2 b}-\frac {\ln \left (\tanh \left (x \right )-1\right )}{2 a +2 b}-\frac {b^{3} \ln \left (a +b \tanh \left (x \right )\right )}{a^{2} \left (a -b \right ) \left (a +b \right )}\)	\(78\)
parallelrisch	\(\frac {-b^{3} \ln \left (a +b \tanh \left (x \right )\right ) \tanh \left (x \right )+\ln \left (1-\tanh \left (x \right )\right ) \tanh \left (x \right ) a^{2} b +\left (a +b \right ) \left (-b \tanh \left (x \right ) \left (a -b \right ) \ln \left (\tanh \left (x \right )\right )+a \left (a x \tanh \left (x \right )-a +b \right )\right )}{\left (a^{4}-a^{2} b^{2}\right ) \tanh \left (x \right )}\)	\(79\)
risch	\(\frac {x}{a +b}+\frac {2 x b}{a^{2}}+\frac {2 x \,b^{3}}{a^{2} \left (a^{2}-b^{2}\right )}-\frac {2}{a \left ({\mathrm e}^{2 x}-1\right )}-\frac {b \ln \left ({\mathrm e}^{2 x}-1\right )}{a^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{2 x}+\frac {a -b}{a +b}\right )}{a^{2} \left (a^{2}-b^{2}\right )}\)	\(98\)

[In]

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

[Out]

-b*ln(tanh(x))/a^2-1/a/tanh(x)+1/(2*a-2*b)*ln(1+tanh(x))-1/(2*a+2*b)*ln(tanh(x)-1)-b^3/a^2/(a-b)/(a+b)*ln(a+b*
tanh(x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (60) = 120\).

Time = 0.28 (sec) , antiderivative size = 271, normalized size of antiderivative = 4.52 \[ \int \frac {\coth ^2(x)}{a+b \tanh (x)} \, dx=-\frac {{\left (a^{3} + a^{2} b\right )} x \cosh \left (x\right )^{2} + 2 \, {\left (a^{3} + a^{2} b\right )} x \cosh \left (x\right ) \sinh \left (x\right ) + {\left (a^{3} + a^{2} b\right )} x \sinh \left (x\right )^{2} - 2 \, a^{3} + 2 \, a b^{2} - {\left (a^{3} + a^{2} b\right )} x - {\left (b^{3} \cosh \left (x\right )^{2} + 2 \, b^{3} \cosh \left (x\right ) \sinh \left (x\right ) + b^{3} \sinh \left (x\right )^{2} - b^{3}\right )} \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b \sinh \left (x\right )\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + {\left (a^{2} b - b^{3} - {\left (a^{2} b - b^{3}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (a^{2} b - b^{3}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (a^{2} b - b^{3}\right )} \sinh \left (x\right )^{2}\right )} \log \left (\frac {2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{4} - a^{2} b^{2} - {\left (a^{4} - a^{2} b^{2}\right )} \cosh \left (x\right )^{2} - 2 \, {\left (a^{4} - a^{2} b^{2}\right )} \cosh \left (x\right ) \sinh \left (x\right ) - {\left (a^{4} - a^{2} b^{2}\right )} \sinh \left (x\right )^{2}} \]

[In]

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

[Out]

-((a^3 + a^2*b)*x*cosh(x)^2 + 2*(a^3 + a^2*b)*x*cosh(x)*sinh(x) + (a^3 + a^2*b)*x*sinh(x)^2 - 2*a^3 + 2*a*b^2
- (a^3 + a^2*b)*x - (b^3*cosh(x)^2 + 2*b^3*cosh(x)*sinh(x) + b^3*sinh(x)^2 - b^3)*log(2*(a*cosh(x) + b*sinh(x)
)/(cosh(x) - sinh(x))) + (a^2*b - b^3 - (a^2*b - b^3)*cosh(x)^2 - 2*(a^2*b - b^3)*cosh(x)*sinh(x) - (a^2*b - b
^3)*sinh(x)^2)*log(2*sinh(x)/(cosh(x) - sinh(x))))/(a^4 - a^2*b^2 - (a^4 - a^2*b^2)*cosh(x)^2 - 2*(a^4 - a^2*b
^2)*cosh(x)*sinh(x) - (a^4 - a^2*b^2)*sinh(x)^2)

Sympy [F]

\[ \int \frac {\coth ^2(x)}{a+b \tanh (x)} \, dx=\int \frac {\coth ^{2}{\left (x \right )}}{a + b \tanh {\left (x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.43 \[ \int \frac {\coth ^2(x)}{a+b \tanh (x)} \, dx=-\frac {b^{3} \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{4} - a^{2} b^{2}} + \frac {x}{a + b} - \frac {b \log \left (e^{\left (-x\right )} + 1\right )}{a^{2}} - \frac {b \log \left (e^{\left (-x\right )} - 1\right )}{a^{2}} + \frac {2}{a e^{\left (-2 \, x\right )} - a} \]

[In]

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

[Out]

-b^3*log(-(a - b)*e^(-2*x) - a - b)/(a^4 - a^2*b^2) + x/(a + b) - b*log(e^(-x) + 1)/a^2 - b*log(e^(-x) - 1)/a^
2 + 2/(a*e^(-2*x) - a)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.25 \[ \int \frac {\coth ^2(x)}{a+b \tanh (x)} \, dx=-\frac {b^{3} \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{4} - a^{2} b^{2}} + \frac {x}{a - b} - \frac {b \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right )}{a^{2}} - \frac {2}{a {\left (e^{\left (2 \, x\right )} - 1\right )}} \]

[In]

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

[Out]

-b^3*log(abs(a*e^(2*x) + b*e^(2*x) + a - b))/(a^4 - a^2*b^2) + x/(a - b) - b*log(abs(e^(2*x) - 1))/a^2 - 2/(a*
(e^(2*x) - 1))

Mupad [B] (verification not implemented)

Time = 1.97 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22 \[ \int \frac {\coth ^2(x)}{a+b \tanh (x)} \, dx=\frac {x}{a-b}-\frac {2}{a\,\left ({\mathrm {e}}^{2\,x}-1\right )}-\frac {b^3\,\ln \left (a-b+a\,{\mathrm {e}}^{2\,x}+b\,{\mathrm {e}}^{2\,x}\right )}{a^4-a^2\,b^2}-\frac {b\,\ln \left ({\mathrm {e}}^{2\,x}-1\right )}{a^2} \]

[In]

int(coth(x)^2/(a + b*tanh(x)),x)

[Out]

x/(a - b) - 2/(a*(exp(2*x) - 1)) - (b^3*log(a - b + a*exp(2*x) + b*exp(2*x)))/(a^4 - a^2*b^2) - (b*log(exp(2*x
) - 1))/a^2