\(\int \frac {\text {csch}^2(x)}{1+\coth (x)} \, dx\) [94]

Optimal result

Integrand size = 11, antiderivative size = 7 \[ \int \frac {\text {csch}^2(x)}{1+\coth (x)} \, dx=-\log (1+\coth (x)) \]

[Out]

-ln(1+coth(x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3568, 31} \[ \int \frac {\text {csch}^2(x)}{1+\coth (x)} \, dx=-\log (\coth (x)+1) \]

[In]

Int[Csch[x]^2/(1 + Coth[x]),x]

[Out]

-Log[1 + Coth[x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3568

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {1}{1+x} \, dx,x,\coth (x)\right ) \\ & = -\log (1+\coth (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.43 \[ \int \frac {\text {csch}^2(x)}{1+\coth (x)} \, dx=-x+\log (\cosh (x))+\log (\tanh (x)) \]

[In]

Integrate[Csch[x]^2/(1 + Coth[x]),x]

[Out]

-x + Log[Cosh[x]] + Log[Tanh[x]]

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 8, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(-\ln \left (1+\coth \left (x \right )\right )\)	\(8\)
default	\(-\ln \left (1+\coth \left (x \right )\right )\)	\(8\)
risch	\(-2 x +\ln \left ({\mathrm e}^{2 x}-1\right )\)	\(12\)

[In]

int(csch(x)^2/(1+coth(x)),x,method=_RETURNVERBOSE)

[Out]

-ln(1+coth(x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18 vs. \(2 (7) = 14\).

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 2.57 \[ \int \frac {\text {csch}^2(x)}{1+\coth (x)} \, dx=-2 \, x + \log \left (\frac {2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) \]

[In]

integrate(csch(x)^2/(1+coth(x)),x, algorithm="fricas")

[Out]

-2*x + log(2*sinh(x)/(cosh(x) - sinh(x)))

Sympy [F]

\[ \int \frac {\text {csch}^2(x)}{1+\coth (x)} \, dx=\int \frac {\operatorname {csch}^{2}{\left (x \right )}}{\coth {\left (x \right )} + 1}\, dx \]

[In]

integrate(csch(x)**2/(1+coth(x)),x)

[Out]

Integral(csch(x)**2/(coth(x) + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 7, normalized size of antiderivative = 1.00 \[ \int \frac {\text {csch}^2(x)}{1+\coth (x)} \, dx=-\log \left (\coth \left (x\right ) + 1\right ) \]

[In]

integrate(csch(x)^2/(1+coth(x)),x, algorithm="maxima")

[Out]

-log(coth(x) + 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.71 \[ \int \frac {\text {csch}^2(x)}{1+\coth (x)} \, dx=-2 \, x + \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \]

[In]

integrate(csch(x)^2/(1+coth(x)),x, algorithm="giac")

[Out]

-2*x + log(abs(e^(2*x) - 1))

Mupad [B] (verification not implemented)

Time = 1.86 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.57 \[ \int \frac {\text {csch}^2(x)}{1+\coth (x)} \, dx=\ln \left ({\mathrm {e}}^{2\,x}-1\right )-2\,x \]

[In]

int(1/(sinh(x)^2*(coth(x) + 1)),x)

[Out]

log(exp(2*x) - 1) - 2*x