\(\int \frac {1}{\sqrt {1+\coth (x)}} \, dx\) [74]

Optimal result

Integrand size = 8, antiderivative size = 32 \[ \int \frac {1}{\sqrt {1+\coth (x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {1+\coth (x)}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{\sqrt {1+\coth (x)}} \]

[Out]

1/2*arctanh(1/2*(1+coth(x))^(1/2)*2^(1/2))*2^(1/2)-1/(1+coth(x))^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3560, 3561, 212} \[ \int \frac {1}{\sqrt {1+\coth (x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {\coth (x)+1}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{\sqrt {\coth (x)+1}} \]

[In]

Int[1/Sqrt[1 + Coth[x]],x]

[Out]

ArcTanh[Sqrt[1 + Coth[x]]/Sqrt[2]]/Sqrt[2] - 1/Sqrt[1 + Coth[x]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3560

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

Rule 3561

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{\sqrt {1+\coth (x)}}+\frac {1}{2} \int \sqrt {1+\coth (x)} \, dx \\ & = -\frac {1}{\sqrt {1+\coth (x)}}+\text {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\sqrt {1+\coth (x)}\right ) \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {1+\coth (x)}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{\sqrt {1+\coth (x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.45 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\sqrt {1+\coth (x)}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1+\coth (x))\right )}{\sqrt {1+\coth (x)}} \]

[In]

Integrate[1/Sqrt[1 + Coth[x]],x]

[Out]

-(Hypergeometric2F1[-1/2, 1, 1/2, (1 + Coth[x])/2]/Sqrt[1 + Coth[x]])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(\frac {\operatorname {arctanh}\left (\frac {\sqrt {1+\coth \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}}{2}-\frac {1}{\sqrt {1+\coth \left (x \right )}}\)	\(27\)
default	\(\frac {\operatorname {arctanh}\left (\frac {\sqrt {1+\coth \left (x \right )}\, \sqrt {2}}{2}\right ) \sqrt {2}}{2}-\frac {1}{\sqrt {1+\coth \left (x \right )}}\)	\(27\)

[In]

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

[Out]

1/2*arctanh(1/2*(1+coth(x))^(1/2)*2^(1/2))*2^(1/2)-1/(1+coth(x))^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (26) = 52\).

Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.66 \[ \int \frac {1}{\sqrt {1+\coth (x)}} \, dx=\frac {{\left (\sqrt {2} \cosh \left (x\right ) + \sqrt {2} \sinh \left (x\right )\right )} \log \left (2 \, \sqrt {2} \sqrt {\frac {\sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} + 2 \, \cosh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sinh \left (x\right )^{2} - 1\right ) - 4 \, \sqrt {\frac {\sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}}}{4 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \]

[In]

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

[Out]

1/4*((sqrt(2)*cosh(x) + sqrt(2)*sinh(x))*log(2*sqrt(2)*sqrt(sinh(x)/(cosh(x) - sinh(x)))*(cosh(x) + sinh(x)) +
2*cosh(x)^2 + 4*cosh(x)*sinh(x) + 2*sinh(x)^2 - 1) - 4*sqrt(sinh(x)/(cosh(x) - sinh(x))))/(cosh(x) + sinh(x))

Sympy [F]

\[ \int \frac {1}{\sqrt {1+\coth (x)}} \, dx=\int \frac {1}{\sqrt {\coth {\left (x \right )} + 1}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(coth(x) + 1), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {1+\coth (x)}} \, dx=\int { \frac {1}{\sqrt {\coth \left (x\right ) + 1}} \,d x } \]

[In]

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

[Out]

integrate(1/sqrt(coth(x) + 1), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (26) = 52\).

Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.06 \[ \int \frac {1}{\sqrt {1+\coth (x)}} \, dx=\frac {\sqrt {2} {\left (\frac {2}{\sqrt {e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}} - \log \left ({\left | 2 \, \sqrt {e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - 2 \, e^{\left (2 \, x\right )} + 1 \right |}\right )\right )}}{4 \, \mathrm {sgn}\left (e^{\left (2 \, x\right )} - 1\right )} \]

[In]

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

[Out]

1/4*sqrt(2)*(2/(sqrt(e^(4*x) - e^(2*x)) - e^(2*x)) - log(abs(2*sqrt(e^(4*x) - e^(2*x)) - 2*e^(2*x) + 1)))/sgn(
e^(2*x) - 1)

Mupad [B] (verification not implemented)

Time = 1.89 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\sqrt {1+\coth (x)}} \, dx=\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {\mathrm {coth}\left (x\right )+1}}{2}\right )}{2}-\frac {1}{\sqrt {\mathrm {coth}\left (x\right )+1}} \]

[In]

int(1/(coth(x) + 1)^(1/2),x)

[Out]

(2^(1/2)*atanh((2^(1/2)*(coth(x) + 1)^(1/2))/2))/2 - 1/(coth(x) + 1)^(1/2)