3.1.17 \(\int \sqrt {1+\text {csch}^2(x)} \, dx\) [17]

3.1.17.1 Optimal result
3.1.17.2 Mathematica [A] (verified)
3.1.17.3 Rubi [A] (verified)
3.1.17.4 Maple [C] (warning: unable to verify)
3.1.17.5 Fricas [A] (verification not implemented)
3.1.17.6 Sympy [F]
3.1.17.7 Maxima [A] (verification not implemented)
3.1.17.8 Giac [B] (verification not implemented)
3.1.17.9 Mupad [F(-1)]

3.1.17.1 Optimal result

Integrand size = 10, antiderivative size = 14 \[ \int \sqrt {1+\text {csch}^2(x)} \, dx=\sqrt {\coth ^2(x)} \log (\sinh (x)) \tanh (x) \]

output
ln(sinh(x))*(coth(x)^2)^(1/2)*tanh(x)
 
3.1.17.2 Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.29 \[ \int \sqrt {1+\text {csch}^2(x)} \, dx=\sqrt {\coth ^2(x)} (\log (\cosh (x))+\log (\tanh (x))) \tanh (x) \]

input
Integrate[Sqrt[1 + Csch[x]^2],x]
 
output
Sqrt[Coth[x]^2]*(Log[Cosh[x]] + Log[Tanh[x]])*Tanh[x]
 
3.1.17.3 Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {3042, 4609, 3042, 4141, 3042, 26, 3956}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \sqrt {\text {csch}^2(x)+1} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {1-\sec \left (\frac {\pi }{2}+i x\right )^2}dx\)

\(\Big \downarrow \) 4609

\(\displaystyle \int \sqrt {\coth ^2(x)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {-\tan \left (\frac {\pi }{2}+i x\right )^2}dx\)

\(\Big \downarrow \) 4141

\(\displaystyle \tanh (x) \sqrt {\coth ^2(x)} \int \coth (x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \tanh (x) \sqrt {\coth ^2(x)} \int -i \tan \left (i x+\frac {\pi }{2}\right )dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \tanh (x) \sqrt {\coth ^2(x)} \int \tan \left (i x+\frac {\pi }{2}\right )dx\)

\(\Big \downarrow \) 3956

\(\displaystyle \tanh (x) \sqrt {\coth ^2(x)} \log (\sinh (x))\)

input
Int[Sqrt[1 + Csch[x]^2],x]
 
output
Sqrt[Coth[x]^2]*Log[Sinh[x]]*Tanh[x]
 

3.1.17.3.1 Defintions of rubi rules used

rule 26
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3956
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d 
*x], x]]/d, x] /; FreeQ[{c, d}, x]
 

rule 4141
Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff 
= FreeFactors[Tan[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Tan[e + f*x]^ 
n)^FracPart[p]/(Tan[e + f*x]/ff)^(n*FracPart[p]))   Int[ActivateTrig[u]*(Ta 
n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] 
 && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / 
; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
 

rule 4609
Int[(u_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A 
ctivateTrig[u*(b*tan[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ 
[a + b, 0]
 
3.1.17.4 Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.40 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21

method result size
default \(-\frac {\operatorname {csgn}\left (\coth \left (x \right )\right ) \left (\ln \left (\coth \left (x \right )-1\right )+\ln \left (1+\coth \left (x \right )\right )\right )}{2}\) \(17\)
risch \(-\frac {\left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 x}\right )^{2}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}\, x}{1+{\mathrm e}^{2 x}}+\frac {\left ({\mathrm e}^{2 x}-1\right ) \sqrt {\frac {\left (1+{\mathrm e}^{2 x}\right )^{2}}{\left ({\mathrm e}^{2 x}-1\right )^{2}}}\, \ln \left ({\mathrm e}^{2 x}-1\right )}{1+{\mathrm e}^{2 x}}\) \(79\)

input
int((1+csch(x)^2)^(1/2),x,method=_RETURNVERBOSE)
 
output
-1/2*csgn(coth(x))*(ln(coth(x)-1)+ln(1+coth(x)))
 
3.1.17.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.29 \[ \int \sqrt {1+\text {csch}^2(x)} \, dx=-x + \log \left (\frac {2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) \]

input
integrate((1+csch(x)^2)^(1/2),x, algorithm="fricas")
 
output
-x + log(2*sinh(x)/(cosh(x) - sinh(x)))
 
3.1.17.6 Sympy [F]

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

input
integrate((1+csch(x)**2)**(1/2),x)
 
output
Integral(sqrt(csch(x)**2 + 1), x)
 
3.1.17.7 Maxima [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.57 \[ \int \sqrt {1+\text {csch}^2(x)} \, dx=-x - \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right ) \]

input
integrate((1+csch(x)^2)^(1/2),x, algorithm="maxima")
 
output
-x - log(e^(-x) + 1) - log(e^(-x) - 1)
 
3.1.17.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 27 vs. \(2 (12) = 24\).

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.93 \[ \int \sqrt {1+\text {csch}^2(x)} \, dx=-x \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) + \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \mathrm {sgn}\left (e^{\left (4 \, x\right )} - 1\right ) \]

input
integrate((1+csch(x)^2)^(1/2),x, algorithm="giac")
 
output
-x*sgn(e^(4*x) - 1) + log(abs(e^(2*x) - 1))*sgn(e^(4*x) - 1)
 
3.1.17.9 Mupad [F(-1)]

Timed out. \[ \int \sqrt {1+\text {csch}^2(x)} \, dx=\int \sqrt {\frac {1}{{\mathrm {sinh}\left (x\right )}^2}+1} \,d x \]

input
int((1/sinh(x)^2 + 1)^(1/2),x)
 
output
int((1/sinh(x)^2 + 1)^(1/2), x)