\(\int \cos ^2(\coth (a+b x)) \, dx\) [227]

Optimal result

Integrand size = 9, antiderivative size = 115 \[ \int \cos ^2(\coth (a+b x)) \, dx=-\frac {\cos (2) \operatorname {CosIntegral}(2-2 \coth (a+b x))}{4 b}+\frac {\cos (2) \operatorname {CosIntegral}(2+2 \coth (a+b x))}{4 b}-\frac {\log (1-\coth (a+b x))}{4 b}+\frac {\log (1+\coth (a+b x))}{4 b}-\frac {\sin (2) \text {Si}(2-2 \coth (a+b x))}{4 b}+\frac {\sin (2) \text {Si}(2+2 \coth (a+b x))}{4 b} \] Output:

-1/4*cos(2)*Ci(2-2*coth(b*x+a))/b+1/4*cos(2)*Ci(2+2*coth(b*x+a))/b-1/4*ln( 
1-coth(b*x+a))/b+1/4*ln(1+coth(b*x+a))/b+1/4*sin(2)*Si(-2+2*coth(b*x+a))/b 
+1/4*sin(2)*Si(2+2*coth(b*x+a))/b
 

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.77 \[ \int \cos ^2(\coth (a+b x)) \, dx=\frac {-\cos (2) \operatorname {CosIntegral}(2-2 \coth (a+b x))+\cos (2) \operatorname {CosIntegral}(2 (1+\coth (a+b x)))-\log (1-\coth (a+b x))+\log (1+\coth (a+b x))-\sin (2) \text {Si}(2-2 \coth (a+b x))+\sin (2) \text {Si}(2 (1+\coth (a+b x)))}{4 b} \] Input:

Integrate[Cos[Coth[a + b*x]]^2,x]
 

Output:

(-(Cos[2]*CosIntegral[2 - 2*Coth[a + b*x]]) + Cos[2]*CosIntegral[2*(1 + Co 
th[a + b*x])] - Log[1 - Coth[a + b*x]] + Log[1 + Coth[a + b*x]] - Sin[2]*S 
inIntegral[2 - 2*Coth[a + b*x]] + Sin[2]*SinIntegral[2*(1 + Coth[a + b*x]) 
])/(4*b)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.88, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4852, 7276, 2009}

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.

\(\displaystyle \int \cos ^2(\coth (a+b x)) \, dx\)

\(\Big \downarrow \) 4852

\(\displaystyle \frac {\int \frac {\cos ^2(\coth (a+b x))}{1-\coth ^2(a+b x)}d\coth (a+b x)}{b}\)

\(\Big \downarrow \) 7276

\(\displaystyle \frac {\int \left (\frac {\cos ^2(\coth (a+b x))}{2 (\coth (a+b x)+1)}-\frac {\cos ^2(\coth (a+b x))}{2 (\coth (a+b x)-1)}\right )d\coth (a+b x)}{b}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {1}{4} \cos (2) \operatorname {CosIntegral}(2-2 \coth (a+b x))+\frac {1}{4} \cos (2) \operatorname {CosIntegral}(2 \coth (a+b x)+2)-\frac {1}{4} \sin (2) \text {Si}(2-2 \coth (a+b x))+\frac {1}{4} \sin (2) \text {Si}(2 \coth (a+b x)+2)-\frac {1}{4} \log (1-\coth (a+b x))+\frac {1}{4} \log (\coth (a+b x)+1)}{b}\)

Input:

Int[Cos[Coth[a + b*x]]^2,x]
 

Output:

(-1/4*(Cos[2]*CosIntegral[2 - 2*Coth[a + b*x]]) + (Cos[2]*CosIntegral[2 + 
2*Coth[a + b*x]])/4 - Log[1 - Coth[a + b*x]]/4 + Log[1 + Coth[a + b*x]]/4 
- (Sin[2]*SinIntegral[2 - 2*Coth[a + b*x]])/4 + (Sin[2]*SinIntegral[2 + 2* 
Coth[a + b*x]])/4)/b
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4852

Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa 
ctors[Cot[v], x]}, -d/Coefficient[v, x, 1]   Subst[Int[SubstFor[1/(1 + d^2* 
x^2), Cot[v]/d, u, x], x], x, Cot[v]/d]], x] /;  !FalseQ[v] && FunctionOfQ[ 
NonfreeFactors[Cot[v], x], u, x, True] && TryPureTanSubst[ActivateTrig[u], 
x]]
 

rule 7276

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {\frac {\operatorname {Si}\left (-2+2 \coth \left (b x +a \right )\right ) \sin \left (2\right )}{4}-\frac {\operatorname {Ci}\left (-2+2 \coth \left (b x +a \right )\right ) \cos \left (2\right )}{4}+\frac {\operatorname {Si}\left (2+2 \coth \left (b x +a \right )\right ) \sin \left (2\right )}{4}+\frac {\operatorname {Ci}\left (2+2 \coth \left (b x +a \right )\right ) \cos \left (2\right )}{4}-\frac {\ln \left (-1+\coth \left (b x +a \right )\right )}{4}+\frac {\ln \left (1+\coth \left (b x +a \right )\right )}{4}}{b}\)	\(88\)
default	\(\frac {\frac {\operatorname {Si}\left (-2+2 \coth \left (b x +a \right )\right ) \sin \left (2\right )}{4}-\frac {\operatorname {Ci}\left (-2+2 \coth \left (b x +a \right )\right ) \cos \left (2\right )}{4}+\frac {\operatorname {Si}\left (2+2 \coth \left (b x +a \right )\right ) \sin \left (2\right )}{4}+\frac {\operatorname {Ci}\left (2+2 \coth \left (b x +a \right )\right ) \cos \left (2\right )}{4}-\frac {\ln \left (-1+\coth \left (b x +a \right )\right )}{4}+\frac {\ln \left (1+\coth \left (b x +a \right )\right )}{4}}{b}\)	\(88\)
risch	\(-\frac {{\mathrm e}^{2 i} \operatorname {expIntegral}_{1}\left (\frac {4 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}+4 i\right )}{8 b}+\frac {{\mathrm e}^{-2 i} \operatorname {expIntegral}_{1}\left (\frac {4 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right )}{8 b}-\frac {{\mathrm e}^{-2 i} \operatorname {expIntegral}_{1}\left (-\frac {4 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}-4 i\right )}{8 b}-\frac {i {\mathrm e}^{2 i} \pi \,\operatorname {csgn}\left (\frac {{\mathrm e}^{-a}}{-{\mathrm e}^{2 b x +a}+{\mathrm e}^{-a}}\right )}{8 b}-\frac {i {\mathrm e}^{2 i} \operatorname {Si}\left (\frac {4 \,{\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right )}{4 b}+\frac {{\mathrm e}^{2 i} \operatorname {expIntegral}_{1}\left (\frac {4 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right )}{8 b}+\frac {\ln \left ({\mathrm e}^{b x}\right )}{2 b}\)	\(214\)

Input:

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

Output:

1/b*(1/4*Si(-2+2*coth(b*x+a))*sin(2)-1/4*Ci(-2+2*coth(b*x+a))*cos(2)+1/4*S 
i(2+2*coth(b*x+a))*sin(2)+1/4*Ci(2+2*coth(b*x+a))*cos(2)-1/4*ln(-1+coth(b* 
x+a))+1/4*ln(1+coth(b*x+a)))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.00 \[ \int \cos ^2(\coth (a+b x)) \, dx=\frac {4 \, b x \cos \left (2\right ) + 4 i \, b x \sin \left (2\right ) + {\left (\cos \left (2\right )^{2} + 2 i \, \cos \left (2\right ) \sin \left (2\right ) - \sin \left (2\right )^{2} + 1\right )} \operatorname {Ci}\left (\frac {2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{\sinh \left (b x + a\right )}\right ) - {\left (\cos \left (2\right )^{2} + 2 i \, \cos \left (2\right ) \sin \left (2\right ) - \sin \left (2\right )^{2} + 1\right )} \operatorname {Ci}\left (\frac {4}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1}\right ) + {\left (-i \, \cos \left (2\right )^{2} + 2 \, \cos \left (2\right ) \sin \left (2\right ) + i \, \sin \left (2\right )^{2} + i\right )} \operatorname {Si}\left (\frac {2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{\sinh \left (b x + a\right )}\right ) + {\left (-i \, \cos \left (2\right )^{2} + 2 \, \cos \left (2\right ) \sin \left (2\right ) + i \, \sin \left (2\right )^{2} + i\right )} \operatorname {Si}\left (\frac {4}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1}\right )}{8 \, {\left (b \cos \left (2\right ) + i \, b \sin \left (2\right )\right )}} \] Input:

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

Output:

1/8*(4*b*x*cos(2) + 4*I*b*x*sin(2) + (cos(2)^2 + 2*I*cos(2)*sin(2) - sin(2 
)^2 + 1)*cos_integral(2*(cosh(b*x + a) + sinh(b*x + a))/sinh(b*x + a)) - ( 
cos(2)^2 + 2*I*cos(2)*sin(2) - sin(2)^2 + 1)*cos_integral(4/(cosh(b*x + a) 
^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1)) + (-I*cos(2)^2 
+ 2*cos(2)*sin(2) + I*sin(2)^2 + I)*sin_integral(2*(cosh(b*x + a) + sinh(b 
*x + a))/sinh(b*x + a)) + (-I*cos(2)^2 + 2*cos(2)*sin(2) + I*sin(2)^2 + I) 
*sin_integral(4/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b* 
x + a)^2 - 1)))/(b*cos(2) + I*b*sin(2))
 

Sympy [F]

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

integrate(cos(coth(b*x+a))**2,x)
 

Output:

Integral(cos(coth(a + b*x))**2, x)
 

Maxima [F]

\[ \int \cos ^2(\coth (a+b x)) \, dx=\int { \cos \left (\coth \left (b x + a\right )\right )^{2} \,d x } \] Input:

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

Output:

1/2*x + 1/2*integrate(cos(2*(e^(2*b*x + 2*a) + 1)/(e^(2*b*x + 2*a) - 1)), 
x)
 

Giac [F]

\[ \int \cos ^2(\coth (a+b x)) \, dx=\int { \cos \left (\coth \left (b x + a\right )\right )^{2} \,d x } \] Input:

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

Output:

integrate(cos(coth(b*x + a))^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \cos ^2(\coth (a+b x)) \, dx=\int {\cos \left (\mathrm {coth}\left (a+b\,x\right )\right )}^2 \,d x \] Input:

int(cos(coth(a + b*x))^2,x)
 

Output:

int(cos(coth(a + b*x))^2, x)
 

Reduce [F]

\[ \int \cos ^2(\coth (a+b x)) \, dx=\int \cos \left (\coth \left (b x +a \right )\right )^{2}d x \] Input:

int(cos(coth(b*x+a))^2,x)
 

Output:

int(cos(coth(a + b*x))**2,x)