\(\int \cos (\tanh (a+b x)) \, dx\) [257]

Optimal result

Integrand size = 7, antiderivative size = 77 \[ \int \cos (\tanh (a+b x)) \, dx=-\frac {\cos (1) \operatorname {CosIntegral}(1-\tanh (a+b x))}{2 b}+\frac {\cos (1) \operatorname {CosIntegral}(1+\tanh (a+b x))}{2 b}-\frac {\sin (1) \text {Si}(1-\tanh (a+b x))}{2 b}+\frac {\sin (1) \text {Si}(1+\tanh (a+b x))}{2 b} \] Output:

-1/2*cos(1)*Ci(1-tanh(b*x+a))/b+1/2*cos(1)*Ci(1+tanh(b*x+a))/b+1/2*sin(1)* 
Si(-1+tanh(b*x+a))/b+1/2*sin(1)*Si(1+tanh(b*x+a))/b
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.81 \[ \int \cos (\tanh (a+b x)) \, dx=\frac {-\cos (1) \operatorname {CosIntegral}(1-\tanh (a+b x))+\cos (1) \operatorname {CosIntegral}(1+\tanh (a+b x))-\sin (1) \text {Si}(1-\tanh (a+b x))+\sin (1) \text {Si}(1+\tanh (a+b x))}{2 b} \] Input:

Integrate[Cos[Tanh[a + b*x]],x]
 

Output:

(-(Cos[1]*CosIntegral[1 - Tanh[a + b*x]]) + Cos[1]*CosIntegral[1 + Tanh[a 
+ b*x]] - Sin[1]*SinIntegral[1 - Tanh[a + b*x]] + Sin[1]*SinIntegral[1 + T 
anh[a + b*x]])/(2*b)
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4853, 3815, 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.

\(\displaystyle \int \cos (\tanh (a+b x)) \, dx\)

\(\Big \downarrow \) 4853

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

\(\Big \downarrow \) 3815

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

\(\Big \downarrow \) 2009

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

Input:

Int[Cos[Tanh[a + b*x]],x]
 

Output:

(-1/2*(Cos[1]*CosIntegral[1 - Tanh[a + b*x]]) + (Cos[1]*CosIntegral[1 + Ta 
nh[a + b*x]])/2 - (Sin[1]*SinIntegral[1 - Tanh[a + b*x]])/2 + (Sin[1]*SinI 
ntegral[1 + Tanh[a + b*x]])/2)/b
 

Defintions of rubi rules used

rule 2009

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

rule 3815

Int[Cos[(c_.) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int 
[ExpandIntegrand[Cos[c + d*x], (a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d}, 
 x] && ILtQ[p, 0] && IGtQ[n, 0] && (EqQ[n, 2] || EqQ[p, -1])
 

rule 4853

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

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {\frac {\operatorname {Si}\left (1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{2}+\frac {\operatorname {Ci}\left (1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{2}+\frac {\operatorname {Si}\left (-1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{2}-\frac {\operatorname {Ci}\left (-1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{2}}{b}\)	\(58\)
default	\(\frac {\frac {\operatorname {Si}\left (1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{2}+\frac {\operatorname {Ci}\left (1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{2}+\frac {\operatorname {Si}\left (-1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{2}-\frac {\operatorname {Ci}\left (-1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{2}}{b}\)	\(58\)
risch	\(\frac {{\mathrm e}^{i} \operatorname {expIntegral}_{1}\left (\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{4 b}-\frac {{\mathrm e}^{-i} \operatorname {expIntegral}_{1}\left (\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}-2 i\right )}{4 b}+\frac {{\mathrm e}^{-i} \operatorname {expIntegral}_{1}\left (-\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{4 b}-\frac {{\mathrm e}^{i} \operatorname {expIntegral}_{1}\left (-\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}+2 i\right )}{4 b}\)	\(112\)

Input:

int(cos(tanh(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

1/b*(1/2*Si(1+tanh(b*x+a))*sin(1)+1/2*Ci(1+tanh(b*x+a))*cos(1)+1/2*Si(-1+t 
anh(b*x+a))*sin(1)-1/2*Ci(-1+tanh(b*x+a))*cos(1))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.81 \[ \int \cos (\tanh (a+b x)) \, dx=\frac {{\left (\cos \left (1\right )^{2} + 2 i \, \cos \left (1\right ) \sin \left (1\right ) - \sin \left (1\right )^{2} + 1\right )} \operatorname {Ci}\left (\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )}{\cosh \left (b x + a\right )}\right ) - {\left (\cos \left (1\right )^{2} + 2 i \, \cos \left (1\right ) \sin \left (1\right ) - \sin \left (1\right )^{2} + 1\right )} \operatorname {Ci}\left (\frac {2}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1}\right ) + {\left (-i \, \cos \left (1\right )^{2} + 2 \, \cos \left (1\right ) \sin \left (1\right ) + i \, \sin \left (1\right )^{2} + i\right )} \operatorname {Si}\left (\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )}{\cosh \left (b x + a\right )}\right ) + {\left (i \, \cos \left (1\right )^{2} - 2 \, \cos \left (1\right ) \sin \left (1\right ) - i \, \sin \left (1\right )^{2} - i\right )} \operatorname {Si}\left (\frac {2}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1}\right )}{4 \, {\left (b \cos \left (1\right ) + i \, b \sin \left (1\right )\right )}} \] Input:

integrate(cos(tanh(b*x+a)),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(cos(tanh(b*x+a)),x)
 

Output:

Integral(cos(tanh(a + b*x)), x)
 

Maxima [F]

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

integrate(cos(tanh(b*x+a)),x, algorithm="maxima")
 

Output:

integrate(cos(tanh(b*x + a)), x)
 

Giac [F]

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

integrate(cos(tanh(b*x+a)),x, algorithm="giac")
 

Output:

integrate(cos(tanh(b*x + a)), x)
 

Mupad [F(-1)]

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

int(cos(tanh(a + b*x)),x)
 

Output:

int(cos(tanh(a + b*x)), x)
 

Reduce [F]

\[ \int \cos (\tanh (a+b x)) \, dx=\int \cos \left (\tanh \left (b x +a \right )\right )d x \] Input:

int(cos(tanh(b*x+a)),x)
 

Output:

int(cos(tanh(a + b*x)),x)