Integrand size = 7, antiderivative size = 77 \[ \int \cos (\coth (a+b x)) \, dx=-\frac {\cos (1) \operatorname {CosIntegral}(1-\coth (a+b x))}{2 b}+\frac {\cos (1) \operatorname {CosIntegral}(1+\coth (a+b x))}{2 b}-\frac {\sin (1) \text {Si}(1-\coth (a+b x))}{2 b}+\frac {\sin (1) \text {Si}(1+\coth (a+b x))}{2 b} \] Output:
-1/2*cos(1)*Ci(1-coth(b*x+a))/b+1/2*cos(1)*Ci(1+coth(b*x+a))/b+1/2*sin(1)* Si(-1+coth(b*x+a))/b+1/2*sin(1)*Si(1+coth(b*x+a))/b
Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.81 \[ \int \cos (\coth (a+b x)) \, dx=-\frac {\cos (1) \operatorname {CosIntegral}(1-\coth (a+b x))-\cos (1) \operatorname {CosIntegral}(1+\coth (a+b x))+\sin (1) \text {Si}(1-\coth (a+b x))-\sin (1) \text {Si}(1+\coth (a+b x))}{2 b} \] Input:
Integrate[Cos[Coth[a + b*x]],x]
Output:
-1/2*(Cos[1]*CosIntegral[1 - Coth[a + b*x]] - Cos[1]*CosIntegral[1 + Coth[ a + b*x]] + Sin[1]*SinIntegral[1 - Coth[a + b*x]] - Sin[1]*SinIntegral[1 + Coth[a + b*x]])/b
Time = 0.34 (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 = {4852, 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.
\(\displaystyle \int \cos (\coth (a+b x)) \, dx\) |
\(\Big \downarrow \) 4852 |
\(\displaystyle \frac {\int \frac {\cos (\coth (a+b x))}{1-\coth ^2(a+b x)}d\coth (a+b x)}{b}\) |
\(\Big \downarrow \) 3815 |
\(\displaystyle \frac {\int \left (\frac {\cos (\coth (a+b x))}{2 (1-\coth (a+b x))}+\frac {\cos (\coth (a+b x))}{2 (\coth (a+b x)+1)}\right )d\coth (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{2} \cos (1) \operatorname {CosIntegral}(1-\coth (a+b x))+\frac {1}{2} \cos (1) \operatorname {CosIntegral}(\coth (a+b x)+1)-\frac {1}{2} \sin (1) \text {Si}(1-\coth (a+b x))+\frac {1}{2} \sin (1) \text {Si}(\coth (a+b x)+1)}{b}\) |
Input:
Int[Cos[Coth[a + b*x]],x]
Output:
(-1/2*(Cos[1]*CosIntegral[1 - Coth[a + b*x]]) + (Cos[1]*CosIntegral[1 + Co th[a + b*x]])/2 - (Sin[1]*SinIntegral[1 - Coth[a + b*x]])/2 + (Sin[1]*SinI ntegral[1 + Coth[a + b*x]])/2)/b
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])
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]]
Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {\frac {\operatorname {Si}\left (-1+\coth \left (b x +a \right )\right ) \sin \left (1\right )}{2}-\frac {\operatorname {Ci}\left (-1+\coth \left (b x +a \right )\right ) \cos \left (1\right )}{2}+\frac {\operatorname {Si}\left (1+\coth \left (b x +a \right )\right ) \sin \left (1\right )}{2}+\frac {\operatorname {Ci}\left (1+\coth \left (b x +a \right )\right ) \cos \left (1\right )}{2}}{b}\) | \(58\) |
default | \(\frac {\frac {\operatorname {Si}\left (-1+\coth \left (b x +a \right )\right ) \sin \left (1\right )}{2}-\frac {\operatorname {Ci}\left (-1+\coth \left (b x +a \right )\right ) \cos \left (1\right )}{2}+\frac {\operatorname {Si}\left (1+\coth \left (b x +a \right )\right ) \sin \left (1\right )}{2}+\frac {\operatorname {Ci}\left (1+\coth \left (b x +a \right )\right ) \cos \left (1\right )}{2}}{b}\) | \(58\) |
risch | \(-\frac {{\mathrm e}^{i} \operatorname {expIntegral}_{1}\left (\frac {2 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}+2 i\right )}{4 b}+\frac {{\mathrm e}^{-i} \operatorname {expIntegral}_{1}\left (\frac {2 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right )}{4 b}-\frac {{\mathrm e}^{-i} \operatorname {expIntegral}_{1}\left (-\frac {2 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}-2 i\right )}{4 b}-\frac {i \pi \,\operatorname {csgn}\left (\frac {{\mathrm e}^{-a}}{-{\mathrm e}^{2 b x +a}+{\mathrm e}^{-a}}\right ) {\mathrm e}^{i}}{4 b}-\frac {i \operatorname {Si}\left (\frac {2 \,{\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right ) {\mathrm e}^{i}}{2 b}+\frac {{\mathrm e}^{i} \operatorname {expIntegral}_{1}\left (\frac {2 i {\mathrm e}^{-a}}{{\mathrm e}^{2 b x +a}-{\mathrm e}^{-a}}\right )}{4 b}\) | \(204\) |
Input:
int(cos(coth(b*x+a)),x,method=_RETURNVERBOSE)
Output:
1/b*(1/2*Si(-1+coth(b*x+a))*sin(1)-1/2*Ci(-1+coth(b*x+a))*cos(1)+1/2*Si(1+ coth(b*x+a))*sin(1)+1/2*Ci(1+coth(b*x+a))*cos(1))
Result contains complex when optimal does not.
Time = 0.12 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.81 \[ \int \cos (\coth (a+b x)) \, dx=\frac {{\left (\cos \left (1\right )^{2} + 2 i \, \cos \left (1\right ) \sin \left (1\right ) - \sin \left (1\right )^{2} + 1\right )} \operatorname {Ci}\left (\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )}{\sinh \left (b x + a\right )}\right ) - {\left (\cos \left (1\right )^{2} + 2 i \, \cos \left (1\right ) \sin \left (1\right ) - \sin \left (1\right )^{2} + 1\right )} \operatorname {Ci}\left (\frac {2}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1}\right ) + {\left (-i \, \cos \left (1\right )^{2} + 2 \, \cos \left (1\right ) \sin \left (1\right ) + i \, \sin \left (1\right )^{2} + i\right )} \operatorname {Si}\left (\frac {\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )}{\sinh \left (b x + a\right )}\right ) + {\left (-i \, \cos \left (1\right )^{2} + 2 \, \cos \left (1\right ) \sin \left (1\right ) + i \, \sin \left (1\right )^{2} + i\right )} \operatorname {Si}\left (\frac {2}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1}\right )}{4 \, {\left (b \cos \left (1\right ) + i \, b \sin \left (1\right )\right )}} \] Input:
integrate(cos(coth(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))/sinh(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))/sinh(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))
\[ \int \cos (\coth (a+b x)) \, dx=\int \cos {\left (\coth {\left (a + b x \right )} \right )}\, dx \] Input:
integrate(cos(coth(b*x+a)),x)
Output:
Integral(cos(coth(a + b*x)), x)
\[ \int \cos (\coth (a+b x)) \, dx=\int { \cos \left (\coth \left (b x + a\right )\right ) \,d x } \] Input:
integrate(cos(coth(b*x+a)),x, algorithm="maxima")
Output:
integrate(cos(coth(b*x + a)), x)
\[ \int \cos (\coth (a+b x)) \, dx=\int { \cos \left (\coth \left (b x + a\right )\right ) \,d x } \] Input:
integrate(cos(coth(b*x+a)),x, algorithm="giac")
Output:
integrate(cos(coth(b*x + a)), x)
Timed out. \[ \int \cos (\coth (a+b x)) \, dx=\int \cos \left (\mathrm {coth}\left (a+b\,x\right )\right ) \,d x \] Input:
int(cos(coth(a + b*x)),x)
Output:
int(cos(coth(a + b*x)), x)
\[ \int \cos (\coth (a+b x)) \, dx=\int \cos \left (\coth \left (b x +a \right )\right )d x \] Input:
int(cos(coth(b*x+a)),x)
Output:
int(cos(coth(a + b*x)),x)