Integrand size = 9, antiderivative size = 115 \[ \int \cos ^2(\tanh (a+b x)) \, dx=-\frac {\cos (2) \operatorname {CosIntegral}(2-2 \tanh (a+b x))}{4 b}+\frac {\cos (2) \operatorname {CosIntegral}(2+2 \tanh (a+b x))}{4 b}-\frac {\log (1-\tanh (a+b x))}{4 b}+\frac {\log (1+\tanh (a+b x))}{4 b}-\frac {\sin (2) \text {Si}(2-2 \tanh (a+b x))}{4 b}+\frac {\sin (2) \text {Si}(2+2 \tanh (a+b x))}{4 b} \]
-1/4*Ci(2-2*tanh(b*x+a))*cos(2)/b+1/4*Ci(2+2*tanh(b*x+a))*cos(2)/b-1/4*ln( 1-tanh(b*x+a))/b+1/4*ln(1+tanh(b*x+a))/b+1/4*Si(-2+2*tanh(b*x+a))*sin(2)/b +1/4*Si(2+2*tanh(b*x+a))*sin(2)/b
Time = 0.57 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.77 \[ \int \cos ^2(\tanh (a+b x)) \, dx=\frac {-\cos (2) \operatorname {CosIntegral}(2-2 \tanh (a+b x))+\cos (2) \operatorname {CosIntegral}(2 (1+\tanh (a+b x)))-\log (1-\tanh (a+b x))+\log (1+\tanh (a+b x))-\sin (2) \text {Si}(2-2 \tanh (a+b x))+\sin (2) \text {Si}(2 (1+\tanh (a+b x)))}{4 b} \]
(-(Cos[2]*CosIntegral[2 - 2*Tanh[a + b*x]]) + Cos[2]*CosIntegral[2*(1 + Ta nh[a + b*x])] - Log[1 - Tanh[a + b*x]] + Log[1 + Tanh[a + b*x]] - Sin[2]*S inIntegral[2 - 2*Tanh[a + b*x]] + Sin[2]*SinIntegral[2*(1 + Tanh[a + b*x]) ])/(4*b)
Time = 0.47 (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 = {4853, 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.
\(\displaystyle \int \cos ^2(\tanh (a+b x)) \, dx\) |
\(\Big \downarrow \) 4853 |
\(\displaystyle \frac {\int \frac {\cos ^2(\tanh (a+b x))}{1-\tanh ^2(a+b x)}d\tanh (a+b x)}{b}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {\int \left (\frac {\cos ^2(\tanh (a+b x))}{2 (\tanh (a+b x)+1)}-\frac {\cos ^2(\tanh (a+b x))}{2 (\tanh (a+b x)-1)}\right )d\tanh (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{4} \cos (2) \operatorname {CosIntegral}(2-2 \tanh (a+b x))+\frac {1}{4} \cos (2) \operatorname {CosIntegral}(2 \tanh (a+b x)+2)-\frac {1}{4} \sin (2) \text {Si}(2-2 \tanh (a+b x))+\frac {1}{4} \sin (2) \text {Si}(2 \tanh (a+b x)+2)-\frac {1}{4} \log (1-\tanh (a+b x))+\frac {1}{4} \log (\tanh (a+b x)+1)}{b}\) |
(-1/4*(Cos[2]*CosIntegral[2 - 2*Tanh[a + b*x]]) + (Cos[2]*CosIntegral[2 + 2*Tanh[a + b*x]])/4 - Log[1 - Tanh[a + b*x]]/4 + Log[1 + Tanh[a + b*x]]/4 - (Sin[2]*SinIntegral[2 - 2*Tanh[a + b*x]])/4 + (Sin[2]*SinIntegral[2 + 2* Tanh[a + b*x]])/4)/b
3.3.45.3.1 Defintions of rubi rules used
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 ]]
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]
Time = 0.35 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.77
method | result | size |
derivativedivides | \(\frac {\frac {\operatorname {Si}\left (-2+2 \tanh \left (b x +a \right )\right ) \sin \left (2\right )}{4}-\frac {\operatorname {Ci}\left (-2+2 \tanh \left (b x +a \right )\right ) \cos \left (2\right )}{4}+\frac {\operatorname {Si}\left (2+2 \tanh \left (b x +a \right )\right ) \sin \left (2\right )}{4}+\frac {\operatorname {Ci}\left (2+2 \tanh \left (b x +a \right )\right ) \cos \left (2\right )}{4}-\frac {\ln \left (-1+\tanh \left (b x +a \right )\right )}{4}+\frac {\ln \left (1+\tanh \left (b x +a \right )\right )}{4}}{b}\) | \(88\) |
default | \(\frac {\frac {\operatorname {Si}\left (-2+2 \tanh \left (b x +a \right )\right ) \sin \left (2\right )}{4}-\frac {\operatorname {Ci}\left (-2+2 \tanh \left (b x +a \right )\right ) \cos \left (2\right )}{4}+\frac {\operatorname {Si}\left (2+2 \tanh \left (b x +a \right )\right ) \sin \left (2\right )}{4}+\frac {\operatorname {Ci}\left (2+2 \tanh \left (b x +a \right )\right ) \cos \left (2\right )}{4}-\frac {\ln \left (-1+\tanh \left (b x +a \right )\right )}{4}+\frac {\ln \left (1+\tanh \left (b x +a \right )\right )}{4}}{b}\) | \(88\) |
risch | \(-\frac {{\mathrm e}^{-2 i} \operatorname {Ei}_{1}\left (\frac {4 i}{1+{\mathrm e}^{2 b x +2 a}}-4 i\right )}{8 b}+\frac {{\mathrm e}^{2 i} \operatorname {Ei}_{1}\left (\frac {4 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{8 b}-\frac {{\mathrm e}^{2 i} \operatorname {Ei}_{1}\left (-\frac {4 i}{1+{\mathrm e}^{2 b x +2 a}}+4 i\right )}{8 b}+\frac {{\mathrm e}^{-2 i} \operatorname {Ei}_{1}\left (-\frac {4 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{8 b}+\frac {x}{2}\) | \(115\) |
1/b*(1/4*Si(-2+2*tanh(b*x+a))*sin(2)-1/4*Ci(-2+2*tanh(b*x+a))*cos(2)+1/4*S i(2+2*tanh(b*x+a))*sin(2)+1/4*Ci(2+2*tanh(b*x+a))*cos(2)-1/4*ln(-1+tanh(b* x+a))+1/4*ln(1+tanh(b*x+a)))
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.00 \[ \int \cos ^2(\tanh (a+b x)) \, dx=\frac {4 \, b x \cos \left (2\right ) + 4 i \, b x \sin \left (2\right ) + {\left (\cos \left (2\right )^{2} + 2 i \, \cos \left (2\right ) \sin \left (2\right ) - \sin \left (2\right )^{2} + 1\right )} \operatorname {Ci}\left (\frac {2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{\cosh \left (b x + a\right )}\right ) - {\left (\cos \left (2\right )^{2} + 2 i \, \cos \left (2\right ) \sin \left (2\right ) - \sin \left (2\right )^{2} + 1\right )} \operatorname {Ci}\left (\frac {4}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1}\right ) + {\left (-i \, \cos \left (2\right )^{2} + 2 \, \cos \left (2\right ) \sin \left (2\right ) + i \, \sin \left (2\right )^{2} + i\right )} \operatorname {Si}\left (\frac {2 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{\cosh \left (b x + a\right )}\right ) + {\left (i \, \cos \left (2\right )^{2} - 2 \, \cos \left (2\right ) \sin \left (2\right ) - i \, \sin \left (2\right )^{2} - i\right )} \operatorname {Si}\left (\frac {4}{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} + 1}\right )}{8 \, {\left (b \cos \left (2\right ) + i \, b \sin \left (2\right )\right )}} \]
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))/cosh(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))/cosh(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))
\[ \int \cos ^2(\tanh (a+b x)) \, dx=\int \cos ^{2}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \]
\[ \int \cos ^2(\tanh (a+b x)) \, dx=\int { \cos \left (\tanh \left (b x + a\right )\right )^{2} \,d x } \]
\[ \int \cos ^2(\tanh (a+b x)) \, dx=\int { \cos \left (\tanh \left (b x + a\right )\right )^{2} \,d x } \]
Timed out. \[ \int \cos ^2(\tanh (a+b x)) \, dx=\int {\cos \left (\mathrm {tanh}\left (a+b\,x\right )\right )}^2 \,d x \]