Integrand size = 9, antiderivative size = 157 \[ \int \cos ^3(\tanh (a+b x)) \, dx=-\frac {\cos (3) \operatorname {CosIntegral}(3-3 \tanh (a+b x))}{8 b}-\frac {3 \cos (1) \operatorname {CosIntegral}(1-\tanh (a+b x))}{8 b}+\frac {3 \cos (1) \operatorname {CosIntegral}(1+\tanh (a+b x))}{8 b}+\frac {\cos (3) \operatorname {CosIntegral}(3+3 \tanh (a+b x))}{8 b}-\frac {\sin (3) \text {Si}(3-3 \tanh (a+b x))}{8 b}-\frac {3 \sin (1) \text {Si}(1-\tanh (a+b x))}{8 b}+\frac {3 \sin (1) \text {Si}(1+\tanh (a+b x))}{8 b}+\frac {\sin (3) \text {Si}(3+3 \tanh (a+b x))}{8 b} \]
-3/8*Ci(1-tanh(b*x+a))*cos(1)/b+3/8*Ci(1+tanh(b*x+a))*cos(1)/b-1/8*Ci(3-3* tanh(b*x+a))*cos(3)/b+1/8*Ci(3+3*tanh(b*x+a))*cos(3)/b+3/8*Si(-1+tanh(b*x+ a))*sin(1)/b+3/8*Si(1+tanh(b*x+a))*sin(1)/b+1/8*Si(-3+3*tanh(b*x+a))*sin(3 )/b+1/8*Si(3+3*tanh(b*x+a))*sin(3)/b
Time = 0.71 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.79 \[ \int \cos ^3(\tanh (a+b x)) \, dx=\frac {-2 \cos (3) \operatorname {CosIntegral}(3-3 \tanh (a+b x))-6 \cos (1) \operatorname {CosIntegral}(1-\tanh (a+b x))+6 \cos (1) \operatorname {CosIntegral}(1+\tanh (a+b x))+2 \cos (3) \operatorname {CosIntegral}(3+3 \tanh (a+b x))-2 \sin (3) \text {Si}(3-3 \tanh (a+b x))-6 \sin (1) \text {Si}(1-\tanh (a+b x))+6 \sin (1) \text {Si}(1+\tanh (a+b x))+2 \sin (3) \text {Si}(3+3 \tanh (a+b x))}{16 b} \]
(-2*Cos[3]*CosIntegral[3 - 3*Tanh[a + b*x]] - 6*Cos[1]*CosIntegral[1 - Tan h[a + b*x]] + 6*Cos[1]*CosIntegral[1 + Tanh[a + b*x]] + 2*Cos[3]*CosIntegr al[3 + 3*Tanh[a + b*x]] - 2*Sin[3]*SinIntegral[3 - 3*Tanh[a + b*x]] - 6*Si n[1]*SinIntegral[1 - Tanh[a + b*x]] + 6*Sin[1]*SinIntegral[1 + Tanh[a + b* x]] + 2*Sin[3]*SinIntegral[3 + 3*Tanh[a + b*x]])/(16*b)
Time = 0.58 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.87, 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.
\(\displaystyle \int \cos ^3(\tanh (a+b x)) \, dx\) |
\(\Big \downarrow \) 4853 |
\(\displaystyle \frac {\int \frac {\cos ^3(\tanh (a+b x))}{1-\tanh ^2(a+b x)}d\tanh (a+b x)}{b}\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \frac {\int \left (\frac {\cos ^3(\tanh (a+b x))}{2 (\tanh (a+b x)+1)}-\frac {\cos ^3(\tanh (a+b x))}{2 (\tanh (a+b x)-1)}\right )d\tanh (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{8} \cos (3) \operatorname {CosIntegral}(3-3 \tanh (a+b x))-\frac {3}{8} \cos (1) \operatorname {CosIntegral}(1-\tanh (a+b x))+\frac {3}{8} \cos (1) \operatorname {CosIntegral}(\tanh (a+b x)+1)+\frac {1}{8} \cos (3) \operatorname {CosIntegral}(3 \tanh (a+b x)+3)-\frac {1}{8} \sin (3) \text {Si}(3-3 \tanh (a+b x))-\frac {3}{8} \sin (1) \text {Si}(1-\tanh (a+b x))+\frac {3}{8} \sin (1) \text {Si}(\tanh (a+b x)+1)+\frac {1}{8} \sin (3) \text {Si}(3 \tanh (a+b x)+3)}{b}\) |
(-1/8*(Cos[3]*CosIntegral[3 - 3*Tanh[a + b*x]]) - (3*Cos[1]*CosIntegral[1 - Tanh[a + b*x]])/8 + (3*Cos[1]*CosIntegral[1 + Tanh[a + b*x]])/8 + (Cos[3 ]*CosIntegral[3 + 3*Tanh[a + b*x]])/8 - (Sin[3]*SinIntegral[3 - 3*Tanh[a + b*x]])/8 - (3*Sin[1]*SinIntegral[1 - Tanh[a + b*x]])/8 + (3*Sin[1]*SinInt egral[1 + Tanh[a + b*x]])/8 + (Sin[3]*SinIntegral[3 + 3*Tanh[a + b*x]])/8) /b
3.3.44.3.1 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 = 0.81 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {\frac {\operatorname {Si}\left (-3+3 \tanh \left (b x +a \right )\right ) \sin \left (3\right )}{8}-\frac {\operatorname {Ci}\left (-3+3 \tanh \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\operatorname {Si}\left (3+3 \tanh \left (b x +a \right )\right ) \sin \left (3\right )}{8}+\frac {\operatorname {Ci}\left (3+3 \tanh \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {3 \,\operatorname {Si}\left (-1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{8}-\frac {3 \,\operatorname {Ci}\left (-1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{8}+\frac {3 \,\operatorname {Si}\left (1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{8}+\frac {3 \,\operatorname {Ci}\left (1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{8}}{b}\) | \(118\) |
default | \(\frac {\frac {\operatorname {Si}\left (-3+3 \tanh \left (b x +a \right )\right ) \sin \left (3\right )}{8}-\frac {\operatorname {Ci}\left (-3+3 \tanh \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {\operatorname {Si}\left (3+3 \tanh \left (b x +a \right )\right ) \sin \left (3\right )}{8}+\frac {\operatorname {Ci}\left (3+3 \tanh \left (b x +a \right )\right ) \cos \left (3\right )}{8}+\frac {3 \,\operatorname {Si}\left (-1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{8}-\frac {3 \,\operatorname {Ci}\left (-1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{8}+\frac {3 \,\operatorname {Si}\left (1+\tanh \left (b x +a \right )\right ) \sin \left (1\right )}{8}+\frac {3 \,\operatorname {Ci}\left (1+\tanh \left (b x +a \right )\right ) \cos \left (1\right )}{8}}{b}\) | \(118\) |
risch | \(-\frac {{\mathrm e}^{3 i} \operatorname {Ei}_{1}\left (-\frac {6 i}{1+{\mathrm e}^{2 b x +2 a}}+6 i\right )}{16 b}+\frac {{\mathrm e}^{3 i} \operatorname {Ei}_{1}\left (\frac {6 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{16 b}+\frac {{\mathrm e}^{-3 i} \operatorname {Ei}_{1}\left (-\frac {6 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{16 b}-\frac {{\mathrm e}^{-3 i} \operatorname {Ei}_{1}\left (\frac {6 i}{1+{\mathrm e}^{2 b x +2 a}}-6 i\right )}{16 b}-\frac {3 \,{\mathrm e}^{-i} \operatorname {Ei}_{1}\left (\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}-2 i\right )}{16 b}+\frac {3 \,{\mathrm e}^{i} \operatorname {Ei}_{1}\left (\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{16 b}-\frac {3 \,{\mathrm e}^{i} \operatorname {Ei}_{1}\left (-\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}+2 i\right )}{16 b}+\frac {3 \,{\mathrm e}^{-i} \operatorname {Ei}_{1}\left (-\frac {2 i}{1+{\mathrm e}^{2 b x +2 a}}\right )}{16 b}\) | \(222\) |
1/b*(1/8*Si(-3+3*tanh(b*x+a))*sin(3)-1/8*Ci(-3+3*tanh(b*x+a))*cos(3)+1/8*S i(3+3*tanh(b*x+a))*sin(3)+1/8*Ci(3+3*tanh(b*x+a))*cos(3)+3/8*Si(-1+tanh(b* x+a))*sin(1)-3/8*Ci(-1+tanh(b*x+a))*cos(1)+3/8*Si(1+tanh(b*x+a))*sin(1)+3/ 8*Ci(1+tanh(b*x+a))*cos(1))
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 698, normalized size of antiderivative = 4.45 \[ \int \cos ^3(\tanh (a+b x)) \, dx=\text {Too large to display} \]
1/16*((cos(3)^2*cos(1) - (cos(1) + I*sin(1))*sin(3)^2 + 2*I*(cos(3)*cos(1) + I*cos(3)*sin(1))*sin(3) + I*(cos(3)^2 + 1)*sin(1) + cos(1))*cos_integra l(3*(cosh(b*x + a) + sinh(b*x + a))/cosh(b*x + a)) - 3*(-2*I*cos(3)*cos(1) *sin(1) + cos(3)*sin(1)^2 - (cos(1)^2 + 1)*cos(3) - I*(cos(1)^2 + 2*I*cos( 1)*sin(1) - sin(1)^2 + 1)*sin(3))*cos_integral((cosh(b*x + a) + sinh(b*x + a))/cosh(b*x + a)) - (cos(3)^2*cos(1) - (cos(1) + I*sin(1))*sin(3)^2 + 2* I*(cos(3)*cos(1) + I*cos(3)*sin(1))*sin(3) + I*(cos(3)^2 + 1)*sin(1) + cos (1))*cos_integral(6/(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sin h(b*x + a)^2 + 1)) - 3*(2*I*cos(3)*cos(1)*sin(1) - cos(3)*sin(1)^2 + (cos( 1)^2 + 1)*cos(3) + I*(cos(1)^2 + 2*I*cos(1)*sin(1) - sin(1)^2 + 1)*sin(3)) *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(3)^2*cos(1) - (-I*cos(1) + sin(1))*sin(3)^2 - 2*I *(I*cos(3)*cos(1) - cos(3)*sin(1))*sin(3) + I*(-I*cos(3)^2 + I)*sin(1) + I *cos(1))*sin_integral(3*(cosh(b*x + a) + sinh(b*x + a))/cosh(b*x + a)) + 3 *(2*cos(3)*cos(1)*sin(1) + I*cos(3)*sin(1)^2 - (I*cos(1)^2 - I)*cos(3) - I *(I*cos(1)^2 - 2*cos(1)*sin(1) - I*sin(1)^2 - I)*sin(3))*sin_integral((cos h(b*x + a) + sinh(b*x + a))/cosh(b*x + a)) + (I*cos(3)^2*cos(1) - (I*cos(1 ) - sin(1))*sin(3)^2 - 2*I*(-I*cos(3)*cos(1) + cos(3)*sin(1))*sin(3) + I*( I*cos(3)^2 - I)*sin(1) - I*cos(1))*sin_integral(6/(cosh(b*x + a)^2 + 2*cos h(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 1)) - 3*(2*cos(3)*cos(1)*s...
\[ \int \cos ^3(\tanh (a+b x)) \, dx=\int \cos ^{3}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \]
\[ \int \cos ^3(\tanh (a+b x)) \, dx=\int { \cos \left (\tanh \left (b x + a\right )\right )^{3} \,d x } \]
\[ \int \cos ^3(\tanh (a+b x)) \, dx=\int { \cos \left (\tanh \left (b x + a\right )\right )^{3} \,d x } \]
Timed out. \[ \int \cos ^3(\tanh (a+b x)) \, dx=\int {\cos \left (\mathrm {tanh}\left (a+b\,x\right )\right )}^3 \,d x \]