Integrand size = 11, antiderivative size = 29 \[ \int \frac {\tanh ^2(x)}{1+\coth (x)} \, dx=\frac {3 x}{2}-\log (\cosh (x))-\frac {3 \tanh (x)}{2}+\frac {\tanh (x)}{2 (1+\coth (x))} \]
Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {\tanh ^2(x)}{1+\coth (x)} \, dx=\frac {1}{4} \left (-\log (1-\tanh (x))+5 \log (1+\tanh (x))+\left (-6+\frac {2}{1+\coth (x)}\right ) \tanh (x)\right ) \]
Result contains complex when optimal does not.
Time = 0.40 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.182, Rules used = {3042, 25, 4035, 3042, 25, 4012, 3042, 26, 4014, 26, 3042, 26, 3956}
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.
\(\displaystyle \int \frac {\tanh ^2(x)}{\coth (x)+1} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{\left (1-i \tan \left (\frac {\pi }{2}+i x\right )\right ) \tan \left (\frac {\pi }{2}+i x\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{\left (1-i \tan \left (i x+\frac {\pi }{2}\right )\right ) \tan \left (i x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 4035 |
\(\displaystyle \frac {1}{2} \int (3-2 \coth (x)) \tanh ^2(x)dx+\frac {\tanh (x)}{2 (\coth (x)+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh (x)}{2 (\coth (x)+1)}+\frac {1}{2} \int -\frac {2 i \tan \left (i x+\frac {\pi }{2}\right )+3}{\tan \left (i x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tanh (x)}{2 (\coth (x)+1)}-\frac {1}{2} \int \frac {2 i \tan \left (i x+\frac {\pi }{2}\right )+3}{\tan \left (i x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {1}{2} (-\int (2-3 \coth (x)) \tanh (x)dx-3 \tanh (x))+\frac {\tanh (x)}{2 (\coth (x)+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh (x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-3 \tanh (x)-\int \frac {i \left (3 i \tan \left (i x+\frac {\pi }{2}\right )+2\right )}{\tan \left (i x+\frac {\pi }{2}\right )}dx\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\tanh (x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-3 \tanh (x)-i \int \frac {3 i \tan \left (i x+\frac {\pi }{2}\right )+2}{\tan \left (i x+\frac {\pi }{2}\right )}dx\right )\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle \frac {\tanh (x)}{2 (\coth (x)+1)}+\frac {1}{2} (-3 \tanh (x)-i (2 \int -i \tanh (x)dx+3 i x))\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\tanh (x)}{2 (\coth (x)+1)}+\frac {1}{2} (-3 \tanh (x)-i (3 i x-2 i \int \tanh (x)dx))\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh (x)}{2 (\coth (x)+1)}+\frac {1}{2} (-3 \tanh (x)-i (3 i x-2 i \int -i \tan (i x)dx))\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\tanh (x)}{2 (\coth (x)+1)}+\frac {1}{2} (-3 \tanh (x)-i (3 i x-2 \int \tan (i x)dx))\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {\tanh (x)}{2 (\coth (x)+1)}+\frac {1}{2} (-3 \tanh (x)-i (3 i x-2 i \log (\cosh (x))))\) |
3.2.25.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ (f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) Int[(a + b*Tan[e + f*x] )^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a , b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 ]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a *d)/(a^2 + b^2) Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && N eQ[a*c + b*d, 0]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-a)*((c + d*Tan[e + f*x])^(n + 1)/(2*f*(b* c - a*d)*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a*(b*c - a*d)) Int[(c + d *Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !GtQ[n, 0]
Time = 0.13 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03
method | result | size |
risch | \(\frac {5 x}{2}+\frac {{\mathrm e}^{-2 x}}{4}+\frac {2}{1+{\mathrm e}^{2 x}}-\ln \left (1+{\mathrm e}^{2 x}\right )\) | \(30\) |
parallelrisch | \(\frac {\left (2+2 \tanh \left (x \right )\right ) \ln \left (1-\tanh \left (x \right )\right )+5 \tanh \left (x \right ) x -2 \tanh \left (x \right )^{2}+5 x +3}{2+2 \tanh \left (x \right )}\) | \(40\) |
default | \(-\frac {2 \tanh \left (\frac {x}{2}\right )}{1+\tanh \left (\frac {x}{2}\right )^{2}}-\ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {1}{\tanh \left (\frac {x}{2}\right )+1}+\frac {5 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}\) | \(65\) |
Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (23) = 46\).
Time = 0.33 (sec) , antiderivative size = 186, normalized size of antiderivative = 6.41 \[ \int \frac {\tanh ^2(x)}{1+\coth (x)} \, dx=\frac {10 \, x \cosh \left (x\right )^{4} + 40 \, x \cosh \left (x\right ) \sinh \left (x\right )^{3} + 10 \, x \sinh \left (x\right )^{4} + {\left (10 \, x + 9\right )} \cosh \left (x\right )^{2} + {\left (60 \, x \cosh \left (x\right )^{2} + 10 \, x + 9\right )} \sinh \left (x\right )^{2} - 4 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + {\left (6 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + \cosh \left (x\right )^{2} + 2 \, {\left (2 \, \cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 2 \, {\left (20 \, x \cosh \left (x\right )^{3} + {\left (10 \, x + 9\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1}{4 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + {\left (6 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + \cosh \left (x\right )^{2} + 2 \, {\left (2 \, \cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right )\right )}} \]
1/4*(10*x*cosh(x)^4 + 40*x*cosh(x)*sinh(x)^3 + 10*x*sinh(x)^4 + (10*x + 9) *cosh(x)^2 + (60*x*cosh(x)^2 + 10*x + 9)*sinh(x)^2 - 4*(cosh(x)^4 + 4*cosh (x)*sinh(x)^3 + sinh(x)^4 + (6*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*(2 *cosh(x)^3 + cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) + 2*(20* x*cosh(x)^3 + (10*x + 9)*cosh(x))*sinh(x) + 1)/(cosh(x)^4 + 4*cosh(x)*sinh (x)^3 + sinh(x)^4 + (6*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*(2*cosh(x) ^3 + cosh(x))*sinh(x))
\[ \int \frac {\tanh ^2(x)}{1+\coth (x)} \, dx=\int \frac {\tanh ^{2}{\left (x \right )}}{\coth {\left (x \right )} + 1}\, dx \]
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\tanh ^2(x)}{1+\coth (x)} \, dx=\frac {1}{2} \, x - \frac {2}{e^{\left (-2 \, x\right )} + 1} + \frac {1}{4} \, e^{\left (-2 \, x\right )} - \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]
Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {\tanh ^2(x)}{1+\coth (x)} \, dx=\frac {5}{2} \, x + \frac {{\left (9 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-2 \, x\right )}}{4 \, {\left (e^{\left (2 \, x\right )} + 1\right )}} - \log \left (e^{\left (2 \, x\right )} + 1\right ) \]
Time = 1.91 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\tanh ^2(x)}{1+\coth (x)} \, dx=\frac {5\,x}{2}-\ln \left ({\mathrm {e}}^{2\,x}+1\right )+\frac {{\mathrm {e}}^{-2\,x}}{4}+\frac {2}{{\mathrm {e}}^{2\,x}+1} \]