Integrand size = 11, antiderivative size = 29 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {3 x}{2}-\frac {3 \coth (x)}{2}-\log (\sinh (x))+\frac {\coth (x)}{2 (1+\tanh (x))} \] Output:
3/2*x-3/2*coth(x)-ln(sinh(x))+coth(x)/(2+2*tanh(x))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.20 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.62 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \left (\coth ^2(x)+\frac {\coth ^4(x)}{1+\coth (x)}-\coth ^3(x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\tanh ^2(x)\right )-2 (\log (\cosh (x))+\log (\tanh (x)))\right ) \] Input:
Integrate[Coth[x]^2/(1 + Tanh[x]),x]
Output:
(Coth[x]^2 + Coth[x]^4/(1 + Coth[x]) - Coth[x]^3*Hypergeometric2F1[-3/2, 1 , -1/2, Tanh[x]^2] - 2*(Log[Cosh[x]] + Log[Tanh[x]]))/2
Result contains complex when optimal does not.
Time = 0.41 (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.
\(\displaystyle \int \frac {\coth ^2(x)}{\tanh (x)+1} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {1}{(1-i \tan (i x)) \tan (i x)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {1}{(1-i \tan (i x)) \tan (i x)^2}dx\) |
\(\Big \downarrow \) 4035 |
\(\displaystyle \frac {1}{2} \int \coth ^2(x) (3-2 \tanh (x))dx+\frac {\coth (x)}{2 (\tanh (x)+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\coth (x)}{2 (\tanh (x)+1)}+\frac {1}{2} \int -\frac {2 i \tan (i x)+3}{\tan (i x)^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\coth (x)}{2 (\tanh (x)+1)}-\frac {1}{2} \int \frac {2 i \tan (i x)+3}{\tan (i x)^2}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {1}{2} (-\int \coth (x) (2-3 \tanh (x))dx-3 \coth (x))+\frac {\coth (x)}{2 (\tanh (x)+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\coth (x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-3 \coth (x)-\int \frac {i (3 i \tan (i x)+2)}{\tan (i x)}dx\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\coth (x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-3 \coth (x)-i \int \frac {3 i \tan (i x)+2}{\tan (i x)}dx\right )\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle \frac {\coth (x)}{2 (\tanh (x)+1)}+\frac {1}{2} (-3 \coth (x)-i (2 \int -i \coth (x)dx+3 i x))\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\coth (x)}{2 (\tanh (x)+1)}+\frac {1}{2} (-3 \coth (x)-i (3 i x-2 i \int \coth (x)dx))\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\coth (x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-3 \coth (x)-i \left (3 i x-2 i \int -i \tan \left (i x+\frac {\pi }{2}\right )dx\right )\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\coth (x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-3 \coth (x)-i \left (3 i x-2 \int \tan \left (i x+\frac {\pi }{2}\right )dx\right )\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {\coth (x)}{2 (\tanh (x)+1)}+\frac {1}{2} (-3 \coth (x)-i (3 i x-2 i \log (\sinh (x))))\) |
Input:
Int[Coth[x]^2/(1 + Tanh[x]),x]
Output:
(-3*Coth[x] - I*((3*I)*x - (2*I)*Log[Sinh[x]]))/2 + Coth[x]/(2*(1 + Tanh[x ]))
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 = 0.46 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03
method | result | size |
risch | \(\frac {5 x}{2}-\frac {{\mathrm e}^{-2 x}}{4}-\frac {2}{{\mathrm e}^{2 x}-1}-\ln \left ({\mathrm e}^{2 x}-1\right )\) | \(30\) |
parallelrisch | \(\frac {\left (2+2 \tanh \left (x \right )\right ) \ln \left (1-\tanh \left (x \right )\right )+\left (-2-2 \tanh \left (x \right )\right ) \ln \left (\tanh \left (x \right )\right )+5 \tanh \left (x \right ) x +5 x -2 \coth \left (x \right )-3}{2+2 \tanh \left (x \right )}\) | \(48\) |
default | \(-\frac {\tanh \left (\frac {x}{2}\right )}{2}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}-\frac {1}{2 \tanh \left (\frac {x}{2}\right )}-\ln \left (\tanh \left (\frac {x}{2}\right )\right )-\frac {1}{\left (1+\tanh \left (\frac {x}{2}\right )\right )^{2}}+\frac {1}{1+\tanh \left (\frac {x}{2}\right )}+\frac {5 \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{2}\) | \(59\) |
Input:
int(coth(x)^2/(1+tanh(x)),x,method=_RETURNVERBOSE)
Output:
5/2*x-1/4*exp(-2*x)-2/(exp(2*x)-1)-ln(exp(2*x)-1)
Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (23) = 46\).
Time = 0.09 (sec) , antiderivative size = 196, normalized size of antiderivative = 6.76 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {10 \, x \cosh \left (x\right )^{4} + 40 \, x \cosh \left (x\right ) \sinh \left (x\right )^{3} + 10 \, x \sinh \left (x\right )^{4} - {\left (10 \, x + 9\right )} \cosh \left (x\right )^{2} + {\left (60 \, x \cosh \left (x\right )^{2} - 10 \, x - 9\right )} \sinh \left (x\right )^{2} - 4 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + {\left (6 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - \cosh \left (x\right )^{2} + 2 \, {\left (2 \, \cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac {2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 2 \, {\left (20 \, x \cosh \left (x\right )^{3} - {\left (10 \, x + 9\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1}{4 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + {\left (6 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - \cosh \left (x\right )^{2} + 2 \, {\left (2 \, \cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right )\right )}} \] Input:
integrate(coth(x)^2/(1+tanh(x)),x, algorithm="fricas")
Output:
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*sinh(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))
\[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\int \frac {\coth ^{2}{\left (x \right )}}{\tanh {\left (x \right )} + 1}\, dx \] Input:
integrate(coth(x)**2/(1+tanh(x)),x)
Output:
Integral(coth(x)**2/(tanh(x) + 1), x)
Time = 0.04 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \, x + \frac {2}{e^{\left (-2 \, x\right )} - 1} - \frac {1}{4} \, e^{\left (-2 \, x\right )} - \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right ) \] Input:
integrate(coth(x)^2/(1+tanh(x)),x, algorithm="maxima")
Output:
1/2*x + 2/(e^(-2*x) - 1) - 1/4*e^(-2*x) - log(e^(-x) + 1) - log(e^(-x) - 1 )
Time = 0.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {5}{2} \, x - \frac {{\left (9 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-2 \, x\right )}}{4 \, {\left (e^{\left (2 \, x\right )} - 1\right )}} - \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \] Input:
integrate(coth(x)^2/(1+tanh(x)),x, algorithm="giac")
Output:
5/2*x - 1/4*(9*e^(2*x) - 1)*e^(-2*x)/(e^(2*x) - 1) - log(abs(e^(2*x) - 1))
Time = 2.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {5\,x}{2}-\ln \left ({\mathrm {e}}^{2\,x}-1\right )-\frac {{\mathrm {e}}^{-2\,x}}{4}-\frac {2}{{\mathrm {e}}^{2\,x}-1} \] Input:
int(coth(x)^2/(tanh(x) + 1),x)
Output:
(5*x)/2 - log(exp(2*x) - 1) - exp(-2*x)/4 - 2/(exp(2*x) - 1)
Time = 0.24 (sec) , antiderivative size = 95, normalized size of antiderivative = 3.28 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {-4 e^{4 x} \mathrm {log}\left (e^{x}-1\right )-4 e^{4 x} \mathrm {log}\left (e^{x}+1\right )+10 e^{4 x} x -9 e^{4 x}+4 e^{2 x} \mathrm {log}\left (e^{x}-1\right )+4 e^{2 x} \mathrm {log}\left (e^{x}+1\right )-10 e^{2 x} x +1}{4 e^{2 x} \left (e^{2 x}-1\right )} \] Input:
int(coth(x)^2/(1+tanh(x)),x)
Output:
( - 4*e**(4*x)*log(e**x - 1) - 4*e**(4*x)*log(e**x + 1) + 10*e**(4*x)*x - 9*e**(4*x) + 4*e**(2*x)*log(e**x - 1) + 4*e**(2*x)*log(e**x + 1) - 10*e**( 2*x)*x + 1)/(4*e**(2*x)*(e**(2*x) - 1))