Integrand size = 11, antiderivative size = 43 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\frac {5 x}{2}-\frac {5 \coth (x)}{2}+\coth ^2(x)-\frac {5 \coth ^3(x)}{6}-2 \log (\sinh (x))+\frac {\coth ^3(x)}{2 (1+\tanh (x))} \] Output:
5/2*x-5/2*coth(x)+coth(x)^2-5/6*coth(x)^3-2*ln(sinh(x))+coth(x)^3/(2+2*tan h(x))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.23 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \left (2 \coth ^2(x)+\coth ^4(x)+\frac {\coth ^6(x)}{1+\coth (x)}-\coth ^5(x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\tanh ^2(x)\right )-4 (\log (\cosh (x))+\log (\tanh (x)))\right ) \] Input:
Integrate[Coth[x]^4/(1 + Tanh[x]),x]
Output:
(2*Coth[x]^2 + Coth[x]^4 + Coth[x]^6/(1 + Coth[x]) - Coth[x]^5*Hypergeomet ric2F1[-5/2, 1, -3/2, Tanh[x]^2] - 4*(Log[Cosh[x]] + Log[Tanh[x]]))/2
Result contains complex when optimal does not.
Time = 0.61 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.56, number of steps used = 20, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.818, Rules used = {3042, 4035, 25, 3042, 4012, 25, 3042, 26, 4012, 26, 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.
| \(\displaystyle \int \frac {\coth ^4(x)}{\tanh (x)+1} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(1-i \tan (i x)) \tan (i x)^4}dx\) |
| \(\Big \downarrow \) 4035 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}-\frac {1}{2} \int -\coth ^4(x) (5-4 \tanh (x))dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \int \coth ^4(x) (5-4 \tanh (x))dx+\frac {\coth ^3(x)}{2 (\tanh (x)+1)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \int \frac {4 i \tan (i x)+5}{\tan (i x)^4}dx\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {1}{2} \left (\int -\coth ^3(x) (4-5 \tanh (x))dx-\frac {5 \coth ^3(x)}{3}\right )+\frac {\coth ^3(x)}{2 (\tanh (x)+1)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} \left (-\int \coth ^3(x) (4-5 \tanh (x))dx-\frac {5}{3} \coth ^3(x)\right )+\frac {\coth ^3(x)}{2 (\tanh (x)+1)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5}{3} \coth ^3(x)-\int -\frac {i (5 i \tan (i x)+4)}{\tan (i x)^3}dx\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \int \frac {5 i \tan (i x)+4}{\tan (i x)^3}dx\right )\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \left (\int -i \coth ^2(x) (5-4 \tanh (x))dx-2 i \coth ^2(x)\right )\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \left (-i \int \coth ^2(x) (5-4 \tanh (x))dx-2 i \coth ^2(x)\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \left (-i \int -\frac {4 i \tan (i x)+5}{\tan (i x)^2}dx-2 i \coth ^2(x)\right )\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \left (i \int \frac {4 i \tan (i x)+5}{\tan (i x)^2}dx-2 i \coth ^2(x)\right )\right )\) |
| \(\Big \downarrow \) 4012 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \left (i (\int \coth (x) (4-5 \tanh (x))dx+5 \coth (x))-2 i \coth ^2(x)\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \left (i \left (5 \coth (x)+\int \frac {i (5 i \tan (i x)+4)}{\tan (i x)}dx\right )-2 i \coth ^2(x)\right )\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \left (i \left (5 \coth (x)+i \int \frac {5 i \tan (i x)+4}{\tan (i x)}dx\right )-2 i \coth ^2(x)\right )\right )\) |
| \(\Big \downarrow \) 4014 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \left (i (5 \coth (x)+i (4 \int -i \coth (x)dx+5 i x))-2 i \coth ^2(x)\right )\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \left (i (5 \coth (x)+i (5 i x-4 i \int \coth (x)dx))-2 i \coth ^2(x)\right )\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \left (i \left (5 \coth (x)+i \left (5 i x-4 i \int -i \tan \left (i x+\frac {\pi }{2}\right )dx\right )\right )-2 i \coth ^2(x)\right )\right )\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \left (i \left (5 \coth (x)+i \left (5 i x-4 \int \tan \left (i x+\frac {\pi }{2}\right )dx\right )\right )-2 i \coth ^2(x)\right )\right )\) |
| \(\Big \downarrow \) 3956 |
| \(\displaystyle \frac {\coth ^3(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5 \coth ^3(x)}{3}+i \left (i (5 \coth (x)+i (5 i x-4 i \log (\sinh (x))))-2 i \coth ^2(x)\right )\right )\) |
Input:
Int[Coth[x]^4/(1 + Tanh[x]),x]
Output:
((-5*Coth[x]^3)/3 + I*((-2*I)*Coth[x]^2 + I*(5*Coth[x] + I*((5*I)*x - (4*I )*Log[Sinh[x]]))))/2 + Coth[x]^3/(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 = 0.54 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02
| method | result | size |
| risch | \(\frac {9 x}{2}-\frac {{\mathrm e}^{-2 x}}{4}-\frac {2 \left (6 \,{\mathrm e}^{4 x}-9 \,{\mathrm e}^{2 x}+7\right )}{3 \left ({\mathrm e}^{2 x}-1\right )^{3}}-2 \ln \left ({\mathrm e}^{2 x}-1\right )\) | \(44\) |
| parallelrisch | \(\frac {\left (12 \tanh \left (x \right )+12\right ) \ln \left (1-\tanh \left (x \right )\right )+\left (-12 \tanh \left (x \right )-12\right ) \ln \left (\tanh \left (x \right )\right )-2 \coth \left (x \right )^{3}+27 \tanh \left (x \right ) x +\coth \left (x \right )^{2}+27 x -9 \coth \left (x \right )-15}{6+6 \tanh \left (x \right )}\) | \(58\) |
| default | \(-\frac {\tanh \left (\frac {x}{2}\right )^{3}}{24}+\frac {\tanh \left (\frac {x}{2}\right )^{2}}{8}-\frac {9 \tanh \left (\frac {x}{2}\right )}{8}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}-\frac {1}{24 \tanh \left (\frac {x}{2}\right )^{3}}+\frac {1}{8 \tanh \left (\frac {x}{2}\right )^{2}}-\frac {9}{8 \tanh \left (\frac {x}{2}\right )}-2 \ln \left (\tanh \left (\frac {x}{2}\right )\right )-\frac {1}{\left (1+\tanh \left (\frac {x}{2}\right )\right )^{2}}+\frac {1}{1+\tanh \left (\frac {x}{2}\right )}+\frac {9 \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{2}\) | \(91\) |
Input:
int(coth(x)^4/(1+tanh(x)),x,method=_RETURNVERBOSE)
Output:
9/2*x-1/4*exp(-2*x)-2/3*(6*exp(4*x)-9*exp(2*x)+7)/(exp(2*x)-1)^3-2*ln(exp( 2*x)-1)
Leaf count of result is larger than twice the leaf count of optimal. 582 vs. \(2 (35) = 70\).
Time = 0.10 (sec) , antiderivative size = 582, normalized size of antiderivative = 13.53 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\text {Too large to display} \] Input:
integrate(coth(x)^4/(1+tanh(x)),x, algorithm="fricas")
Output:
1/12*(54*x*cosh(x)^8 + 432*x*cosh(x)*sinh(x)^7 + 54*x*sinh(x)^8 - 3*(54*x + 17)*cosh(x)^6 + 3*(504*x*cosh(x)^2 - 54*x - 17)*sinh(x)^6 + 18*(168*x*co sh(x)^3 - (54*x + 17)*cosh(x))*sinh(x)^5 + 81*(2*x + 1)*cosh(x)^4 + 9*(420 *x*cosh(x)^4 - 5*(54*x + 17)*cosh(x)^2 + 18*x + 9)*sinh(x)^4 + 12*(252*x*c osh(x)^5 - 5*(54*x + 17)*cosh(x)^3 + 27*(2*x + 1)*cosh(x))*sinh(x)^3 - (54 *x + 65)*cosh(x)^2 + (1512*x*cosh(x)^6 - 45*(54*x + 17)*cosh(x)^4 + 486*(2 *x + 1)*cosh(x)^2 - 54*x - 65)*sinh(x)^2 - 24*(cosh(x)^8 + 8*cosh(x)*sinh( x)^7 + sinh(x)^8 + (28*cosh(x)^2 - 3)*sinh(x)^6 - 3*cosh(x)^6 + 2*(28*cosh (x)^3 - 9*cosh(x))*sinh(x)^5 + (70*cosh(x)^4 - 45*cosh(x)^2 + 3)*sinh(x)^4 + 3*cosh(x)^4 + 4*(14*cosh(x)^5 - 15*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + ( 28*cosh(x)^6 - 45*cosh(x)^4 + 18*cosh(x)^2 - 1)*sinh(x)^2 - cosh(x)^2 + 2* (4*cosh(x)^7 - 9*cosh(x)^5 + 6*cosh(x)^3 - cosh(x))*sinh(x))*log(2*sinh(x) /(cosh(x) - sinh(x))) + 2*(216*x*cosh(x)^7 - 9*(54*x + 17)*cosh(x)^5 + 162 *(2*x + 1)*cosh(x)^3 - (54*x + 65)*cosh(x))*sinh(x) + 3)/(cosh(x)^8 + 8*co sh(x)*sinh(x)^7 + sinh(x)^8 + (28*cosh(x)^2 - 3)*sinh(x)^6 - 3*cosh(x)^6 + 2*(28*cosh(x)^3 - 9*cosh(x))*sinh(x)^5 + (70*cosh(x)^4 - 45*cosh(x)^2 + 3 )*sinh(x)^4 + 3*cosh(x)^4 + 4*(14*cosh(x)^5 - 15*cosh(x)^3 + 3*cosh(x))*si nh(x)^3 + (28*cosh(x)^6 - 45*cosh(x)^4 + 18*cosh(x)^2 - 1)*sinh(x)^2 - cos h(x)^2 + 2*(4*cosh(x)^7 - 9*cosh(x)^5 + 6*cosh(x)^3 - cosh(x))*sinh(x))
\[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\int \frac {\coth ^{4}{\left (x \right )}}{\tanh {\left (x \right )} + 1}\, dx \] Input:
integrate(coth(x)**4/(1+tanh(x)),x)
Output:
Integral(coth(x)**4/(tanh(x) + 1), x)
Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.49 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \, x - \frac {2 \, {\left (15 \, e^{\left (-2 \, x\right )} - 12 \, e^{\left (-4 \, x\right )} - 7\right )}}{3 \, {\left (3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1\right )}} - \frac {1}{4} \, e^{\left (-2 \, x\right )} - 2 \, \log \left (e^{\left (-x\right )} + 1\right ) - 2 \, \log \left (e^{\left (-x\right )} - 1\right ) \] Input:
integrate(coth(x)^4/(1+tanh(x)),x, algorithm="maxima")
Output:
1/2*x - 2/3*(15*e^(-2*x) - 12*e^(-4*x) - 7)/(3*e^(-2*x) - 3*e^(-4*x) + e^( -6*x) - 1) - 1/4*e^(-2*x) - 2*log(e^(-x) + 1) - 2*log(e^(-x) - 1)
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.12 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\frac {9}{2} \, x - \frac {{\left (51 \, e^{\left (6 \, x\right )} - 81 \, e^{\left (4 \, x\right )} + 65 \, e^{\left (2 \, x\right )} - 3\right )} e^{\left (-2 \, x\right )}}{12 \, {\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} - 2 \, \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \] Input:
integrate(coth(x)^4/(1+tanh(x)),x, algorithm="giac")
Output:
9/2*x - 1/12*(51*e^(6*x) - 81*e^(4*x) + 65*e^(2*x) - 3)*e^(-2*x)/(e^(2*x) - 1)^3 - 2*log(abs(e^(2*x) - 1))
Time = 2.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.60 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\frac {9\,x}{2}-2\,\ln \left ({\mathrm {e}}^{2\,x}-1\right )-\frac {{\mathrm {e}}^{-2\,x}}{4}-\frac {8}{3\,\left (3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1\right )}-\frac {2}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}-\frac {4}{{\mathrm {e}}^{2\,x}-1} \] Input:
int(coth(x)^4/(tanh(x) + 1),x)
Output:
(9*x)/2 - 2*log(exp(2*x) - 1) - exp(-2*x)/4 - 8/(3*(3*exp(2*x) - 3*exp(4*x ) + exp(6*x) - 1)) - 2/(exp(4*x) - 2*exp(2*x) + 1) - 4/(exp(2*x) - 1)
Time = 0.24 (sec) , antiderivative size = 191, normalized size of antiderivative = 4.44 \[ \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx=\frac {-24 e^{8 x} \mathrm {log}\left (e^{x}-1\right )-24 e^{8 x} \mathrm {log}\left (e^{x}+1\right )+54 e^{8 x} x -17 e^{8 x}+72 e^{6 x} \mathrm {log}\left (e^{x}-1\right )+72 e^{6 x} \mathrm {log}\left (e^{x}+1\right )-162 e^{6 x} x -72 e^{4 x} \mathrm {log}\left (e^{x}-1\right )-72 e^{4 x} \mathrm {log}\left (e^{x}+1\right )+162 e^{4 x} x +30 e^{4 x}+24 e^{2 x} \mathrm {log}\left (e^{x}-1\right )+24 e^{2 x} \mathrm {log}\left (e^{x}+1\right )-54 e^{2 x} x -48 e^{2 x}+3}{12 e^{2 x} \left (e^{6 x}-3 e^{4 x}+3 e^{2 x}-1\right )} \] Input:
int(coth(x)^4/(1+tanh(x)),x)
Output:
( - 24*e**(8*x)*log(e**x - 1) - 24*e**(8*x)*log(e**x + 1) + 54*e**(8*x)*x - 17*e**(8*x) + 72*e**(6*x)*log(e**x - 1) + 72*e**(6*x)*log(e**x + 1) - 16 2*e**(6*x)*x - 72*e**(4*x)*log(e**x - 1) - 72*e**(4*x)*log(e**x + 1) + 162 *e**(4*x)*x + 30*e**(4*x) + 24*e**(2*x)*log(e**x - 1) + 24*e**(2*x)*log(e* *x + 1) - 54*e**(2*x)*x - 48*e**(2*x) + 3)/(12*e**(2*x)*(e**(6*x) - 3*e**( 4*x) + 3*e**(2*x) - 1))