Integrand size = 11, antiderivative size = 43 \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\frac {5 x}{2}-2 \log (\cosh (x))-\frac {5 \tanh (x)}{2}+\tanh ^2(x)-\frac {5 \tanh ^3(x)}{6}+\frac {\tanh ^3(x)}{2 (1+\coth (x))} \]
Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.93 \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\frac {5}{2} \text {arctanh}(\tanh (x))-2 \log (\cosh (x))-\frac {5 \tanh (x)}{2}+\tanh ^2(x)+\left (-\frac {5}{6}+\frac {1}{2+2 \coth (x)}\right ) \tanh ^3(x) \]
(5*ArcTanh[Tanh[x]])/2 - 2*Log[Cosh[x]] - (5*Tanh[x])/2 + Tanh[x]^2 + (-5/ 6 + (2 + 2*Coth[x])^(-1))*Tanh[x]^3
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.
\(\displaystyle \int \frac {\tanh ^4(x)}{\coth (x)+1} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (1-i \tan \left (\frac {\pi }{2}+i x\right )\right ) \tan \left (\frac {\pi }{2}+i x\right )^4}dx\) |
\(\Big \downarrow \) 4035 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}-\frac {1}{2} \int -\left ((5-4 \coth (x)) \tanh ^4(x)\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \int (5-4 \coth (x)) \tanh ^4(x)dx+\frac {\tanh ^3(x)}{2 (\coth (x)+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \int \frac {4 i \tan \left (i x+\frac {\pi }{2}\right )+5}{\tan \left (i x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {1}{2} \left (\int -\left ((4-5 \coth (x)) \tanh ^3(x)\right )dx-\frac {5 \tanh ^3(x)}{3}\right )+\frac {\tanh ^3(x)}{2 (\coth (x)+1)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (-\int (4-5 \coth (x)) \tanh ^3(x)dx-\frac {5}{3} \tanh ^3(x)\right )+\frac {\tanh ^3(x)}{2 (\coth (x)+1)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5}{3} \tanh ^3(x)-\int -\frac {i \left (5 i \tan \left (i x+\frac {\pi }{2}\right )+4\right )}{\tan \left (i x+\frac {\pi }{2}\right )^3}dx\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \int \frac {5 i \tan \left (i x+\frac {\pi }{2}\right )+4}{\tan \left (i x+\frac {\pi }{2}\right )^3}dx\right )\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \left (\int -i (5-4 \coth (x)) \tanh ^2(x)dx-2 i \tanh ^2(x)\right )\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \left (-i \int (5-4 \coth (x)) \tanh ^2(x)dx-2 i \tanh ^2(x)\right )\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \left (-i \int -\frac {4 i \tan \left (i x+\frac {\pi }{2}\right )+5}{\tan \left (i x+\frac {\pi }{2}\right )^2}dx-2 i \tanh ^2(x)\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \left (i \int \frac {4 i \tan \left (i x+\frac {\pi }{2}\right )+5}{\tan \left (i x+\frac {\pi }{2}\right )^2}dx-2 i \tanh ^2(x)\right )\right )\) |
\(\Big \downarrow \) 4012 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \left (i (\int (4-5 \coth (x)) \tanh (x)dx+5 \tanh (x))-2 i \tanh ^2(x)\right )\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \left (i \left (5 \tanh (x)+\int \frac {i \left (5 i \tan \left (i x+\frac {\pi }{2}\right )+4\right )}{\tan \left (i x+\frac {\pi }{2}\right )}dx\right )-2 i \tanh ^2(x)\right )\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \left (i \left (5 \tanh (x)+i \int \frac {5 i \tan \left (i x+\frac {\pi }{2}\right )+4}{\tan \left (i x+\frac {\pi }{2}\right )}dx\right )-2 i \tanh ^2(x)\right )\right )\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \left (i (5 \tanh (x)+i (4 \int -i \tanh (x)dx+5 i x))-2 i \tanh ^2(x)\right )\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \left (i (5 \tanh (x)+i (5 i x-4 i \int \tanh (x)dx))-2 i \tanh ^2(x)\right )\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \left (i (5 \tanh (x)+i (5 i x-4 i \int -i \tan (i x)dx))-2 i \tanh ^2(x)\right )\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \left (i (5 \tanh (x)+i (5 i x-4 \int \tan (i x)dx))-2 i \tanh ^2(x)\right )\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {\tanh ^3(x)}{2 (\coth (x)+1)}+\frac {1}{2} \left (-\frac {5 \tanh ^3(x)}{3}+i \left (i (5 \tanh (x)+i (5 i x-4 i \log (\cosh (x))))-2 i \tanh ^2(x)\right )\right )\) |
Tanh[x]^3/(2*(1 + Coth[x])) + ((-5*Tanh[x]^3)/3 + I*((-2*I)*Tanh[x]^2 + I* (I*((5*I)*x - (4*I)*Log[Cosh[x]]) + 5*Tanh[x])))/2
3.2.23.3.1 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 = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.02
method | result | size |
risch | \(\frac {9 x}{2}+\frac {{\mathrm e}^{-2 x}}{4}+\frac {4 \,{\mathrm e}^{4 x}+6 \,{\mathrm e}^{2 x}+\frac {14}{3}}{\left (1+{\mathrm e}^{2 x}\right )^{3}}-2 \ln \left (1+{\mathrm e}^{2 x}\right )\) | \(44\) |
parallelrisch | \(\frac {\left (12 \tanh \left (x \right )+12\right ) \ln \left (1-\tanh \left (x \right )\right )-2 \tanh \left (x \right )^{4}+\tanh \left (x \right )^{3}+27 \tanh \left (x \right ) x -9 \tanh \left (x \right )^{2}+27 x +15}{6+6 \tanh \left (x \right )}\) | \(50\) |
default | \(\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {1}{\tanh \left (\frac {x}{2}\right )+1}+\frac {9 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}-\frac {4 \left (\tanh \left (\frac {x}{2}\right )^{5}-\frac {\tanh \left (\frac {x}{2}\right )^{4}}{2}+\frac {8 \tanh \left (\frac {x}{2}\right )^{3}}{3}-\frac {\tanh \left (\frac {x}{2}\right )^{2}}{2}+\tanh \left (\frac {x}{2}\right )\right )}{\left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )^{3}}-2 \ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}\) | \(96\) |
Leaf count of result is larger than twice the leaf count of optimal. 571 vs. \(2 (35) = 70\).
Time = 0.26 (sec) , antiderivative size = 571, normalized size of antiderivative = 13.28 \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\text {Too large to display} \]
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*cosh(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))
\[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\int \frac {\tanh ^{4}{\left (x \right )}}{\coth {\left (x \right )} + 1}\, dx \]
Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\frac {1}{2} \, x - \frac {2 \, {\left (15 \, e^{\left (-2 \, x\right )} + 12 \, e^{\left (-4 \, x\right )} + 7\right )}}{3 \, {\left (3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1\right )}} + \frac {1}{4} \, e^{\left (-2 \, x\right )} - 2 \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]
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^(-2*x) + 1)
Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.09 \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\frac {9}{2} \, x + \frac {{\left (51 \, e^{\left (6 \, x\right )} + 81 \, e^{\left (4 \, x\right )} + 65 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (-2 \, x\right )}}{12 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} - 2 \, \log \left (e^{\left (2 \, x\right )} + 1\right ) \]
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(e^(2*x) + 1)
Time = 1.96 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.60 \[ \int \frac {\tanh ^4(x)}{1+\coth (x)} \, dx=\frac {9\,x}{2}-2\,\ln \left ({\mathrm {e}}^{2\,x}+1\right )+\frac {{\mathrm {e}}^{-2\,x}}{4}+\frac {8}{3\,\left (3\,{\mathrm {e}}^{2\,x}+3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}+1\right )}-\frac {2}{2\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{4\,x}+1}+\frac {4}{{\mathrm {e}}^{2\,x}+1} \]