Integrand size = 11, antiderivative size = 43 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {5 x}{2}-2 \log (\cosh (x))-\frac {5 \tanh (x)}{2}+\tanh ^2(x)-\frac {5 \tanh ^3(x)}{6}+\frac {\tanh ^4(x)}{2 (1+\tanh (x))} \]
Time = 0.13 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.21 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {-12 \log (\cosh (x))-3 (5+4 \log (\cosh (x))) \tanh (x)-9 \tanh ^2(x)+\tanh ^3(x)-2 \tanh ^4(x)+15 \text {arctanh}(\tanh (x)) (1+\tanh (x))}{6 (1+\tanh (x))} \]
(-12*Log[Cosh[x]] - 3*(5 + 4*Log[Cosh[x]])*Tanh[x] - 9*Tanh[x]^2 + Tanh[x] ^3 - 2*Tanh[x]^4 + 15*ArcTanh[Tanh[x]]*(1 + Tanh[x]))/(6*(1 + Tanh[x]))
Result contains complex when optimal does not.
Time = 0.51 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.63, number of steps used = 21, number of rules used = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.909, Rules used = {3042, 26, 4033, 26, 3042, 26, 4011, 25, 26, 3042, 25, 4011, 26, 26, 3042, 26, 4008, 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.
\(\displaystyle \int \frac {\tanh ^5(x)}{\tanh (x)+1} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \tan (i x)^5}{1-i \tan (i x)}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\tan (i x)^5}{1-i \tan (i x)}dx\) |
\(\Big \downarrow \) 4033 |
\(\displaystyle -i \left (\frac {1}{2} \int -i (4-5 \tanh (x)) \tanh ^3(x)dx+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}-\frac {1}{2} i \int (4-5 \tanh (x)) \tanh ^3(x)dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}-\frac {1}{2} i \int i (5 i \tan (i x)+4) \tan (i x)^3dx\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} \int (5 i \tan (i x)+4) \tan (i x)^3dx+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle -i \left (\frac {1}{2} \left (\int -\left ((4 i \tanh (x)-5 i) \tanh ^2(x)\right )dx-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -i \left (\frac {1}{2} \left (-\int -i (5-4 \tanh (x)) \tanh ^2(x)dx-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} \left (i \int (5-4 \tanh (x)) \tanh ^2(x)dx-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (\frac {1}{2} \left (i \int -\left ((4 i \tan (i x)+5) \tan (i x)^2\right )dx-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -i \left (\frac {1}{2} \left (-i \int (4 i \tan (i x)+5) \tan (i x)^2dx-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle -i \left (\frac {1}{2} \left (-i \left (-2 \tanh ^2(x)+\int i (5 i \tanh (x)-4 i) \tanh (x)dx\right )-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} \left (-i \left (-2 \tanh ^2(x)+i \int -i (4-5 \tanh (x)) \tanh (x)dx\right )-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} \left (-i \left (\int (4-5 \tanh (x)) \tanh (x)dx-2 \tanh ^2(x)\right )-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (\frac {1}{2} \left (-i \left (-2 \tanh ^2(x)+\int -i (5 i \tan (i x)+4) \tan (i x)dx\right )-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} \left (-i \left (-2 \tanh ^2(x)-i \int (5 i \tan (i x)+4) \tan (i x)dx\right )-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle -i \left (\frac {1}{2} \left (-i \left (-2 \tanh ^2(x)-i (4 \int i \tanh (x)dx-5 i x+5 i \tanh (x))\right )-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} \left (-i \left (-2 \tanh ^2(x)-i (4 i \int \tanh (x)dx-5 i x+5 i \tanh (x))\right )-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \left (\frac {1}{2} \left (-i \left (-2 \tanh ^2(x)-i (4 i \int -i \tan (i x)dx-5 i x+5 i \tanh (x))\right )-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \left (\frac {1}{2} \left (-i \left (-2 \tanh ^2(x)-i (4 \int \tan (i x)dx-5 i x+5 i \tanh (x))\right )-\frac {5}{3} i \tanh ^3(x)\right )+\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}\right )\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -i \left (\frac {i \tanh ^4(x)}{2 (\tanh (x)+1)}+\frac {1}{2} \left (-\frac {5}{3} i \tanh ^3(x)-i \left (-2 \tanh ^2(x)-i (-5 i x+5 i \tanh (x)+4 i \log (\cosh (x)))\right )\right )\right )\) |
(-I)*(((I/2)*Tanh[x]^4)/(1 + Tanh[x]) + (((-5*I)/3)*Tanh[x]^3 - I*((-I)*(( -5*I)*x + (4*I)*Log[Cosh[x]] + (5*I)*Tanh[x]) - 2*Tanh[x]^2))/2)
3.2.15.3.1 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_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(a*c - b*d)*x, x] + (Simp[b*d*(Tan[e + f*x]/f), x] + Simp[(b*c + a*d) Int[Tan[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*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] && GtQ[m, 0]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((c + d*Tan[e + f*x])^(n - 1)/( 2*a*f*(a + b*Tan[e + f*x]))), x] + Simp[1/(2*a^2) Int[(c + d*Tan[e + f*x] )^(n - 2)*Simp[a*c^2 + a*d^2*(n - 1) - b*c*d*n - d*(a*c*(n - 2) + b*d*n)*Ta n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[n, 1]
Time = 0.06 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.93
method | result | size |
derivativedivides | \(-\frac {\tanh \left (x \right )^{3}}{3}+\frac {\tanh \left (x \right )^{2}}{2}-2 \tanh \left (x \right )+\frac {1}{2+2 \tanh \left (x \right )}+\frac {9 \ln \left (1+\tanh \left (x \right )\right )}{4}-\frac {\ln \left (\tanh \left (x \right )-1\right )}{4}\) | \(40\) |
default | \(-\frac {\tanh \left (x \right )^{3}}{3}+\frac {\tanh \left (x \right )^{2}}{2}-2 \tanh \left (x \right )+\frac {1}{2+2 \tanh \left (x \right )}+\frac {9 \ln \left (1+\tanh \left (x \right )\right )}{4}-\frac {\ln \left (\tanh \left (x \right )-1\right )}{4}\) | \(40\) |
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 | \(-\frac {2 \tanh \left (x \right )^{4}-15-\tanh \left (x \right )^{3}-12 \ln \left (1-\tanh \left (x \right )\right ) \tanh \left (x \right )-27 \tanh \left (x \right ) x +9 \tanh \left (x \right )^{2}-12 \ln \left (1-\tanh \left (x \right )\right )-27 x}{6 \left (1+\tanh \left (x \right )\right )}\) | \(57\) |
Leaf count of result is larger than twice the leaf count of optimal. 571 vs. \(2 (35) = 70\).
Time = 0.25 (sec) , antiderivative size = 571, normalized size of antiderivative = 13.28 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\text {Too large to display} \]
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))
Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (42) = 84\).
Time = 0.21 (sec) , antiderivative size = 104, normalized size of antiderivative = 2.42 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {3 x \tanh {\left (x \right )}}{6 \tanh {\left (x \right )} + 6} + \frac {3 x}{6 \tanh {\left (x \right )} + 6} + \frac {12 \log {\left (\tanh {\left (x \right )} + 1 \right )} \tanh {\left (x \right )}}{6 \tanh {\left (x \right )} + 6} + \frac {12 \log {\left (\tanh {\left (x \right )} + 1 \right )}}{6 \tanh {\left (x \right )} + 6} - \frac {2 \tanh ^{4}{\left (x \right )}}{6 \tanh {\left (x \right )} + 6} + \frac {\tanh ^{3}{\left (x \right )}}{6 \tanh {\left (x \right )} + 6} - \frac {9 \tanh ^{2}{\left (x \right )}}{6 \tanh {\left (x \right )} + 6} + \frac {15}{6 \tanh {\left (x \right )} + 6} \]
3*x*tanh(x)/(6*tanh(x) + 6) + 3*x/(6*tanh(x) + 6) + 12*log(tanh(x) + 1)*ta nh(x)/(6*tanh(x) + 6) + 12*log(tanh(x) + 1)/(6*tanh(x) + 6) - 2*tanh(x)**4 /(6*tanh(x) + 6) + tanh(x)**3/(6*tanh(x) + 6) - 9*tanh(x)**2/(6*tanh(x) + 6) + 15/(6*tanh(x) + 6)
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \, x - \frac {2 \, {\left (15 \, e^{\left (-2 \, x\right )} + 12 \, e^{\left (-4 \, x\right )} + 7\right )}}{3 \, {\left (3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1\right )}} + \frac {1}{4} \, e^{\left (-2 \, x\right )} - 2 \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]
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 = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.09 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {9}{2} \, x + \frac {{\left (51 \, e^{\left (6 \, x\right )} + 81 \, e^{\left (4 \, x\right )} + 65 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (-2 \, x\right )}}{12 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} - 2 \, \log \left (e^{\left (2 \, x\right )} + 1\right ) \]
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 = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {x}{2}+2\,\ln \left (\mathrm {tanh}\left (x\right )+1\right )-2\,\mathrm {tanh}\left (x\right )+\frac {{\mathrm {tanh}\left (x\right )}^2}{2}-\frac {{\mathrm {tanh}\left (x\right )}^3}{3}+\frac {1}{2\,\left (\mathrm {tanh}\left (x\right )+1\right )} \]