\(\int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx\) [115]

Optimal result

Integrand size = 11, antiderivative size = 43 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {5 x}{2}-2 \log (\cosh (x))-\frac {5 \tanh (x)}{2}+\tanh ^2(x)-\frac {5 \tanh ^3(x)}{6}+\frac {\tanh ^4(x)}{2 (1+\tanh (x))} \]

[Out]

5/2*x-2*ln(cosh(x))-5/2*tanh(x)+tanh(x)^2-5/6*tanh(x)^3+1/2*tanh(x)^4/(1+tanh(x))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3631, 3609, 3606, 3556} \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {5 x}{2}+\frac {\tanh ^4(x)}{2 (\tanh (x)+1)}-\frac {5 \tanh ^3(x)}{6}+\tanh ^2(x)-\frac {5 \tanh (x)}{2}-2 \log (\cosh (x)) \]

[In]

Int[Tanh[x]^5/(1 + Tanh[x]),x]

[Out]

(5*x)/2 - 2*Log[Cosh[x]] - (5*Tanh[x])/2 + Tanh[x]^2 - (5*Tanh[x]^3)/6 + Tanh[x]^4/(2*(1 + Tanh[x]))

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3606

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(a*c - b
*d)*x, x] + (Dist[b*c + a*d, Int[Tan[e + f*x], x], x] + Simp[b*d*(Tan[e + f*x]/f), x]) /; FreeQ[{a, b, c, d, e
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

Rule 3609

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]

Rule 3631

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] + Dist[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)*Tan[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\tanh ^4(x)}{2 (1+\tanh (x))}-\frac {1}{2} \int (4-5 \tanh (x)) \tanh ^3(x) \, dx \\ & = -\frac {5}{6} \tanh ^3(x)+\frac {\tanh ^4(x)}{2 (1+\tanh (x))}+\frac {1}{2} i \int (-5 i+4 i \tanh (x)) \tanh ^2(x) \, dx \\ & = \tanh ^2(x)-\frac {5 \tanh ^3(x)}{6}+\frac {\tanh ^4(x)}{2 (1+\tanh (x))}+\frac {1}{2} \int \tanh (x) (-4+5 \tanh (x)) \, dx \\ & = \frac {5 x}{2}-\frac {5 \tanh (x)}{2}+\tanh ^2(x)-\frac {5 \tanh ^3(x)}{6}+\frac {\tanh ^4(x)}{2 (1+\tanh (x))}-2 \int \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))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.21 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {-12 \log (\cosh (x))-3 (5+4 \log (\cosh (x))) \tanh (x)-9 \tanh ^2(x)+\tanh ^3(x)-2 \tanh ^4(x)+15 \text {arctanh}(\tanh (x)) (1+\tanh (x))}{6 (1+\tanh (x))} \]

[In]

Integrate[Tanh[x]^5/(1 + Tanh[x]),x]

[Out]

(-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]))

Maple [A] (verified)

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	\(-\frac {2 \tanh \left (x \right )^{4}-15-\tanh \left (x \right )^{3}-12 \ln \left (1-\tanh \left (x \right )\right ) \tanh \left (x \right )-27 \tanh \left (x \right ) x +9 \tanh \left (x \right )^{2}-12 \ln \left (1-\tanh \left (x \right )\right )-27 x}{6 \left (1+\tanh \left (x \right )\right )}\)	\(57\)

[In]

int(tanh(x)^5/(1+tanh(x)),x,method=_RETURNVERBOSE)

[Out]

-1/3*tanh(x)^3+1/2*tanh(x)^2-2*tanh(x)+1/2/(1+tanh(x))+9/4*ln(1+tanh(x))-1/4*ln(tanh(x)-1)

Fricas [B] (verification not implemented)

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 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\text {Too large to display} \]

[In]

integrate(tanh(x)^5/(1+tanh(x)),x, algorithm="fricas")

[Out]

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*cosh(x)^3 + (54*x + 17)*cosh(x))*sinh(x)^5 + 81*(2*x + 1)*cosh(x)^4 + 9*(4
20*x*cosh(x)^4 + 5*(54*x + 17)*cosh(x)^2 + 18*x + 9)*sinh(x)^4 + 12*(252*x*cosh(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 + 48
6*(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)*sin
h(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*cosh(x)*sinh(x)^7 + sinh(x)^8 + (28*cosh(x)^2 + 3)*sinh(x)^6 + 3*c
osh(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)*sin
h(x)^2 + cosh(x)^2 + 2*(4*cosh(x)^7 + 9*cosh(x)^5 + 6*cosh(x)^3 + cosh(x))*sinh(x))

Sympy [B] (verification not implemented)

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 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {3 x \tanh {\left (x \right )}}{6 \tanh {\left (x \right )} + 6} + \frac {3 x}{6 \tanh {\left (x \right )} + 6} + \frac {12 \log {\left (\tanh {\left (x \right )} + 1 \right )} \tanh {\left (x \right )}}{6 \tanh {\left (x \right )} + 6} + \frac {12 \log {\left (\tanh {\left (x \right )} + 1 \right )}}{6 \tanh {\left (x \right )} + 6} - \frac {2 \tanh ^{4}{\left (x \right )}}{6 \tanh {\left (x \right )} + 6} + \frac {\tanh ^{3}{\left (x \right )}}{6 \tanh {\left (x \right )} + 6} - \frac {9 \tanh ^{2}{\left (x \right )}}{6 \tanh {\left (x \right )} + 6} + \frac {15}{6 \tanh {\left (x \right )} + 6} \]

[In]

integrate(tanh(x)**5/(1+tanh(x)),x)

[Out]

3*x*tanh(x)/(6*tanh(x) + 6) + 3*x/(6*tanh(x) + 6) + 12*log(tanh(x) + 1)*tanh(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)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.28 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \, x - \frac {2 \, {\left (15 \, e^{\left (-2 \, x\right )} + 12 \, e^{\left (-4 \, x\right )} + 7\right )}}{3 \, {\left (3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1\right )}} + \frac {1}{4} \, e^{\left (-2 \, x\right )} - 2 \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]

[In]

integrate(tanh(x)^5/(1+tanh(x)),x, algorithm="maxima")

[Out]

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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.09 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {9}{2} \, x + \frac {{\left (51 \, e^{\left (6 \, x\right )} + 81 \, e^{\left (4 \, x\right )} + 65 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (-2 \, x\right )}}{12 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} - 2 \, \log \left (e^{\left (2 \, x\right )} + 1\right ) \]

[In]

integrate(tanh(x)^5/(1+tanh(x)),x, algorithm="giac")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {\tanh ^5(x)}{1+\tanh (x)} \, dx=\frac {x}{2}+2\,\ln \left (\mathrm {tanh}\left (x\right )+1\right )-2\,\mathrm {tanh}\left (x\right )+\frac {{\mathrm {tanh}\left (x\right )}^2}{2}-\frac {{\mathrm {tanh}\left (x\right )}^3}{3}+\frac {1}{2\,\left (\mathrm {tanh}\left (x\right )+1\right )} \]

[In]

int(tanh(x)^5/(tanh(x) + 1),x)

[Out]

x/2 + 2*log(tanh(x) + 1) - 2*tanh(x) + tanh(x)^2/2 - tanh(x)^3/3 + 1/(2*(tanh(x) + 1))