\(\int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx\) [116]

Optimal result

Integrand size = 11, antiderivative size = 37 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=-\frac {3 x}{2}+2 \log (\cosh (x))+\frac {3 \tanh (x)}{2}-\tanh ^2(x)+\frac {\tanh ^3(x)}{2 (1+\tanh (x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3631, 3609, 3606, 3556} \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=-\frac {3 x}{2}+\frac {\tanh ^3(x)}{2 (\tanh (x)+1)}-\tanh ^2(x)+\frac {3 \tanh (x)}{2}+2 \log (\cosh (x)) \]

[In]

Int[Tanh[x]^4/(1 + Tanh[x]),x]

[Out]

(-3*x)/2 + 2*Log[Cosh[x]] + (3*Tanh[x])/2 - Tanh[x]^2 + Tanh[x]^3/(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 ^3(x)}{2 (1+\tanh (x))}-\frac {1}{2} \int (3-4 \tanh (x)) \tanh ^2(x) \, dx \\ & = -\tanh ^2(x)+\frac {\tanh ^3(x)}{2 (1+\tanh (x))}+\frac {1}{2} i \int (-4 i+3 i \tanh (x)) \tanh (x) \, dx \\ & = -\frac {3 x}{2}+\frac {3 \tanh (x)}{2}-\tanh ^2(x)+\frac {\tanh ^3(x)}{2 (1+\tanh (x))}+2 \int \tanh (x) \, dx \\ & = -\frac {3 x}{2}+2 \log (\cosh (x))+\frac {3 \tanh (x)}{2}-\tanh ^2(x)+\frac {\tanh ^3(x)}{2 (1+\tanh (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.22 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=\frac {4 \log (\cosh (x))+(3+4 \log (\cosh (x))) \tanh (x)+\tanh ^2(x)-\tanh ^3(x)-3 \text {arctanh}(\tanh (x)) (1+\tanh (x))}{2 (1+\tanh (x))} \]

[In]

Integrate[Tanh[x]^4/(1 + Tanh[x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81

method	result	size
risch	\(-\frac {7 x}{2}-\frac {{\mathrm e}^{-2 x}}{4}-\frac {2}{\left (1+{\mathrm e}^{2 x}\right )^{2}}+2 \ln \left (1+{\mathrm e}^{2 x}\right )\)	\(30\)
derivativedivides	\(-\frac {\tanh \left (x \right )^{2}}{2}+\tanh \left (x \right )-\frac {\ln \left (\tanh \left (x \right )-1\right )}{4}-\frac {1}{2 \left (1+\tanh \left (x \right )\right )}-\frac {7 \ln \left (1+\tanh \left (x \right )\right )}{4}\)	\(32\)
default	\(-\frac {\tanh \left (x \right )^{2}}{2}+\tanh \left (x \right )-\frac {\ln \left (\tanh \left (x \right )-1\right )}{4}-\frac {1}{2 \left (1+\tanh \left (x \right )\right )}-\frac {7 \ln \left (1+\tanh \left (x \right )\right )}{4}\)	\(32\)
parallelrisch	\(-\frac {3+\tanh \left (x \right )^{3}+4 \ln \left (1-\tanh \left (x \right )\right ) \tanh \left (x \right )+7 \tanh \left (x \right ) x -\tanh \left (x \right )^{2}+4 \ln \left (1-\tanh \left (x \right )\right )+7 x}{2 \left (1+\tanh \left (x \right )\right )}\)	\(49\)

[In]

int(tanh(x)^4/(1+tanh(x)),x,method=_RETURNVERBOSE)

[Out]

-7/2*x-1/4*exp(-2*x)-2/(1+exp(2*x))^2+2*ln(1+exp(2*x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (31) = 62\).

Time = 0.28 (sec) , antiderivative size = 354, normalized size of antiderivative = 9.57 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=-\frac {14 \, x \cosh \left (x\right )^{6} + 84 \, x \cosh \left (x\right ) \sinh \left (x\right )^{5} + 14 \, x \sinh \left (x\right )^{6} + {\left (28 \, x + 1\right )} \cosh \left (x\right )^{4} + {\left (210 \, x \cosh \left (x\right )^{2} + 28 \, x + 1\right )} \sinh \left (x\right )^{4} + 4 \, {\left (70 \, x \cosh \left (x\right )^{3} + {\left (28 \, x + 1\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + 2 \, {\left (7 \, x + 5\right )} \cosh \left (x\right )^{2} + 2 \, {\left (105 \, x \cosh \left (x\right )^{4} + 3 \, {\left (28 \, x + 1\right )} \cosh \left (x\right )^{2} + 7 \, x + 5\right )} \sinh \left (x\right )^{2} - 8 \, {\left (\cosh \left (x\right )^{6} + 6 \, \cosh \left (x\right ) \sinh \left (x\right )^{5} + \sinh \left (x\right )^{6} + {\left (15 \, \cosh \left (x\right )^{2} + 2\right )} \sinh \left (x\right )^{4} + 2 \, \cosh \left (x\right )^{4} + 4 \, {\left (5 \, \cosh \left (x\right )^{3} + 2 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + {\left (15 \, \cosh \left (x\right )^{4} + 12 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + \cosh \left (x\right )^{2} + 2 \, {\left (3 \, \cosh \left (x\right )^{5} + 4 \, \cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 4 \, {\left (21 \, x \cosh \left (x\right )^{5} + {\left (28 \, x + 1\right )} \cosh \left (x\right )^{3} + {\left (7 \, x + 5\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1}{4 \, {\left (\cosh \left (x\right )^{6} + 6 \, \cosh \left (x\right ) \sinh \left (x\right )^{5} + \sinh \left (x\right )^{6} + {\left (15 \, \cosh \left (x\right )^{2} + 2\right )} \sinh \left (x\right )^{4} + 2 \, \cosh \left (x\right )^{4} + 4 \, {\left (5 \, \cosh \left (x\right )^{3} + 2 \, \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} + {\left (15 \, \cosh \left (x\right )^{4} + 12 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + \cosh \left (x\right )^{2} + 2 \, {\left (3 \, \cosh \left (x\right )^{5} + 4 \, \cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right )\right )}} \]

[In]

integrate(tanh(x)^4/(1+tanh(x)),x, algorithm="fricas")

[Out]

-1/4*(14*x*cosh(x)^6 + 84*x*cosh(x)*sinh(x)^5 + 14*x*sinh(x)^6 + (28*x + 1)*cosh(x)^4 + (210*x*cosh(x)^2 + 28*
x + 1)*sinh(x)^4 + 4*(70*x*cosh(x)^3 + (28*x + 1)*cosh(x))*sinh(x)^3 + 2*(7*x + 5)*cosh(x)^2 + 2*(105*x*cosh(x
)^4 + 3*(28*x + 1)*cosh(x)^2 + 7*x + 5)*sinh(x)^2 - 8*(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + (15*cosh(
x)^2 + 2)*sinh(x)^4 + 2*cosh(x)^4 + 4*(5*cosh(x)^3 + 2*cosh(x))*sinh(x)^3 + (15*cosh(x)^4 + 12*cosh(x)^2 + 1)*
sinh(x)^2 + cosh(x)^2 + 2*(3*cosh(x)^5 + 4*cosh(x)^3 + cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) +
4*(21*x*cosh(x)^5 + (28*x + 1)*cosh(x)^3 + (7*x + 5)*cosh(x))*sinh(x) + 1)/(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 +
sinh(x)^6 + (15*cosh(x)^2 + 2)*sinh(x)^4 + 2*cosh(x)^4 + 4*(5*cosh(x)^3 + 2*cosh(x))*sinh(x)^3 + (15*cosh(x)^4
+ 12*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*(3*cosh(x)^5 + 4*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. 85 vs. \(2 (34) = 68\).

Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.30 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=\frac {x \tanh {\left (x \right )}}{2 \tanh {\left (x \right )} + 2} + \frac {x}{2 \tanh {\left (x \right )} + 2} - \frac {4 \log {\left (\tanh {\left (x \right )} + 1 \right )} \tanh {\left (x \right )}}{2 \tanh {\left (x \right )} + 2} - \frac {4 \log {\left (\tanh {\left (x \right )} + 1 \right )}}{2 \tanh {\left (x \right )} + 2} - \frac {\tanh ^{3}{\left (x \right )}}{2 \tanh {\left (x \right )} + 2} + \frac {\tanh ^{2}{\left (x \right )}}{2 \tanh {\left (x \right )} + 2} - \frac {3}{2 \tanh {\left (x \right )} + 2} \]

[In]

integrate(tanh(x)**4/(1+tanh(x)),x)

[Out]

x*tanh(x)/(2*tanh(x) + 2) + x/(2*tanh(x) + 2) - 4*log(tanh(x) + 1)*tanh(x)/(2*tanh(x) + 2) - 4*log(tanh(x) + 1
)/(2*tanh(x) + 2) - tanh(x)**3/(2*tanh(x) + 2) + tanh(x)**2/(2*tanh(x) + 2) - 3/(2*tanh(x) + 2)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.16 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \, x + \frac {2 \, {\left (2 \, e^{\left (-2 \, x\right )} + 1\right )}}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} - \frac {1}{4} \, e^{\left (-2 \, x\right )} + 2 \, \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]

[In]

integrate(tanh(x)^4/(1+tanh(x)),x, algorithm="maxima")

[Out]

1/2*x + 2*(2*e^(-2*x) + 1)/(2*e^(-2*x) + e^(-4*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 = 39, normalized size of antiderivative = 1.05 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=-\frac {7}{2} \, x - \frac {{\left (e^{\left (4 \, x\right )} + 10 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-2 \, x\right )}}{4 \, {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} + 2 \, \log \left (e^{\left (2 \, x\right )} + 1\right ) \]

[In]

integrate(tanh(x)^4/(1+tanh(x)),x, algorithm="giac")

[Out]

-7/2*x - 1/4*(e^(4*x) + 10*e^(2*x) + 1)*e^(-2*x)/(e^(2*x) + 1)^2 + 2*log(e^(2*x) + 1)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int \frac {\tanh ^4(x)}{1+\tanh (x)} \, dx=\frac {x}{2}-2\,\ln \left (\mathrm {tanh}\left (x\right )+1\right )+\mathrm {tanh}\left (x\right )-\frac {{\mathrm {tanh}\left (x\right )}^2}{2}-\frac {1}{2\,\left (\mathrm {tanh}\left (x\right )+1\right )} \]

[In]

int(tanh(x)^4/(tanh(x) + 1),x)

[Out]

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