\(\int \frac {\tanh ^3(x)}{1+\tanh (x)} \, dx\) [117]

Optimal result

Integrand size = 11, antiderivative size = 31 \[ \int \frac {\tanh ^3(x)}{1+\tanh (x)} \, dx=\frac {3 x}{2}-\log (\cosh (x))-\frac {3 \tanh (x)}{2}+\frac {\tanh ^2(x)}{2 (1+\tanh (x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3631, 3606, 3556} \[ \int \frac {\tanh ^3(x)}{1+\tanh (x)} \, dx=\frac {3 x}{2}+\frac {\tanh ^2(x)}{2 (\tanh (x)+1)}-\frac {3 \tanh (x)}{2}-\log (\cosh (x)) \]

[In]

Int[Tanh[x]^3/(1 + Tanh[x]),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {\tanh ^3(x)}{1+\tanh (x)} \, dx=-\frac {2 \log (\cosh (x))+(3+2 \log (\cosh (x))) \tanh (x)+2 \tanh ^2(x)-3 \text {arctanh}(\tanh (x)) (1+\tanh (x))}{2 (1+\tanh (x))} \]

[In]

Integrate[Tanh[x]^3/(1 + Tanh[x]),x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(-\tanh \left (x \right )-\frac {\ln \left (\tanh \left (x \right )-1\right )}{4}+\frac {1}{2+2 \tanh \left (x \right )}+\frac {5 \ln \left (1+\tanh \left (x \right )\right )}{4}\)	\(28\)
default	\(-\tanh \left (x \right )-\frac {\ln \left (\tanh \left (x \right )-1\right )}{4}+\frac {1}{2+2 \tanh \left (x \right )}+\frac {5 \ln \left (1+\tanh \left (x \right )\right )}{4}\)	\(28\)
risch	\(\frac {5 x}{2}+\frac {{\mathrm e}^{-2 x}}{4}+\frac {2}{1+{\mathrm e}^{2 x}}-\ln \left (1+{\mathrm e}^{2 x}\right )\)	\(30\)
parallelrisch	\(-\frac {-3-2 \ln \left (1-\tanh \left (x \right )\right ) \tanh \left (x \right )-5 \tanh \left (x \right ) x +2 \tanh \left (x \right )^{2}-2 \ln \left (1-\tanh \left (x \right )\right )-5 x}{2 \left (1+\tanh \left (x \right )\right )}\)	\(45\)

[In]

int(tanh(x)^3/(1+tanh(x)),x,method=_RETURNVERBOSE)

[Out]

-tanh(x)-1/4*ln(tanh(x)-1)+1/2/(1+tanh(x))+5/4*ln(1+tanh(x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (25) = 50\).

Time = 0.26 (sec) , antiderivative size = 186, normalized size of antiderivative = 6.00 \[ \int \frac {\tanh ^3(x)}{1+\tanh (x)} \, dx=\frac {10 \, x \cosh \left (x\right )^{4} + 40 \, x \cosh \left (x\right ) \sinh \left (x\right )^{3} + 10 \, x \sinh \left (x\right )^{4} + {\left (10 \, x + 9\right )} \cosh \left (x\right )^{2} + {\left (60 \, x \cosh \left (x\right )^{2} + 10 \, x + 9\right )} \sinh \left (x\right )^{2} - 4 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + {\left (6 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + \cosh \left (x\right )^{2} + 2 \, {\left (2 \, \cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 2 \, {\left (20 \, x \cosh \left (x\right )^{3} + {\left (10 \, x + 9\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1}{4 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + {\left (6 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + \cosh \left (x\right )^{2} + 2 \, {\left (2 \, \cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right )\right )}} \]

[In]

integrate(tanh(x)^3/(1+tanh(x)),x, algorithm="fricas")

[Out]

1/4*(10*x*cosh(x)^4 + 40*x*cosh(x)*sinh(x)^3 + 10*x*sinh(x)^4 + (10*x + 9)*cosh(x)^2 + (60*x*cosh(x)^2 + 10*x
+ 9)*sinh(x)^2 - 4*(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + (6*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*
(2*cosh(x)^3 + cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))) + 2*(20*x*cosh(x)^3 + (10*x + 9)*cosh(x))*
sinh(x) + 1)/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + (6*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*(2*cos
h(x)^3 + cosh(x))*sinh(x))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (27) = 54\).

Time = 0.18 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.42 \[ \int \frac {\tanh ^3(x)}{1+\tanh (x)} \, dx=\frac {x \tanh {\left (x \right )}}{2 \tanh {\left (x \right )} + 2} + \frac {x}{2 \tanh {\left (x \right )} + 2} + \frac {2 \log {\left (\tanh {\left (x \right )} + 1 \right )} \tanh {\left (x \right )}}{2 \tanh {\left (x \right )} + 2} + \frac {2 \log {\left (\tanh {\left (x \right )} + 1 \right )}}{2 \tanh {\left (x \right )} + 2} - \frac {2 \tanh ^{2}{\left (x \right )}}{2 \tanh {\left (x \right )} + 2} + \frac {3}{2 \tanh {\left (x \right )} + 2} \]

[In]

integrate(tanh(x)**3/(1+tanh(x)),x)

[Out]

x*tanh(x)/(2*tanh(x) + 2) + x/(2*tanh(x) + 2) + 2*log(tanh(x) + 1)*tanh(x)/(2*tanh(x) + 2) + 2*log(tanh(x) + 1
)/(2*tanh(x) + 2) - 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 = 29, normalized size of antiderivative = 0.94 \[ \int \frac {\tanh ^3(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \, x - \frac {2}{e^{\left (-2 \, x\right )} + 1} + \frac {1}{4} \, e^{\left (-2 \, x\right )} - \log \left (e^{\left (-2 \, x\right )} + 1\right ) \]

[In]

integrate(tanh(x)^3/(1+tanh(x)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13 \[ \int \frac {\tanh ^3(x)}{1+\tanh (x)} \, dx=\frac {5}{2} \, x + \frac {{\left (9 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-2 \, x\right )}}{4 \, {\left (e^{\left (2 \, x\right )} + 1\right )}} - \log \left (e^{\left (2 \, x\right )} + 1\right ) \]

[In]

integrate(tanh(x)^3/(1+tanh(x)),x, algorithm="giac")

[Out]

5/2*x + 1/4*(9*e^(2*x) + 1)*e^(-2*x)/(e^(2*x) + 1) - log(e^(2*x) + 1)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {\tanh ^3(x)}{1+\tanh (x)} \, dx=\frac {x}{2}+\ln \left (\mathrm {tanh}\left (x\right )+1\right )-\mathrm {tanh}\left (x\right )+\frac {1}{2\,\left (\mathrm {tanh}\left (x\right )+1\right )} \]

[In]

int(tanh(x)^3/(tanh(x) + 1),x)

[Out]

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