3.116 \(\int \frac{\tanh ^4(x)}{1+\tanh (x)} \, dx\)

Optimal. Leaf size=37 \[ -\frac{3 x}{2}+\frac{\tanh ^3(x)}{2 (\tanh (x)+1)}-\tanh ^2(x)+\frac{3 \tanh (x)}{2}+2 \log (\cosh (x)) \]

[Out]

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

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Rubi [A]  time = 0.0657397, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {3550, 3528, 3525, 3475} \[ -\frac{3 x}{2}+\frac{\tanh ^3(x)}{2 (\tanh (x)+1)}-\tanh ^2(x)+\frac{3 \tanh (x)}{2}+2 \log (\cosh (x)) \]

Antiderivative was successfully verified.

[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 3550

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]

Rule 3528

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 3525

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 3475

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

Rubi steps

\begin{align*} \int \frac{\tanh ^4(x)}{1+\tanh (x)} \, dx &=\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]  time = 0.0516263, size = 33, normalized size = 0.89 \[ \frac{1}{4} \left (-6 x+\sinh (2 x)-\cosh (2 x)+4 \tanh (x)+2 \text{sech}^2(x)+8 \log (\cosh (x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^4/(1 + Tanh[x]),x]

[Out]

(-6*x - Cosh[2*x] + 8*Log[Cosh[x]] + 2*Sech[x]^2 + Sinh[2*x] + 4*Tanh[x])/4

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Maple [A]  time = 0.019, size = 32, normalized size = 0.9 \begin{align*} -{\frac{ \left ( \tanh \left ( x \right ) \right ) ^{2}}{2}}+\tanh \left ( x \right ) -{\frac{1}{2+2\,\tanh \left ( x \right ) }}-{\frac{7\,\ln \left ( 1+\tanh \left ( x \right ) \right ) }{4}}-{\frac{\ln \left ( \tanh \left ( x \right ) -1 \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^4/(1+tanh(x)),x)

[Out]

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

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Maxima [A]  time = 1.76215, size = 58, normalized size = 1.57 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B]  time = 2.32173, size = 1168, normalized size = 31.57 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [B]  time = 0.538901, size = 85, normalized size = 2.3 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.27143, size = 53, normalized size = 1.43 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)