3.125 \(\int \frac{\tanh ^2(x)}{1+\coth (x)} \, dx\)

Optimal. Leaf size=29 \[ \frac{3 x}{2}-\frac{3 \tanh (x)}{2}-\log (\cosh (x))+\frac{\tanh (x)}{2 (\coth (x)+1)} \]

[Out]

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

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Rubi [A]  time = 0.0742931, antiderivative size = 29, 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 = {3552, 3529, 3531, 3475} \[ \frac{3 x}{2}-\frac{3 \tanh (x)}{2}-\log (\cosh (x))+\frac{\tanh (x)}{2 (\coth (x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2/(1 + Coth[x]),x]

[Out]

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

Rule 3552

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(a
*(c + d*Tan[e + f*x])^(n + 1))/(2*f*(b*c - a*d)*(a + b*Tan[e + f*x])), x] + Dist[1/(2*a*(b*c - a*d)), Int[(c +
d*Tan[e + f*x])^n*Simp[b*c + a*d*(n - 1) - b*d*n*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x]
&& NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !GtQ[n, 0]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*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] && NeQ[a*c + b*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 ^2(x)}{1+\coth (x)} \, dx &=\frac{\tanh (x)}{2 (1+\coth (x))}-\frac{1}{2} \int (-3+2 \coth (x)) \tanh ^2(x) \, dx\\ &=-\frac{3 \tanh (x)}{2}+\frac{\tanh (x)}{2 (1+\coth (x))}-\frac{1}{2} i \int (-2 i+3 i \coth (x)) \tanh (x) \, dx\\ &=\frac{3 x}{2}-\frac{3 \tanh (x)}{2}+\frac{\tanh (x)}{2 (1+\coth (x))}-\int \tanh (x) \, dx\\ &=\frac{3 x}{2}-\log (\cosh (x))-\frac{3 \tanh (x)}{2}+\frac{\tanh (x)}{2 (1+\coth (x))}\\ \end{align*}

Mathematica [A]  time = 0.0456504, size = 27, normalized size = 0.93 \[ \frac{1}{4} (6 x-\sinh (2 x)+\cosh (2 x)-4 \tanh (x)-4 \log (\cosh (x))) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^2/(1 + Coth[x]),x]

[Out]

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

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Maple [B]  time = 0.035, size = 65, normalized size = 2.2 \begin{align*} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}- \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}+{\frac{5}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-2\,{\frac{\tanh \left ( x/2 \right ) }{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1}}-\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^2/(1+coth(x)),x)

[Out]

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

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Maxima [A]  time = 1.67494, size = 39, normalized size = 1.34 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^2/(1+coth(x)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.69727, size = 632, normalized size = 21.79 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^2/(1+coth(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))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{2}{\left (x \right )}}{\coth{\left (x \right )} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)**2/(1+coth(x)),x)

[Out]

Integral(tanh(x)**2/(coth(x) + 1), x)

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Giac [A]  time = 1.14485, size = 47, normalized size = 1.62 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tanh(x)^2/(1+coth(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)