3.124 \(\int \frac{\tanh ^3(x)}{1+\coth (x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0991369, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {3552, 3529, 3531, 3475} \[ -\frac{3 x}{2}-\tanh ^2(x)+\frac{3 \tanh (x)}{2}+2 \log (\cosh (x))+\frac{\tanh ^2(x)}{2 (\coth (x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^3/(1 + Coth[x]),x]

[Out]

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

Mathematica [A]  time = 0.0532405, 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]^3/(1 + Coth[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 [B]  time = 0.037, size = 80, 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{7}{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{ \left ( \tanh \left ( x/2 \right ) \right ) ^{3}- \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+\tanh \left ( x/2 \right ) }{ \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+2\,\ln \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^3/(1+coth(x)),x)

[Out]

-1/(tanh(1/2*x)+1)^2+1/(tanh(1/2*x)+1)-7/2*ln(tanh(1/2*x)+1)-1/2*ln(tanh(1/2*x)-1)+2*(tanh(1/2*x)^3-tanh(1/2*x
)^2+tanh(1/2*x))/(tanh(1/2*x)^2+1)^2+2*ln(tanh(1/2*x)^2+1)

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Maxima [A]  time = 1.54025, 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)^3/(1+coth(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 = 2.64764, 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)^3/(1+coth(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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tanh ^{3}{\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)**3/(1+coth(x)),x)

[Out]

Integral(tanh(x)**3/(coth(x) + 1), x)

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Giac [A]  time = 1.11696, 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)^3/(1+coth(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)