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

Optimal. Leaf size=43 \[ \frac{5 x}{2}-\frac{5 \tanh ^3(x)}{6}+\tanh ^2(x)-\frac{5 \tanh (x)}{2}-2 \log (\cosh (x))+\frac{\tanh ^3(x)}{2 (\coth (x)+1)} \]

[Out]

(5*x)/2 - 2*Log[Cosh[x]] - (5*Tanh[x])/2 + Tanh[x]^2 - (5*Tanh[x]^3)/6 + Tanh[x]^3/(2*(1 + Coth[x]))

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Rubi [A]  time = 0.114614, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {3552, 3529, 3531, 3475} \[ \frac{5 x}{2}-\frac{5 \tanh ^3(x)}{6}+\tanh ^2(x)-\frac{5 \tanh (x)}{2}-2 \log (\cosh (x))+\frac{\tanh ^3(x)}{2 (\coth (x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^4/(1 + Coth[x]),x]

[Out]

(5*x)/2 - 2*Log[Cosh[x]] - (5*Tanh[x])/2 + Tanh[x]^2 - (5*Tanh[x]^3)/6 + Tanh[x]^3/(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 ^4(x)}{1+\coth (x)} \, dx &=\frac{\tanh ^3(x)}{2 (1+\coth (x))}-\frac{1}{2} \int (-5+4 \coth (x)) \tanh ^4(x) \, dx\\ &=-\frac{5}{6} \tanh ^3(x)+\frac{\tanh ^3(x)}{2 (1+\coth (x))}-\frac{1}{2} i \int (-4 i+5 i \coth (x)) \tanh ^3(x) \, dx\\ &=\tanh ^2(x)-\frac{5 \tanh ^3(x)}{6}+\frac{\tanh ^3(x)}{2 (1+\coth (x))}+\frac{1}{2} \int (5-4 \coth (x)) \tanh ^2(x) \, dx\\ &=-\frac{5 \tanh (x)}{2}+\tanh ^2(x)-\frac{5 \tanh ^3(x)}{6}+\frac{\tanh ^3(x)}{2 (1+\coth (x))}+\frac{1}{2} i \int (4 i-5 i \coth (x)) \tanh (x) \, dx\\ &=\frac{5 x}{2}-\frac{5 \tanh (x)}{2}+\tanh ^2(x)-\frac{5 \tanh ^3(x)}{6}+\frac{\tanh ^3(x)}{2 (1+\coth (x))}-2 \int \tanh (x) \, dx\\ &=\frac{5 x}{2}-2 \log (\cosh (x))-\frac{5 \tanh (x)}{2}+\tanh ^2(x)-\frac{5 \tanh ^3(x)}{6}+\frac{\tanh ^3(x)}{2 (1+\coth (x))}\\ \end{align*}

Mathematica [A]  time = 0.095612, size = 40, normalized size = 0.93 \[ \frac{1}{12} \left (30 x-3 \sinh (2 x)+3 \cosh (2 x)-28 \tanh (x)-24 \log (\cosh (x))+(4 \tanh (x)-6) \text{sech}^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tanh[x]^4/(1 + Coth[x]),x]

[Out]

(30*x + 3*Cosh[2*x] - 24*Log[Cosh[x]] - 3*Sinh[2*x] - 28*Tanh[x] + Sech[x]^2*(-6 + 4*Tanh[x]))/12

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Maple [B]  time = 0.043, size = 96, 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{9}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{2}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-4\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{5}-1/2\, \left ( \tanh \left ( x/2 \right ) \right ) ^{4}+8/3\, \left ( \tanh \left ( x/2 \right ) \right ) ^{3}-1/2\, \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+\tanh \left ( x/2 \right ) }{ \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-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)^4/(1+coth(x)),x)

[Out]

1/(tanh(1/2*x)+1)^2-1/(tanh(1/2*x)+1)+9/2*ln(tanh(1/2*x)+1)-1/2*ln(tanh(1/2*x)-1)-4*(tanh(1/2*x)^5-1/2*tanh(1/
2*x)^4+8/3*tanh(1/2*x)^3-1/2*tanh(1/2*x)^2+tanh(1/2*x))/(tanh(1/2*x)^2+1)^3-2*ln(tanh(1/2*x)^2+1)

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Maxima [A]  time = 1.70061, size = 74, normalized size = 1.72 \begin{align*} \frac{1}{2} \, x - \frac{2 \,{\left (15 \, e^{\left (-2 \, x\right )} + 12 \, e^{\left (-4 \, x\right )} + 7\right )}}{3 \,{\left (3 \, e^{\left (-2 \, x\right )} + 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} + 1\right )}} + \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+coth(x)),x, algorithm="maxima")

[Out]

1/2*x - 2/3*(15*e^(-2*x) + 12*e^(-4*x) + 7)/(3*e^(-2*x) + 3*e^(-4*x) + e^(-6*x) + 1) + 1/4*e^(-2*x) - 2*log(e^
(-2*x) + 1)

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Fricas [B]  time = 2.6535, size = 1897, normalized size = 44.12 \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+coth(x)),x, algorithm="fricas")

[Out]

1/12*(54*x*cosh(x)^8 + 432*x*cosh(x)*sinh(x)^7 + 54*x*sinh(x)^8 + 3*(54*x + 17)*cosh(x)^6 + 3*(504*x*cosh(x)^2
+ 54*x + 17)*sinh(x)^6 + 18*(168*x*cosh(x)^3 + (54*x + 17)*cosh(x))*sinh(x)^5 + 81*(2*x + 1)*cosh(x)^4 + 9*(4
20*x*cosh(x)^4 + 5*(54*x + 17)*cosh(x)^2 + 18*x + 9)*sinh(x)^4 + 12*(252*x*cosh(x)^5 + 5*(54*x + 17)*cosh(x)^3
+ 27*(2*x + 1)*cosh(x))*sinh(x)^3 + (54*x + 65)*cosh(x)^2 + (1512*x*cosh(x)^6 + 45*(54*x + 17)*cosh(x)^4 + 48
6*(2*x + 1)*cosh(x)^2 + 54*x + 65)*sinh(x)^2 - 24*(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + (28*cosh(x)^2
+ 3)*sinh(x)^6 + 3*cosh(x)^6 + 2*(28*cosh(x)^3 + 9*cosh(x))*sinh(x)^5 + (70*cosh(x)^4 + 45*cosh(x)^2 + 3)*sin
h(x)^4 + 3*cosh(x)^4 + 4*(14*cosh(x)^5 + 15*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + (28*cosh(x)^6 + 45*cosh(x)^4 +
18*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*(4*cosh(x)^7 + 9*cosh(x)^5 + 6*cosh(x)^3 + cosh(x))*sinh(x))*log(2
*cosh(x)/(cosh(x) - sinh(x))) + 2*(216*x*cosh(x)^7 + 9*(54*x + 17)*cosh(x)^5 + 162*(2*x + 1)*cosh(x)^3 + (54*x
+ 65)*cosh(x))*sinh(x) + 3)/(cosh(x)^8 + 8*cosh(x)*sinh(x)^7 + sinh(x)^8 + (28*cosh(x)^2 + 3)*sinh(x)^6 + 3*c
osh(x)^6 + 2*(28*cosh(x)^3 + 9*cosh(x))*sinh(x)^5 + (70*cosh(x)^4 + 45*cosh(x)^2 + 3)*sinh(x)^4 + 3*cosh(x)^4
+ 4*(14*cosh(x)^5 + 15*cosh(x)^3 + 3*cosh(x))*sinh(x)^3 + (28*cosh(x)^6 + 45*cosh(x)^4 + 18*cosh(x)^2 + 1)*sin
h(x)^2 + cosh(x)^2 + 2*(4*cosh(x)^7 + 9*cosh(x)^5 + 6*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 ^{4}{\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)**4/(1+coth(x)),x)

[Out]

Integral(tanh(x)**4/(coth(x) + 1), x)

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Giac [A]  time = 1.11034, size = 63, normalized size = 1.47 \begin{align*} \frac{9}{2} \, x + \frac{{\left (51 \, e^{\left (6 \, x\right )} + 81 \, e^{\left (4 \, x\right )} + 65 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (-2 \, x\right )}}{12 \,{\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} - 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+coth(x)),x, algorithm="giac")

[Out]

9/2*x + 1/12*(51*e^(6*x) + 81*e^(4*x) + 65*e^(2*x) + 3)*e^(-2*x)/(e^(2*x) + 1)^3 - 2*log(e^(2*x) + 1)