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

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

[Out]

5/2*x-5/2*coth(x)+coth(x)^2-5/6*coth(x)^3-2*ln(sinh(x))+1/2*coth(x)^3/(1+tanh(x))

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Rubi [A]  time = 0.10, antiderivative size = 43, normalized size of antiderivative = 1.00, 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} \[ \frac {5 x}{2}-\frac {5 \coth ^3(x)}{6}+\coth ^2(x)-\frac {5 \coth (x)}{2}-2 \log (\sinh (x))+\frac {\coth ^3(x)}{2 (\tanh (x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^4/(1 + Tanh[x]),x]

[Out]

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

Rule 3475

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

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 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]

Rubi steps

\begin {align*} \int \frac {\coth ^4(x)}{1+\tanh (x)} \, dx &=\frac {\coth ^3(x)}{2 (1+\tanh (x))}-\frac {1}{2} \int \coth ^4(x) (-5+4 \tanh (x)) \, dx\\ &=-\frac {5}{6} \coth ^3(x)+\frac {\coth ^3(x)}{2 (1+\tanh (x))}-\frac {1}{2} i \int \coth ^3(x) (-4 i+5 i \tanh (x)) \, dx\\ &=\coth ^2(x)-\frac {5 \coth ^3(x)}{6}+\frac {\coth ^3(x)}{2 (1+\tanh (x))}+\frac {1}{2} \int \coth ^2(x) (5-4 \tanh (x)) \, dx\\ &=-\frac {5 \coth (x)}{2}+\coth ^2(x)-\frac {5 \coth ^3(x)}{6}+\frac {\coth ^3(x)}{2 (1+\tanh (x))}+\frac {1}{2} i \int \coth (x) (4 i-5 i \tanh (x)) \, dx\\ &=\frac {5 x}{2}-\frac {5 \coth (x)}{2}+\coth ^2(x)-\frac {5 \coth ^3(x)}{6}+\frac {\coth ^3(x)}{2 (1+\tanh (x))}-2 \int \coth (x) \, dx\\ &=\frac {5 x}{2}-\frac {5 \coth (x)}{2}+\coth ^2(x)-\frac {5 \coth ^3(x)}{6}-2 \log (\sinh (x))+\frac {\coth ^3(x)}{2 (1+\tanh (x))}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 42, normalized size = 0.98 \[ \frac {1}{12} \left (-3 \cosh (2 x)-4 \coth (x) \left (\text {csch}^2(x)+7\right )+3 \left (10 x+\sinh (2 x)+2 \text {csch}^2(x)-8 \log (\sinh (x))\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^4/(1 + Tanh[x]),x]

[Out]

(-3*Cosh[2*x] - 4*Coth[x]*(7 + Csch[x]^2) + 3*(10*x + 2*Csch[x]^2 - 8*Log[Sinh[x]] + Sinh[2*x]))/12

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fricas [B]  time = 0.50, size = 582, normalized size = 13.53 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^4/(1+tanh(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
*sinh(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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giac [A]  time = 0.14, size = 48, normalized size = 1.12 \[ \frac {9}{2} \, x - \frac {{\left (51 \, e^{\left (6 \, x\right )} - 81 \, e^{\left (4 \, x\right )} + 65 \, e^{\left (2 \, x\right )} - 3\right )} e^{\left (-2 \, x\right )}}{12 \, {\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} - 2 \, \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^4/(1+tanh(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(abs(e^(2*x) - 1))

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maple [B]  time = 0.10, size = 91, normalized size = 2.12 \[ -\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{24}+\frac {\left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{8}-\frac {9 \tanh \left (\frac {x}{2}\right )}{8}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}-\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{\tanh \left (\frac {x}{2}\right )+1}+\frac {9 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}-\frac {1}{24 \tanh \left (\frac {x}{2}\right )^{3}}+\frac {1}{8 \tanh \left (\frac {x}{2}\right )^{2}}-\frac {9}{8 \tanh \left (\frac {x}{2}\right )}-2 \ln \left (\tanh \left (\frac {x}{2}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*tanh(1/2*x)^3+1/8*tanh(1/2*x)^2-9/8*tanh(1/2*x)-1/2*ln(tanh(1/2*x)-1)-1/(tanh(1/2*x)+1)^2+1/(tanh(1/2*x)
+1)+9/2*ln(tanh(1/2*x)+1)-1/24/tanh(1/2*x)^3+1/8/tanh(1/2*x)^2-9/8/tanh(1/2*x)-2*ln(tanh(1/2*x))

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maxima [A]  time = 0.31, size = 64, normalized size = 1.49 \[ \frac {1}{2} \, x - \frac {2 \, {\left (15 \, e^{\left (-2 \, x\right )} - 12 \, e^{\left (-4 \, x\right )} - 7\right )}}{3 \, {\left (3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1\right )}} - \frac {1}{4} \, e^{\left (-2 \, x\right )} - 2 \, \log \left (e^{\left (-x\right )} + 1\right ) - 2 \, \log \left (e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 1.10, size = 69, normalized size = 1.60 \[ \frac {9\,x}{2}-2\,\ln \left ({\mathrm {e}}^{2\,x}-1\right )-\frac {{\mathrm {e}}^{-2\,x}}{4}-\frac {8}{3\,\left (3\,{\mathrm {e}}^{2\,x}-3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{6\,x}-1\right )}-\frac {2}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1}-\frac {4}{{\mathrm {e}}^{2\,x}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^4/(tanh(x) + 1),x)

[Out]

(9*x)/2 - 2*log(exp(2*x) - 1) - exp(-2*x)/4 - 8/(3*(3*exp(2*x) - 3*exp(4*x) + exp(6*x) - 1)) - 2/(exp(4*x) - 2
*exp(2*x) + 1) - 4/(exp(2*x) - 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{4}{\relax (x )}}{\tanh {\relax (x )} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**4/(1+tanh(x)),x)

[Out]

Integral(coth(x)**4/(tanh(x) + 1), x)

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