3.76 \(\int \frac {\text {csch}^4(x)}{1+\tanh (x)} \, dx\)

Optimal. Leaf size=17 \[ \frac {\coth ^2(x)}{2}-\frac {\coth ^3(x)}{3} \]

[Out]

1/2*coth(x)^2-1/3*coth(x)^3

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Rubi [A]  time = 0.04, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3516, 848, 43} \[ \frac {\coth ^2(x)}{2}-\frac {\coth ^3(x)}{3} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^4/(1 + Tanh[x]),x]

[Out]

Coth[x]^2/2 - Coth[x]^3/3

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 3516

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b/f, Subst[Int
[(x^m*(a + x)^n)/(b^2 + x^2)^(m/2 + 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[m/
2]

Rubi steps

\begin {align*} \int \frac {\text {csch}^4(x)}{1+\tanh (x)} \, dx &=-\operatorname {Subst}\left (\int \frac {-1+x^2}{x^4 (1+x)} \, dx,x,\tanh (x)\right )\\ &=-\operatorname {Subst}\left (\int \frac {-1+x}{x^4} \, dx,x,\tanh (x)\right )\\ &=-\operatorname {Subst}\left (\int \left (-\frac {1}{x^4}+\frac {1}{x^3}\right ) \, dx,x,\tanh (x)\right )\\ &=\frac {\coth ^2(x)}{2}-\frac {\coth ^3(x)}{3}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 20, normalized size = 1.18 \[ -\frac {1}{6} \text {csch}(x) (2 \cosh (x)+(2 \coth (x)-3) \text {csch}(x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^4/(1 + Tanh[x]),x]

[Out]

-1/6*(Csch[x]*(2*Cosh[x] + (-3 + 2*Coth[x])*Csch[x]))

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fricas [B]  time = 0.52, size = 84, normalized size = 4.94 \[ -\frac {4 \, {\left (2 \, \cosh \relax (x) + \sinh \relax (x)\right )}}{3 \, {\left (\cosh \relax (x)^{5} + 5 \, \cosh \relax (x) \sinh \relax (x)^{4} + \sinh \relax (x)^{5} + {\left (10 \, \cosh \relax (x)^{2} - 3\right )} \sinh \relax (x)^{3} - 3 \, \cosh \relax (x)^{3} + {\left (10 \, \cosh \relax (x)^{3} - 9 \, \cosh \relax (x)\right )} \sinh \relax (x)^{2} + {\left (5 \, \cosh \relax (x)^{4} - 9 \, \cosh \relax (x)^{2} + 4\right )} \sinh \relax (x) + 2 \, \cosh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(1+tanh(x)),x, algorithm="fricas")

[Out]

-4/3*(2*cosh(x) + sinh(x))/(cosh(x)^5 + 5*cosh(x)*sinh(x)^4 + sinh(x)^5 + (10*cosh(x)^2 - 3)*sinh(x)^3 - 3*cos
h(x)^3 + (10*cosh(x)^3 - 9*cosh(x))*sinh(x)^2 + (5*cosh(x)^4 - 9*cosh(x)^2 + 4)*sinh(x) + 2*cosh(x))

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giac [A]  time = 0.11, size = 18, normalized size = 1.06 \[ -\frac {2 \, {\left (3 \, e^{\left (2 \, x\right )} + 1\right )}}{3 \, {\left (e^{\left (2 \, x\right )} - 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(1+tanh(x)),x, algorithm="giac")

[Out]

-2/3*(3*e^(2*x) + 1)/(e^(2*x) - 1)^3

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maple [B]  time = 0.10, size = 48, normalized size = 2.82 \[ -\frac {\left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{24}+\frac {\left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{8}-\frac {\tanh \left (\frac {x}{2}\right )}{8}-\frac {1}{24 \tanh \left (\frac {x}{2}\right )^{3}}-\frac {1}{8 \tanh \left (\frac {x}{2}\right )}+\frac {1}{8 \tanh \left (\frac {x}{2}\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.30, size = 75, normalized size = 4.41 \[ -\frac {2 \, e^{\left (-2 \, x\right )}}{3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1} + \frac {4 \, e^{\left (-4 \, x\right )}}{3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1} + \frac {2}{3 \, {\left (3 \, e^{\left (-2 \, x\right )} - 3 \, e^{\left (-4 \, x\right )} + e^{\left (-6 \, x\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)^4/(1+tanh(x)),x, algorithm="maxima")

[Out]

-2*e^(-2*x)/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6*x) - 1) + 4*e^(-4*x)/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6*x) - 1) + 2
/3/(3*e^(-2*x) - 3*e^(-4*x) + e^(-6*x) - 1)

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mupad [B]  time = 1.06, size = 18, normalized size = 1.06 \[ -\frac {2\,\left (3\,{\mathrm {e}}^{2\,x}+1\right )}{3\,{\left ({\mathrm {e}}^{2\,x}-1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sinh(x)^4*(tanh(x) + 1)),x)

[Out]

-(2*(3*exp(2*x) + 1))/(3*(exp(2*x) - 1)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csch(x)**4/(tanh(x) + 1), x)

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