3.63 \(\int \frac{1}{1-\cosh ^4(x)} \, dx\)

Optimal. Leaf size=25 \[ \frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{2 \sqrt{2}}+\frac{\coth (x)}{2} \]

[Out]

ArcTanh[Tanh[x]/Sqrt[2]]/(2*Sqrt[2]) + Coth[x]/2

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Rubi [A]  time = 0.0169727, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3209, 388, 206} \[ \frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{2 \sqrt{2}}+\frac{\coth (x)}{2} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Cosh[x]^4)^(-1),x]

[Out]

ArcTanh[Tanh[x]/Sqrt[2]]/(2*Sqrt[2]) + Coth[x]/2

Rule 3209

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dis
t[ff/f, Subst[Int[(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(2*p + 1), x], x, Tan[e + f*x]/ff], x
]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[p]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.10027, size = 24, normalized size = 0.96 \[ \frac{1}{4} \left (\sqrt{2} \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )+2 \coth (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cosh[x]^4)^(-1),x]

[Out]

(Sqrt[2]*ArcTanh[Tanh[x]/Sqrt[2]] + 2*Coth[x])/4

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Maple [B]  time = 0.017, size = 100, normalized size = 4. \begin{align*}{\frac{1}{4}\tanh \left ({\frac{x}{2}} \right ) }+{\frac{1}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{\sqrt{2}}{16}\ln \left ({ \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}} \right ) }-{\frac{\sqrt{2}}{16}\ln \left ({ \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-cosh(x)^4),x)

[Out]

1/4*tanh(1/2*x)+1/4/tanh(1/2*x)+1/16*2^(1/2)*ln((tanh(1/2*x)^2+2^(1/2)*tanh(1/2*x)+1)/(tanh(1/2*x)^2-2^(1/2)*t
anh(1/2*x)+1))-1/16*2^(1/2)*ln((tanh(1/2*x)^2-2^(1/2)*tanh(1/2*x)+1)/(tanh(1/2*x)^2+2^(1/2)*tanh(1/2*x)+1))

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Maxima [B]  time = 1.68216, size = 61, normalized size = 2.44 \begin{align*} -\frac{1}{8} \, \sqrt{2} \log \left (-\frac{2 \, \sqrt{2} - e^{\left (-2 \, x\right )} - 3}{2 \, \sqrt{2} + e^{\left (-2 \, x\right )} + 3}\right ) - \frac{1}{e^{\left (-2 \, x\right )} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(x)^4),x, algorithm="maxima")

[Out]

-1/8*sqrt(2)*log(-(2*sqrt(2) - e^(-2*x) - 3)/(2*sqrt(2) + e^(-2*x) + 3)) - 1/(e^(-2*x) - 1)

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Fricas [B]  time = 2.1641, size = 382, normalized size = 15.28 \begin{align*} \frac{{\left (\sqrt{2} \cosh \left (x\right )^{2} + 2 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt{2} \sinh \left (x\right )^{2} - \sqrt{2}\right )} \log \left (-\frac{3 \,{\left (2 \, \sqrt{2} - 3\right )} \cosh \left (x\right )^{2} - 4 \,{\left (3 \, \sqrt{2} - 4\right )} \cosh \left (x\right ) \sinh \left (x\right ) + 3 \,{\left (2 \, \sqrt{2} - 3\right )} \sinh \left (x\right )^{2} + 2 \, \sqrt{2} - 3}{\cosh \left (x\right )^{2} + \sinh \left (x\right )^{2} + 3}\right ) + 8}{8 \,{\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(x)^4),x, algorithm="fricas")

[Out]

1/8*((sqrt(2)*cosh(x)^2 + 2*sqrt(2)*cosh(x)*sinh(x) + sqrt(2)*sinh(x)^2 - sqrt(2))*log(-(3*(2*sqrt(2) - 3)*cos
h(x)^2 - 4*(3*sqrt(2) - 4)*cosh(x)*sinh(x) + 3*(2*sqrt(2) - 3)*sinh(x)^2 + 2*sqrt(2) - 3)/(cosh(x)^2 + sinh(x)
^2 + 3)) + 8)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)

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Sympy [B]  time = 3.71029, size = 75, normalized size = 3. \begin{align*} - \frac{\sqrt{2} \log{\left (4 \tanh ^{2}{\left (\frac{x}{2} \right )} - 4 \sqrt{2} \tanh{\left (\frac{x}{2} \right )} + 4 \right )}}{8} + \frac{\sqrt{2} \log{\left (4 \tanh ^{2}{\left (\frac{x}{2} \right )} + 4 \sqrt{2} \tanh{\left (\frac{x}{2} \right )} + 4 \right )}}{8} + \frac{\tanh{\left (\frac{x}{2} \right )}}{4} + \frac{1}{4 \tanh{\left (\frac{x}{2} \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(x)**4),x)

[Out]

-sqrt(2)*log(4*tanh(x/2)**2 - 4*sqrt(2)*tanh(x/2) + 4)/8 + sqrt(2)*log(4*tanh(x/2)**2 + 4*sqrt(2)*tanh(x/2) +
4)/8 + tanh(x/2)/4 + 1/(4*tanh(x/2))

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Giac [B]  time = 1.24074, size = 58, normalized size = 2.32 \begin{align*} \frac{1}{8} \, \sqrt{2} \log \left (-\frac{2 \, \sqrt{2} - e^{\left (2 \, x\right )} - 3}{2 \, \sqrt{2} + e^{\left (2 \, x\right )} + 3}\right ) + \frac{1}{e^{\left (2 \, x\right )} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(x)^4),x, algorithm="giac")

[Out]

1/8*sqrt(2)*log(-(2*sqrt(2) - e^(2*x) - 3)/(2*sqrt(2) + e^(2*x) + 3)) + 1/(e^(2*x) - 1)