3.75 \(\int \frac{1}{1-\cosh ^8(x)} \, dx\)

Optimal. Leaf size=69 \[ \frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{1-i}}\right )}{4 \sqrt{1-i}}+\frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{1+i}}\right )}{4 \sqrt{1+i}}+\frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{4 \sqrt{2}}+\frac{\coth (x)}{4} \]

[Out]

ArcTanh[Tanh[x]/Sqrt[1 - I]]/(4*Sqrt[1 - I]) + ArcTanh[Tanh[x]/Sqrt[1 + I]]/(4*Sqrt[1 + I]) + ArcTanh[Tanh[x]/
Sqrt[2]]/(4*Sqrt[2]) + Coth[x]/4

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Rubi [A]  time = 0.0774158, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {3211, 3181, 206, 3175, 3767, 8} \[ \frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{1-i}}\right )}{4 \sqrt{1-i}}+\frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{1+i}}\right )}{4 \sqrt{1+i}}+\frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{4 \sqrt{2}}+\frac{\coth (x)}{4} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Cosh[x]^8)^(-1),x]

[Out]

ArcTanh[Tanh[x]/Sqrt[1 - I]]/(4*Sqrt[1 - I]) + ArcTanh[Tanh[x]/Sqrt[1 + I]]/(4*Sqrt[1 + I]) + ArcTanh[Tanh[x]/
Sqrt[2]]/(4*Sqrt[2]) + Coth[x]/4

Rule 3211

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(-1), x_Symbol] :> Module[{k}, Dist[2/(a*n), Sum[Int[1/(1 - Si
n[e + f*x]^2/((-1)^((4*k)/n)*Rt[-(a/b), n/2])), x], {k, 1, n/2}], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[n/
2]

Rule 3181

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

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])

Rule 3175

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[a^p, Int[ActivateTrig[u*cos[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{1}{1-\cosh ^8(x)} \, dx &=\frac{1}{4} \int \frac{1}{1-\cosh ^2(x)} \, dx+\frac{1}{4} \int \frac{1}{1-i \cosh ^2(x)} \, dx+\frac{1}{4} \int \frac{1}{1+i \cosh ^2(x)} \, dx+\frac{1}{4} \int \frac{1}{1+\cosh ^2(x)} \, dx\\ &=-\left (\frac{1}{4} \int \text{csch}^2(x) \, dx\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1-2 x^2} \, dx,x,\coth (x)\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1-(1+i) x^2} \, dx,x,\coth (x)\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1-(1-i) x^2} \, dx,x,\coth (x)\right )\\ &=\frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{1-i}}\right )}{4 \sqrt{1-i}}+\frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{1+i}}\right )}{4 \sqrt{1+i}}+\frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{4 \sqrt{2}}+\frac{1}{4} i \operatorname{Subst}(\int 1 \, dx,x,-i \coth (x))\\ &=\frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{1-i}}\right )}{4 \sqrt{1-i}}+\frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{1+i}}\right )}{4 \sqrt{1+i}}+\frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{4 \sqrt{2}}+\frac{\coth (x)}{4}\\ \end{align*}

Mathematica [A]  time = 0.433931, size = 64, normalized size = 0.93 \[ \frac{1}{8} \left (\frac{2 \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{1-i}}\right )}{\sqrt{1-i}}+\frac{2 \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{1+i}}\right )}{\sqrt{1+i}}+\sqrt{2} \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )+2 \coth (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cosh[x]^8)^(-1),x]

[Out]

((2*ArcTanh[Tanh[x]/Sqrt[1 - I]])/Sqrt[1 - I] + (2*ArcTanh[Tanh[x]/Sqrt[1 + I]])/Sqrt[1 + I] + Sqrt[2]*ArcTanh
[Tanh[x]/Sqrt[2]] + 2*Coth[x])/8

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Maple [C]  time = 0.032, size = 136, normalized size = 2. \begin{align*}{\frac{1}{8}\tanh \left ({\frac{x}{2}} \right ) }+{\frac{1}{8} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{1}{8}\sum _{{\it \_R}={\it RootOf} \left ( 2\,{{\it \_Z}}^{4}-2\,{{\it \_Z}}^{2}+1 \right ) }{\it \_R}\,\ln \left ( 2\,{\it \_R}\,\tanh \left ( x/2 \right ) + \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }+{\frac{\sqrt{2}}{32}\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}}{32}\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)^8),x)

[Out]

1/8*tanh(1/2*x)+1/8/tanh(1/2*x)+1/8*sum(_R*ln(2*_R*tanh(1/2*x)+tanh(1/2*x)^2+1),_R=RootOf(2*_Z^4-2*_Z^2+1))+1/
32*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))-1/32*2^(1/2)*ln((ta
nh(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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{16} \, \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}{2 \,{\left (e^{\left (2 \, x\right )} - 1\right )}} + 8 \, \int \frac{e^{\left (4 \, x\right )}}{e^{\left (8 \, x\right )} + 4 \, e^{\left (6 \, x\right )} + 22 \, e^{\left (4 \, x\right )} + 4 \, e^{\left (2 \, x\right )} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(x)^8),x, algorithm="maxima")

[Out]

1/16*sqrt(2)*log(-(2*sqrt(2) - e^(2*x) - 3)/(2*sqrt(2) + e^(2*x) + 3)) + 1/2/(e^(2*x) - 1) + 8*integrate(e^(4*
x)/(e^(8*x) + 4*e^(6*x) + 22*e^(4*x) + 4*e^(2*x) + 1), x)

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Fricas [B]  time = 3.06715, size = 2198, normalized size = 31.86 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(x)^8),x, algorithm="fricas")

[Out]

1/32*(4*(2^(1/4)*e^(2*x) - 2^(1/4))*sqrt(-2*sqrt(2) + 4)*arctan(1/14*(sqrt(2)*(5*sqrt(2) + 6) + 8*sqrt(2) + 4)
*e^(2*x) - 1/28*(2*sqrt(2)*(5*sqrt(2) + 6) - (2^(3/4)*(8*sqrt(2) + 11) + 2*2^(1/4)*(5*sqrt(2) + 6))*sqrt(-2*sq
rt(2) + 4) + 16*sqrt(2) + 8)*sqrt((2^(3/4)*e^(2*x) + 2^(1/4)*(3*sqrt(2) + 4))*sqrt(-2*sqrt(2) + 4) + 4*sqrt(2)
+ e^(4*x) + 2*e^(2*x) + 5) + 1/14*sqrt(2)*(3*sqrt(2) - 2) - 1/28*((2^(3/4)*(8*sqrt(2) + 11) + 2*2^(1/4)*(5*sq
rt(2) + 6))*e^(2*x) + 2^(3/4)*(2*sqrt(2) + 1) + 2*2^(1/4)*(3*sqrt(2) - 2))*sqrt(-2*sqrt(2) + 4) + 1/7*sqrt(2)
- 3/7) + 4*(2^(1/4)*e^(2*x) - 2^(1/4))*sqrt(-2*sqrt(2) + 4)*arctan(-1/14*(sqrt(2)*(5*sqrt(2) + 6) + 8*sqrt(2)
+ 4)*e^(2*x) + 1/28*(2*sqrt(2)*(5*sqrt(2) + 6) + (2^(3/4)*(8*sqrt(2) + 11) + 2*2^(1/4)*(5*sqrt(2) + 6))*sqrt(-
2*sqrt(2) + 4) + 16*sqrt(2) + 8)*sqrt(-(2^(3/4)*e^(2*x) + 2^(1/4)*(3*sqrt(2) + 4))*sqrt(-2*sqrt(2) + 4) + 4*sq
rt(2) + e^(4*x) + 2*e^(2*x) + 5) - 1/14*sqrt(2)*(3*sqrt(2) - 2) - 1/28*((2^(3/4)*(8*sqrt(2) + 11) + 2*2^(1/4)*
(5*sqrt(2) + 6))*e^(2*x) + 2^(3/4)*(2*sqrt(2) + 1) + 2*2^(1/4)*(3*sqrt(2) - 2))*sqrt(-2*sqrt(2) + 4) - 1/7*sqr
t(2) + 3/7) - (2^(1/4)*(sqrt(2) + 1)*e^(2*x) - 2^(1/4)*(sqrt(2) + 1))*sqrt(-2*sqrt(2) + 4)*log((2^(3/4)*e^(2*x
) + 2^(1/4)*(3*sqrt(2) + 4))*sqrt(-2*sqrt(2) + 4) + 4*sqrt(2) + e^(4*x) + 2*e^(2*x) + 5) + (2^(1/4)*(sqrt(2) +
1)*e^(2*x) - 2^(1/4)*(sqrt(2) + 1))*sqrt(-2*sqrt(2) + 4)*log(-(2^(3/4)*e^(2*x) + 2^(1/4)*(3*sqrt(2) + 4))*sqr
t(-2*sqrt(2) + 4) + 4*sqrt(2) + e^(4*x) + 2*e^(2*x) + 5) + 2*(sqrt(2)*e^(2*x) - sqrt(2))*log(-(2*(2*sqrt(2) -
3)*e^(2*x) + 12*sqrt(2) - e^(4*x) - 17)/(e^(4*x) + 6*e^(2*x) + 1)) + 16)/(e^(2*x) - 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(x)**8),x)

[Out]

Timed out

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Giac [A]  time = 1.54299, size = 61, normalized size = 0.88 \begin{align*} \frac{1}{16} \, \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}{2 \,{\left (e^{\left (2 \, x\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cosh(x)^8),x, algorithm="giac")

[Out]

1/16*sqrt(2)*log(-(2*sqrt(2) - e^(2*x) - 3)/(2*sqrt(2) + e^(2*x) + 3)) + 1/2/(e^(2*x) - 1)