3.275 \(\int \frac{1}{1-\sinh ^8(x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0772041, 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} \[ \frac{\tanh ^{-1}\left (\sqrt{1-i} \tanh (x)\right )}{4 \sqrt{1-i}}+\frac{\tanh ^{-1}\left (\sqrt{1+i} \tanh (x)\right )}{4 \sqrt{1+i}}+\frac{\tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )}{4 \sqrt{2}}+\frac{\tanh (x)}{4} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Sinh[x]^8)^(-1),x]

[Out]

ArcTanh[Sqrt[1 - I]*Tanh[x]]/(4*Sqrt[1 - I]) + ArcTanh[Sqrt[1 + I]*Tanh[x]]/(4*Sqrt[1 + I]) + ArcTanh[Sqrt[2]*
Tanh[x]]/(4*Sqrt[2]) + Tanh[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-\sinh ^8(x)} \, dx &=\frac{1}{4} \int \frac{1}{1-\sinh ^2(x)} \, dx+\frac{1}{4} \int \frac{1}{1-i \sinh ^2(x)} \, dx+\frac{1}{4} \int \frac{1}{1+i \sinh ^2(x)} \, dx+\frac{1}{4} \int \frac{1}{1+\sinh ^2(x)} \, dx\\ &=\frac{1}{4} \int \text{sech}^2(x) \, dx+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1-2 x^2} \, dx,x,\tanh (x)\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1-(1+i) x^2} \, dx,x,\tanh (x)\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1-(1-i) x^2} \, dx,x,\tanh (x)\right )\\ &=\frac{\tanh ^{-1}\left (\sqrt{1-i} \tanh (x)\right )}{4 \sqrt{1-i}}+\frac{\tanh ^{-1}\left (\sqrt{1+i} \tanh (x)\right )}{4 \sqrt{1+i}}+\frac{\tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )}{4 \sqrt{2}}+\frac{1}{4} i \operatorname{Subst}(\int 1 \, dx,x,-i \tanh (x))\\ &=\frac{\tanh ^{-1}\left (\sqrt{1-i} \tanh (x)\right )}{4 \sqrt{1-i}}+\frac{\tanh ^{-1}\left (\sqrt{1+i} \tanh (x)\right )}{4 \sqrt{1+i}}+\frac{\tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )}{4 \sqrt{2}}+\frac{\tanh (x)}{4}\\ \end{align*}

Mathematica [A]  time = 0.482848, size = 64, normalized size = 0.93 \[ \frac{1}{8} \left (\frac{2 \tanh ^{-1}\left (\sqrt{1-i} \tanh (x)\right )}{\sqrt{1-i}}+\frac{2 \tanh ^{-1}\left (\sqrt{1+i} \tanh (x)\right )}{\sqrt{1+i}}+\sqrt{2} \tanh ^{-1}\left (\sqrt{2} \tanh (x)\right )+2 \tanh (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Sinh[x]^8)^(-1),x]

[Out]

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

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Maple [C]  time = 0.033, size = 99, normalized size = 1.4 \begin{align*}{\frac{1}{8}\sum _{{\it \_R}={\it RootOf} \left ( 2\,{{\it \_Z}}^{4}-2\,{{\it \_Z}}^{2}+1 \right ) }{\it \_R}\,\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+ \left ( -4\,{{\it \_R}}^{3}+4\,{\it \_R} \right ) \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{1}{2}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-1}}+{\frac{\sqrt{2}}{8}{\it Artanh} \left ({\frac{\sqrt{2}}{4} \left ( 2\,\tanh \left ( x/2 \right ) -2 \right ) } \right ) }+{\frac{\sqrt{2}}{8}{\it Artanh} \left ({\frac{\sqrt{2}}{4} \left ( 2\,\tanh \left ( x/2 \right ) +2 \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-sinh(x)^8),x)

[Out]

1/8*sum(_R*ln(tanh(1/2*x)^2+(-4*_R^3+4*_R)*tanh(1/2*x)+1),_R=RootOf(2*_Z^4-2*_Z^2+1))+1/2*tanh(1/2*x)/(tanh(1/
2*x)^2+1)+1/8*2^(1/2)*arctanh(1/4*(2*tanh(1/2*x)-2)*2^(1/2))+1/8*2^(1/2)*arctanh(1/4*(2*tanh(1/2*x)+2)*2^(1/2)
)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{16} \, \sqrt{2} \log \left (-\frac{\sqrt{2} - e^{x} + 1}{\sqrt{2} + e^{x} - 1}\right ) + \frac{1}{16} \, \sqrt{2} \log \left (-\frac{\sqrt{2} - e^{x} - 1}{\sqrt{2} + e^{x} + 1}\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-sinh(x)^8),x, algorithm="maxima")

[Out]

-1/16*sqrt(2)*log(-(sqrt(2) - e^x + 1)/(sqrt(2) + e^x - 1)) + 1/16*sqrt(2)*log(-(sqrt(2) - e^x - 1)/(sqrt(2) +
e^x + 1)) - 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 = 1.85785, 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-sinh(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*s
qrt(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*
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*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*s
qrt(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*sq
rt(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))*sq
rt(-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-sinh(x)**8),x)

[Out]

Timed out

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Giac [A]  time = 1.38856, size = 65, normalized size = 0.94 \begin{align*} -\frac{1}{16} \, \sqrt{2} \log \left (\frac{{\left | -4 \, \sqrt{2} + 2 \, e^{\left (2 \, x\right )} - 6 \right |}}{{\left | 4 \, \sqrt{2} + 2 \, e^{\left (2 \, x\right )} - 6 \right |}}\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-sinh(x)^8),x, algorithm="giac")

[Out]

-1/16*sqrt(2)*log(abs(-4*sqrt(2) + 2*e^(2*x) - 6)/abs(4*sqrt(2) + 2*e^(2*x) - 6)) - 1/2/(e^(2*x) + 1)