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]

1/4*arctanh((1-I)^(1/2)*tanh(x))/(1-I)^(1/2)+1/4*arctanh((1+I)^(1/2)*tanh(x))/(1+I)^(1/2)+1/8*arctanh(2^(1/2)*
tanh(x))*2^(1/2)+1/4*tanh(x)

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Rubi [A]  time = 0.08, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, 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 8

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

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*}

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Mathematica [A]  time = 0.50, 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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fricas [B]  time = 1.55, size = 708, normalized size = 10.26 \[ \text {result too large to display} \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [C]  time = 0.05, size = 99, normalized size = 1.43 \[ \frac {\sqrt {2}\, \arctanh \left (\frac {\left (2 \tanh \left (\frac {x}{2}\right )-2\right ) \sqrt {2}}{4}\right )}{8}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (2 \textit {\_Z}^{4}-2 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+\left (-4 \textit {\_R}^{3}+4 \textit {\_R} \right ) \tanh \left (\frac {x}{2}\right )+1\right )\right )}{8}+\frac {\sqrt {2}\, \arctanh \left (\frac {\left (2 \tanh \left (\frac {x}{2}\right )+2\right ) \sqrt {2}}{4}\right )}{8}+\frac {\tanh \left (\frac {x}{2}\right )}{2 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*2^(1/2)*arctanh(1/4*(2*tanh(1/2*x)-2)*2^(1/2))+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))+1/8*2^(1/2)*arctanh(1/4*(2*tanh(1/2*x)+2)*2^(1/2))+1/2*tanh(1/2*x)/(tanh(1/2*x)^2+1
)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{16} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - e^{x} + 1}{\sqrt {2} + e^{x} - 1}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - e^{x} - 1}{\sqrt {2} + e^{x} + 1}\right ) - \frac {1}{2 \, {\left (e^{\left (2 \, x\right )} + 1\right )}} + 8 \, \int \frac {e^{\left (4 \, x\right )}}{e^{\left (8 \, x\right )} - 4 \, e^{\left (6 \, x\right )} + 22 \, e^{\left (4 \, x\right )} - 4 \, e^{\left (2 \, x\right )} + 1}\,{d x} \]

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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mupad [B]  time = 4.81, size = 273, normalized size = 3.96 \[ \frac {\sqrt {2}\,\ln \left (582732658686033920\,{\mathrm {e}}^{2\,x}-70697326355677184\,\sqrt {2}+412054214575915008\,\sqrt {2}\,{\mathrm {e}}^{2\,x}-99981117754441728\right )}{16}-\frac {\sqrt {2}\,\ln \left (582732658686033920\,{\mathrm {e}}^{2\,x}+70697326355677184\,\sqrt {2}-412054214575915008\,\sqrt {2}\,{\mathrm {e}}^{2\,x}-99981117754441728\right )}{16}-\frac {1}{2\,\left ({\mathrm {e}}^{2\,x}+1\right )}-\frac {\sqrt {2}\,\sqrt {1-\mathrm {i}}\,\ln \left ({\mathrm {e}}^{2\,x}\,\left (155613434002538496+429723297714798592{}\mathrm {i}\right )+\sqrt {2}\,\sqrt {1-\mathrm {i}}\,\left (-54684829282729984+21956972328779776{}\mathrm {i}\right )+\sqrt {2}\,\sqrt {1-\mathrm {i}}\,{\mathrm {e}}^{2\,x}\,\left (12296353929494528-271474128182050816{}\mathrm {i}\right )+70836483296067584-69311013991743488{}\mathrm {i}\right )}{16}+\frac {\sqrt {2}\,\sqrt {1-\mathrm {i}}\,\ln \left ({\mathrm {e}}^{2\,x}\,\left (155613434002538496+429723297714798592{}\mathrm {i}\right )+\sqrt {2}\,\sqrt {1-\mathrm {i}}\,\left (54684829282729984-21956972328779776{}\mathrm {i}\right )+\sqrt {2}\,\sqrt {1-\mathrm {i}}\,{\mathrm {e}}^{2\,x}\,\left (-12296353929494528+271474128182050816{}\mathrm {i}\right )+70836483296067584-69311013991743488{}\mathrm {i}\right )}{16}-\frac {\sqrt {2}\,\sqrt {1+1{}\mathrm {i}}\,\ln \left ({\mathrm {e}}^{2\,x}\,\left (155613434002538496-429723297714798592{}\mathrm {i}\right )+\sqrt {2}\,\sqrt {1+1{}\mathrm {i}}\,\left (-54684829282729984-21956972328779776{}\mathrm {i}\right )+\sqrt {2}\,\sqrt {1+1{}\mathrm {i}}\,{\mathrm {e}}^{2\,x}\,\left (12296353929494528+271474128182050816{}\mathrm {i}\right )+70836483296067584+69311013991743488{}\mathrm {i}\right )}{16}+\frac {\sqrt {2}\,\sqrt {1+1{}\mathrm {i}}\,\ln \left ({\mathrm {e}}^{2\,x}\,\left (155613434002538496-429723297714798592{}\mathrm {i}\right )+\sqrt {2}\,\sqrt {1+1{}\mathrm {i}}\,\left (54684829282729984+21956972328779776{}\mathrm {i}\right )+\sqrt {2}\,\sqrt {1+1{}\mathrm {i}}\,{\mathrm {e}}^{2\,x}\,\left (-12296353929494528-271474128182050816{}\mathrm {i}\right )+70836483296067584+69311013991743488{}\mathrm {i}\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(sinh(x)^8 - 1),x)

[Out]

(2^(1/2)*log(582732658686033920*exp(2*x) - 70697326355677184*2^(1/2) + 412054214575915008*2^(1/2)*exp(2*x) - 9
9981117754441728))/16 - (2^(1/2)*log(582732658686033920*exp(2*x) + 70697326355677184*2^(1/2) - 412054214575915
008*2^(1/2)*exp(2*x) - 99981117754441728))/16 - 1/(2*(exp(2*x) + 1)) - (2^(1/2)*(1 - 1i)^(1/2)*log(exp(2*x)*(1
55613434002538496 + 429723297714798592i) - 2^(1/2)*(1 - 1i)^(1/2)*(54684829282729984 - 21956972328779776i) + 2
^(1/2)*(1 - 1i)^(1/2)*exp(2*x)*(12296353929494528 - 271474128182050816i) + (70836483296067584 - 69311013991743
488i)))/16 + (2^(1/2)*(1 - 1i)^(1/2)*log(exp(2*x)*(155613434002538496 + 429723297714798592i) + 2^(1/2)*(1 - 1i
)^(1/2)*(54684829282729984 - 21956972328779776i) - 2^(1/2)*(1 - 1i)^(1/2)*exp(2*x)*(12296353929494528 - 271474
128182050816i) + (70836483296067584 - 69311013991743488i)))/16 - (2^(1/2)*(1 + 1i)^(1/2)*log(exp(2*x)*(1556134
34002538496 - 429723297714798592i) - 2^(1/2)*(1 + 1i)^(1/2)*(54684829282729984 + 21956972328779776i) + 2^(1/2)
*(1 + 1i)^(1/2)*exp(2*x)*(12296353929494528 + 271474128182050816i) + (70836483296067584 + 69311013991743488i))
)/16 + (2^(1/2)*(1 + 1i)^(1/2)*log(exp(2*x)*(155613434002538496 - 429723297714798592i) + 2^(1/2)*(1 + 1i)^(1/2
)*(54684829282729984 + 21956972328779776i) - 2^(1/2)*(1 + 1i)^(1/2)*exp(2*x)*(12296353929494528 + 271474128182
050816i) + (70836483296067584 + 69311013991743488i)))/16

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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