3.269 \(\int \frac {1}{a+b \sinh ^8(x)} \, dx\)

Optimal. Leaf size=245 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}-\sqrt [4]{b}} \tanh (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}-\sqrt [4]{b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}} \tanh (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}} \tanh (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}+\sqrt [4]{b}} \tanh (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+\sqrt [4]{b}}} \]

[Out]

-1/4*arctanh(((-a)^(1/4)-b^(1/4))^(1/2)*tanh(x)/(-a)^(1/8))/(-a)^(7/8)/((-a)^(1/4)-b^(1/4))^(1/2)-1/4*arctanh(
((-a)^(1/4)-I*b^(1/4))^(1/2)*tanh(x)/(-a)^(1/8))/(-a)^(7/8)/((-a)^(1/4)-I*b^(1/4))^(1/2)-1/4*arctanh(((-a)^(1/
4)+I*b^(1/4))^(1/2)*tanh(x)/(-a)^(1/8))/(-a)^(7/8)/((-a)^(1/4)+I*b^(1/4))^(1/2)-1/4*arctanh(((-a)^(1/4)+b^(1/4
))^(1/2)*tanh(x)/(-a)^(1/8))/(-a)^(7/8)/((-a)^(1/4)+b^(1/4))^(1/2)

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Rubi [A]  time = 0.52, antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3211, 3181, 206} \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}} \tanh (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}} \tanh (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}+\sqrt [4]{b}} \tanh (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+\sqrt [4]{b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a \sqrt [4]{b}+(-a)^{5/4}} \tanh (x)}{(-a)^{5/8}}\right )}{4 (-a)^{3/8} \sqrt {a \sqrt [4]{b}+(-a)^{5/4}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sinh[x]^8)^(-1),x]

[Out]

-ArcTanh[(Sqrt[(-a)^(1/4) - I*b^(1/4)]*Tanh[x])/(-a)^(1/8)]/(4*(-a)^(7/8)*Sqrt[(-a)^(1/4) - I*b^(1/4)]) - ArcT
anh[(Sqrt[(-a)^(1/4) + I*b^(1/4)]*Tanh[x])/(-a)^(1/8)]/(4*(-a)^(7/8)*Sqrt[(-a)^(1/4) + I*b^(1/4)]) - ArcTanh[(
Sqrt[(-a)^(1/4) + b^(1/4)]*Tanh[x])/(-a)^(1/8)]/(4*(-a)^(7/8)*Sqrt[(-a)^(1/4) + b^(1/4)]) - ArcTanh[(Sqrt[(-a)
^(5/4) + a*b^(1/4)]*Tanh[x])/(-a)^(5/8)]/(4*(-a)^(3/8)*Sqrt[(-a)^(5/4) + a*b^(1/4)])

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 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]

Rubi steps

\begin {align*} \int \frac {1}{a+b \sinh ^8(x)} \, dx &=\frac {\int \frac {1}{1-\frac {\sqrt [4]{b} \sinh ^2(x)}{\sqrt [4]{-a}}} \, dx}{4 a}+\frac {\int \frac {1}{1-\frac {i \sqrt [4]{b} \sinh ^2(x)}{\sqrt [4]{-a}}} \, dx}{4 a}+\frac {\int \frac {1}{1+\frac {i \sqrt [4]{b} \sinh ^2(x)}{\sqrt [4]{-a}}} \, dx}{4 a}+\frac {\int \frac {1}{1+\frac {\sqrt [4]{b} \sinh ^2(x)}{\sqrt [4]{-a}}} \, dx}{4 a}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{1-\left (1-\frac {\sqrt [4]{b}}{\sqrt [4]{-a}}\right ) x^2} \, dx,x,\tanh (x)\right )}{4 a}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-\left (1-\frac {i \sqrt [4]{b}}{\sqrt [4]{-a}}\right ) x^2} \, dx,x,\tanh (x)\right )}{4 a}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-\left (1+\frac {i \sqrt [4]{b}}{\sqrt [4]{-a}}\right ) x^2} \, dx,x,\tanh (x)\right )}{4 a}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-\left (1+\frac {\sqrt [4]{b}}{\sqrt [4]{-a}}\right ) x^2} \, dx,x,\tanh (x)\right )}{4 a}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}} \tanh (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}-i \sqrt [4]{b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}} \tanh (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+i \sqrt [4]{b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {\sqrt [4]{-a}+\sqrt [4]{b}} \tanh (x)}{\sqrt [8]{-a}}\right )}{4 (-a)^{7/8} \sqrt {\sqrt [4]{-a}+\sqrt [4]{b}}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {(-a)^{5/4}+a \sqrt [4]{b}} \tanh (x)}{(-a)^{5/8}}\right )}{4 (-a)^{3/8} \sqrt {(-a)^{5/4}+a \sqrt [4]{b}}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 160, normalized size = 0.65 \[ 16 \text {RootSum}\left [\text {$\#$1}^8 b-8 \text {$\#$1}^7 b+28 \text {$\#$1}^6 b-56 \text {$\#$1}^5 b+256 \text {$\#$1}^4 a+70 \text {$\#$1}^4 b-56 \text {$\#$1}^3 b+28 \text {$\#$1}^2 b-8 \text {$\#$1} b+b\& ,\frac {\text {$\#$1}^3 x+\text {$\#$1}^3 \log (-\text {$\#$1} \sinh (x)+\text {$\#$1} \cosh (x)-\sinh (x)-\cosh (x))}{\text {$\#$1}^7 b-7 \text {$\#$1}^6 b+21 \text {$\#$1}^5 b-35 \text {$\#$1}^4 b+128 \text {$\#$1}^3 a+35 \text {$\#$1}^3 b-21 \text {$\#$1}^2 b+7 \text {$\#$1} b-b}\& \right ] \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sinh[x]^8)^(-1),x]

[Out]

16*RootSum[b - 8*b*#1 + 28*b*#1^2 - 56*b*#1^3 + 256*a*#1^4 + 70*b*#1^4 - 56*b*#1^5 + 28*b*#1^6 - 8*b*#1^7 + b*
#1^8 & , (x*#1^3 + Log[-Cosh[x] - Sinh[x] + Cosh[x]*#1 - Sinh[x]*#1]*#1^3)/(-b + 7*b*#1 - 21*b*#1^2 + 128*a*#1
^3 + 35*b*#1^3 - 35*b*#1^4 + 21*b*#1^5 - 7*b*#1^6 + b*#1^7) & ]

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fricas [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/(a+b*sinh(x)^8),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.80, size = 1, normalized size = 0.00 \[ 0 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(x)^8),x, algorithm="giac")

[Out]

0

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maple [C]  time = 0.06, size = 162, normalized size = 0.66 \[ \frac {\left (\munderset {\textit {\_R} =\RootOf \left (a \,\textit {\_Z}^{16}-8 a \,\textit {\_Z}^{14}+28 a \,\textit {\_Z}^{12}-56 a \,\textit {\_Z}^{10}+\left (70 a +256 b \right ) \textit {\_Z}^{8}-56 a \,\textit {\_Z}^{6}+28 a \,\textit {\_Z}^{4}-8 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (-\textit {\_R}^{14}+7 \textit {\_R}^{12}-21 \textit {\_R}^{10}+35 \textit {\_R}^{8}-35 \textit {\_R}^{6}+21 \textit {\_R}^{4}-7 \textit {\_R}^{2}+1\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{15} a -7 \textit {\_R}^{13} a +21 \textit {\_R}^{11} a -35 \textit {\_R}^{9} a +35 \textit {\_R}^{7} a +128 \textit {\_R}^{7} b -21 \textit {\_R}^{5} a +7 \textit {\_R}^{3} a -\textit {\_R} a}\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sinh(x)^8),x)

[Out]

1/8*sum((-_R^14+7*_R^12-21*_R^10+35*_R^8-35*_R^6+21*_R^4-7*_R^2+1)/(_R^15*a-7*_R^13*a+21*_R^11*a-35*_R^9*a+35*
_R^7*a+128*_R^7*b-21*_R^5*a+7*_R^3*a-_R*a)*ln(tanh(1/2*x)-_R),_R=RootOf(a*_Z^16-8*a*_Z^14+28*a*_Z^12-56*a*_Z^1
0+(70*a+256*b)*_Z^8-56*a*_Z^6+28*a*_Z^4-8*a*_Z^2+a))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{b \sinh \relax (x)^{8} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(x)^8),x, algorithm="maxima")

[Out]

integrate(1/(b*sinh(x)^8 + a), x)

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sinh(x)^8),x)

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{a + b \sinh ^{8}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sinh(x)**8),x)

[Out]

Integral(1/(a + b*sinh(x)**8), x)

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