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]

-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)^(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) + 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)])

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Rubi [A]  time = 0.516823, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, 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 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])

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*}

Mathematica [C]  time = 0.245581, size = 160, normalized size = 0.65 \[ 16 \text{RootSum}\left [256 \text{$\#$1}^4 a+\text{$\#$1}^8 b-8 \text{$\#$1}^7 b+28 \text{$\#$1}^6 b-56 \text{$\#$1}^5 b+70 \text{$\#$1}^4 b-56 \text{$\#$1}^3 b+28 \text{$\#$1}^2 b-8 \text{$\#$1} b+b\& ,\frac{\text{$\#$1}^3 x+\text{$\#$1}^3 \log (-\text{$\#$1} \sinh (x)+\text{$\#$1} \cosh (x)-\sinh (x)-\cosh (x))}{128 \text{$\#$1}^3 a+\text{$\#$1}^7 b-7 \text{$\#$1}^6 b+21 \text{$\#$1}^5 b-35 \text{$\#$1}^4 b+35 \text{$\#$1}^3 b-21 \text{$\#$1}^2 b+7 \text{$\#$1} b-b}\& \right ] \]

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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Maple [C]  time = 0.033, size = 162, normalized size = 0.7 \begin{align*}{\frac{1}{8}\sum _{{\it \_R}={\it RootOf} \left ( a{{\it \_Z}}^{16}-8\,a{{\it \_Z}}^{14}+28\,a{{\it \_Z}}^{12}-56\,a{{\it \_Z}}^{10}+ \left ( 70\,a+256\,b \right ){{\it \_Z}}^{8}-56\,a{{\it \_Z}}^{6}+28\,a{{\it \_Z}}^{4}-8\,a{{\it \_Z}}^{2}+a \right ) }{\frac{-{{\it \_R}}^{14}+7\,{{\it \_R}}^{12}-21\,{{\it \_R}}^{10}+35\,{{\it \_R}}^{8}-35\,{{\it \_R}}^{6}+21\,{{\it \_R}}^{4}-7\,{{\it \_R}}^{2}+1}{{{\it \_R}}^{15}a-7\,{{\it \_R}}^{13}a+21\,{{\it \_R}}^{11}a-35\,{{\it \_R}}^{9}a+35\,{{\it \_R}}^{7}a+128\,{{\it \_R}}^{7}b-21\,{{\it \_R}}^{5}a+7\,{{\it \_R}}^{3}a-{\it \_R}\,a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -{\it \_R} \right ) }} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{b \sinh \left (x\right )^{8} + a}\,{d x} \end{align*}

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

[Out]

Timed out

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

[Out]

Timed out

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Giac [A]  time = 1.90444, size = 1, normalized size = 0. \begin{align*} 0 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0