3.257 \(\int \frac{1}{1+\sin ^8(x)} \, dx\)

Optimal. Leaf size=218 \[ \frac{1}{8} \left (\sqrt{1+\sqrt{4-2 \sqrt{2}}}+\sqrt{2+2 \sqrt [4]{2}+2 \sqrt{1+\sqrt{2}}+2 \sqrt{2+\sqrt{2}}}+\sqrt{1+\sqrt{4+2 \sqrt{2}}}\right ) \left (x-\tan ^{-1}(\tan (x))\right )+\frac{\tan ^{-1}\left (\sqrt{1-\sqrt [4]{-1}} \tan (x)\right )}{4 \sqrt{1-\sqrt [4]{-1}}}+\frac{\tan ^{-1}\left (\sqrt{1+\sqrt [4]{-1}} \tan (x)\right )}{4 \sqrt{1+\sqrt [4]{-1}}}+\frac{\tan ^{-1}\left (\sqrt{1-(-1)^{3/4}} \tan (x)\right )}{4 \sqrt{1-(-1)^{3/4}}}+\frac{\tan ^{-1}\left (\sqrt{1+(-1)^{3/4}} \tan (x)\right )}{4 \sqrt{1+(-1)^{3/4}}} \]

[Out]

((Sqrt[1 + Sqrt[4 - 2*Sqrt[2]]] + Sqrt[2 + 2*2^(1/4) + 2*Sqrt[1 + Sqrt[2]] + 2*Sqrt[2 + Sqrt[2]]] + Sqrt[1 + S
qrt[4 + 2*Sqrt[2]]])*(x - ArcTan[Tan[x]]))/8 + ArcTan[Sqrt[1 - (-1)^(1/4)]*Tan[x]]/(4*Sqrt[1 - (-1)^(1/4)]) +
ArcTan[Sqrt[1 + (-1)^(1/4)]*Tan[x]]/(4*Sqrt[1 + (-1)^(1/4)]) + ArcTan[Sqrt[1 - (-1)^(3/4)]*Tan[x]]/(4*Sqrt[1 -
(-1)^(3/4)]) + ArcTan[Sqrt[1 + (-1)^(3/4)]*Tan[x]]/(4*Sqrt[1 + (-1)^(3/4)])

________________________________________________________________________________________

Rubi [A]  time = 0.200197, antiderivative size = 129, normalized size of antiderivative = 0.59, number of steps used = 9, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {3211, 3181, 203} \[ \frac{\tan ^{-1}\left (\sqrt{1-\sqrt [4]{-1}} \tan (x)\right )}{4 \sqrt{1-\sqrt [4]{-1}}}+\frac{\tan ^{-1}\left (\sqrt{1+\sqrt [4]{-1}} \tan (x)\right )}{4 \sqrt{1+\sqrt [4]{-1}}}+\frac{\tan ^{-1}\left (\sqrt{1-(-1)^{3/4}} \tan (x)\right )}{4 \sqrt{1-(-1)^{3/4}}}+\frac{\tan ^{-1}\left (\sqrt{1+(-1)^{3/4}} \tan (x)\right )}{4 \sqrt{1+(-1)^{3/4}}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Sin[x]^8)^(-1),x]

[Out]

ArcTan[Sqrt[1 - (-1)^(1/4)]*Tan[x]]/(4*Sqrt[1 - (-1)^(1/4)]) + ArcTan[Sqrt[1 + (-1)^(1/4)]*Tan[x]]/(4*Sqrt[1 +
(-1)^(1/4)]) + ArcTan[Sqrt[1 - (-1)^(3/4)]*Tan[x]]/(4*Sqrt[1 - (-1)^(3/4)]) + ArcTan[Sqrt[1 + (-1)^(3/4)]*Tan
[x]]/(4*Sqrt[1 + (-1)^(3/4)])

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1}{1+\sin ^8(x)} \, dx &=\frac{1}{4} \int \frac{1}{1-\sqrt [4]{-1} \sin ^2(x)} \, dx+\frac{1}{4} \int \frac{1}{1+\sqrt [4]{-1} \sin ^2(x)} \, dx+\frac{1}{4} \int \frac{1}{1-(-1)^{3/4} \sin ^2(x)} \, dx+\frac{1}{4} \int \frac{1}{1+(-1)^{3/4} \sin ^2(x)} \, dx\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+\left (1-\sqrt [4]{-1}\right ) x^2} \, dx,x,\tan (x)\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+\left (1+\sqrt [4]{-1}\right ) x^2} \, dx,x,\tan (x)\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+\left (1-(-1)^{3/4}\right ) x^2} \, dx,x,\tan (x)\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+\left (1+(-1)^{3/4}\right ) x^2} \, dx,x,\tan (x)\right )\\ &=\frac{\tan ^{-1}\left (\sqrt{1-\sqrt [4]{-1}} \tan (x)\right )}{4 \sqrt{1-\sqrt [4]{-1}}}+\frac{\tan ^{-1}\left (\sqrt{1+\sqrt [4]{-1}} \tan (x)\right )}{4 \sqrt{1+\sqrt [4]{-1}}}+\frac{\tan ^{-1}\left (\sqrt{1-(-1)^{3/4}} \tan (x)\right )}{4 \sqrt{1-(-1)^{3/4}}}+\frac{\tan ^{-1}\left (\sqrt{1+(-1)^{3/4}} \tan (x)\right )}{4 \sqrt{1+(-1)^{3/4}}}\\ \end{align*}

Mathematica [C]  time = 0.147735, size = 141, normalized size = 0.65 \[ 8 \text{RootSum}\left [\text{$\#$1}^8-8 \text{$\#$1}^7+28 \text{$\#$1}^6-56 \text{$\#$1}^5+326 \text{$\#$1}^4-56 \text{$\#$1}^3+28 \text{$\#$1}^2-8 \text{$\#$1}+1\& ,\frac{2 \text{$\#$1}^3 \tan ^{-1}\left (\frac{\sin (2 x)}{\cos (2 x)-\text{$\#$1}}\right )-i \text{$\#$1}^3 \log \left (\text{$\#$1}^2-2 \text{$\#$1} \cos (2 x)+1\right )}{\text{$\#$1}^7-7 \text{$\#$1}^6+21 \text{$\#$1}^5-35 \text{$\#$1}^4+163 \text{$\#$1}^3-21 \text{$\#$1}^2+7 \text{$\#$1}-1}\& \right ] \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Sin[x]^8)^(-1),x]

[Out]

8*RootSum[1 - 8*#1 + 28*#1^2 - 56*#1^3 + 326*#1^4 - 56*#1^5 + 28*#1^6 - 8*#1^7 + #1^8 & , (2*ArcTan[Sin[2*x]/(
Cos[2*x] - #1)]*#1^3 - I*Log[1 - 2*Cos[2*x]*#1 + #1^2]*#1^3)/(-1 + 7*#1 - 21*#1^2 + 163*#1^3 - 35*#1^4 + 21*#1
^5 - 7*#1^6 + #1^7) & ]

________________________________________________________________________________________

Maple [C]  time = 0.066, size = 71, normalized size = 0.3 \begin{align*}{\frac{1}{8}\sum _{{\it \_R}={\it RootOf} \left ( 2\,{{\it \_Z}}^{8}+4\,{{\it \_Z}}^{6}+6\,{{\it \_Z}}^{4}+4\,{{\it \_Z}}^{2}+1 \right ) }{\frac{ \left ({{\it \_R}}^{6}+3\,{{\it \_R}}^{4}+3\,{{\it \_R}}^{2}+1 \right ) \ln \left ( \tan \left ( x \right ) -{\it \_R} \right ) }{2\,{{\it \_R}}^{7}+3\,{{\it \_R}}^{5}+3\,{{\it \_R}}^{3}+{\it \_R}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+sin(x)^8),x)

[Out]

1/8*sum((_R^6+3*_R^4+3*_R^2+1)/(2*_R^7+3*_R^5+3*_R^3+_R)*ln(tan(x)-_R),_R=RootOf(2*_Z^8+4*_Z^6+6*_Z^4+4*_Z^2+1
))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sin \left (x\right )^{8} + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+sin(x)^8),x, algorithm="maxima")

[Out]

integrate(1/(sin(x)^8 + 1), x)

________________________________________________________________________________________

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/(1+sin(x)^8),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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+sin(x)**8),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+sin(x)^8),x, algorithm="giac")

[Out]

Exception raised: TypeError