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]

1/4*arctan((1-(-1)^(1/4))^(1/2)*tan(x))/(1-(-1)^(1/4))^(1/2)+1/4*arctan((1+(-1)^(1/4))^(1/2)*tan(x))/(1+(-1)^(
1/4))^(1/2)+1/4*arctan((1-(-1)^(3/4))^(1/2)*tan(x))/(1-(-1)^(3/4))^(1/2)+1/4*arctan((1+(-1)^(3/4))^(1/2)*tan(x
))/(1+(-1)^(3/4))^(1/2)+1/8*(x-arctan(tan(x)))*((1+(4-2*2^(1/2))^(1/2))^(1/2)+(2+2*2^(1/4)+2*(1+2^(1/2))^(1/2)
+2*(2+2^(1/2))^(1/2))^(1/2)+(1+(4+2*2^(1/2))^(1/2))^(1/2))

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Rubi [A]  time = 0.20, 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 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])

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}{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*}

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Mathematica [C]  time = 0.15, 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) & ]

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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/(1+sin(x)^8),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [C]  time = 0.20, size = 71, normalized size = 0.33 \[ \frac {\left (\munderset {\textit {\_R} =\RootOf \left (2 \textit {\_Z}^{8}+4 \textit {\_Z}^{6}+6 \textit {\_Z}^{4}+4 \textit {\_Z}^{2}+1\right )}{\sum }\frac {\left (\textit {\_R}^{6}+3 \textit {\_R}^{4}+3 \textit {\_R}^{2}+1\right ) \ln \left (\tan \relax (x )-\textit {\_R} \right )}{2 \textit {\_R}^{7}+3 \textit {\_R}^{5}+3 \textit {\_R}^{3}+\textit {\_R}}\right )}{8} \]

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
))

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

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)

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mupad [B]  time = 14.92, size = 945, normalized size = 4.33 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^8 + 1),x)

[Out]

atan((tan(x)*((- 2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*1i)/(256*((3*2^(1/2)*(- 2*2^(1/2) - 3)^(1/2))/2048 -
(- 2*2^(1/2) - 3)^(1/2)/512)) - (2^(1/2)*tan(x)*((- 2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*1i)/(256*((3*2^(1/
2)*(- 2*2^(1/2) - 3)^(1/2))/2048 - (- 2*2^(1/2) - 3)^(1/2)/512)) - (tan(x)*(- 2*2^(1/2) - 3)^(1/2)*((- 2*2^(1/
2) - 3)^(1/2)/128 - 1/128)^(1/2)*7i)/(256*((3*2^(1/2)*(- 2*2^(1/2) - 3)^(1/2))/2048 - (- 2*2^(1/2) - 3)^(1/2)/
512)) + (2^(1/2)*tan(x)*(- 2*2^(1/2) - 3)^(1/2)*((- 2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*5i)/(256*((3*2^(1/
2)*(- 2*2^(1/2) - 3)^(1/2))/2048 - (- 2*2^(1/2) - 3)^(1/2)/512)))*((- 2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*
2i - atan((tan(x)*(- (- 2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*1i)/(256*((3*2^(1/2)*(- 2*2^(1/2) - 3)^(1/2))/
2048 - (- 2*2^(1/2) - 3)^(1/2)/512)) - (2^(1/2)*tan(x)*(- (- 2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*1i)/(256*
((3*2^(1/2)*(- 2*2^(1/2) - 3)^(1/2))/2048 - (- 2*2^(1/2) - 3)^(1/2)/512)) + (tan(x)*(- 2*2^(1/2) - 3)^(1/2)*(-
(- 2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*7i)/(256*((3*2^(1/2)*(- 2*2^(1/2) - 3)^(1/2))/2048 - (- 2*2^(1/2)
- 3)^(1/2)/512)) - (2^(1/2)*tan(x)*(- 2*2^(1/2) - 3)^(1/2)*(- (- 2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*5i)/(
256*((3*2^(1/2)*(- 2*2^(1/2) - 3)^(1/2))/2048 - (- 2*2^(1/2) - 3)^(1/2)/512)))*(- (- 2*2^(1/2) - 3)^(1/2)/128
- 1/128)^(1/2)*2i + atan((tan(x)*(- (2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*1i)/(256*((3*2^(1/2)*(2*2^(1/2) -
3)^(1/2))/2048 + (2*2^(1/2) - 3)^(1/2)/512)) + (2^(1/2)*tan(x)*(- (2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*1i
)/(256*((3*2^(1/2)*(2*2^(1/2) - 3)^(1/2))/2048 + (2*2^(1/2) - 3)^(1/2)/512)) + (tan(x)*(2*2^(1/2) - 3)^(1/2)*(
- (2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*7i)/(256*((3*2^(1/2)*(2*2^(1/2) - 3)^(1/2))/2048 + (2*2^(1/2) - 3)^
(1/2)/512)) + (2^(1/2)*tan(x)*(2*2^(1/2) - 3)^(1/2)*(- (2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*5i)/(256*((3*2
^(1/2)*(2*2^(1/2) - 3)^(1/2))/2048 + (2*2^(1/2) - 3)^(1/2)/512)))*(- (2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*
2i - atan((tan(x)*((2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*1i)/(256*((3*2^(1/2)*(2*2^(1/2) - 3)^(1/2))/2048 +
(2*2^(1/2) - 3)^(1/2)/512)) + (2^(1/2)*tan(x)*((2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*1i)/(256*((3*2^(1/2)*
(2*2^(1/2) - 3)^(1/2))/2048 + (2*2^(1/2) - 3)^(1/2)/512)) - (tan(x)*(2*2^(1/2) - 3)^(1/2)*((2*2^(1/2) - 3)^(1/
2)/128 - 1/128)^(1/2)*7i)/(256*((3*2^(1/2)*(2*2^(1/2) - 3)^(1/2))/2048 + (2*2^(1/2) - 3)^(1/2)/512)) - (2^(1/2
)*tan(x)*(2*2^(1/2) - 3)^(1/2)*((2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*5i)/(256*((3*2^(1/2)*(2*2^(1/2) - 3)^
(1/2))/2048 + (2*2^(1/2) - 3)^(1/2)/512)))*((2*2^(1/2) - 3)^(1/2)/128 - 1/128)^(1/2)*2i

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+sin(x)**8),x)

[Out]

Integral(1/(sin(x)**8 + 1), x)

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