3.85 \(\int \frac {1}{1-\cos ^8(x)} \, dx\)

Optimal. Leaf size=89 \[ \frac {x}{4 \sqrt {2}}-\frac {\cot (x)}{4}-\frac {\tan ^{-1}\left (\sqrt {1-i} \cot (x)\right )}{4 \sqrt {1-i}}-\frac {\tan ^{-1}\left (\sqrt {1+i} \cot (x)\right )}{4 \sqrt {1+i}}-\frac {\tan ^{-1}\left (\frac {\sin (x) \cos (x)}{\cos ^2(x)+\sqrt {2}+1}\right )}{4 \sqrt {2}} \]

[Out]

-1/4*cot(x)-1/4*arctan(cot(x)*(1-I)^(1/2))/(1-I)^(1/2)-1/4*arctan(cot(x)*(1+I)^(1/2))/(1+I)^(1/2)+1/8*x*2^(1/2
)-1/8*arctan(cos(x)*sin(x)/(1+cos(x)^2+2^(1/2)))*2^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 89, 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, 203, 3175, 3767, 8} \[ \frac {x}{4 \sqrt {2}}-\frac {\cot (x)}{4}-\frac {\tan ^{-1}\left (\sqrt {1-i} \cot (x)\right )}{4 \sqrt {1-i}}-\frac {\tan ^{-1}\left (\sqrt {1+i} \cot (x)\right )}{4 \sqrt {1+i}}-\frac {\tan ^{-1}\left (\frac {\sin (x) \cos (x)}{\cos ^2(x)+\sqrt {2}+1}\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Cos[x]^8)^(-1),x]

[Out]

x/(4*Sqrt[2]) - ArcTan[Sqrt[1 - I]*Cot[x]]/(4*Sqrt[1 - I]) - ArcTan[Sqrt[1 + I]*Cot[x]]/(4*Sqrt[1 + I]) - ArcT
an[(Cos[x]*Sin[x])/(1 + Sqrt[2] + Cos[x]^2)]/(4*Sqrt[2]) - Cot[x]/4

Rule 8

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

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-\cos ^8(x)} \, dx &=\frac {1}{4} \int \frac {1}{1-\cos ^2(x)} \, dx+\frac {1}{4} \int \frac {1}{1-i \cos ^2(x)} \, dx+\frac {1}{4} \int \frac {1}{1+i \cos ^2(x)} \, dx+\frac {1}{4} \int \frac {1}{1+\cos ^2(x)} \, dx\\ &=\frac {1}{4} \int \csc ^2(x) \, dx-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+(1-i) x^2} \, dx,x,\cot (x)\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+(1+i) x^2} \, dx,x,\cot (x)\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\cot (x)\right )\\ &=\frac {x}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {1-i} \cot (x)\right )}{4 \sqrt {1-i}}-\frac {\tan ^{-1}\left (\sqrt {1+i} \cot (x)\right )}{4 \sqrt {1+i}}-\frac {\tan ^{-1}\left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\cos ^2(x)}\right )}{4 \sqrt {2}}-\frac {1}{4} \operatorname {Subst}(\int 1 \, dx,x,\cot (x))\\ &=\frac {x}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {1-i} \cot (x)\right )}{4 \sqrt {1-i}}-\frac {\tan ^{-1}\left (\sqrt {1+i} \cot (x)\right )}{4 \sqrt {1+i}}-\frac {\tan ^{-1}\left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\cos ^2(x)}\right )}{4 \sqrt {2}}-\frac {\cot (x)}{4}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 64, normalized size = 0.72 \[ \frac {1}{8} \left (\frac {2 \tan ^{-1}\left (\frac {\tan (x)}{\sqrt {1-i}}\right )}{\sqrt {1-i}}+\frac {2 \tan ^{-1}\left (\frac {\tan (x)}{\sqrt {1+i}}\right )}{\sqrt {1+i}}+\sqrt {2} \tan ^{-1}\left (\frac {\tan (x)}{\sqrt {2}}\right )-2 \cot (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Cos[x]^8)^(-1),x]

[Out]

((2*ArcTan[Tan[x]/Sqrt[1 - I]])/Sqrt[1 - I] + (2*ArcTan[Tan[x]/Sqrt[1 + I]])/Sqrt[1 + I] + Sqrt[2]*ArcTan[Tan[
x]/Sqrt[2]] - 2*Cot[x])/8

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fricas [B]  time = 63.66, size = 3963, normalized size = 44.53 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(x)^8),x, algorithm="fricas")

[Out]

-1/64*(2*(2^(1/4)*cos(x)^2 - 2^(1/4))*sqrt(2*sqrt(2) + 4)*arctan(1/4*(32*(sqrt(2)*(3*sqrt(2) + 2) - 2*sqrt(2)
- 6)*cos(x)^16 - 16*(sqrt(2)*(19*sqrt(2) + 22) - 8*sqrt(2) - 52)*cos(x)^14 + 32*(sqrt(2)*(8*sqrt(2) + 19) + 2*
sqrt(2) - 37)*cos(x)^12 + 16*(2*sqrt(2)*(4*sqrt(2) - 13) - 22*sqrt(2) + 39)*cos(x)^10 - 8*(sqrt(2)*(41*sqrt(2)
- 10) - 42*sqrt(2) - 2)*cos(x)^8 + 4*(sqrt(2)*(49*sqrt(2) + 6) - 32*sqrt(2) - 32)*cos(x)^6 - 8*(sqrt(2)*(6*sq
rt(2) + 1) - 2*sqrt(2) - 5)*cos(x)^4 + 2*(8*(2^(3/4)*(2*sqrt(2) - 1) - 2*2^(1/4)*(3*sqrt(2) + 2))*cos(x)^15 -
8*(2^(3/4)*(3*sqrt(2) + 2) - 4*2^(1/4)*(4*sqrt(2) + 5))*cos(x)^13 - 4*(2*2^(3/4)*(3*sqrt(2) - 10) + 2^(1/4)*(1
9*sqrt(2) + 58))*cos(x)^11 + 4*(6*2^(3/4)*(3*sqrt(2) - 4) - 2^(1/4)*(19*sqrt(2) - 32))*cos(x)^9 - 2*(2^(3/4)*(
28*sqrt(2) - 27) - 4*2^(1/4)*(15*sqrt(2) - 2))*cos(x)^7 + 2*(2^(3/4)*(9*sqrt(2) - 8) - 2*2^(1/4)*(15*sqrt(2) +
2))*cos(x)^5 - (2*2^(3/4)*(sqrt(2) - 1) - 2^(1/4)*(13*sqrt(2) + 2))*cos(x)^3 - 2^(3/4)*cos(x))*sqrt(2*sqrt(2)
+ 4)*sin(x) + 4*cos(x)^2 + (16*(sqrt(2)*(5*sqrt(2) - 6) - 8*sqrt(2) + 4)*cos(x)^14 - 56*(sqrt(2)*(5*sqrt(2) -
6) - 8*sqrt(2) + 4)*cos(x)^12 + 8*(sqrt(2)*(49*sqrt(2) - 62) - 76*sqrt(2) + 54)*cos(x)^10 - 40*(sqrt(2)*(7*sq
rt(2) - 10) - 10*sqrt(2) + 13)*cos(x)^8 + 4*(sqrt(2)*(27*sqrt(2) - 46) - 32*sqrt(2) + 92)*cos(x)^6 - 2*(11*sqr
t(2)*(sqrt(2) - 2) - 8*sqrt(2) + 72)*cos(x)^4 + 2*(sqrt(2)*(sqrt(2) - 2) + 14)*cos(x)^2 + (8*(2^(3/4)*(8*sqrt(
2) - 11) - 2*2^(1/4)*(5*sqrt(2) - 6))*cos(x)^13 - 24*(2^(3/4)*(8*sqrt(2) - 11) - 2*2^(1/4)*(5*sqrt(2) - 6))*co
s(x)^11 + 4*(2*2^(3/4)*(28*sqrt(2) - 39) - 2^(1/4)*(73*sqrt(2) - 94))*cos(x)^9 - 8*(2^(3/4)*(16*sqrt(2) - 23)
- 2^(1/4)*(23*sqrt(2) - 34))*cos(x)^7 + 2*(9*2^(3/4)*(2*sqrt(2) - 3) - 8*2^(1/4)*(4*sqrt(2) - 7))*cos(x)^5 - 2
*(2^(3/4)*(2*sqrt(2) - 3) - 6*2^(1/4)*(sqrt(2) - 2))*cos(x)^3 - 2^(1/4)*(sqrt(2) - 2)*cos(x))*sqrt(2*sqrt(2) +
4)*sin(x) - 2)*sqrt(-4*(4*sqrt(2) - 5)*cos(x)^4 + 16*(sqrt(2) - 1)*cos(x)^2 + 4*(2^(1/4)*(3*sqrt(2) - 4)*cos(
x)^3 - 2^(1/4)*(sqrt(2) - 2)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) + 4))/(112*cos(x)^16 - 448*cos(x)^14 + 608*cos
(x)^12 - 256*cos(x)^10 - 152*cos(x)^8 + 208*cos(x)^6 - 88*cos(x)^4 + 16*cos(x)^2 - 1)) - 2*(2^(1/4)*cos(x)^2 -
2^(1/4))*sqrt(2*sqrt(2) + 4)*arctan(-1/4*(32*(sqrt(2)*(3*sqrt(2) + 2) - 2*sqrt(2) - 6)*cos(x)^16 - 16*(sqrt(2
)*(19*sqrt(2) + 22) - 8*sqrt(2) - 52)*cos(x)^14 + 32*(sqrt(2)*(8*sqrt(2) + 19) + 2*sqrt(2) - 37)*cos(x)^12 + 1
6*(2*sqrt(2)*(4*sqrt(2) - 13) - 22*sqrt(2) + 39)*cos(x)^10 - 8*(sqrt(2)*(41*sqrt(2) - 10) - 42*sqrt(2) - 2)*co
s(x)^8 + 4*(sqrt(2)*(49*sqrt(2) + 6) - 32*sqrt(2) - 32)*cos(x)^6 - 8*(sqrt(2)*(6*sqrt(2) + 1) - 2*sqrt(2) - 5)
*cos(x)^4 + 2*(8*(2^(3/4)*(2*sqrt(2) - 1) - 2*2^(1/4)*(3*sqrt(2) + 2))*cos(x)^15 - 8*(2^(3/4)*(3*sqrt(2) + 2)
- 4*2^(1/4)*(4*sqrt(2) + 5))*cos(x)^13 - 4*(2*2^(3/4)*(3*sqrt(2) - 10) + 2^(1/4)*(19*sqrt(2) + 58))*cos(x)^11
+ 4*(6*2^(3/4)*(3*sqrt(2) - 4) - 2^(1/4)*(19*sqrt(2) - 32))*cos(x)^9 - 2*(2^(3/4)*(28*sqrt(2) - 27) - 4*2^(1/4
)*(15*sqrt(2) - 2))*cos(x)^7 + 2*(2^(3/4)*(9*sqrt(2) - 8) - 2*2^(1/4)*(15*sqrt(2) + 2))*cos(x)^5 - (2*2^(3/4)*
(sqrt(2) - 1) - 2^(1/4)*(13*sqrt(2) + 2))*cos(x)^3 - 2^(3/4)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) + 4*cos(x)^2 -
(16*(sqrt(2)*(5*sqrt(2) - 6) - 8*sqrt(2) + 4)*cos(x)^14 - 56*(sqrt(2)*(5*sqrt(2) - 6) - 8*sqrt(2) + 4)*cos(x)
^12 + 8*(sqrt(2)*(49*sqrt(2) - 62) - 76*sqrt(2) + 54)*cos(x)^10 - 40*(sqrt(2)*(7*sqrt(2) - 10) - 10*sqrt(2) +
13)*cos(x)^8 + 4*(sqrt(2)*(27*sqrt(2) - 46) - 32*sqrt(2) + 92)*cos(x)^6 - 2*(11*sqrt(2)*(sqrt(2) - 2) - 8*sqrt
(2) + 72)*cos(x)^4 + 2*(sqrt(2)*(sqrt(2) - 2) + 14)*cos(x)^2 + (8*(2^(3/4)*(8*sqrt(2) - 11) - 2*2^(1/4)*(5*sqr
t(2) - 6))*cos(x)^13 - 24*(2^(3/4)*(8*sqrt(2) - 11) - 2*2^(1/4)*(5*sqrt(2) - 6))*cos(x)^11 + 4*(2*2^(3/4)*(28*
sqrt(2) - 39) - 2^(1/4)*(73*sqrt(2) - 94))*cos(x)^9 - 8*(2^(3/4)*(16*sqrt(2) - 23) - 2^(1/4)*(23*sqrt(2) - 34)
)*cos(x)^7 + 2*(9*2^(3/4)*(2*sqrt(2) - 3) - 8*2^(1/4)*(4*sqrt(2) - 7))*cos(x)^5 - 2*(2^(3/4)*(2*sqrt(2) - 3) -
6*2^(1/4)*(sqrt(2) - 2))*cos(x)^3 - 2^(1/4)*(sqrt(2) - 2)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) - 2)*sqrt(-4*(4*
sqrt(2) - 5)*cos(x)^4 + 16*(sqrt(2) - 1)*cos(x)^2 + 4*(2^(1/4)*(3*sqrt(2) - 4)*cos(x)^3 - 2^(1/4)*(sqrt(2) - 2
)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) + 4))/(112*cos(x)^16 - 448*cos(x)^14 + 608*cos(x)^12 - 256*cos(x)^10 - 15
2*cos(x)^8 + 208*cos(x)^6 - 88*cos(x)^4 + 16*cos(x)^2 - 1)) + 2*(2^(1/4)*cos(x)^2 - 2^(1/4))*sqrt(2*sqrt(2) +
4)*arctan(-1/4*(32*(sqrt(2)*(3*sqrt(2) + 2) - 2*sqrt(2) - 6)*cos(x)^16 - 16*(sqrt(2)*(19*sqrt(2) + 22) - 8*sqr
t(2) - 52)*cos(x)^14 + 32*(sqrt(2)*(8*sqrt(2) + 19) + 2*sqrt(2) - 37)*cos(x)^12 + 16*(2*sqrt(2)*(4*sqrt(2) - 1
3) - 22*sqrt(2) + 39)*cos(x)^10 - 8*(sqrt(2)*(41*sqrt(2) - 10) - 42*sqrt(2) - 2)*cos(x)^8 + 4*(sqrt(2)*(49*sqr
t(2) + 6) - 32*sqrt(2) - 32)*cos(x)^6 - 8*(sqrt(2)*(6*sqrt(2) + 1) - 2*sqrt(2) - 5)*cos(x)^4 - 2*(8*(2^(3/4)*(
2*sqrt(2) - 1) - 2*2^(1/4)*(3*sqrt(2) + 2))*cos(x)^15 - 8*(2^(3/4)*(3*sqrt(2) + 2) - 4*2^(1/4)*(4*sqrt(2) + 5)
)*cos(x)^13 - 4*(2*2^(3/4)*(3*sqrt(2) - 10) + 2^(1/4)*(19*sqrt(2) + 58))*cos(x)^11 + 4*(6*2^(3/4)*(3*sqrt(2) -
4) - 2^(1/4)*(19*sqrt(2) - 32))*cos(x)^9 - 2*(2^(3/4)*(28*sqrt(2) - 27) - 4*2^(1/4)*(15*sqrt(2) - 2))*cos(x)^
7 + 2*(2^(3/4)*(9*sqrt(2) - 8) - 2*2^(1/4)*(15*sqrt(2) + 2))*cos(x)^5 - (2*2^(3/4)*(sqrt(2) - 1) - 2^(1/4)*(13
*sqrt(2) + 2))*cos(x)^3 - 2^(3/4)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) + 4*cos(x)^2 + (16*(sqrt(2)*(5*sqrt(2) -
6) - 8*sqrt(2) + 4)*cos(x)^14 - 56*(sqrt(2)*(5*sqrt(2) - 6) - 8*sqrt(2) + 4)*cos(x)^12 + 8*(sqrt(2)*(49*sqrt(2
) - 62) - 76*sqrt(2) + 54)*cos(x)^10 - 40*(sqrt(2)*(7*sqrt(2) - 10) - 10*sqrt(2) + 13)*cos(x)^8 + 4*(sqrt(2)*(
27*sqrt(2) - 46) - 32*sqrt(2) + 92)*cos(x)^6 - 2*(11*sqrt(2)*(sqrt(2) - 2) - 8*sqrt(2) + 72)*cos(x)^4 + 2*(sqr
t(2)*(sqrt(2) - 2) + 14)*cos(x)^2 - (8*(2^(3/4)*(8*sqrt(2) - 11) - 2*2^(1/4)*(5*sqrt(2) - 6))*cos(x)^13 - 24*(
2^(3/4)*(8*sqrt(2) - 11) - 2*2^(1/4)*(5*sqrt(2) - 6))*cos(x)^11 + 4*(2*2^(3/4)*(28*sqrt(2) - 39) - 2^(1/4)*(73
*sqrt(2) - 94))*cos(x)^9 - 8*(2^(3/4)*(16*sqrt(2) - 23) - 2^(1/4)*(23*sqrt(2) - 34))*cos(x)^7 + 2*(9*2^(3/4)*(
2*sqrt(2) - 3) - 8*2^(1/4)*(4*sqrt(2) - 7))*cos(x)^5 - 2*(2^(3/4)*(2*sqrt(2) - 3) - 6*2^(1/4)*(sqrt(2) - 2))*c
os(x)^3 - 2^(1/4)*(sqrt(2) - 2)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) - 2)*sqrt(-4*(4*sqrt(2) - 5)*cos(x)^4 + 16*
(sqrt(2) - 1)*cos(x)^2 - 4*(2^(1/4)*(3*sqrt(2) - 4)*cos(x)^3 - 2^(1/4)*(sqrt(2) - 2)*cos(x))*sqrt(2*sqrt(2) +
4)*sin(x) + 4))/(112*cos(x)^16 - 448*cos(x)^14 + 608*cos(x)^12 - 256*cos(x)^10 - 152*cos(x)^8 + 208*cos(x)^6 -
88*cos(x)^4 + 16*cos(x)^2 - 1)) - 2*(2^(1/4)*cos(x)^2 - 2^(1/4))*sqrt(2*sqrt(2) + 4)*arctan(1/4*(32*(sqrt(2)*
(3*sqrt(2) + 2) - 2*sqrt(2) - 6)*cos(x)^16 - 16*(sqrt(2)*(19*sqrt(2) + 22) - 8*sqrt(2) - 52)*cos(x)^14 + 32*(s
qrt(2)*(8*sqrt(2) + 19) + 2*sqrt(2) - 37)*cos(x)^12 + 16*(2*sqrt(2)*(4*sqrt(2) - 13) - 22*sqrt(2) + 39)*cos(x)
^10 - 8*(sqrt(2)*(41*sqrt(2) - 10) - 42*sqrt(2) - 2)*cos(x)^8 + 4*(sqrt(2)*(49*sqrt(2) + 6) - 32*sqrt(2) - 32)
*cos(x)^6 - 8*(sqrt(2)*(6*sqrt(2) + 1) - 2*sqrt(2) - 5)*cos(x)^4 - 2*(8*(2^(3/4)*(2*sqrt(2) - 1) - 2*2^(1/4)*(
3*sqrt(2) + 2))*cos(x)^15 - 8*(2^(3/4)*(3*sqrt(2) + 2) - 4*2^(1/4)*(4*sqrt(2) + 5))*cos(x)^13 - 4*(2*2^(3/4)*(
3*sqrt(2) - 10) + 2^(1/4)*(19*sqrt(2) + 58))*cos(x)^11 + 4*(6*2^(3/4)*(3*sqrt(2) - 4) - 2^(1/4)*(19*sqrt(2) -
32))*cos(x)^9 - 2*(2^(3/4)*(28*sqrt(2) - 27) - 4*2^(1/4)*(15*sqrt(2) - 2))*cos(x)^7 + 2*(2^(3/4)*(9*sqrt(2) -
8) - 2*2^(1/4)*(15*sqrt(2) + 2))*cos(x)^5 - (2*2^(3/4)*(sqrt(2) - 1) - 2^(1/4)*(13*sqrt(2) + 2))*cos(x)^3 - 2^
(3/4)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) + 4*cos(x)^2 - (16*(sqrt(2)*(5*sqrt(2) - 6) - 8*sqrt(2) + 4)*cos(x)^1
4 - 56*(sqrt(2)*(5*sqrt(2) - 6) - 8*sqrt(2) + 4)*cos(x)^12 + 8*(sqrt(2)*(49*sqrt(2) - 62) - 76*sqrt(2) + 54)*c
os(x)^10 - 40*(sqrt(2)*(7*sqrt(2) - 10) - 10*sqrt(2) + 13)*cos(x)^8 + 4*(sqrt(2)*(27*sqrt(2) - 46) - 32*sqrt(2
) + 92)*cos(x)^6 - 2*(11*sqrt(2)*(sqrt(2) - 2) - 8*sqrt(2) + 72)*cos(x)^4 + 2*(sqrt(2)*(sqrt(2) - 2) + 14)*cos
(x)^2 - (8*(2^(3/4)*(8*sqrt(2) - 11) - 2*2^(1/4)*(5*sqrt(2) - 6))*cos(x)^13 - 24*(2^(3/4)*(8*sqrt(2) - 11) - 2
*2^(1/4)*(5*sqrt(2) - 6))*cos(x)^11 + 4*(2*2^(3/4)*(28*sqrt(2) - 39) - 2^(1/4)*(73*sqrt(2) - 94))*cos(x)^9 - 8
*(2^(3/4)*(16*sqrt(2) - 23) - 2^(1/4)*(23*sqrt(2) - 34))*cos(x)^7 + 2*(9*2^(3/4)*(2*sqrt(2) - 3) - 8*2^(1/4)*(
4*sqrt(2) - 7))*cos(x)^5 - 2*(2^(3/4)*(2*sqrt(2) - 3) - 6*2^(1/4)*(sqrt(2) - 2))*cos(x)^3 - 2^(1/4)*(sqrt(2) -
2)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) - 2)*sqrt(-4*(4*sqrt(2) - 5)*cos(x)^4 + 16*(sqrt(2) - 1)*cos(x)^2 - 4*(
2^(1/4)*(3*sqrt(2) - 4)*cos(x)^3 - 2^(1/4)*(sqrt(2) - 2)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) + 4))/(112*cos(x)^
16 - 448*cos(x)^14 + 608*cos(x)^12 - 256*cos(x)^10 - 152*cos(x)^8 + 208*cos(x)^6 - 88*cos(x)^4 + 16*cos(x)^2 -
1)) + (2^(1/4)*(sqrt(2) - 1)*cos(x)^2 - 2^(1/4)*(sqrt(2) - 1))*sqrt(2*sqrt(2) + 4)*log(-(4*sqrt(2) - 5)*cos(x
)^4 + 4*(sqrt(2) - 1)*cos(x)^2 + (2^(1/4)*(3*sqrt(2) - 4)*cos(x)^3 - 2^(1/4)*(sqrt(2) - 2)*cos(x))*sqrt(2*sqrt
(2) + 4)*sin(x) + 1) - (2^(1/4)*(sqrt(2) - 1)*cos(x)^2 - 2^(1/4)*(sqrt(2) - 1))*sqrt(2*sqrt(2) + 4)*log(-(4*sq
rt(2) - 5)*cos(x)^4 + 4*(sqrt(2) - 1)*cos(x)^2 - (2^(1/4)*(3*sqrt(2) - 4)*cos(x)^3 - 2^(1/4)*(sqrt(2) - 2)*cos
(x))*sqrt(2*sqrt(2) + 4)*sin(x) + 1) + 4*(sqrt(2)*cos(x)^2 - sqrt(2))*arctan(1/4*(3*sqrt(2)*cos(x)^2 - sqrt(2)
)/(cos(x)*sin(x))) - 16*cos(x)*sin(x))/(cos(x)^2 - 1)

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giac [B]  time = 0.88, size = 222, normalized size = 2.49 \[ \frac {1}{8} \, \sqrt {2} {\left (x + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, x\right ) - \sin \left (2 \, x\right )}{\sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2} - \cos \left (2 \, x\right ) + 1}\right )\right )} + \frac {1}{8} \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} + 2 \, \tan \relax (x)\right )}}{2 \, \sqrt {\sqrt {2} + 2}}\right )\right )} \sqrt {\sqrt {2} + 1} + \frac {1}{8} \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} - 2 \, \tan \relax (x)\right )}}{2 \, \sqrt {\sqrt {2} + 2}}\right )\right )} \sqrt {\sqrt {2} + 1} - \frac {1}{16} \, \sqrt {\sqrt {2} - 1} \log \left (\tan \relax (x)^{2} + 2^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \relax (x) + \sqrt {2}\right ) + \frac {1}{16} \, \sqrt {\sqrt {2} - 1} \log \left (\tan \relax (x)^{2} - 2^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \relax (x) + \sqrt {2}\right ) - \frac {1}{4 \, \tan \relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*(x + arctan(-(sqrt(2)*sin(2*x) - sin(2*x))/(sqrt(2)*cos(2*x) + sqrt(2) - cos(2*x) + 1))) + 1/8*(pi
*floor(x/pi + 1/2) + arctan(1/2*2^(3/4)*(2^(1/4)*sqrt(-sqrt(2) + 2) + 2*tan(x))/sqrt(sqrt(2) + 2)))*sqrt(sqrt(
2) + 1) + 1/8*(pi*floor(x/pi + 1/2) + arctan(-1/2*2^(3/4)*(2^(1/4)*sqrt(-sqrt(2) + 2) - 2*tan(x))/sqrt(sqrt(2)
+ 2)))*sqrt(sqrt(2) + 1) - 1/16*sqrt(sqrt(2) - 1)*log(tan(x)^2 + 2^(1/4)*sqrt(-sqrt(2) + 2)*tan(x) + sqrt(2))
+ 1/16*sqrt(sqrt(2) - 1)*log(tan(x)^2 - 2^(1/4)*sqrt(-sqrt(2) + 2)*tan(x) + sqrt(2)) - 1/4/tan(x)

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maple [B]  time = 0.11, size = 246, normalized size = 2.76 \[ \frac {\arctan \left (\frac {\tan \relax (x ) \sqrt {2}}{2}\right ) \sqrt {2}}{8}+\frac {\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}\, \ln \left (\tan ^{2}\relax (x )-\tan \relax (x ) \sqrt {-2+2 \sqrt {2}}+\sqrt {2}\right )}{32}+\frac {\arctan \left (\frac {2 \tan \relax (x )-\sqrt {-2+2 \sqrt {2}}}{\sqrt {2 \sqrt {2}+2}}\right ) \sqrt {2}}{8 \sqrt {2 \sqrt {2}+2}}+\frac {\arctan \left (\frac {2 \tan \relax (x )-\sqrt {-2+2 \sqrt {2}}}{\sqrt {2 \sqrt {2}+2}}\right )}{4 \sqrt {2 \sqrt {2}+2}}-\frac {\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}\, \ln \left (\tan ^{2}\relax (x )+\tan \relax (x ) \sqrt {-2+2 \sqrt {2}}+\sqrt {2}\right )}{32}+\frac {\arctan \left (\frac {2 \tan \relax (x )+\sqrt {-2+2 \sqrt {2}}}{\sqrt {2 \sqrt {2}+2}}\right ) \sqrt {2}}{8 \sqrt {2 \sqrt {2}+2}}+\frac {\arctan \left (\frac {2 \tan \relax (x )+\sqrt {-2+2 \sqrt {2}}}{\sqrt {2 \sqrt {2}+2}}\right )}{4 \sqrt {2 \sqrt {2}+2}}-\frac {1}{4 \tan \relax (x )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*arctan(1/2*tan(x)*2^(1/2))*2^(1/2)+1/32*2^(1/2)*(-2+2*2^(1/2))^(1/2)*ln(tan(x)^2-tan(x)*(-2+2*2^(1/2))^(1/
2)+2^(1/2))+1/8/(2*2^(1/2)+2)^(1/2)*arctan((2*tan(x)-(-2+2*2^(1/2))^(1/2))/(2*2^(1/2)+2)^(1/2))*2^(1/2)+1/4/(2
*2^(1/2)+2)^(1/2)*arctan((2*tan(x)-(-2+2*2^(1/2))^(1/2))/(2*2^(1/2)+2)^(1/2))-1/32*2^(1/2)*(-2+2*2^(1/2))^(1/2
)*ln(tan(x)^2+tan(x)*(-2+2*2^(1/2))^(1/2)+2^(1/2))+1/8/(2*2^(1/2)+2)^(1/2)*arctan((2*tan(x)+(-2+2*2^(1/2))^(1/
2))/(2*2^(1/2)+2)^(1/2))*2^(1/2)+1/4/(2*2^(1/2)+2)^(1/2)*arctan((2*tan(x)+(-2+2*2^(1/2))^(1/2))/(2*2^(1/2)+2)^
(1/2))-1/4/tan(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\frac {1}{2} \, {\left (\sqrt {2} \cos \left (2 \, x\right )^{2} + \sqrt {2} \sin \left (2 \, x\right )^{2} - 2 \, \sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2}\right )} {\left (2 \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} + 2 \, \tan \relax (x)\right )}}{2 \, \sqrt {\sqrt {2} + 2}}\right )\right )} \sqrt {\sqrt {2} + 1} + 2 \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (-\frac {2^{\frac {3}{4}} {\left (2^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} - 2 \, \tan \relax (x)\right )}}{2 \, \sqrt {\sqrt {2} + 2}}\right )\right )} \sqrt {\sqrt {2} + 1} - \sqrt {\sqrt {2} - 1} \log \left (\tan \relax (x)^{2} + 2^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \relax (x) + \sqrt {2}\right ) + \sqrt {\sqrt {2} - 1} \log \left (\tan \relax (x)^{2} - 2^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \relax (x) + \sqrt {2}\right )\right )} + {\left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} - 2 \, \cos \left (2 \, x\right ) + 1\right )} \arctan \left (\frac {4 \, \sqrt {2} \sin \left (2 \, x\right )}{2 \, {\left (2 \, \sqrt {2} + 3\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 12 \, \sqrt {2} + 17}, \frac {\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 6 \, \cos \left (2 \, x\right ) + 1}{2 \, {\left (2 \, \sqrt {2} + 3\right )} \cos \left (2 \, x\right ) + \cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 12 \, \sqrt {2} + 17}\right ) - 4 \, \sqrt {2} \sin \left (2 \, x\right )}{8 \, {\left (\sqrt {2} \cos \left (2 \, x\right )^{2} + \sqrt {2} \sin \left (2 \, x\right )^{2} - 2 \, \sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-cos(x)^8),x, algorithm="maxima")

[Out]

1/8*((cos(2*x)^2 + sin(2*x)^2 - 2*cos(2*x) + 1)*arctan2(4*sqrt(2)*sin(2*x)/(2*(2*sqrt(2) + 3)*cos(2*x) + cos(2
*x)^2 + sin(2*x)^2 + 12*sqrt(2) + 17), (cos(2*x)^2 + sin(2*x)^2 + 6*cos(2*x) + 1)/(2*(2*sqrt(2) + 3)*cos(2*x)
+ cos(2*x)^2 + sin(2*x)^2 + 12*sqrt(2) + 17)) + 64*(sqrt(2)*cos(2*x)^2 + sqrt(2)*sin(2*x)^2 - 2*sqrt(2)*cos(2*
x) + sqrt(2))*integrate(((4*cos(2*x) + 1)*cos(4*x) + cos(8*x)*cos(4*x) + 4*cos(6*x)*cos(4*x) + 22*cos(4*x)^2 +
sin(8*x)*sin(4*x) + 4*sin(6*x)*sin(4*x) + 22*sin(4*x)^2 + 4*sin(4*x)*sin(2*x))/(2*(4*cos(6*x) + 22*cos(4*x) +
4*cos(2*x) + 1)*cos(8*x) + cos(8*x)^2 + 8*(22*cos(4*x) + 4*cos(2*x) + 1)*cos(6*x) + 16*cos(6*x)^2 + 44*(4*cos
(2*x) + 1)*cos(4*x) + 484*cos(4*x)^2 + 16*cos(2*x)^2 + 4*(2*sin(6*x) + 11*sin(4*x) + 2*sin(2*x))*sin(8*x) + si
n(8*x)^2 + 16*(11*sin(4*x) + 2*sin(2*x))*sin(6*x) + 16*sin(6*x)^2 + 484*sin(4*x)^2 + 176*sin(4*x)*sin(2*x) + 1
6*sin(2*x)^2 + 8*cos(2*x) + 1), x) - 4*sqrt(2)*sin(2*x))/(sqrt(2)*cos(2*x)^2 + sqrt(2)*sin(2*x)^2 - 2*sqrt(2)*
cos(2*x) + sqrt(2))

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mupad [B]  time = 2.27, size = 241, normalized size = 2.71 \[ \frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\relax (x)}{2}\right )}{8}-\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\relax (x)\,\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}\,1{}\mathrm {i}}{2\,\left (16\,\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}-\frac {1}{16}\right )}+\frac {\sqrt {2}\,\mathrm {tan}\relax (x)\,\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,1{}\mathrm {i}}{2\,\left (16\,\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}-\frac {1}{16}\right )}\right )\,\left (\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}\,2{}\mathrm {i}-\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,2{}\mathrm {i}\right )+\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tan}\relax (x)\,\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}\,1{}\mathrm {i}}{2\,\left (16\,\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}+\frac {1}{16}\right )}-\frac {\sqrt {2}\,\mathrm {tan}\relax (x)\,\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,1{}\mathrm {i}}{2\,\left (16\,\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}+\frac {1}{16}\right )}\right )\,\left (\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}\,2{}\mathrm {i}+\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,2{}\mathrm {i}\right )-\frac {1}{4\,\mathrm {tan}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(cos(x)^8 - 1),x)

[Out]

atan((2^(1/2)*tan(x)*(- 2^(1/2)/256 - 1/256)^(1/2)*1i)/(2*(16*(2^(1/2)/256 - 1/256)^(1/2)*(- 2^(1/2)/256 - 1/2
56)^(1/2) + 1/16)) - (2^(1/2)*tan(x)*(2^(1/2)/256 - 1/256)^(1/2)*1i)/(2*(16*(2^(1/2)/256 - 1/256)^(1/2)*(- 2^(
1/2)/256 - 1/256)^(1/2) + 1/16)))*((- 2^(1/2)/256 - 1/256)^(1/2)*2i + (2^(1/2)/256 - 1/256)^(1/2)*2i) - atan((
2^(1/2)*tan(x)*(- 2^(1/2)/256 - 1/256)^(1/2)*1i)/(2*(16*(2^(1/2)/256 - 1/256)^(1/2)*(- 2^(1/2)/256 - 1/256)^(1
/2) - 1/16)) + (2^(1/2)*tan(x)*(2^(1/2)/256 - 1/256)^(1/2)*1i)/(2*(16*(2^(1/2)/256 - 1/256)^(1/2)*(- 2^(1/2)/2
56 - 1/256)^(1/2) - 1/16)))*((- 2^(1/2)/256 - 1/256)^(1/2)*2i - (2^(1/2)/256 - 1/256)^(1/2)*2i) - 1/(4*tan(x))
+ (2^(1/2)*atan((2^(1/2)*tan(x))/2))/8

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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-cos(x)**8),x)

[Out]

Timed out

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