3.260 \(\int \frac {1}{1-\sin ^8(x)} \, dx\)

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

[Out]

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

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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} \[ \frac {x}{4 \sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {1-i} \tan (x)\right )}{4 \sqrt {1-i}}+\frac {\tan ^{-1}\left (\sqrt {1+i} \tan (x)\right )}{4 \sqrt {1+i}}+\frac {\tan (x)}{4}+\frac {\tan ^{-1}\left (\frac {\sin (x) \cos (x)}{\sin ^2(x)+\sqrt {2}+1}\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Sin[x]^8)^(-1),x]

[Out]

x/(4*Sqrt[2]) + ArcTan[(Cos[x]*Sin[x])/(1 + Sqrt[2] + Sin[x]^2)]/(4*Sqrt[2]) + ArcTan[Sqrt[1 - I]*Tan[x]]/(4*S
qrt[1 - I]) + ArcTan[Sqrt[1 + I]*Tan[x]]/(4*Sqrt[1 + I]) + Tan[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-\sin ^8(x)} \, dx &=\frac {1}{4} \int \frac {1}{1-\sin ^2(x)} \, dx+\frac {1}{4} \int \frac {1}{1-i \sin ^2(x)} \, dx+\frac {1}{4} \int \frac {1}{1+i \sin ^2(x)} \, dx+\frac {1}{4} \int \frac {1}{1+\sin ^2(x)} \, dx\\ &=\frac {1}{4} \int \sec ^2(x) \, dx+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+(1-i) x^2} \, dx,x,\tan (x)\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+(1+i) x^2} \, dx,x,\tan (x)\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\tan (x)\right )\\ &=\frac {x}{4 \sqrt {2}}+\frac {\tan ^{-1}\left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\sin ^2(x)}\right )}{4 \sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {1-i} \tan (x)\right )}{4 \sqrt {1-i}}+\frac {\tan ^{-1}\left (\sqrt {1+i} \tan (x)\right )}{4 \sqrt {1+i}}-\frac {1}{4} \operatorname {Subst}(\int 1 \, dx,x,-\tan (x))\\ &=\frac {x}{4 \sqrt {2}}+\frac {\tan ^{-1}\left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\sin ^2(x)}\right )}{4 \sqrt {2}}+\frac {\tan ^{-1}\left (\sqrt {1-i} \tan (x)\right )}{4 \sqrt {1-i}}+\frac {\tan ^{-1}\left (\sqrt {1+i} \tan (x)\right )}{4 \sqrt {1+i}}+\frac {\tan (x)}{4}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 64, normalized size = 0.72 \[ \frac {1}{8} \left (\frac {2 \tan ^{-1}\left (\sqrt {1-i} \tan (x)\right )}{\sqrt {1-i}}+\frac {2 \tan ^{-1}\left (\sqrt {1+i} \tan (x)\right )}{\sqrt {1+i}}+\sqrt {2} \tan ^{-1}\left (\sqrt {2} \tan (x)\right )+2 \tan (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Sin[x]^8)^(-1),x]

[Out]

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

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fricas [B]  time = 33.49, size = 3884, normalized size = 43.64 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(2^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 1)*cos(x)*log(-(4*sqrt(2) - 5)*cos(x)^4 + 2*(2*sqrt(2) - 3)*cos(
x)^2 + (2^(1/4)*(3*sqrt(2) - 4)*cos(x)^3 - 2*2^(1/4)*(sqrt(2) - 1)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) + 2) - 2
^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 1)*cos(x)*log(-(4*sqrt(2) - 5)*cos(x)^4 + 2*(2*sqrt(2) - 3)*cos(x)^2 - (
2^(1/4)*(3*sqrt(2) - 4)*cos(x)^3 - 2*2^(1/4)*(sqrt(2) - 1)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) + 2) + 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)*(29*sqrt
(2) + 10) - 24*sqrt(2) - 44)*cos(x)^14 + 16*(sqrt(2)*(51*sqrt(2) - 4) - 52*sqrt(2) - 46)*cos(x)^12 - 16*(sqrt(
2)*(41*sqrt(2) - 36) - 54*sqrt(2) + 15)*cos(x)^10 + 8*(sqrt(2)*(29*sqrt(2) - 90) - 58*sqrt(2) + 132)*cos(x)^8
- 4*(sqrt(2)*(5*sqrt(2) - 98) - 32*sqrt(2) + 216)*cos(x)^6 - 4*(sqrt(2)*(sqrt(2) + 24) + 4*sqrt(2) - 82)*cos(x
)^4 + 4*(2*sqrt(2) - 15)*cos(x)^2 + 2*(8*(2^(3/4)*(2*sqrt(2) - 1) - 2*2^(1/4)*(3*sqrt(2) + 2))*cos(x)^15 - 8*(
2^(3/4)*(11*sqrt(2) - 9) - 2*2^(1/4)*(13*sqrt(2) + 4))*cos(x)^13 + 4*(2*2^(3/4)*(21*sqrt(2) - 23) - 2^(1/4)*(7
9*sqrt(2) - 14))*cos(x)^11 - 8*(2^(3/4)*(19*sqrt(2) - 27) - 2^(1/4)*(27*sqrt(2) - 31))*cos(x)^9 + 2*(2^(3/4)*(
36*sqrt(2) - 65) - 32*2^(1/4)*(sqrt(2) - 4))*cos(x)^7 - 2*(2^(3/4)*(9*sqrt(2) - 19) - 2*2^(1/4)*(sqrt(2) - 30)
)*cos(x)^5 + (2*2^(3/4)*(sqrt(2) - 2) + 2^(1/4)*(sqrt(2) + 26))*cos(x)^3 - 2*2^(1/4)*cos(x))*sqrt(2*sqrt(2) +
4)*sin(x) + (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*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*sq
rt(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*sqr
t(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 + 8*(2*sqrt(2) - 3)*cos(x)^2 + 4*(2^(1/4)*(3*sqrt(2) - 4)*cos(x)^3 - 2*2^(1/4
)*(sqrt(2) - 1)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) + 8) + 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))*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)*(29*sqrt(2) + 10) - 2
4*sqrt(2) - 44)*cos(x)^14 + 16*(sqrt(2)*(51*sqrt(2) - 4) - 52*sqrt(2) - 46)*cos(x)^12 - 16*(sqrt(2)*(41*sqrt(2
) - 36) - 54*sqrt(2) + 15)*cos(x)^10 + 8*(sqrt(2)*(29*sqrt(2) - 90) - 58*sqrt(2) + 132)*cos(x)^8 - 4*(sqrt(2)*
(5*sqrt(2) - 98) - 32*sqrt(2) + 216)*cos(x)^6 - 4*(sqrt(2)*(sqrt(2) + 24) + 4*sqrt(2) - 82)*cos(x)^4 + 4*(2*sq
rt(2) - 15)*cos(x)^2 + 2*(8*(2^(3/4)*(2*sqrt(2) - 1) - 2*2^(1/4)*(3*sqrt(2) + 2))*cos(x)^15 - 8*(2^(3/4)*(11*s
qrt(2) - 9) - 2*2^(1/4)*(13*sqrt(2) + 4))*cos(x)^13 + 4*(2*2^(3/4)*(21*sqrt(2) - 23) - 2^(1/4)*(79*sqrt(2) - 1
4))*cos(x)^11 - 8*(2^(3/4)*(19*sqrt(2) - 27) - 2^(1/4)*(27*sqrt(2) - 31))*cos(x)^9 + 2*(2^(3/4)*(36*sqrt(2) -
65) - 32*2^(1/4)*(sqrt(2) - 4))*cos(x)^7 - 2*(2^(3/4)*(9*sqrt(2) - 19) - 2*2^(1/4)*(sqrt(2) - 30))*cos(x)^5 +
(2*2^(3/4)*(sqrt(2) - 2) + 2^(1/4)*(sqrt(2) + 26))*cos(x)^3 - 2*2^(1/4)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) - (
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)^1
2 + 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*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*sq
rt(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*sq
rt(2) - 5)*cos(x)^4 + 8*(2*sqrt(2) - 3)*cos(x)^2 + 4*(2^(1/4)*(3*sqrt(2) - 4)*cos(x)^3 - 2*2^(1/4)*(sqrt(2) -
1)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) + 8) + 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))*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)*(29*sqrt(2) + 10) - 24*sqrt(2) - 4
4)*cos(x)^14 + 16*(sqrt(2)*(51*sqrt(2) - 4) - 52*sqrt(2) - 46)*cos(x)^12 - 16*(sqrt(2)*(41*sqrt(2) - 36) - 54*
sqrt(2) + 15)*cos(x)^10 + 8*(sqrt(2)*(29*sqrt(2) - 90) - 58*sqrt(2) + 132)*cos(x)^8 - 4*(sqrt(2)*(5*sqrt(2) -
98) - 32*sqrt(2) + 216)*cos(x)^6 - 4*(sqrt(2)*(sqrt(2) + 24) + 4*sqrt(2) - 82)*cos(x)^4 + 4*(2*sqrt(2) - 15)*c
os(x)^2 - 2*(8*(2^(3/4)*(2*sqrt(2) - 1) - 2*2^(1/4)*(3*sqrt(2) + 2))*cos(x)^15 - 8*(2^(3/4)*(11*sqrt(2) - 9) -
2*2^(1/4)*(13*sqrt(2) + 4))*cos(x)^13 + 4*(2*2^(3/4)*(21*sqrt(2) - 23) - 2^(1/4)*(79*sqrt(2) - 14))*cos(x)^11
- 8*(2^(3/4)*(19*sqrt(2) - 27) - 2^(1/4)*(27*sqrt(2) - 31))*cos(x)^9 + 2*(2^(3/4)*(36*sqrt(2) - 65) - 32*2^(1
/4)*(sqrt(2) - 4))*cos(x)^7 - 2*(2^(3/4)*(9*sqrt(2) - 19) - 2*2^(1/4)*(sqrt(2) - 30))*cos(x)^5 + (2*2^(3/4)*(s
qrt(2) - 2) + 2^(1/4)*(sqrt(2) + 26))*cos(x)^3 - 2*2^(1/4)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) + (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*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)*(sqr
t(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)*co
s(x)^4 + 8*(2*sqrt(2) - 3)*cos(x)^2 - 4*(2^(1/4)*(3*sqrt(2) - 4)*cos(x)^3 - 2*2^(1/4)*(sqrt(2) - 1)*cos(x))*sq
rt(2*sqrt(2) + 4)*sin(x) + 8) + 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))*cos(x) - 2*2^(1/4)*sqrt(2*sqrt(2) + 4)*arctan(1/4*(32*(sqr
t(2)*(3*sqrt(2) + 2) - 2*sqrt(2) - 6)*cos(x)^16 - 16*(sqrt(2)*(29*sqrt(2) + 10) - 24*sqrt(2) - 44)*cos(x)^14 +
16*(sqrt(2)*(51*sqrt(2) - 4) - 52*sqrt(2) - 46)*cos(x)^12 - 16*(sqrt(2)*(41*sqrt(2) - 36) - 54*sqrt(2) + 15)*
cos(x)^10 + 8*(sqrt(2)*(29*sqrt(2) - 90) - 58*sqrt(2) + 132)*cos(x)^8 - 4*(sqrt(2)*(5*sqrt(2) - 98) - 32*sqrt(
2) + 216)*cos(x)^6 - 4*(sqrt(2)*(sqrt(2) + 24) + 4*sqrt(2) - 82)*cos(x)^4 + 4*(2*sqrt(2) - 15)*cos(x)^2 - 2*(8
*(2^(3/4)*(2*sqrt(2) - 1) - 2*2^(1/4)*(3*sqrt(2) + 2))*cos(x)^15 - 8*(2^(3/4)*(11*sqrt(2) - 9) - 2*2^(1/4)*(13
*sqrt(2) + 4))*cos(x)^13 + 4*(2*2^(3/4)*(21*sqrt(2) - 23) - 2^(1/4)*(79*sqrt(2) - 14))*cos(x)^11 - 8*(2^(3/4)*
(19*sqrt(2) - 27) - 2^(1/4)*(27*sqrt(2) - 31))*cos(x)^9 + 2*(2^(3/4)*(36*sqrt(2) - 65) - 32*2^(1/4)*(sqrt(2) -
4))*cos(x)^7 - 2*(2^(3/4)*(9*sqrt(2) - 19) - 2*2^(1/4)*(sqrt(2) - 30))*cos(x)^5 + (2*2^(3/4)*(sqrt(2) - 2) +
2^(1/4)*(sqrt(2) + 26))*cos(x)^3 - 2*2^(1/4)*cos(x))*sqrt(2*sqrt(2) + 4)*sin(x) - (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*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*s
qrt(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 + 8*(2*
sqrt(2) - 3)*cos(x)^2 - 4*(2^(1/4)*(3*sqrt(2) - 4)*cos(x)^3 - 2*2^(1/4)*(sqrt(2) - 1)*cos(x))*sqrt(2*sqrt(2) +
4)*sin(x) + 8) + 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))*cos(x) + 4*sqrt(2)*arctan(1/4*(3*sqrt(2)*cos(x)^2 - 2*sqrt(2))/(cos(x)*s
in(x)))*cos(x) - 16*sin(x))/cos(x)

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giac [B]  time = 0.35, size = 220, normalized size = 2.47 \[ \frac {1}{8} \, \sqrt {2} {\left (x + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, x\right ) - 2 \, \sin \left (2 \, x\right )}{\sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2} - 2 \, \cos \left (2 \, x\right ) + 2}\right )\right )} + \frac {1}{8} \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} + 2 \, \tan \relax (x)\right )}}{\sqrt {\sqrt {2} + 2}}\right )\right )} \sqrt {\sqrt {2} + 1} + \frac {1}{8} \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (-\frac {2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} - 2 \, \tan \relax (x)\right )}}{\sqrt {\sqrt {2} + 2}}\right )\right )} \sqrt {\sqrt {2} + 1} + \frac {1}{16} \, \sqrt {\sqrt {2} - 1} \log \left (\tan \relax (x)^{2} + \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \relax (x) + \sqrt {\frac {1}{2}}\right ) - \frac {1}{16} \, \sqrt {\sqrt {2} - 1} \log \left (\tan \relax (x)^{2} - \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \relax (x) + \sqrt {\frac {1}{2}}\right ) + \frac {1}{4} \, \tan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.22, size = 255, normalized size = 2.87 \[ \frac {\tan \relax (x )}{4}-\frac {\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}\, \ln \left (-\sqrt {-2+2 \sqrt {2}}\, \sqrt {2}\, \tan \relax (x )+2 \left (\tan ^{2}\relax (x )\right )+\sqrt {2}\right )}{32}+\frac {\arctan \left (\frac {-\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}+4 \tan \relax (x )}{2 \sqrt {1+\sqrt {2}}}\right ) \sqrt {2}}{8 \sqrt {1+\sqrt {2}}}+\frac {\arctan \left (\frac {-\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}+4 \tan \relax (x )}{2 \sqrt {1+\sqrt {2}}}\right )}{8 \sqrt {1+\sqrt {2}}}+\frac {\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}\, \ln \left (\sqrt {2}+2 \left (\tan ^{2}\relax (x )\right )+\sqrt {-2+2 \sqrt {2}}\, \sqrt {2}\, \tan \relax (x )\right )}{32}+\frac {\arctan \left (\frac {4 \tan \relax (x )+\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}}{2 \sqrt {1+\sqrt {2}}}\right ) \sqrt {2}}{8 \sqrt {1+\sqrt {2}}}+\frac {\arctan \left (\frac {4 \tan \relax (x )+\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}}{2 \sqrt {1+\sqrt {2}}}\right )}{8 \sqrt {1+\sqrt {2}}}+\frac {\arctan \left (\sqrt {2}\, \tan \relax (x )\right ) \sqrt {2}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right )} {\left (2 \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} + 2 \, \tan \relax (x)\right )}}{\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 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} - 2 \, \tan \relax (x)\right )}}{\sqrt {\sqrt {2} + 2}}\right )\right )} \sqrt {\sqrt {2} + 1} + \sqrt {\sqrt {2} - 1} \log \left (\tan \relax (x)^{2} + \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \relax (x) + \sqrt {\frac {1}{2}}\right ) - \sqrt {\sqrt {2} - 1} \log \left (\tan \relax (x)^{2} - \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \relax (x) + \sqrt {\frac {1}{2}}\right )\right )} + {\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 )} \arctan \left (\frac {2 \, \sqrt {2} \sin \relax (x)}{2 \, {\left (\sqrt {2} + 1\right )} \cos \relax (x) + \cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \sqrt {2} + 3}, \frac {\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \cos \relax (x) - 1}{2 \, {\left (\sqrt {2} + 1\right )} \cos \relax (x) + \cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \sqrt {2} + 3}\right ) - {\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 )} \arctan \left (\frac {2 \, \sqrt {2} \sin \relax (x)}{2 \, {\left (\sqrt {2} - 1\right )} \cos \relax (x) + \cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \sqrt {2} + 3}, \frac {\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \cos \relax (x) - 1}{2 \, {\left (\sqrt {2} - 1\right )} \cos \relax (x) + \cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \sqrt {2} + 3}\right ) + 8 \, \sin \left (2 \, x\right )}{16 \, {\left (\cos \left (2 \, x\right )^{2} + \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((sqrt(2)*cos(2*x)^2 + sqrt(2)*sin(2*x)^2 + 2*sqrt(2)*cos(2*x) + sqrt(2))*arctan2(2*sqrt(2)*sin(x)/(2*(sq
rt(2) + 1)*cos(x) + cos(x)^2 + sin(x)^2 + 2*sqrt(2) + 3), (cos(x)^2 + sin(x)^2 + 2*cos(x) - 1)/(2*(sqrt(2) + 1
)*cos(x) + cos(x)^2 + sin(x)^2 + 2*sqrt(2) + 3)) - (sqrt(2)*cos(2*x)^2 + sqrt(2)*sin(2*x)^2 + 2*sqrt(2)*cos(2*
x) + sqrt(2))*arctan2(2*sqrt(2)*sin(x)/(2*(sqrt(2) - 1)*cos(x) + cos(x)^2 + sin(x)^2 - 2*sqrt(2) + 3), (cos(x)
^2 + sin(x)^2 - 2*cos(x) - 1)/(2*(sqrt(2) - 1)*cos(x) + cos(x)^2 + sin(x)^2 - 2*sqrt(2) + 3)) + 128*(cos(2*x)^
2 + sin(2*x)^2 + 2*cos(2*x) + 1)*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) - sin(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*si
n(4*x)*sin(2*x) - 16*sin(2*x)^2 + 8*cos(2*x) - 1), x) + 8*sin(2*x))/(cos(2*x)^2 + sin(2*x)^2 + 2*cos(2*x) + 1)

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mupad [B]  time = 14.04, size = 141, normalized size = 1.58 \[ \frac {\mathrm {tan}\relax (x)}{4}+\mathrm {atan}\left (\sqrt {2}\,\mathrm {tan}\relax (x)\,\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}\,8{}\mathrm {i}-\sqrt {2}\,\mathrm {tan}\relax (x)\,\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,8{}\mathrm {i}\right )\,\left (\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}\,2{}\mathrm {i}+\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,2{}\mathrm {i}\right )+\mathrm {atan}\left (\sqrt {2}\,\mathrm {tan}\relax (x)\,\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}\,8{}\mathrm {i}+\sqrt {2}\,\mathrm {tan}\relax (x)\,\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,8{}\mathrm {i}\right )\,\left (\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}\,2{}\mathrm {i}-\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,2{}\mathrm {i}\right )+\frac {\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\mathrm {tan}\relax (x)\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(sin(x)^8 - 1),x)

[Out]

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

[Out]

Timed out

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