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]

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

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Rubi [A]  time = 0.0770086, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, 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 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])

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 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]

Rule 8

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

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*}

Mathematica [A]  time = 0.166119, 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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Maple [B]  time = 0.078, size = 255, normalized size = 2.9 \begin{align*}{\frac{\tan \left ( x \right ) }{4}}+{\frac{\arctan \left ( \sqrt{2}\tan \left ( x \right ) \right ) \sqrt{2}}{8}}+{\frac{\sqrt{2}\sqrt{-2+2\,\sqrt{2}}\ln \left ( \sqrt{2}+\sqrt{-2+2\,\sqrt{2}}\sqrt{2}\tan \left ( x \right ) +2\, \left ( \tan \left ( x \right ) \right ) ^{2} \right ) }{32}}+{\frac{\sqrt{2}}{8\,\sqrt{1+\sqrt{2}}}\arctan \left ({\frac{\sqrt{2}\sqrt{-2+2\,\sqrt{2}}+4\,\tan \left ( x \right ) }{2\,\sqrt{1+\sqrt{2}}}} \right ) }+{\frac{1}{8\,\sqrt{1+\sqrt{2}}}\arctan \left ({\frac{\sqrt{2}\sqrt{-2+2\,\sqrt{2}}+4\,\tan \left ( x \right ) }{2\,\sqrt{1+\sqrt{2}}}} \right ) }-{\frac{\sqrt{2}\sqrt{-2+2\,\sqrt{2}}\ln \left ( -\sqrt{-2+2\,\sqrt{2}}\sqrt{2}\tan \left ( x \right ) +2\, \left ( \tan \left ( x \right ) \right ) ^{2}+\sqrt{2} \right ) }{32}}+{\frac{\sqrt{2}}{8\,\sqrt{1+\sqrt{2}}}\arctan \left ({\frac{-\sqrt{2}\sqrt{-2+2\,\sqrt{2}}+4\,\tan \left ( x \right ) }{2\,\sqrt{1+\sqrt{2}}}} \right ) }+{\frac{1}{8\,\sqrt{1+\sqrt{2}}}\arctan \left ({\frac{-\sqrt{2}\sqrt{-2+2\,\sqrt{2}}+4\,\tan \left ( x \right ) }{2\,\sqrt{1+\sqrt{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [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="maxima")

[Out]

Timed out

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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/(1-sin(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/(1-sin(x)**8),x)

[Out]

Timed out

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Giac [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="giac")

[Out]

integrate(-1/(sin(x)^8 - 1), x)