3.259 \(\int \frac{1}{1-\sin ^6(x)} \, dx\)

Optimal. Leaf size=71 \[ \frac{\tan ^{-1}\left (\sqrt{1+\sqrt [3]{-1}} \tan (x)\right )}{3 \sqrt{1+\sqrt [3]{-1}}}+\frac{\tan ^{-1}\left (\sqrt{1-(-1)^{2/3}} \tan (x)\right )}{3 \sqrt{1-(-1)^{2/3}}}+\frac{\tan (x)}{3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.135382, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{\tan ^{-1}\left (\sqrt{1+\sqrt [3]{-1}} \tan (x)\right )}{3 \sqrt{1+\sqrt [3]{-1}}}+\frac{\tan ^{-1}\left (\sqrt{1-(-1)^{2/3}} \tan (x)\right )}{3 \sqrt{1-(-1)^{2/3}}}+\frac{\tan (x)}{3} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Sin[x]^6)^(-1),x]

[Out]

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

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 ^6(x)} \, dx &=\frac{1}{3} \int \frac{1}{1-\sin ^2(x)} \, dx+\frac{1}{3} \int \frac{1}{1+\sqrt [3]{-1} \sin ^2(x)} \, dx+\frac{1}{3} \int \frac{1}{1-(-1)^{2/3} \sin ^2(x)} \, dx\\ &=\frac{1}{3} \int \sec ^2(x) \, dx+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{1+\left (1+\sqrt [3]{-1}\right ) x^2} \, dx,x,\tan (x)\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{1+\left (1-(-1)^{2/3}\right ) x^2} \, dx,x,\tan (x)\right )\\ &=\frac{\tan ^{-1}\left (\sqrt{1+\sqrt [3]{-1}} \tan (x)\right )}{3 \sqrt{1+\sqrt [3]{-1}}}+\frac{\tan ^{-1}\left (\sqrt{1-(-1)^{2/3}} \tan (x)\right )}{3 \sqrt{1-(-1)^{2/3}}}-\frac{1}{3} \operatorname{Subst}(\int 1 \, dx,x,-\tan (x))\\ &=\frac{\tan ^{-1}\left (\sqrt{1+\sqrt [3]{-1}} \tan (x)\right )}{3 \sqrt{1+\sqrt [3]{-1}}}+\frac{\tan ^{-1}\left (\sqrt{1-(-1)^{2/3}} \tan (x)\right )}{3 \sqrt{1-(-1)^{2/3}}}+\frac{\tan (x)}{3}\\ \end{align*}

Mathematica [C]  time = 0.276495, size = 117, normalized size = 1.65 \[ \frac{\cos (x) (-8 \cos (2 x)+\cos (4 x)+15) \left (-6 \sin (x)+i \sqrt [4]{-3} \left (\sqrt{3}+3 i\right ) \cos (x) \tan ^{-1}\left (\frac{1}{2} \sqrt [4]{-\frac{1}{3}} \left (\sqrt{3}-3 i\right ) \tan (x)\right )+\sqrt [4]{-3} \left (\sqrt{3}-3 i\right ) \cos (x) \tan ^{-1}\left (\frac{(-1)^{3/4} \left (\sqrt{3}+3 i\right ) \tan (x)}{2 \sqrt [4]{3}}\right )\right )}{144 \left (\sin ^6(x)-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Sin[x]^6)^(-1),x]

[Out]

(Cos[x]*(15 - 8*Cos[2*x] + Cos[4*x])*(I*(-3)^(1/4)*(3*I + Sqrt[3])*ArcTan[((-1/3)^(1/4)*(-3*I + Sqrt[3])*Tan[x
])/2]*Cos[x] + (-3)^(1/4)*(-3*I + Sqrt[3])*ArcTan[((-1)^(3/4)*(3*I + Sqrt[3])*Tan[x])/(2*3^(1/4))]*Cos[x] - 6*
Sin[x]))/(144*(-1 + Sin[x]^6))

________________________________________________________________________________________

Maple [B]  time = 0.139, size = 255, normalized size = 3.6 \begin{align*}{\frac{\tan \left ( x \right ) }{3}}+{\frac{\sqrt{3}\sqrt{2\,\sqrt{3}-3}\ln \left ( \sqrt{3}+\sqrt{2\,\sqrt{3}-3}\sqrt{3}\tan \left ( x \right ) +3\, \left ( \tan \left ( x \right ) \right ) ^{2} \right ) }{36}}+{\frac{\sqrt{3}}{3\,\sqrt{6\,\sqrt{3}+9}}\arctan \left ({\frac{\sqrt{2\,\sqrt{3}-3}\sqrt{3}+6\,\tan \left ( x \right ) }{\sqrt{6\,\sqrt{3}+9}}} \right ) }+{\frac{1}{2\,\sqrt{6\,\sqrt{3}+9}}\arctan \left ({\frac{\sqrt{2\,\sqrt{3}-3}\sqrt{3}+6\,\tan \left ( x \right ) }{\sqrt{6\,\sqrt{3}+9}}} \right ) }-{\frac{\sqrt{3}\sqrt{2\,\sqrt{3}-3}\ln \left ( -\sqrt{2\,\sqrt{3}-3}\sqrt{3}\tan \left ( x \right ) +3\, \left ( \tan \left ( x \right ) \right ) ^{2}+\sqrt{3} \right ) }{36}}+{\frac{\sqrt{3}}{3\,\sqrt{6\,\sqrt{3}+9}}\arctan \left ({\frac{-\sqrt{2\,\sqrt{3}-3}\sqrt{3}+6\,\tan \left ( x \right ) }{\sqrt{6\,\sqrt{3}+9}}} \right ) }+{\frac{1}{2\,\sqrt{6\,\sqrt{3}+9}}\arctan \left ({\frac{-\sqrt{2\,\sqrt{3}-3}\sqrt{3}+6\,\tan \left ( x \right ) }{\sqrt{6\,\sqrt{3}+9}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-sin(x)^6),x)

[Out]

1/3*tan(x)+1/36*3^(1/2)*(2*3^(1/2)-3)^(1/2)*ln(3^(1/2)+(2*3^(1/2)-3)^(1/2)*3^(1/2)*tan(x)+3*tan(x)^2)+1/3/(6*3
^(1/2)+9)^(1/2)*arctan(((2*3^(1/2)-3)^(1/2)*3^(1/2)+6*tan(x))/(6*3^(1/2)+9)^(1/2))*3^(1/2)+1/2/(6*3^(1/2)+9)^(
1/2)*arctan(((2*3^(1/2)-3)^(1/2)*3^(1/2)+6*tan(x))/(6*3^(1/2)+9)^(1/2))-1/36*3^(1/2)*(2*3^(1/2)-3)^(1/2)*ln(-(
2*3^(1/2)-3)^(1/2)*3^(1/2)*tan(x)+3*tan(x)^2+3^(1/2))+1/3/(6*3^(1/2)+9)^(1/2)*arctan((-(2*3^(1/2)-3)^(1/2)*3^(
1/2)+6*tan(x))/(6*3^(1/2)+9)^(1/2))*3^(1/2)+1/2/(6*3^(1/2)+9)^(1/2)*arctan((-(2*3^(1/2)-3)^(1/2)*3^(1/2)+6*tan
(x))/(6*3^(1/2)+9)^(1/2))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-sin(x)^6),x, algorithm="maxima")

[Out]

-1/3*(3*(cos(2*x)^2 + sin(2*x)^2 + 2*cos(2*x) + 1)*integrate(-4/3*((cos(6*x) - 10*cos(4*x) + cos(2*x))*cos(8*x
) + (110*cos(4*x) - 16*cos(2*x) + 1)*cos(6*x) - 8*cos(6*x)^2 + 10*(11*cos(2*x) - 1)*cos(4*x) - 300*cos(4*x)^2
- 8*cos(2*x)^2 + (sin(6*x) - 10*sin(4*x) + sin(2*x))*sin(8*x) + 2*(55*sin(4*x) - 8*sin(2*x))*sin(6*x) - 8*sin(
6*x)^2 - 300*sin(4*x)^2 + 110*sin(4*x)*sin(2*x) - 8*sin(2*x)^2 + cos(2*x))/(2*(8*cos(6*x) - 30*cos(4*x) + 8*co
s(2*x) - 1)*cos(8*x) - cos(8*x)^2 + 16*(30*cos(4*x) - 8*cos(2*x) + 1)*cos(6*x) - 64*cos(6*x)^2 + 60*(8*cos(2*x
) - 1)*cos(4*x) - 900*cos(4*x)^2 - 64*cos(2*x)^2 + 4*(4*sin(6*x) - 15*sin(4*x) + 4*sin(2*x))*sin(8*x) - sin(8*
x)^2 + 32*(15*sin(4*x) - 4*sin(2*x))*sin(6*x) - 64*sin(6*x)^2 - 900*sin(4*x)^2 + 480*sin(4*x)*sin(2*x) - 64*si
n(2*x)^2 + 16*cos(2*x) - 1), x) - 2*sin(2*x))/(cos(2*x)^2 + sin(2*x)^2 + 2*cos(2*x) + 1)

________________________________________________________________________________________

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)^6),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)**6),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{1}{\sin \left (x\right )^{6} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-sin(x)^6),x, algorithm="giac")

[Out]

integrate(-1/(sin(x)^6 - 1), x)