\(\int \frac {1}{1-\sin ^6(x)} \, dx\) [259]

Optimal result

Integrand size = 10, antiderivative size = 71 \[ \int \frac {1}{1-\sin ^6(x)} \, dx=\frac {\arctan \left (\sqrt {1+\sqrt [3]{-1}} \tan (x)\right )}{3 \sqrt {1+\sqrt [3]{-1}}}+\frac {\arctan \left (\sqrt {1-(-1)^{2/3}} \tan (x)\right )}{3 \sqrt {1-(-1)^{2/3}}}+\frac {\tan (x)}{3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3290, 3260, 209, 3254, 3852, 8} \[ \int \frac {1}{1-\sin ^6(x)} \, dx=\frac {\arctan \left (\sqrt {1+\sqrt [3]{-1}} \tan (x)\right )}{3 \sqrt {1+\sqrt [3]{-1}}}+\frac {\arctan \left (\sqrt {1-(-1)^{2/3}} \tan (x)\right )}{3 \sqrt {1-(-1)^{2/3}}}+\frac {\tan (x)}{3} \]

[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 8

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3254

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 3260

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 3290

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 3852

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.58 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.65 \[ \int \frac {1}{1-\sin ^6(x)} \, dx=\frac {\cos (x) (15-8 \cos (2 x)+\cos (4 x)) \left (i \sqrt [4]{-3} \left (3 i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \sqrt [4]{-\frac {1}{3}} \left (-3 i+\sqrt {3}\right ) \tan (x)\right ) \cos (x)+\sqrt [4]{-3} \left (-3 i+\sqrt {3}\right ) \arctan \left (\frac {(-1)^{3/4} \left (3 i+\sqrt {3}\right ) \tan (x)}{2 \sqrt [4]{3}}\right ) \cos (x)-6 \sin (x)\right )}{144 \left (-1+\sin ^6(x)\right )} \]

[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 [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.72 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {2 i}{3 \left ({\mathrm e}^{2 i x}+1\right )}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (3888 \textit {\_Z}^{4}+108 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}-1296 i \textit {\_R}^{3}+216 \textit {\_R}^{2}+1\right )\right )\)	\(51\)
default	\(\frac {\tan \left (x \right )}{3}+\frac {\sqrt {3}\, \left (\frac {\sqrt {2 \sqrt {3}-3}\, \ln \left (\sqrt {3}+\sqrt {2 \sqrt {3}-3}\, \sqrt {3}\, \tan \left (x \right )+3 \left (\tan ^{2}\left (x \right )\right )\right )}{6}+\frac {2 \left (-\frac {\left (2 \sqrt {3}-3\right ) \sqrt {3}}{6}+2\right ) \arctan \left (\frac {\sqrt {2 \sqrt {3}-3}\, \sqrt {3}+6 \tan \left (x \right )}{\sqrt {6 \sqrt {3}+9}}\right )}{\sqrt {6 \sqrt {3}+9}}\right )}{6}+\frac {\sqrt {3}\, \left (-\frac {\sqrt {2 \sqrt {3}-3}\, \ln \left (-\sqrt {2 \sqrt {3}-3}\, \sqrt {3}\, \tan \left (x \right )+3 \left (\tan ^{2}\left (x \right )\right )+\sqrt {3}\right )}{6}+\frac {2 \left (-\frac {\left (2 \sqrt {3}-3\right ) \sqrt {3}}{6}+2\right ) \arctan \left (\frac {-\sqrt {2 \sqrt {3}-3}\, \sqrt {3}+6 \tan \left (x \right )}{\sqrt {6 \sqrt {3}+9}}\right )}{\sqrt {6 \sqrt {3}+9}}\right )}{6}\)	\(202\)

[In]

int(1/(1-sin(x)^6),x,method=_RETURNVERBOSE)

[Out]

2/3*I/(exp(2*I*x)+1)+sum(_R*ln(exp(2*I*x)-1296*I*_R^3+216*_R^2+1),_R=RootOf(3888*_Z^4+108*_Z^2+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.32 (sec) , antiderivative size = 239, normalized size of antiderivative = 3.37 \[ \int \frac {1}{1-\sin ^6(x)} \, dx=-\frac {\sqrt {6} \sqrt {i \, \sqrt {3} - 3} \cos \left (x\right ) \log \left (\sqrt {6} {\left (i \, \sqrt {3} - 3\right )}^{\frac {3}{2}} \cos \left (x\right ) \sin \left (x\right ) - 6 \, {\left (-i \, \sqrt {3} + 2\right )} \cos \left (x\right )^{2} - 3 i \, \sqrt {3} + 9\right ) - \sqrt {6} \sqrt {i \, \sqrt {3} - 3} \cos \left (x\right ) \log \left (\sqrt {6} \sqrt {i \, \sqrt {3} - 3} {\left (-i \, \sqrt {3} + 3\right )} \cos \left (x\right ) \sin \left (x\right ) - 6 \, {\left (-i \, \sqrt {3} + 2\right )} \cos \left (x\right )^{2} - 3 i \, \sqrt {3} + 9\right ) + \sqrt {6} \sqrt {-i \, \sqrt {3} - 3} \cos \left (x\right ) \log \left (\sqrt {6} {\left (i \, \sqrt {3} + 3\right )} \sqrt {-i \, \sqrt {3} - 3} \cos \left (x\right ) \sin \left (x\right ) - 6 \, {\left (-i \, \sqrt {3} - 2\right )} \cos \left (x\right )^{2} - 3 i \, \sqrt {3} - 9\right ) - \sqrt {6} \sqrt {-i \, \sqrt {3} - 3} \cos \left (x\right ) \log \left (\sqrt {6} {\left (-i \, \sqrt {3} - 3\right )}^{\frac {3}{2}} \cos \left (x\right ) \sin \left (x\right ) - 6 \, {\left (-i \, \sqrt {3} - 2\right )} \cos \left (x\right )^{2} - 3 i \, \sqrt {3} - 9\right ) - 24 \, \sin \left (x\right )}{72 \, \cos \left (x\right )} \]

[In]

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

[Out]

-1/72*(sqrt(6)*sqrt(I*sqrt(3) - 3)*cos(x)*log(sqrt(6)*(I*sqrt(3) - 3)^(3/2)*cos(x)*sin(x) - 6*(-I*sqrt(3) + 2)
*cos(x)^2 - 3*I*sqrt(3) + 9) - sqrt(6)*sqrt(I*sqrt(3) - 3)*cos(x)*log(sqrt(6)*sqrt(I*sqrt(3) - 3)*(-I*sqrt(3)
+ 3)*cos(x)*sin(x) - 6*(-I*sqrt(3) + 2)*cos(x)^2 - 3*I*sqrt(3) + 9) + sqrt(6)*sqrt(-I*sqrt(3) - 3)*cos(x)*log(
sqrt(6)*(I*sqrt(3) + 3)*sqrt(-I*sqrt(3) - 3)*cos(x)*sin(x) - 6*(-I*sqrt(3) - 2)*cos(x)^2 - 3*I*sqrt(3) - 9) -
sqrt(6)*sqrt(-I*sqrt(3) - 3)*cos(x)*log(sqrt(6)*(-I*sqrt(3) - 3)^(3/2)*cos(x)*sin(x) - 6*(-I*sqrt(3) - 2)*cos(
x)^2 - 3*I*sqrt(3) - 9) - 24*sin(x))/cos(x)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{1-\sin ^6(x)} \, dx=\text {Timed out} \]

[In]

integrate(1/(1-sin(x)**6),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{1-\sin ^6(x)} \, dx=\int { -\frac {1}{\sin \left (x\right )^{6} - 1} \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (49) = 98\).

Time = 0.35 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.77 \[ \int \frac {1}{1-\sin ^6(x)} \, dx=\frac {1}{18} \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor - \arctan \left (-\frac {3 \, \left (\frac {1}{3}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (\sqrt {6} - \sqrt {2}\right )} + 4 \, \tan \left (x\right )\right )}}{\sqrt {6} + \sqrt {2}}\right )\right )} \sqrt {6 \, \sqrt {3} + 9} + \frac {1}{18} \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (-\frac {3 \, \left (\frac {1}{3}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{3}\right )^{\frac {1}{4}} {\left (\sqrt {6} - \sqrt {2}\right )} - 4 \, \tan \left (x\right )\right )}}{\sqrt {6} + \sqrt {2}}\right )\right )} \sqrt {6 \, \sqrt {3} + 9} + \frac {1}{36} \, \sqrt {6 \, \sqrt {3} - 9} \log \left (\frac {1}{2} \, {\left (\sqrt {6} \left (\frac {1}{3}\right )^{\frac {1}{4}} - \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}}\right )} \tan \left (x\right ) + \tan \left (x\right )^{2} + \sqrt {\frac {1}{3}}\right ) - \frac {1}{36} \, \sqrt {6 \, \sqrt {3} - 9} \log \left (-\frac {1}{2} \, {\left (\sqrt {6} \left (\frac {1}{3}\right )^{\frac {1}{4}} - \sqrt {2} \left (\frac {1}{3}\right )^{\frac {1}{4}}\right )} \tan \left (x\right ) + \tan \left (x\right )^{2} + \sqrt {\frac {1}{3}}\right ) + \frac {1}{3} \, \tan \left (x\right ) \]

[In]

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

[Out]

1/18*(pi*floor(x/pi + 1/2) - arctan(-3*(1/3)^(3/4)*((1/3)^(1/4)*(sqrt(6) - sqrt(2)) + 4*tan(x))/(sqrt(6) + sqr
t(2))))*sqrt(6*sqrt(3) + 9) + 1/18*(pi*floor(x/pi + 1/2) + arctan(-3*(1/3)^(3/4)*((1/3)^(1/4)*(sqrt(6) - sqrt(
2)) - 4*tan(x))/(sqrt(6) + sqrt(2))))*sqrt(6*sqrt(3) + 9) + 1/36*sqrt(6*sqrt(3) - 9)*log(1/2*(sqrt(6)*(1/3)^(1
/4) - sqrt(2)*(1/3)^(1/4))*tan(x) + tan(x)^2 + sqrt(1/3)) - 1/36*sqrt(6*sqrt(3) - 9)*log(-1/2*(sqrt(6)*(1/3)^(
1/4) - sqrt(2)*(1/3)^(1/4))*tan(x) + tan(x)^2 + sqrt(1/3)) + 1/3*tan(x)

Mupad [B] (verification not implemented)

Time = 13.56 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.39 \[ \int \frac {1}{1-\sin ^6(x)} \, dx=\frac {\mathrm {tan}\left (x\right )}{3}-\frac {\sqrt {6}\,\mathrm {atan}\left (3^{1/4}\,\sqrt {6}\,\mathrm {tan}\left (x\right )\,\left (\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+3^{3/4}\,\sqrt {6}\,\mathrm {tan}\left (x\right )\,\left (\frac {1}{12}+\frac {1}{12}{}\mathrm {i}\right )\right )\,\left (3^{1/4}\,\left (1+1{}\mathrm {i}\right )+3^{3/4}\,\left (-1+1{}\mathrm {i}\right )\right )\,1{}\mathrm {i}}{36}+\frac {\sqrt {6}\,\mathrm {atan}\left (3^{1/4}\,\sqrt {6}\,\mathrm {tan}\left (x\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+3^{3/4}\,\sqrt {6}\,\mathrm {tan}\left (x\right )\,\left (\frac {1}{12}-\frac {1}{12}{}\mathrm {i}\right )\right )\,\left (3^{1/4}\,\left (1-\mathrm {i}\right )+3^{3/4}\,\left (-1-\mathrm {i}\right )\right )\,1{}\mathrm {i}}{36} \]

[In]

int(-1/(sin(x)^6 - 1),x)

[Out]

tan(x)/3 - (6^(1/2)*atan(3^(1/4)*6^(1/2)*tan(x)*(1/4 - 1i/4) + 3^(3/4)*6^(1/2)*tan(x)*(1/12 + 1i/12))*(3^(1/4)
*(1 + 1i) - 3^(3/4)*(1 - 1i))*1i)/36 + (6^(1/2)*atan(3^(1/4)*6^(1/2)*tan(x)*(1/4 + 1i/4) + 3^(3/4)*6^(1/2)*tan
(x)*(1/12 - 1i/12))*(3^(1/4)*(1 - 1i) - 3^(3/4)*(1 + 1i))*1i)/36