Integrand size = 10, antiderivative size = 89 \[ \int \frac {1}{1-\sin ^8(x)} \, dx=\frac {x}{4 \sqrt {2}}+\frac {\arctan \left (\frac {\cos (x) \sin (x)}{1+\sqrt {2}+\sin ^2(x)}\right )}{4 \sqrt {2}}+\frac {\arctan \left (\sqrt {1-i} \tan (x)\right )}{4 \sqrt {1-i}}+\frac {\arctan \left (\sqrt {1+i} \tan (x)\right )}{4 \sqrt {1+i}}+\frac {\tan (x)}{4} \]
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)
Time = 3.40 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.72 \[ \int \frac {1}{1-\sin ^8(x)} \, dx=\frac {1}{8} \left (\frac {2 \arctan \left (\sqrt {1-i} \tan (x)\right )}{\sqrt {1-i}}+\frac {2 \arctan \left (\sqrt {1+i} \tan (x)\right )}{\sqrt {1+i}}+\sqrt {2} \arctan \left (\sqrt {2} \tan (x)\right )+2 \tan (x)\right ) \]
((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
Time = 0.42 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.900, Rules used = {3042, 3690, 3042, 3654, 3042, 3660, 216, 4254, 24}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{1-\sin ^8(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{1-\sin (x)^8}dx\) |
| \(\Big \downarrow \) 3690 |
| \(\displaystyle \frac {1}{4} \int \frac {1}{1-\sin ^2(x)}dx+\frac {1}{4} \int \frac {1}{1-i \sin ^2(x)}dx+\frac {1}{4} \int \frac {1}{i \sin ^2(x)+1}dx+\frac {1}{4} \int \frac {1}{\sin ^2(x)+1}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {1}{1-\sin (x)^2}dx+\frac {1}{4} \int \frac {1}{1-i \sin (x)^2}dx+\frac {1}{4} \int \frac {1}{i \sin (x)^2+1}dx+\frac {1}{4} \int \frac {1}{\sin (x)^2+1}dx\) |
| \(\Big \downarrow \) 3654 |
| \(\displaystyle \frac {1}{4} \int \frac {1}{1-i \sin (x)^2}dx+\frac {1}{4} \int \frac {1}{i \sin (x)^2+1}dx+\frac {1}{4} \int \frac {1}{\sin (x)^2+1}dx+\frac {1}{4} \int \sec ^2(x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {1}{1-i \sin (x)^2}dx+\frac {1}{4} \int \frac {1}{i \sin (x)^2+1}dx+\frac {1}{4} \int \frac {1}{\sin (x)^2+1}dx+\frac {1}{4} \int \csc \left (x+\frac {\pi }{2}\right )^2dx\) |
| \(\Big \downarrow \) 3660 |
| \(\displaystyle \frac {1}{4} \int \frac {1}{(1-i) \tan ^2(x)+1}d\tan (x)+\frac {1}{4} \int \frac {1}{(1+i) \tan ^2(x)+1}d\tan (x)+\frac {1}{4} \int \frac {1}{2 \tan ^2(x)+1}d\tan (x)+\frac {1}{4} \int \csc \left (x+\frac {\pi }{2}\right )^2dx\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{4} \int \csc \left (x+\frac {\pi }{2}\right )^2dx+\frac {\arctan \left (\sqrt {1-i} \tan (x)\right )}{4 \sqrt {1-i}}+\frac {\arctan \left (\sqrt {1+i} \tan (x)\right )}{4 \sqrt {1+i}}+\frac {\arctan \left (\sqrt {2} \tan (x)\right )}{4 \sqrt {2}}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle -\frac {1}{4} \int 1d(-\tan (x))+\frac {\arctan \left (\sqrt {1-i} \tan (x)\right )}{4 \sqrt {1-i}}+\frac {\arctan \left (\sqrt {1+i} \tan (x)\right )}{4 \sqrt {1+i}}+\frac {\arctan \left (\sqrt {2} \tan (x)\right )}{4 \sqrt {2}}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\arctan \left (\sqrt {1-i} \tan (x)\right )}{4 \sqrt {1-i}}+\frac {\arctan \left (\sqrt {1+i} \tan (x)\right )}{4 \sqrt {1+i}}+\frac {\arctan \left (\sqrt {2} \tan (x)\right )}{4 \sqrt {2}}+\frac {\tan (x)}{4}\) |
ArcTan[Sqrt[1 - I]*Tan[x]]/(4*Sqrt[1 - I]) + ArcTan[Sqrt[1 + I]*Tan[x]]/(4 *Sqrt[1 + I]) + ArcTan[Sqrt[2]*Tan[x]]/(4*Sqrt[2]) + Tan[x]/4
3.3.60.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[ 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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(-1), x_Symbol] :> Module[{ k}, Simp[2/(a*n) Sum[Int[1/(1 - Sin[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]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (65 ) = 130\).
Time = 5.77 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.98
| method | result | size |
| risch | \(\frac {i}{2 \,{\mathrm e}^{2 i x}+2}+\frac {\sqrt {-2-2 i}\, \ln \left ({\mathrm e}^{2 i x}+i \sqrt {-2-2 i}-\sqrt {-2-2 i}-1-2 i\right )}{16}-\frac {\sqrt {-2-2 i}\, \ln \left ({\mathrm e}^{2 i x}-i \sqrt {-2-2 i}+\sqrt {-2-2 i}-1-2 i\right )}{16}+\frac {\sqrt {-2+2 i}\, \ln \left ({\mathrm e}^{2 i x}+i \sqrt {-2+2 i}+\sqrt {-2+2 i}-1+2 i\right )}{16}-\frac {\sqrt {-2+2 i}\, \ln \left ({\mathrm e}^{2 i x}-i \sqrt {-2+2 i}-\sqrt {-2+2 i}-1+2 i\right )}{16}+\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-2 \sqrt {2}-3\right )}{16}-\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+2 \sqrt {2}-3\right )}{16}\) | \(176\) |
| default | \(\frac {\tan \left (x \right )}{4}+\frac {\sqrt {2}\, \left (-\frac {\sqrt {-2+2 \sqrt {2}}\, \ln \left (-\sqrt {-2+2 \sqrt {2}}\, \sqrt {2}\, \tan \left (x \right )+2 \left (\tan ^{2}\left (x \right )\right )+\sqrt {2}\right )}{4}+\frac {\left (-\frac {\left (-2+2 \sqrt {2}\right ) \sqrt {2}}{4}+2\right ) \arctan \left (\frac {-\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}+4 \tan \left (x \right )}{2 \sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}\right )}{8}+\frac {\sqrt {2}\, \left (\frac {\sqrt {-2+2 \sqrt {2}}\, \ln \left (\sqrt {2}+\sqrt {-2+2 \sqrt {2}}\, \sqrt {2}\, \tan \left (x \right )+2 \left (\tan ^{2}\left (x \right )\right )\right )}{4}+\frac {\left (-\frac {\left (-2+2 \sqrt {2}\right ) \sqrt {2}}{4}+2\right ) \arctan \left (\frac {\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}+4 \tan \left (x \right )}{2 \sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}\right )}{8}+\frac {\arctan \left (\sqrt {2}\, \tan \left (x \right )\right ) \sqrt {2}}{8}\) | \(206\) |
1/2*I/(exp(2*I*x)+1)+1/16*(-2-2*I)^(1/2)*ln(exp(2*I*x)+I*(-2-2*I)^(1/2)-(- 2-2*I)^(1/2)-1-2*I)-1/16*(-2-2*I)^(1/2)*ln(exp(2*I*x)-I*(-2-2*I)^(1/2)+(-2 -2*I)^(1/2)-1-2*I)+1/16*(-2+2*I)^(1/2)*ln(exp(2*I*x)+I*(-2+2*I)^(1/2)+(-2+ 2*I)^(1/2)-1+2*I)-1/16*(-2+2*I)^(1/2)*ln(exp(2*I*x)-I*(-2+2*I)^(1/2)-(-2+2 *I)^(1/2)-1+2*I)+1/16*I*2^(1/2)*ln(exp(2*I*x)-2*2^(1/2)-3)-1/16*I*2^(1/2)* ln(exp(2*I*x)+2*2^(1/2)-3)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (57) = 114\).
Time = 0.32 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.87 \[ \int \frac {1}{1-\sin ^8(x)} \, dx=\frac {\sqrt {2} \sqrt {i - 1} \cos \left (x\right ) \log \left (-\left (i - 1\right ) \, \sqrt {2} \sqrt {i - 1} \cos \left (x\right ) \sin \left (x\right ) + \left (2 i - 1\right ) \, \cos \left (x\right )^{2} - i + 1\right ) - \sqrt {2} \sqrt {i - 1} \cos \left (x\right ) \log \left (\left (i - 1\right ) \, \sqrt {2} \sqrt {i - 1} \cos \left (x\right ) \sin \left (x\right ) + \left (2 i - 1\right ) \, \cos \left (x\right )^{2} - i + 1\right ) - \sqrt {2} \sqrt {-i - 1} \cos \left (x\right ) \log \left (\left (i + 1\right ) \, \sqrt {2} \sqrt {-i - 1} \cos \left (x\right ) \sin \left (x\right ) + \left (2 i + 1\right ) \, \cos \left (x\right )^{2} - i - 1\right ) + \sqrt {2} \sqrt {-i - 1} \cos \left (x\right ) \log \left (-\left (i + 1\right ) \, \sqrt {2} \sqrt {-i - 1} \cos \left (x\right ) \sin \left (x\right ) + \left (2 i + 1\right ) \, \cos \left (x\right )^{2} - i - 1\right ) - 2 \, \sqrt {2} \arctan \left (\frac {3 \, \sqrt {2} \cos \left (x\right )^{2} - 2 \, \sqrt {2}}{4 \, \cos \left (x\right ) \sin \left (x\right )}\right ) \cos \left (x\right ) + 8 \, \sin \left (x\right )}{32 \, \cos \left (x\right )} \]
1/32*(sqrt(2)*sqrt(I - 1)*cos(x)*log(-(I - 1)*sqrt(2)*sqrt(I - 1)*cos(x)*s in(x) + (2*I - 1)*cos(x)^2 - I + 1) - sqrt(2)*sqrt(I - 1)*cos(x)*log((I - 1)*sqrt(2)*sqrt(I - 1)*cos(x)*sin(x) + (2*I - 1)*cos(x)^2 - I + 1) - sqrt( 2)*sqrt(-I - 1)*cos(x)*log((I + 1)*sqrt(2)*sqrt(-I - 1)*cos(x)*sin(x) + (2 *I + 1)*cos(x)^2 - I - 1) + sqrt(2)*sqrt(-I - 1)*cos(x)*log(-(I + 1)*sqrt( 2)*sqrt(-I - 1)*cos(x)*sin(x) + (2*I + 1)*cos(x)^2 - I - 1) - 2*sqrt(2)*ar ctan(1/4*(3*sqrt(2)*cos(x)^2 - 2*sqrt(2))/(cos(x)*sin(x)))*cos(x) + 8*sin( x))/cos(x)
Timed out. \[ \int \frac {1}{1-\sin ^8(x)} \, dx=\text {Timed out} \]
\[ \int \frac {1}{1-\sin ^8(x)} \, dx=\int { -\frac {1}{\sin \left (x\right )^{8} - 1} \,d x } \]
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*(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)) - (sqrt(2)*cos(2*x)^2 + sqr t(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*sin(4*x)*sin (2*x) - 16*sin(2*x)^2 + 8*cos(2*x) - 1), x) + 8*sin(2*x))/(cos(2*x)^2 + si n(2*x)^2 + 2*cos(2*x) + 1)
Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (57) = 114\).
Time = 0.53 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.47 \[ \int \frac {1}{1-\sin ^8(x)} \, dx=\frac {1}{8} \, \sqrt {2} {\left (x + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, x\right ) - 2 \, \sin \left (2 \, x\right )}{\sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2} - 2 \, \cos \left (2 \, x\right ) + 2}\right )\right )} + \frac {1}{8} \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} + 2 \, \tan \left (x\right )\right )}}{\sqrt {\sqrt {2} + 2}}\right )\right )} \sqrt {\sqrt {2} + 1} + \frac {1}{8} \, {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (-\frac {2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} - 2 \, \tan \left (x\right )\right )}}{\sqrt {\sqrt {2} + 2}}\right )\right )} \sqrt {\sqrt {2} + 1} + \frac {1}{16} \, \sqrt {\sqrt {2} - 1} \log \left (\tan \left (x\right )^{2} + \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \left (x\right ) + \sqrt {\frac {1}{2}}\right ) - \frac {1}{16} \, \sqrt {\sqrt {2} - 1} \log \left (\tan \left (x\right )^{2} - \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \left (x\right ) + \sqrt {\frac {1}{2}}\right ) + \frac {1}{4} \, \tan \left (x\right ) \]
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)))*sq rt(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)*t an(x) + sqrt(1/2)) - 1/16*sqrt(sqrt(2) - 1)*log(tan(x)^2 - (1/2)^(1/4)*sqr t(-sqrt(2) + 2)*tan(x) + sqrt(1/2)) + 1/4*tan(x)
Time = 13.79 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.58 \[ \int \frac {1}{1-\sin ^8(x)} \, dx=\frac {\mathrm {tan}\left (x\right )}{4}+\mathrm {atan}\left (\sqrt {2}\,\mathrm {tan}\left (x\right )\,\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}\,8{}\mathrm {i}-\sqrt {2}\,\mathrm {tan}\left (x\right )\,\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,8{}\mathrm {i}\right )\,\left (\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}\,2{}\mathrm {i}+\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,2{}\mathrm {i}\right )+\mathrm {atan}\left (\sqrt {2}\,\mathrm {tan}\left (x\right )\,\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}\,8{}\mathrm {i}+\sqrt {2}\,\mathrm {tan}\left (x\right )\,\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,8{}\mathrm {i}\right )\,\left (\sqrt {-\frac {\sqrt {2}}{256}-\frac {1}{256}}\,2{}\mathrm {i}-\sqrt {\frac {\sqrt {2}}{256}-\frac {1}{256}}\,2{}\mathrm {i}\right )+\frac {\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\mathrm {tan}\left (x\right )\right )}{8} \]
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)*8i)*((- 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