\(\int \frac {1}{a-a \sin ^8(x)} \, dx\) [57]

Optimal result
Mathematica [C] (verified)
Rubi [C] (verified)
Maple [C] (verified)
Fricas [B] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [B] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 11, antiderivative size = 174 \[ \int \frac {1}{a-a \sin ^8(x)} \, dx=-\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {-1+\sqrt {2}}-2 \tan (x)}{\sqrt {1+\sqrt {2}}}\right )}{8 a}+\frac {\arctan \left (\sqrt {2} \tan (x)\right )}{4 \sqrt {2} a}+\frac {\sqrt {1+\sqrt {2}} \arctan \left (\frac {\sqrt {-1+\sqrt {2}}+2 \tan (x)}{\sqrt {1+\sqrt {2}}}\right )}{8 a}+\frac {\sqrt {-1+\sqrt {2}} \text {arctanh}\left (\frac {\sqrt {2 \left (-1+\sqrt {2}\right )} \tan (x)}{1+\sqrt {2} \tan ^2(x)}\right )}{8 a}+\frac {\tan (x)}{4 a} \] Output:

-1/8*(1+2^(1/2))^(1/2)*arctan(((2^(1/2)-1)^(1/2)-2*tan(x))/(1+2^(1/2))^(1/ 
2))/a+1/8*arctan(tan(x)*2^(1/2))*2^(1/2)/a+1/8*(1+2^(1/2))^(1/2)*arctan((( 
2^(1/2)-1)^(1/2)+2*tan(x))/(1+2^(1/2))^(1/2))/a+1/8*(2^(1/2)-1)^(1/2)*arct 
anh((-2+2*2^(1/2))^(1/2)*tan(x)/(1+2^(1/2)*tan(x)^2))/a+1/4*tan(x)/a
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 5.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.39 \[ \int \frac {1}{a-a \sin ^8(x)} \, dx=\frac {\frac {2 \arctan \left (\sqrt {1-i} \tan (x)\right )}{\sqrt {1-i}}+\frac {2 \arctan \left (\sqrt {1+i} \tan (x)\right )}{\sqrt {1+i}}+\sqrt {2} \arctan \left (\sqrt {2} \tan (x)\right )+2 \tan (x)}{8 a} \] Input:

Integrate[(a - a*Sin[x]^8)^(-1),x]
 

Output:

((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*a)
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.45 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.47, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.818, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{a-a \sin (x)^8}dx\)

\(\Big \downarrow \) 3690

\(\displaystyle \frac {\int \frac {1}{1-\sin ^2(x)}dx}{4 a}+\frac {\int \frac {1}{1-i \sin ^2(x)}dx}{4 a}+\frac {\int \frac {1}{i \sin ^2(x)+1}dx}{4 a}+\frac {\int \frac {1}{\sin ^2(x)+1}dx}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{1-\sin (x)^2}dx}{4 a}+\frac {\int \frac {1}{1-i \sin (x)^2}dx}{4 a}+\frac {\int \frac {1}{i \sin (x)^2+1}dx}{4 a}+\frac {\int \frac {1}{\sin (x)^2+1}dx}{4 a}\)

\(\Big \downarrow \) 3654

\(\displaystyle \frac {\int \frac {1}{1-i \sin (x)^2}dx}{4 a}+\frac {\int \frac {1}{i \sin (x)^2+1}dx}{4 a}+\frac {\int \frac {1}{\sin (x)^2+1}dx}{4 a}+\frac {\int \sec ^2(x)dx}{4 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{1-i \sin (x)^2}dx}{4 a}+\frac {\int \frac {1}{i \sin (x)^2+1}dx}{4 a}+\frac {\int \frac {1}{\sin (x)^2+1}dx}{4 a}+\frac {\int \csc \left (x+\frac {\pi }{2}\right )^2dx}{4 a}\)

\(\Big \downarrow \) 3660

\(\displaystyle \frac {\int \frac {1}{(1-i) \tan ^2(x)+1}d\tan (x)}{4 a}+\frac {\int \frac {1}{(1+i) \tan ^2(x)+1}d\tan (x)}{4 a}+\frac {\int \frac {1}{2 \tan ^2(x)+1}d\tan (x)}{4 a}+\frac {\int \csc \left (x+\frac {\pi }{2}\right )^2dx}{4 a}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\int \csc \left (x+\frac {\pi }{2}\right )^2dx}{4 a}+\frac {\arctan \left (\sqrt {1-i} \tan (x)\right )}{4 \sqrt {1-i} a}+\frac {\arctan \left (\sqrt {1+i} \tan (x)\right )}{4 \sqrt {1+i} a}+\frac {\arctan \left (\sqrt {2} \tan (x)\right )}{4 \sqrt {2} a}\)

\(\Big \downarrow \) 4254

\(\displaystyle -\frac {\int 1d(-\tan (x))}{4 a}+\frac {\arctan \left (\sqrt {1-i} \tan (x)\right )}{4 \sqrt {1-i} a}+\frac {\arctan \left (\sqrt {1+i} \tan (x)\right )}{4 \sqrt {1+i} a}+\frac {\arctan \left (\sqrt {2} \tan (x)\right )}{4 \sqrt {2} a}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\arctan \left (\sqrt {1-i} \tan (x)\right )}{4 \sqrt {1-i} a}+\frac {\arctan \left (\sqrt {1+i} \tan (x)\right )}{4 \sqrt {1+i} a}+\frac {\arctan \left (\sqrt {2} \tan (x)\right )}{4 \sqrt {2} a}+\frac {\tan (x)}{4 a}\)

Input:

Int[(a - a*Sin[x]^8)^(-1),x]
 

Output:

ArcTan[Sqrt[1 - I]*Tan[x]]/(4*Sqrt[1 - I]*a) + ArcTan[Sqrt[1 + I]*Tan[x]]/ 
(4*Sqrt[1 + I]*a) + ArcTan[Sqrt[2]*Tan[x]]/(4*Sqrt[2]*a) + Tan[x]/(4*a)
 

Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 216
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])
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3654
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]
 

rule 3660
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]
 

rule 3690
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]
 

rule 4254
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]
 
Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.73 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.13

method result size
risch \(\frac {i}{2 a \left ({\mathrm e}^{2 i x}+1\right )}+\frac {\sqrt {-2-2 i}\, \ln \left ({\mathrm e}^{2 i x}+i \sqrt {-2-2 i}-\sqrt {-2-2 i}-1-2 i\right )}{16 a}-\frac {\sqrt {-2-2 i}\, \ln \left ({\mathrm e}^{2 i x}-i \sqrt {-2-2 i}+\sqrt {-2-2 i}-1-2 i\right )}{16 a}+\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-2 \sqrt {2}-3\right )}{16 a}-\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+2 \sqrt {2}-3\right )}{16 a}+\frac {\sqrt {-2+2 i}\, \ln \left ({\mathrm e}^{2 i x}+i \sqrt {-2+2 i}+\sqrt {-2+2 i}-1+2 i\right )}{16 a}-\frac {\sqrt {-2+2 i}\, \ln \left ({\mathrm e}^{2 i x}-i \sqrt {-2+2 i}-\sqrt {-2+2 i}-1+2 i\right )}{16 a}\) \(197\)
default \(\frac {\frac {\tan \left (x \right )}{4}+\frac {\sqrt {2}\, \left (\frac {\sqrt {-2+2 \sqrt {2}}\, \ln \left (\sqrt {2}+\sqrt {-2+2 \sqrt {2}}\, \sqrt {2}\, \tan \left (x \right )+2 \tan \left (x \right )^{2}\right )}{4}+\frac {\left (-\frac {\left (-2+2 \sqrt {2}\right ) \sqrt {2}}{4}+2\right ) \arctan \left (\frac {\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}+4 \tan \left (x \right )}{2 \sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}\right )}{8}+\frac {\sqrt {2}\, \left (-\frac {\sqrt {-2+2 \sqrt {2}}\, \ln \left (-\sqrt {-2+2 \sqrt {2}}\, \sqrt {2}\, \tan \left (x \right )+2 \tan \left (x \right )^{2}+\sqrt {2}\right )}{4}+\frac {\left (-\frac {\left (-2+2 \sqrt {2}\right ) \sqrt {2}}{4}+2\right ) \arctan \left (\frac {-\sqrt {2}\, \sqrt {-2+2 \sqrt {2}}+4 \tan \left (x \right )}{2 \sqrt {1+\sqrt {2}}}\right )}{\sqrt {1+\sqrt {2}}}\right )}{8}+\frac {\sqrt {2}\, \arctan \left (\tan \left (x \right ) \sqrt {2}\right )}{8}}{a}\) \(210\)

Input:

int(1/(a-a*sin(x)^8),x,method=_RETURNVERBOSE)
 

Output:

1/2*I/a/(exp(2*I*x)+1)+1/16*(-2-2*I)^(1/2)/a*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)/a*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)/a*ln(exp(2*I*x)-2*2^(1/2)-3)-1/16* 
I*2^(1/2)/a*ln(exp(2*I*x)+2*2^(1/2)-3)+1/16*(-2+2*I)^(1/2)/a*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)/a*ln(exp(2*I*x 
)-I*(-2+2*I)^(1/2)-(-2+2*I)^(1/2)-1+2*I)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 474 vs. \(2 (126) = 252\).

Time = 0.16 (sec) , antiderivative size = 474, normalized size of antiderivative = 2.72 \[ \int \frac {1}{a-a \sin ^8(x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/16*(sqrt(1/2)*a*sqrt(-(a^2*sqrt(-1/a^4) + 1)/a^2)*cos(x)*log(1/2*sqrt(1 
/2)*(a^3*sqrt(-1/a^4)*cos(x)*sin(x) + a*cos(x)*sin(x))*sqrt(-(a^2*sqrt(-1/ 
a^4) + 1)/a^2) + 1/4*cos(x)^2 + 1/4*(2*a^2*cos(x)^2 - a^2)*sqrt(-1/a^4) - 
1/4) - sqrt(1/2)*a*sqrt(-(a^2*sqrt(-1/a^4) + 1)/a^2)*cos(x)*log(-1/2*sqrt( 
1/2)*(a^3*sqrt(-1/a^4)*cos(x)*sin(x) + a*cos(x)*sin(x))*sqrt(-(a^2*sqrt(-1 
/a^4) + 1)/a^2) + 1/4*cos(x)^2 + 1/4*(2*a^2*cos(x)^2 - a^2)*sqrt(-1/a^4) - 
 1/4) + sqrt(1/2)*a*sqrt((a^2*sqrt(-1/a^4) - 1)/a^2)*cos(x)*log(1/2*sqrt(1 
/2)*(a^3*sqrt(-1/a^4)*cos(x)*sin(x) - a*cos(x)*sin(x))*sqrt((a^2*sqrt(-1/a 
^4) - 1)/a^2) - 1/4*cos(x)^2 + 1/4*(2*a^2*cos(x)^2 - a^2)*sqrt(-1/a^4) + 1 
/4) - sqrt(1/2)*a*sqrt((a^2*sqrt(-1/a^4) - 1)/a^2)*cos(x)*log(-1/2*sqrt(1/ 
2)*(a^3*sqrt(-1/a^4)*cos(x)*sin(x) - a*cos(x)*sin(x))*sqrt((a^2*sqrt(-1/a^ 
4) - 1)/a^2) - 1/4*cos(x)^2 + 1/4*(2*a^2*cos(x)^2 - a^2)*sqrt(-1/a^4) + 1/ 
4) + sqrt(2)*arctan(1/4*(3*sqrt(2)*cos(x)^2 - 2*sqrt(2))/(cos(x)*sin(x)))* 
cos(x) - 4*sin(x))/(a*cos(x))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{a-a \sin ^8(x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{a-a \sin ^8(x)} \, dx=\int { -\frac {1}{a \sin \left (x\right )^{8} - a} \,d x } \] Input:

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

Output:

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*(a*cos(2*x)^2 + a*sin(2*x)^2 + 2*a*cos(2*x) + a)*int 
egrate(-((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))/(a*cos(8*x)^2 + 16*a*cos(6*x)^2 + 484*a*cos(4*x) 
^2 + 16*a*cos(2*x)^2 + a*sin(8*x)^2 + 16*a*sin(6*x)^2 + 484*a*sin(4*x)^2 - 
 176*a*sin(4*x)*sin(2*x) + 16*a*sin(2*x)^2 - 2*(4*a*cos(6*x) - 22*a*cos(4* 
x) + 4*a*cos(2*x) - a)*cos(8*x) - 8*(22*a*cos(4*x) - 4*a*cos(2*x) + a)*cos 
(6*x) - 44*(4*a*cos(2*x) - a)*cos(4*x) - 8*a*cos(2*x) - 4*(2*a*sin(6*x) - 
11*a*sin(4*x) + 2*a*sin(2*x))*sin(8*x) - 16*(11*a*sin(4*x) - 2*a*sin(2*x)) 
*sin(6*x) + a), x) + 8*sin(2*x))/(a*cos(2*x)^2 + a*sin(2*x)^2 + 2*a*cos(2* 
x) + a)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (126) = 252\).

Time = 0.55 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.84 \[ \int \frac {1}{a-a \sin ^8(x)} \, dx=\frac {\sqrt {2} {\left (x + \arctan \left (-\frac {\sqrt {2} \sin \left (2 \, x\right ) - 2 \, \sin \left (2 \, x\right )}{\sqrt {2} \cos \left (2 \, x\right ) + \sqrt {2} - 2 \, \cos \left (2 \, x\right ) + 2}\right )\right )}}{8 \, a} + \frac {\tan \left (x\right )}{4 \, a} + \frac {{\left (a^{2} \sqrt {2 \, \sqrt {2} + 2} + a \sqrt {2 \, \sqrt {2} - 2} {\left | a \right |}\right )} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} + 2 \, \tan \left (x\right )\right )}}{\sqrt {\sqrt {2} + 2}}\right )\right )}}{16 \, a^{3}} + \frac {{\left (a^{2} \sqrt {2 \, \sqrt {2} + 2} + a \sqrt {2 \, \sqrt {2} - 2} {\left | a \right |}\right )} {\left (\pi \left \lfloor \frac {x}{\pi } + \frac {1}{2} \right \rfloor + \arctan \left (-\frac {2 \, \left (\frac {1}{2}\right )^{\frac {3}{4}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} - 2 \, \tan \left (x\right )\right )}}{\sqrt {\sqrt {2} + 2}}\right )\right )}}{16 \, a^{3}} - \frac {{\left (a^{2} \sqrt {2 \, \sqrt {2} - 2} - a \sqrt {2 \, \sqrt {2} + 2} {\left | a \right |}\right )} \log \left (\tan \left (x\right )^{2} + \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \left (x\right ) + \sqrt {\frac {1}{2}}\right )}{32 \, a^{3}} + \frac {{\left (a^{2} \sqrt {2 \, \sqrt {2} - 2} - a \sqrt {2 \, \sqrt {2} + 2} {\left | a \right |}\right )} \log \left (\tan \left (x\right )^{2} - \left (\frac {1}{2}\right )^{\frac {1}{4}} \sqrt {-\sqrt {2} + 2} \tan \left (x\right ) + \sqrt {\frac {1}{2}}\right )}{32 \, a^{3}} \] Input:

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

Output:

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)))/a + 1/4*tan(x)/a + 1/16*(a^2*sqrt(2*sqrt(2) 
 + 2) + a*sqrt(2*sqrt(2) - 2)*abs(a))*(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)))/a 
^3 + 1/16*(a^2*sqrt(2*sqrt(2) + 2) + a*sqrt(2*sqrt(2) - 2)*abs(a))*(pi*flo 
or(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)))/a^3 - 1/32*(a^2*sqrt(2*sqrt(2) - 2) - a*sqrt( 
2*sqrt(2) + 2)*abs(a))*log(tan(x)^2 + (1/2)^(1/4)*sqrt(-sqrt(2) + 2)*tan(x 
) + sqrt(1/2))/a^3 + 1/32*(a^2*sqrt(2*sqrt(2) - 2) - a*sqrt(2*sqrt(2) + 2) 
*abs(a))*log(tan(x)^2 - (1/2)^(1/4)*sqrt(-sqrt(2) + 2)*tan(x) + sqrt(1/2)) 
/a^3
 

Mupad [B] (verification not implemented)

Time = 37.50 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.47 \[ \int \frac {1}{a-a \sin ^8(x)} \, dx=\frac {\mathrm {tan}\left (x\right )}{4\,a}+\frac {\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,\mathrm {tan}\left (x\right )\right )}{8\,a}+\mathrm {atan}\left (a\,\mathrm {tan}\left (x\right )\,\sqrt {\frac {-\frac {1}{128}-\frac {1}{128}{}\mathrm {i}}{a^2}}\,\left (8+8{}\mathrm {i}\right )\right )\,\sqrt {\frac {-\frac {1}{128}-\frac {1}{128}{}\mathrm {i}}{a^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (a\,\mathrm {tan}\left (x\right )\,\sqrt {\frac {-\frac {1}{128}+\frac {1}{128}{}\mathrm {i}}{a^2}}\,\left (8-8{}\mathrm {i}\right )\right )\,\sqrt {\frac {-\frac {1}{128}+\frac {1}{128}{}\mathrm {i}}{a^2}}\,2{}\mathrm {i} \] Input:

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

Output:

tan(x)/(4*a) + atan(a*tan(x)*((- 1/128 - 1i/128)/a^2)^(1/2)*(8 + 8i))*((- 
1/128 - 1i/128)/a^2)^(1/2)*2i - atan(a*tan(x)*((- 1/128 + 1i/128)/a^2)^(1/ 
2)*(8 - 8i))*((- 1/128 + 1i/128)/a^2)^(1/2)*2i + (2^(1/2)*atan(2^(1/2)*tan 
(x)))/(8*a)
 

Reduce [F]

\[ \int \frac {1}{a-a \sin ^8(x)} \, dx=-\frac {\int \frac {1}{\sin \left (x \right )^{8}-1}d x}{a} \] Input:

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

Output:

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