Integrand size = 23, antiderivative size = 163 \[ \int \frac {\sin ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right ) x}{16 b^4}+\frac {a^{7/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{b^4 \sqrt {a+b} d}-\frac {\left (8 a^2-6 a b+5 b^2\right ) \cos (c+d x) \sin (c+d x)}{16 b^3 d}+\frac {(6 a-5 b) \cos (c+d x) \sin ^3(c+d x)}{24 b^2 d}-\frac {\cos (c+d x) \sin ^5(c+d x)}{6 b d} \]
-1/16*(16*a^3-8*a^2*b+6*a*b^2-5*b^3)*x/b^4-1/16*(8*a^2-6*a*b+5*b^2)*cos(d* x+c)*sin(d*x+c)/b^3/d+1/24*(6*a-5*b)*cos(d*x+c)*sin(d*x+c)^3/b^2/d-1/6*cos (d*x+c)*sin(d*x+c)^5/b/d+a^(7/2)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/b^ 4/d/(a+b)^(1/2)
Time = 1.17 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.82 \[ \int \frac {\sin ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {12 \left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right ) (c+d x)-\frac {192 a^{7/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{\sqrt {a+b}}+3 b \left (16 a^2-16 a b+15 b^2\right ) \sin (2 (c+d x))+3 (2 a-3 b) b^2 \sin (4 (c+d x))+b^3 \sin (6 (c+d x))}{192 b^4 d} \]
-1/192*(12*(16*a^3 - 8*a^2*b + 6*a*b^2 - 5*b^3)*(c + d*x) - (192*a^(7/2)*A rcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/Sqrt[a + b] + 3*b*(16*a^2 - 16* a*b + 15*b^2)*Sin[2*(c + d*x)] + 3*(2*a - 3*b)*b^2*Sin[4*(c + d*x)] + b^3* Sin[6*(c + d*x)])/(b^4*d)
Time = 0.48 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.24, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 3666, 372, 440, 27, 440, 397, 216, 218}
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.
| \(\displaystyle \int \frac {\sin ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^8}{a+b \sin (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3666 |
| \(\displaystyle \frac {\int \frac {\tan ^8(c+d x)}{\left (\tan ^2(c+d x)+1\right )^4 \left ((a+b) \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{d}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\int \frac {\tan ^4(c+d x) \left (5 a-(a-5 b) \tan ^2(c+d x)\right )}{\left (\tan ^2(c+d x)+1\right )^3 \left ((a+b) \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{6 b}-\frac {\tan ^5(c+d x)}{6 b \left (\tan ^2(c+d x)+1\right )^3}}{d}\) |
| \(\Big \downarrow \) 440 |
| \(\displaystyle \frac {\frac {\frac {(6 a-5 b) \tan ^3(c+d x)}{4 b \left (\tan ^2(c+d x)+1\right )^2}-\frac {\int \frac {3 \tan ^2(c+d x) \left (a (6 a-5 b)-\left (2 a^2-b a+5 b^2\right ) \tan ^2(c+d x)\right )}{\left (\tan ^2(c+d x)+1\right )^2 \left ((a+b) \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{4 b}}{6 b}-\frac {\tan ^5(c+d x)}{6 b \left (\tan ^2(c+d x)+1\right )^3}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {(6 a-5 b) \tan ^3(c+d x)}{4 b \left (\tan ^2(c+d x)+1\right )^2}-\frac {3 \int \frac {\tan ^2(c+d x) \left (a (6 a-5 b)-\left (2 a^2-b a+5 b^2\right ) \tan ^2(c+d x)\right )}{\left (\tan ^2(c+d x)+1\right )^2 \left ((a+b) \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{4 b}}{6 b}-\frac {\tan ^5(c+d x)}{6 b \left (\tan ^2(c+d x)+1\right )^3}}{d}\) |
| \(\Big \downarrow \) 440 |
| \(\displaystyle \frac {\frac {\frac {(6 a-5 b) \tan ^3(c+d x)}{4 b \left (\tan ^2(c+d x)+1\right )^2}-\frac {3 \left (\frac {\left (8 a^2-6 a b+5 b^2\right ) \tan (c+d x)}{2 b \left (\tan ^2(c+d x)+1\right )}-\frac {\int \frac {a \left (8 a^2-6 b a+5 b^2\right )-\left (8 a^3-2 b a^2+b^2 a-5 b^3\right ) \tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \left ((a+b) \tan ^2(c+d x)+a\right )}d\tan (c+d x)}{2 b}\right )}{4 b}}{6 b}-\frac {\tan ^5(c+d x)}{6 b \left (\tan ^2(c+d x)+1\right )^3}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {(6 a-5 b) \tan ^3(c+d x)}{4 b \left (\tan ^2(c+d x)+1\right )^2}-\frac {3 \left (\frac {\left (8 a^2-6 a b+5 b^2\right ) \tan (c+d x)}{2 b \left (\tan ^2(c+d x)+1\right )}-\frac {\frac {16 a^4 \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{b}-\frac {\left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right ) \int \frac {1}{\tan ^2(c+d x)+1}d\tan (c+d x)}{b}}{2 b}\right )}{4 b}}{6 b}-\frac {\tan ^5(c+d x)}{6 b \left (\tan ^2(c+d x)+1\right )^3}}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {(6 a-5 b) \tan ^3(c+d x)}{4 b \left (\tan ^2(c+d x)+1\right )^2}-\frac {3 \left (\frac {\left (8 a^2-6 a b+5 b^2\right ) \tan (c+d x)}{2 b \left (\tan ^2(c+d x)+1\right )}-\frac {\frac {16 a^4 \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{b}-\frac {\left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right ) \arctan (\tan (c+d x))}{b}}{2 b}\right )}{4 b}}{6 b}-\frac {\tan ^5(c+d x)}{6 b \left (\tan ^2(c+d x)+1\right )^3}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {(6 a-5 b) \tan ^3(c+d x)}{4 b \left (\tan ^2(c+d x)+1\right )^2}-\frac {3 \left (\frac {\left (8 a^2-6 a b+5 b^2\right ) \tan (c+d x)}{2 b \left (\tan ^2(c+d x)+1\right )}-\frac {\frac {16 a^{7/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{b \sqrt {a+b}}-\frac {\left (16 a^3-8 a^2 b+6 a b^2-5 b^3\right ) \arctan (\tan (c+d x))}{b}}{2 b}\right )}{4 b}}{6 b}-\frac {\tan ^5(c+d x)}{6 b \left (\tan ^2(c+d x)+1\right )^3}}{d}\) |
(-1/6*Tan[c + d*x]^5/(b*(1 + Tan[c + d*x]^2)^3) + (((6*a - 5*b)*Tan[c + d* x]^3)/(4*b*(1 + Tan[c + d*x]^2)^2) - (3*(-1/2*(-(((16*a^3 - 8*a^2*b + 6*a* b^2 - 5*b^3)*ArcTan[Tan[c + d*x]])/b) + (16*a^(7/2)*ArcTan[(Sqrt[a + b]*Ta n[c + d*x])/Sqrt[a]])/(b*Sqrt[a + b]))/b + ((8*a^2 - 6*a*b + 5*b^2)*Tan[c + d*x])/(2*b*(1 + Tan[c + d*x]^2))))/(4*b))/(6*b))/d
3.1.85.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ g^2/(2*b*(b*c - a*d)*(p + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c *f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && LtQ[p, -1] && GtQ[m, 1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1 )/f Subst[Int[x^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1)) , x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] & & IntegerQ[p]
Time = 1.63 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.03
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{4} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{b^{4} \sqrt {a \left (a +b \right )}}-\frac {\frac {\left (\frac {1}{2} a^{2} b -\frac {5}{8} a \,b^{2}+\frac {11}{16} b^{3}\right ) \left (\tan ^{5}\left (d x +c \right )\right )+\left (a^{2} b -a \,b^{2}+\frac {5}{6} b^{3}\right ) \left (\tan ^{3}\left (d x +c \right )\right )+\left (\frac {1}{2} a^{2} b -\frac {3}{8} a \,b^{2}+\frac {5}{16} b^{3}\right ) \tan \left (d x +c \right )}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}+\frac {\left (16 a^{3}-8 a^{2} b +6 a \,b^{2}-5 b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{16}}{b^{4}}}{d}\) | \(168\) |
| default | \(\frac {\frac {a^{4} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{b^{4} \sqrt {a \left (a +b \right )}}-\frac {\frac {\left (\frac {1}{2} a^{2} b -\frac {5}{8} a \,b^{2}+\frac {11}{16} b^{3}\right ) \left (\tan ^{5}\left (d x +c \right )\right )+\left (a^{2} b -a \,b^{2}+\frac {5}{6} b^{3}\right ) \left (\tan ^{3}\left (d x +c \right )\right )+\left (\frac {1}{2} a^{2} b -\frac {3}{8} a \,b^{2}+\frac {5}{16} b^{3}\right ) \tan \left (d x +c \right )}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}+\frac {\left (16 a^{3}-8 a^{2} b +6 a \,b^{2}-5 b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{16}}{b^{4}}}{d}\) | \(168\) |
| risch | \(-\frac {x \,a^{3}}{b^{4}}+\frac {x \,a^{2}}{2 b^{3}}-\frac {3 a x}{8 b^{2}}+\frac {5 x}{16 b}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 b^{3} d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a}{8 b^{2} d}+\frac {15 i {\mathrm e}^{2 i \left (d x +c \right )}}{128 b d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 b^{3} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a}{8 b^{2} d}-\frac {15 i {\mathrm e}^{-2 i \left (d x +c \right )}}{128 b d}+\frac {\sqrt {-a \left (a +b \right )}\, a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 \left (a +b \right ) d \,b^{4}}-\frac {\sqrt {-a \left (a +b \right )}\, a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 \left (a +b \right ) d \,b^{4}}-\frac {\sin \left (6 d x +6 c \right )}{192 d b}-\frac {\sin \left (4 d x +4 c \right ) a}{32 b^{2} d}+\frac {3 \sin \left (4 d x +4 c \right )}{64 b d}\) | \(314\) |
1/d*(1/b^4*a^4/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2))-1/ b^4*(((1/2*a^2*b-5/8*a*b^2+11/16*b^3)*tan(d*x+c)^5+(a^2*b-a*b^2+5/6*b^3)*t an(d*x+c)^3+(1/2*a^2*b-3/8*a*b^2+5/16*b^3)*tan(d*x+c))/(1+tan(d*x+c)^2)^3+ 1/16*(16*a^3-8*a^2*b+6*a*b^2-5*b^3)*arctan(tan(d*x+c))))
Time = 0.33 (sec) , antiderivative size = 453, normalized size of antiderivative = 2.78 \[ \int \frac {\sin ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [\frac {12 \, a^{3} \sqrt {-\frac {a}{a + b}} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 3 \, {\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} d x - {\left (8 \, b^{3} \cos \left (d x + c\right )^{5} + 2 \, {\left (6 \, a b^{2} - 13 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{2} b - 10 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, b^{4} d}, -\frac {24 \, a^{3} \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) + 3 \, {\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} d x + {\left (8 \, b^{3} \cos \left (d x + c\right )^{5} + 2 \, {\left (6 \, a b^{2} - 13 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{2} b - 10 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, b^{4} d}\right ] \]
[1/48*(12*a^3*sqrt(-a/(a + b))*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 - 4*((2*a^2 + 3*a*b + b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^2)*cos(d*x + c))*sqrt(-a/(a + b))*sin(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)) - 3*(16*a^3 - 8*a^2*b + 6*a*b^2 - 5*b^3)*d*x - (8*b^3* cos(d*x + c)^5 + 2*(6*a*b^2 - 13*b^3)*cos(d*x + c)^3 + 3*(8*a^2*b - 10*a*b ^2 + 11*b^3)*cos(d*x + c))*sin(d*x + c))/(b^4*d), -1/48*(24*a^3*sqrt(a/(a + b))*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt(a/(a + b))/(a*cos (d*x + c)*sin(d*x + c))) + 3*(16*a^3 - 8*a^2*b + 6*a*b^2 - 5*b^3)*d*x + (8 *b^3*cos(d*x + c)^5 + 2*(6*a*b^2 - 13*b^3)*cos(d*x + c)^3 + 3*(8*a^2*b - 1 0*a*b^2 + 11*b^3)*cos(d*x + c))*sin(d*x + c))/(b^4*d)]
Timed out. \[ \int \frac {\sin ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\text {Timed out} \]
Time = 0.33 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.18 \[ \int \frac {\sin ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {48 \, a^{4} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} b^{4}} - \frac {3 \, {\left (8 \, a^{2} - 10 \, a b + 11 \, b^{2}\right )} \tan \left (d x + c\right )^{5} + 8 \, {\left (6 \, a^{2} - 6 \, a b + 5 \, b^{2}\right )} \tan \left (d x + c\right )^{3} + 3 \, {\left (8 \, a^{2} - 6 \, a b + 5 \, b^{2}\right )} \tan \left (d x + c\right )}{b^{3} \tan \left (d x + c\right )^{6} + 3 \, b^{3} \tan \left (d x + c\right )^{4} + 3 \, b^{3} \tan \left (d x + c\right )^{2} + b^{3}} - \frac {3 \, {\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} {\left (d x + c\right )}}{b^{4}}}{48 \, d} \]
1/48*(48*a^4*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)*a) *b^4) - (3*(8*a^2 - 10*a*b + 11*b^2)*tan(d*x + c)^5 + 8*(6*a^2 - 6*a*b + 5 *b^2)*tan(d*x + c)^3 + 3*(8*a^2 - 6*a*b + 5*b^2)*tan(d*x + c))/(b^3*tan(d* x + c)^6 + 3*b^3*tan(d*x + c)^4 + 3*b^3*tan(d*x + c)^2 + b^3) - 3*(16*a^3 - 8*a^2*b + 6*a*b^2 - 5*b^3)*(d*x + c)/b^4)/d
Time = 0.45 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.43 \[ \int \frac {\sin ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {48 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} a^{4}}{\sqrt {a^{2} + a b} b^{4}} - \frac {3 \, {\left (16 \, a^{3} - 8 \, a^{2} b + 6 \, a b^{2} - 5 \, b^{3}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {24 \, a^{2} \tan \left (d x + c\right )^{5} - 30 \, a b \tan \left (d x + c\right )^{5} + 33 \, b^{2} \tan \left (d x + c\right )^{5} + 48 \, a^{2} \tan \left (d x + c\right )^{3} - 48 \, a b \tan \left (d x + c\right )^{3} + 40 \, b^{2} \tan \left (d x + c\right )^{3} + 24 \, a^{2} \tan \left (d x + c\right ) - 18 \, a b \tan \left (d x + c\right ) + 15 \, b^{2} \tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )}^{3} b^{3}}}{48 \, d} \]
1/48*(48*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b)))*a^4/(sqrt(a^2 + a*b)*b^4) - 3*(16 *a^3 - 8*a^2*b + 6*a*b^2 - 5*b^3)*(d*x + c)/b^4 - (24*a^2*tan(d*x + c)^5 - 30*a*b*tan(d*x + c)^5 + 33*b^2*tan(d*x + c)^5 + 48*a^2*tan(d*x + c)^3 - 4 8*a*b*tan(d*x + c)^3 + 40*b^2*tan(d*x + c)^3 + 24*a^2*tan(d*x + c) - 18*a* b*tan(d*x + c) + 15*b^2*tan(d*x + c))/((tan(d*x + c)^2 + 1)^3*b^3))/d
Time = 15.31 (sec) , antiderivative size = 2244, normalized size of antiderivative = 13.77 \[ \int \frac {\sin ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\text {Too large to display} \]
(atan(((((tan(c + d*x)*(15*a*b^8 + 768*a^8*b + 512*a^9 + 25*b^9 + 11*a^2*b ^7 - 63*a^3*b^6 - 224*a^4*b^5 - 140*a^5*b^4 + 256*a^7*b^2))/(128*b^6) - (( ((5*a*b^12)/4 + a^2*b^11 + (a^3*b^10)/4 + (5*a^4*b^9)/2 + 2*a^5*b^8)/b^9 - (tan(c + d*x)*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i)*(4096*a*b^10 + 102 4*b^11 + 5120*a^2*b^9 + 2048*a^3*b^8))/(4096*b^10))*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i))/(32*b^4))*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i)*1i) /(32*b^4) + (((tan(c + d*x)*(15*a*b^8 + 768*a^8*b + 512*a^9 + 25*b^9 + 11* a^2*b^7 - 63*a^3*b^6 - 224*a^4*b^5 - 140*a^5*b^4 + 256*a^7*b^2))/(128*b^6) + ((((5*a*b^12)/4 + a^2*b^11 + (a^3*b^10)/4 + (5*a^4*b^9)/2 + 2*a^5*b^8)/ b^9 + (tan(c + d*x)*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i)*(4096*a*b^10 + 1024*b^11 + 5120*a^2*b^9 + 2048*a^3*b^8))/(4096*b^10))*(a*b^2*6i - a^2*b *8i + a^3*16i - b^3*5i))/(32*b^4))*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i )*1i)/(32*b^4))/(((a^10*b)/4 + a^11 + (25*a^4*b^7)/128 - (5*a^5*b^6)/64 + (21*a^6*b^5)/128 - (21*a^7*b^4)/32 - (15*a^8*b^3)/32 - (a^9*b^2)/8)/b^9 - (((tan(c + d*x)*(15*a*b^8 + 768*a^8*b + 512*a^9 + 25*b^9 + 11*a^2*b^7 - 63 *a^3*b^6 - 224*a^4*b^5 - 140*a^5*b^4 + 256*a^7*b^2))/(128*b^6) - ((((5*a*b ^12)/4 + a^2*b^11 + (a^3*b^10)/4 + (5*a^4*b^9)/2 + 2*a^5*b^8)/b^9 - (tan(c + d*x)*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i)*(4096*a*b^10 + 1024*b^11 + 5120*a^2*b^9 + 2048*a^3*b^8))/(4096*b^10))*(a*b^2*6i - a^2*b*8i + a^3*16 i - b^3*5i))/(32*b^4))*(a*b^2*6i - a^2*b*8i + a^3*16i - b^3*5i))/(32*b^...