Integrand size = 23, antiderivative size = 117 \[ \int \frac {\cot ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {(a+b)^{7/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{9/2} d}+\frac {(a+b)^3 \cot (c+d x)}{a^4 d}-\frac {(a+b)^2 \cot ^3(c+d x)}{3 a^3 d}+\frac {(a+b) \cot ^5(c+d x)}{5 a^2 d}-\frac {\cot ^7(c+d x)}{7 a d} \]
(a+b)^(7/2)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/a^(9/2)/d+(a+b)^3*cot(d *x+c)/a^4/d-1/3*(a+b)^2*cot(d*x+c)^3/a^3/d+1/5*(a+b)*cot(d*x+c)^5/a^2/d-1/ 7*cot(d*x+c)^7/a/d
Time = 3.25 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.15 \[ \int \frac {\cot ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {(a+b)^{7/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{9/2} d}+\frac {\cot (c+d x) \left (176 a^3+406 a^2 b+350 a b^2+105 b^3-a \left (122 a^2+112 a b+35 b^2\right ) \csc ^2(c+d x)+3 a^2 (22 a+7 b) \csc ^4(c+d x)-15 a^3 \csc ^6(c+d x)\right )}{105 a^4 d} \]
((a + b)^(7/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(9/2)*d) + ( Cot[c + d*x]*(176*a^3 + 406*a^2*b + 350*a*b^2 + 105*b^3 - a*(122*a^2 + 112 *a*b + 35*b^2)*Csc[c + d*x]^2 + 3*a^2*(22*a + 7*b)*Csc[c + d*x]^4 - 15*a^3 *Csc[c + d*x]^6))/(105*a^4*d)
Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3674, 264, 264, 264, 264, 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.
\(\displaystyle \int \frac {\cot ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^8 \left (a+b \sin (c+d x)^2\right )}dx\) |
\(\Big \downarrow \) 3674 |
\(\displaystyle \frac {\int \frac {\cot ^8(c+d x)}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {-\frac {(a+b) \int \frac {\cot ^6(c+d x)}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{a}-\frac {\cot ^7(c+d x)}{7 a}}{d}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {-\frac {(a+b) \left (-\frac {(a+b) \int \frac {\cot ^4(c+d x)}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{a}-\frac {\cot ^5(c+d x)}{5 a}\right )}{a}-\frac {\cot ^7(c+d x)}{7 a}}{d}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {-\frac {(a+b) \left (-\frac {(a+b) \left (-\frac {(a+b) \int \frac {\cot ^2(c+d x)}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{a}-\frac {\cot ^3(c+d x)}{3 a}\right )}{a}-\frac {\cot ^5(c+d x)}{5 a}\right )}{a}-\frac {\cot ^7(c+d x)}{7 a}}{d}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {-\frac {(a+b) \left (-\frac {(a+b) \left (-\frac {(a+b) \left (-\frac {(a+b) \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{a}-\frac {\cot (c+d x)}{a}\right )}{a}-\frac {\cot ^3(c+d x)}{3 a}\right )}{a}-\frac {\cot ^5(c+d x)}{5 a}\right )}{a}-\frac {\cot ^7(c+d x)}{7 a}}{d}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {-\frac {(a+b) \left (-\frac {(a+b) \left (-\frac {(a+b) \left (-\frac {\sqrt {a+b} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\cot (c+d x)}{a}\right )}{a}-\frac {\cot ^3(c+d x)}{3 a}\right )}{a}-\frac {\cot ^5(c+d x)}{5 a}\right )}{a}-\frac {\cot ^7(c+d x)}{7 a}}{d}\) |
(-1/7*Cot[c + d*x]^7/a - ((a + b)*(-1/5*Cot[c + d*x]^5/a - ((a + b)*(-1/3* Cot[c + d*x]^3/a - ((a + b)*(-((Sqrt[a + b]*ArcTan[(Sqrt[a + b]*Tan[c + d* x])/Sqrt[a]])/a^(3/2)) - Cot[c + d*x]/a))/a))/a))/a)/d
3.5.56.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*((d_.)*tan[(e_.) + (f_.) *(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[f f/f Subst[Int[(d*ff*x)^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(p + 1) ), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m}, x] && IntegerQ [p]
Time = 10.11 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.32
method | result | size |
derivativedivides | \(\frac {-\frac {-a -b}{5 a^{2} \tan \left (d x +c \right )^{5}}-\frac {-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}}{a^{4} \tan \left (d x +c \right )}-\frac {1}{7 a \tan \left (d x +c \right )^{7}}-\frac {a^{2}+2 a b +b^{2}}{3 a^{3} \tan \left (d x +c \right )^{3}}+\frac {\left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{4} \sqrt {a \left (a +b \right )}}}{d}\) | \(155\) |
default | \(\frac {-\frac {-a -b}{5 a^{2} \tan \left (d x +c \right )^{5}}-\frac {-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}}{a^{4} \tan \left (d x +c \right )}-\frac {1}{7 a \tan \left (d x +c \right )^{7}}-\frac {a^{2}+2 a b +b^{2}}{3 a^{3} \tan \left (d x +c \right )^{3}}+\frac {\left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{4} \sqrt {a \left (a +b \right )}}}{d}\) | \(155\) |
risch | \(\frac {2 i \left (406 a^{2} b +350 a \,b^{2}+105 b^{3}+176 a^{3}+420 a \,b^{2} {\mathrm e}^{12 i \left (d x +c \right )}+5040 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-2212 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}+630 a^{2} b \,{\mathrm e}^{12 i \left (d x +c \right )}-2030 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+5586 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-2100 b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-3080 a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-630 b^{3} {\mathrm e}^{10 i \left (d x +c \right )}+3080 a^{3} {\mathrm e}^{8 i \left (d x +c \right )}-1260 a^{3} {\mathrm e}^{10 i \left (d x +c \right )}-630 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+105 b^{3} {\mathrm e}^{12 i \left (d x +c \right )}-812 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2436 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+1575 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+420 a^{3} {\mathrm e}^{12 i \left (d x +c \right )}+1575 b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-2940 a^{2} b \,{\mathrm e}^{10 i \left (d x +c \right )}-7840 a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}+6370 a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+5390 a \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-2310 a \,b^{2} {\mathrm e}^{10 i \left (d x +c \right )}-6860 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}\right )}{105 d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{7}}+\frac {\sqrt {-a \left (a +b \right )}\, \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 a^{2} d}+\frac {3 \sqrt {-a \left (a +b \right )}\, \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 ) b}{2 a^{3} d}+\frac {3 \sqrt {-a \left (a +b \right )}\, \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 ) b^{2}}{2 a^{4} d}+\frac {\sqrt {-a \left (a +b \right )}\, \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 ) b^{3}}{2 a^{5} d}-\frac {\sqrt {-a \left (a +b \right )}\, \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 a^{2} d}-\frac {3 \sqrt {-a \left (a +b \right )}\, \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 ) b}{2 a^{3} d}-\frac {3 \sqrt {-a \left (a +b \right )}\, \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 ) b^{2}}{2 a^{4} d}-\frac {\sqrt {-a \left (a +b \right )}\, \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 ) b^{3}}{2 a^{5} d}\) | \(797\) |
1/d*(-1/5*(-a-b)/a^2/tan(d*x+c)^5-1/a^4*(-a^3-3*a^2*b-3*a*b^2-b^3)/tan(d*x +c)-1/7/a/tan(d*x+c)^7-1/3/a^3*(a^2+2*a*b+b^2)/tan(d*x+c)^3+(a^4+4*a^3*b+6 *a^2*b^2+4*a*b^3+b^4)/a^4/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b) )^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 367 vs. \(2 (103) = 206\).
Time = 0.34 (sec) , antiderivative size = 834, normalized size of antiderivative = 7.13 \[ \int \frac {\cot ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [\frac {4 \, {\left (176 \, a^{3} + 406 \, a^{2} b + 350 \, a b^{2} + 105 \, b^{3}\right )} \cos \left (d x + c\right )^{7} - 28 \, {\left (58 \, a^{3} + 158 \, a^{2} b + 145 \, a b^{2} + 45 \, b^{3}\right )} \cos \left (d x + c\right )^{5} + 140 \, {\left (10 \, a^{3} + 29 \, a^{2} b + 28 \, a b^{2} + 9 \, b^{3}\right )} \cos \left (d x + c\right )^{3} + 105 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3} + 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-\frac {a + b}{a}} \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} + a b\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + a b\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a + b}{a}} \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 ) \sin \left (d x + c\right ) - 420 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )}{420 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} - 3 \, a^{4} d \cos \left (d x + c\right )^{4} + 3 \, a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d\right )} \sin \left (d x + c\right )}, \frac {2 \, {\left (176 \, a^{3} + 406 \, a^{2} b + 350 \, a b^{2} + 105 \, b^{3}\right )} \cos \left (d x + c\right )^{7} - 14 \, {\left (58 \, a^{3} + 158 \, a^{2} b + 145 \, a b^{2} + 45 \, b^{3}\right )} \cos \left (d x + c\right )^{5} + 70 \, {\left (10 \, a^{3} + 29 \, a^{2} b + 28 \, a b^{2} + 9 \, b^{3}\right )} \cos \left (d x + c\right )^{3} - 105 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3} + 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a + b}{a}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a + b}{a}}}{2 \, {\left (a + b\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 210 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{6} - 3 \, a^{4} d \cos \left (d x + c\right )^{4} + 3 \, a^{4} d \cos \left (d x + c\right )^{2} - a^{4} d\right )} \sin \left (d x + c\right )}\right ] \]
[1/420*(4*(176*a^3 + 406*a^2*b + 350*a*b^2 + 105*b^3)*cos(d*x + c)^7 - 28* (58*a^3 + 158*a^2*b + 145*a*b^2 + 45*b^3)*cos(d*x + c)^5 + 140*(10*a^3 + 2 9*a^2*b + 28*a*b^2 + 9*b^3)*cos(d*x + c)^3 + 105*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c)^6 - 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c)^4 - a^3 - 3*a^2*b - 3*a*b^2 - b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d* x + c)^2)*sqrt(-(a + b)/a)*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 + a*b)*cos(d*x + c)^3 - (a ^2 + a*b)*cos(d*x + c))*sqrt(-(a + b)/a)*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))* sin(d*x + c) - 420*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c))/((a^4*d*c os(d*x + c)^6 - 3*a^4*d*cos(d*x + c)^4 + 3*a^4*d*cos(d*x + c)^2 - a^4*d)*s in(d*x + c)), 1/210*(2*(176*a^3 + 406*a^2*b + 350*a*b^2 + 105*b^3)*cos(d*x + c)^7 - 14*(58*a^3 + 158*a^2*b + 145*a*b^2 + 45*b^3)*cos(d*x + c)^5 + 70 *(10*a^3 + 29*a^2*b + 28*a*b^2 + 9*b^3)*cos(d*x + c)^3 - 105*((a^3 + 3*a^2 *b + 3*a*b^2 + b^3)*cos(d*x + c)^6 - 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos (d*x + c)^4 - a^3 - 3*a^2*b - 3*a*b^2 - b^3 + 3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c)^2)*sqrt((a + b)/a)*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt((a + b)/a)/((a + b)*cos(d*x + c)*sin(d*x + c)))*sin(d*x + c ) - 210*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(d*x + c))/((a^4*d*cos(d*x + c) ^6 - 3*a^4*d*cos(d*x + c)^4 + 3*a^4*d*cos(d*x + c)^2 - a^4*d)*sin(d*x +...
\[ \int \frac {\cot ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\cot ^{8}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \]
Time = 0.33 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.32 \[ \int \frac {\cot ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {105 \, {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a^{4}} + \frac {105 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (d x + c\right )^{6} - 35 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \tan \left (d x + c\right )^{4} - 15 \, a^{3} + 21 \, {\left (a^{3} + a^{2} b\right )} \tan \left (d x + c\right )^{2}}{a^{4} \tan \left (d x + c\right )^{7}}}{105 \, d} \]
1/105*(105*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*arctan((a + b)*tan( d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)*a)*a^4) + (105*(a^3 + 3*a^2*b + 3* a*b^2 + b^3)*tan(d*x + c)^6 - 35*(a^3 + 2*a^2*b + a*b^2)*tan(d*x + c)^4 - 15*a^3 + 21*(a^3 + a^2*b)*tan(d*x + c)^2)/(a^4*tan(d*x + c)^7))/d
Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (103) = 206\).
Time = 0.59 (sec) , antiderivative size = 238, normalized size of antiderivative = 2.03 \[ \int \frac {\cot ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {105 \, {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )}}{\sqrt {a^{2} + a b} a^{4}} + \frac {105 \, a^{3} \tan \left (d x + c\right )^{6} + 315 \, a^{2} b \tan \left (d x + c\right )^{6} + 315 \, a b^{2} \tan \left (d x + c\right )^{6} + 105 \, b^{3} \tan \left (d x + c\right )^{6} - 35 \, a^{3} \tan \left (d x + c\right )^{4} - 70 \, a^{2} b \tan \left (d x + c\right )^{4} - 35 \, a b^{2} \tan \left (d x + c\right )^{4} + 21 \, a^{3} \tan \left (d x + c\right )^{2} + 21 \, a^{2} b \tan \left (d x + c\right )^{2} - 15 \, a^{3}}{a^{4} \tan \left (d x + c\right )^{7}}}{105 \, d} \]
1/105*(105*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*(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)))/(sqrt(a^2 + a*b)*a^4) + (105*a^3*tan(d*x + c)^6 + 315*a^2*b*t an(d*x + c)^6 + 315*a*b^2*tan(d*x + c)^6 + 105*b^3*tan(d*x + c)^6 - 35*a^3 *tan(d*x + c)^4 - 70*a^2*b*tan(d*x + c)^4 - 35*a*b^2*tan(d*x + c)^4 + 21*a ^3*tan(d*x + c)^2 + 21*a^2*b*tan(d*x + c)^2 - 15*a^3)/(a^4*tan(d*x + c)^7) )/d
Time = 16.78 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.85 \[ \int \frac {\cot ^8(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\sqrt {a+b}}{\sqrt {a}}\right )\,{\left (a+b\right )}^{7/2}}{a^{9/2}\,d}-\frac {\frac {1}{7\,a}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a+b\right )}{5\,a^2}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^4\,{\left (a+b\right )}^2}{3\,a^3}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^6\,{\left (a+b\right )}^3}{a^4}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^7} \]