Integrand size = 21, antiderivative size = 221 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {2 b \left (3 a^2-5 b^2\right ) \csc (c+d x)}{a^6 d}+\frac {\left (a^2-3 b^2\right ) \csc ^2(c+d x)}{a^5 d}+\frac {b \csc ^3(c+d x)}{a^4 d}-\frac {\csc ^4(c+d x)}{4 a^3 d}+\frac {\left (a^4-12 a^2 b^2+15 b^4\right ) \log (\sin (c+d x))}{a^7 d}-\frac {\left (a^4-12 a^2 b^2+15 b^4\right ) \log (a+b \sin (c+d x))}{a^7 d}+\frac {\left (a^2-b^2\right )^2}{2 a^5 d (a+b \sin (c+d x))^2}+\frac {a^4-6 a^2 b^2+5 b^4}{a^6 d (a+b \sin (c+d x))} \]
-2*b*(3*a^2-5*b^2)*csc(d*x+c)/a^6/d+(a^2-3*b^2)*csc(d*x+c)^2/a^5/d+b*csc(d *x+c)^3/a^4/d-1/4*csc(d*x+c)^4/a^3/d+(a^4-12*a^2*b^2+15*b^4)*ln(sin(d*x+c) )/a^7/d-(a^4-12*a^2*b^2+15*b^4)*ln(a+b*sin(d*x+c))/a^7/d+1/2*(a^2-b^2)^2/a ^5/d/(a+b*sin(d*x+c))^2+(a^4-6*a^2*b^2+5*b^4)/a^6/d/(a+b*sin(d*x+c))
Time = 4.83 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.88 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {-8 a b \left (3 a^2-5 b^2\right ) \csc (c+d x)+4 a^2 \left (a^2-3 b^2\right ) \csc ^2(c+d x)+4 a^3 b \csc ^3(c+d x)-a^4 \csc ^4(c+d x)+4 \left (a^4-12 a^2 b^2+15 b^4\right ) \log (\sin (c+d x))-4 \left (a^4-12 a^2 b^2+15 b^4\right ) \log (a+b \sin (c+d x))+\frac {2 \left (a^3-a b^2\right )^2}{(a+b \sin (c+d x))^2}+\frac {4 a \left (a^4-6 a^2 b^2+5 b^4\right )}{a+b \sin (c+d x)}}{4 a^7 d} \]
(-8*a*b*(3*a^2 - 5*b^2)*Csc[c + d*x] + 4*a^2*(a^2 - 3*b^2)*Csc[c + d*x]^2 + 4*a^3*b*Csc[c + d*x]^3 - a^4*Csc[c + d*x]^4 + 4*(a^4 - 12*a^2*b^2 + 15*b ^4)*Log[Sin[c + d*x]] - 4*(a^4 - 12*a^2*b^2 + 15*b^4)*Log[a + b*Sin[c + d* x]] + (2*(a^3 - a*b^2)^2)/(a + b*Sin[c + d*x])^2 + (4*a*(a^4 - 6*a^2*b^2 + 5*b^4))/(a + b*Sin[c + d*x]))/(4*a^7*d)
Time = 0.44 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3200, 522, 2009}
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.
| \(\displaystyle \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (c+d x)^5 (a+b \sin (c+d x))^3}dx\) |
| \(\Big \downarrow \) 3200 |
| \(\displaystyle \frac {\int \frac {\csc ^5(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b^5 (a+b \sin (c+d x))^3}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (\frac {\csc ^5(c+d x)}{a^3 b}-\frac {3 \csc ^4(c+d x)}{a^4}+\frac {2 \left (3 b^2-a^2\right ) \csc ^3(c+d x)}{a^5 b}+\frac {2 \left (3 a^2 b^2-5 b^4\right ) \csc ^2(c+d x)}{a^6 b^2}+\frac {\left (a^4-12 b^2 a^2+15 b^4\right ) \csc (c+d x)}{a^7 b}+\frac {-a^4+12 b^2 a^2-15 b^4}{a^7 (a+b \sin (c+d x))}+\frac {-a^4+6 b^2 a^2-5 b^4}{a^6 (a+b \sin (c+d x))^2}-\frac {\left (a^2-b^2\right )^2}{a^5 (a+b \sin (c+d x))^3}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b \csc ^3(c+d x)}{a^4}-\frac {\csc ^4(c+d x)}{4 a^3}-\frac {2 b \left (3 a^2-5 b^2\right ) \csc (c+d x)}{a^6}+\frac {\left (a^2-b^2\right )^2}{2 a^5 (a+b \sin (c+d x))^2}+\frac {\left (a^2-3 b^2\right ) \csc ^2(c+d x)}{a^5}+\frac {\left (a^4-12 a^2 b^2+15 b^4\right ) \log (b \sin (c+d x))}{a^7}-\frac {\left (a^4-12 a^2 b^2+15 b^4\right ) \log (a+b \sin (c+d x))}{a^7}+\frac {a^4-6 a^2 b^2+5 b^4}{a^6 (a+b \sin (c+d x))}}{d}\) |
((-2*b*(3*a^2 - 5*b^2)*Csc[c + d*x])/a^6 + ((a^2 - 3*b^2)*Csc[c + d*x]^2)/ a^5 + (b*Csc[c + d*x]^3)/a^4 - Csc[c + d*x]^4/(4*a^3) + ((a^4 - 12*a^2*b^2 + 15*b^4)*Log[b*Sin[c + d*x]])/a^7 - ((a^4 - 12*a^2*b^2 + 15*b^4)*Log[a + b*Sin[c + d*x]])/a^7 + (a^2 - b^2)^2/(2*a^5*(a + b*Sin[c + d*x])^2) + (a^ 4 - 6*a^2*b^2 + 5*b^4)/(a^6*(a + b*Sin[c + d*x])))/d
3.2.97.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Time = 10.75 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {-\frac {\left (a^{4}-12 a^{2} b^{2}+15 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{7}}+\frac {a^{4}-6 a^{2} b^{2}+5 b^{4}}{a^{6} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{2 a^{5} \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {1}{4 a^{3} \sin \left (d x +c \right )^{4}}-\frac {-2 a^{2}+6 b^{2}}{2 a^{5} \sin \left (d x +c \right )^{2}}+\frac {\left (a^{4}-12 a^{2} b^{2}+15 b^{4}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{7}}+\frac {b}{a^{4} \sin \left (d x +c \right )^{3}}-\frac {2 b \left (3 a^{2}-5 b^{2}\right )}{a^{6} \sin \left (d x +c \right )}}{d}\) | \(207\) |
| default | \(\frac {-\frac {\left (a^{4}-12 a^{2} b^{2}+15 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{7}}+\frac {a^{4}-6 a^{2} b^{2}+5 b^{4}}{a^{6} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{2 a^{5} \left (a +b \sin \left (d x +c \right )\right )^{2}}-\frac {1}{4 a^{3} \sin \left (d x +c \right )^{4}}-\frac {-2 a^{2}+6 b^{2}}{2 a^{5} \sin \left (d x +c \right )^{2}}+\frac {\left (a^{4}-12 a^{2} b^{2}+15 b^{4}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{7}}+\frac {b}{a^{4} \sin \left (d x +c \right )^{3}}-\frac {2 b \left (3 a^{2}-5 b^{2}\right )}{a^{6} \sin \left (d x +c \right )}}{d}\) | \(207\) |
| risch | \(\frac {2 i \left (-12 b^{3} a^{2} {\mathrm e}^{11 i \left (d x +c \right )}+30 b \,a^{4} {\mathrm e}^{5 i \left (d x +c \right )}+11 b \,a^{4} {\mathrm e}^{9 i \left (d x +c \right )}-60 b^{3} a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+40 b^{3} a^{2} {\mathrm e}^{9 i \left (d x +c \right )}+60 b^{3} a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-b \,a^{4} {\mathrm e}^{i \left (d x +c \right )}+26 i a^{5} {\mathrm e}^{6 i \left (d x +c \right )}-20 i a^{5} {\mathrm e}^{8 i \left (d x +c \right )}-20 i a^{5} {\mathrm e}^{4 i \left (d x +c \right )}+3 i a^{5} {\mathrm e}^{10 i \left (d x +c \right )}+3 i a^{5} {\mathrm e}^{2 i \left (d x +c \right )}-180 i b^{4} a \,{\mathrm e}^{4 i \left (d x +c \right )}-36 i a^{3} b^{2} {\mathrm e}^{10 i \left (d x +c \right )}+45 i b^{4} a \,{\mathrm e}^{10 i \left (d x +c \right )}-180 i b^{4} a \,{\mathrm e}^{8 i \left (d x +c \right )}-36 i a^{3} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+45 i b^{4} a \,{\mathrm e}^{2 i \left (d x +c \right )}-236 i a^{3} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+270 i b^{4} a \,{\mathrm e}^{6 i \left (d x +c \right )}+154 i a^{3} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+154 i a^{3} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-150 b^{5} {\mathrm e}^{5 i \left (d x +c \right )}+75 b^{5} {\mathrm e}^{3 i \left (d x +c \right )}-75 b^{5} {\mathrm e}^{9 i \left (d x +c \right )}+150 b^{5} {\mathrm e}^{7 i \left (d x +c \right )}+15 b^{5} {\mathrm e}^{11 i \left (d x +c \right )}-15 b^{5} {\mathrm e}^{i \left (d x +c \right )}+12 b^{3} a^{2} {\mathrm e}^{i \left (d x +c \right )}+b \,a^{4} {\mathrm e}^{11 i \left (d x +c \right )}-40 b^{3} a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-11 b \,a^{4} {\mathrm e}^{3 i \left (d x +c \right )}-30 b \,a^{4} {\mathrm e}^{7 i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right )^{2} d \,a^{6}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{a^{3} d}+\frac {12 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) b^{2}}{a^{5} d}-\frac {15 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) b^{4}}{a^{7} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{3} d}-\frac {12 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{a^{5} d}+\frac {15 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{4}}{a^{7} d}\) | \(754\) |
1/d*(-(a^4-12*a^2*b^2+15*b^4)/a^7*ln(a+b*sin(d*x+c))+(a^4-6*a^2*b^2+5*b^4) /a^6/(a+b*sin(d*x+c))+1/2*(a^4-2*a^2*b^2+b^4)/a^5/(a+b*sin(d*x+c))^2-1/4/a ^3/sin(d*x+c)^4-1/2*(-2*a^2+6*b^2)/a^5/sin(d*x+c)^2+(a^4-12*a^2*b^2+15*b^4 )/a^7*ln(sin(d*x+c))+1/a^4*b/sin(d*x+c)^3-2*b*(3*a^2-5*b^2)/a^6/sin(d*x+c) )
Leaf count of result is larger than twice the leaf count of optimal. 754 vs. \(2 (217) = 434\).
Time = 0.35 (sec) , antiderivative size = 754, normalized size of antiderivative = 3.41 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {9 \, a^{6} - 77 \, a^{4} b^{2} + 90 \, a^{2} b^{4} + 6 \, {\left (a^{6} - 12 \, a^{4} b^{2} + 15 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{4} - {\left (16 \, a^{6} - 149 \, a^{4} b^{2} + 180 \, a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (a^{4} b^{2} - 12 \, a^{2} b^{4} + 15 \, b^{6}\right )} \cos \left (d x + c\right )^{6} - a^{6} + 11 \, a^{4} b^{2} - 3 \, a^{2} b^{4} - 15 \, b^{6} - {\left (a^{6} - 9 \, a^{4} b^{2} - 21 \, a^{2} b^{4} + 45 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, a^{6} - 21 \, a^{4} b^{2} - 6 \, a^{2} b^{4} + 45 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} b - 12 \, a^{3} b^{3} + 15 \, a b^{5} + {\left (a^{5} b - 12 \, a^{3} b^{3} + 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} b - 12 \, a^{3} b^{3} + 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 4 \, {\left ({\left (a^{4} b^{2} - 12 \, a^{2} b^{4} + 15 \, b^{6}\right )} \cos \left (d x + c\right )^{6} - a^{6} + 11 \, a^{4} b^{2} - 3 \, a^{2} b^{4} - 15 \, b^{6} - {\left (a^{6} - 9 \, a^{4} b^{2} - 21 \, a^{2} b^{4} + 45 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, a^{6} - 21 \, a^{4} b^{2} - 6 \, a^{2} b^{4} + 45 \, b^{6}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{5} b - 12 \, a^{3} b^{3} + 15 \, a b^{5} + {\left (a^{5} b - 12 \, a^{3} b^{3} + 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} b - 12 \, a^{3} b^{3} + 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 2 \, {\left (5 \, a^{5} b + 14 \, a^{3} b^{3} - 30 \, a b^{5} - 2 \, {\left (a^{5} b - 12 \, a^{3} b^{3} + 15 \, a b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (2 \, a^{5} b + 19 \, a^{3} b^{3} - 30 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{4 \, {\left (a^{7} b^{2} d \cos \left (d x + c\right )^{6} - {\left (a^{9} + 3 \, a^{7} b^{2}\right )} d \cos \left (d x + c\right )^{4} + {\left (2 \, a^{9} + 3 \, a^{7} b^{2}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{9} + a^{7} b^{2}\right )} d - 2 \, {\left (a^{8} b d \cos \left (d x + c\right )^{4} - 2 \, a^{8} b d \cos \left (d x + c\right )^{2} + a^{8} b d\right )} \sin \left (d x + c\right )\right )}} \]
-1/4*(9*a^6 - 77*a^4*b^2 + 90*a^2*b^4 + 6*(a^6 - 12*a^4*b^2 + 15*a^2*b^4)* cos(d*x + c)^4 - (16*a^6 - 149*a^4*b^2 + 180*a^2*b^4)*cos(d*x + c)^2 + 4*( (a^4*b^2 - 12*a^2*b^4 + 15*b^6)*cos(d*x + c)^6 - a^6 + 11*a^4*b^2 - 3*a^2* b^4 - 15*b^6 - (a^6 - 9*a^4*b^2 - 21*a^2*b^4 + 45*b^6)*cos(d*x + c)^4 + (2 *a^6 - 21*a^4*b^2 - 6*a^2*b^4 + 45*b^6)*cos(d*x + c)^2 - 2*(a^5*b - 12*a^3 *b^3 + 15*a*b^5 + (a^5*b - 12*a^3*b^3 + 15*a*b^5)*cos(d*x + c)^4 - 2*(a^5* b - 12*a^3*b^3 + 15*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(b*sin(d*x + c ) + a) - 4*((a^4*b^2 - 12*a^2*b^4 + 15*b^6)*cos(d*x + c)^6 - a^6 + 11*a^4* b^2 - 3*a^2*b^4 - 15*b^6 - (a^6 - 9*a^4*b^2 - 21*a^2*b^4 + 45*b^6)*cos(d*x + c)^4 + (2*a^6 - 21*a^4*b^2 - 6*a^2*b^4 + 45*b^6)*cos(d*x + c)^2 - 2*(a^ 5*b - 12*a^3*b^3 + 15*a*b^5 + (a^5*b - 12*a^3*b^3 + 15*a*b^5)*cos(d*x + c) ^4 - 2*(a^5*b - 12*a^3*b^3 + 15*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))*log(- 1/2*sin(d*x + c)) - 2*(5*a^5*b + 14*a^3*b^3 - 30*a*b^5 - 2*(a^5*b - 12*a^3 *b^3 + 15*a*b^5)*cos(d*x + c)^4 - 2*(2*a^5*b + 19*a^3*b^3 - 30*a*b^5)*cos( d*x + c)^2)*sin(d*x + c))/(a^7*b^2*d*cos(d*x + c)^6 - (a^9 + 3*a^7*b^2)*d* cos(d*x + c)^4 + (2*a^9 + 3*a^7*b^2)*d*cos(d*x + c)^2 - (a^9 + a^7*b^2)*d - 2*(a^8*b*d*cos(d*x + c)^4 - 2*a^8*b*d*cos(d*x + c)^2 + a^8*b*d)*sin(d*x + c))
\[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int \frac {\cot ^{5}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{3}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.07 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {2 \, a^{4} b \sin \left (d x + c\right ) + 4 \, {\left (a^{4} b - 12 \, a^{2} b^{3} + 15 \, b^{5}\right )} \sin \left (d x + c\right )^{5} - a^{5} + 6 \, {\left (a^{5} - 12 \, a^{3} b^{2} + 15 \, a b^{4}\right )} \sin \left (d x + c\right )^{4} - 4 \, {\left (4 \, a^{4} b - 5 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )^{3} + {\left (4 \, a^{5} - 5 \, a^{3} b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{6} b^{2} \sin \left (d x + c\right )^{6} + 2 \, a^{7} b \sin \left (d x + c\right )^{5} + a^{8} \sin \left (d x + c\right )^{4}} - \frac {4 \, {\left (a^{4} - 12 \, a^{2} b^{2} + 15 \, b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{7}} + \frac {4 \, {\left (a^{4} - 12 \, a^{2} b^{2} + 15 \, b^{4}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{7}}}{4 \, d} \]
1/4*((2*a^4*b*sin(d*x + c) + 4*(a^4*b - 12*a^2*b^3 + 15*b^5)*sin(d*x + c)^ 5 - a^5 + 6*(a^5 - 12*a^3*b^2 + 15*a*b^4)*sin(d*x + c)^4 - 4*(4*a^4*b - 5* a^2*b^3)*sin(d*x + c)^3 + (4*a^5 - 5*a^3*b^2)*sin(d*x + c)^2)/(a^6*b^2*sin (d*x + c)^6 + 2*a^7*b*sin(d*x + c)^5 + a^8*sin(d*x + c)^4) - 4*(a^4 - 12*a ^2*b^2 + 15*b^4)*log(b*sin(d*x + c) + a)/a^7 + 4*(a^4 - 12*a^2*b^2 + 15*b^ 4)*log(sin(d*x + c))/a^7)/d
Time = 0.44 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.48 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {\frac {12 \, {\left (a^{4} - 12 \, a^{2} b^{2} + 15 \, b^{4}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{7}} - \frac {12 \, {\left (a^{4} b - 12 \, a^{2} b^{3} + 15 \, b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{7} b} + \frac {6 \, {\left (3 \, a^{4} b^{2} \sin \left (d x + c\right )^{2} - 36 \, a^{2} b^{4} \sin \left (d x + c\right )^{2} + 45 \, b^{6} \sin \left (d x + c\right )^{2} + 8 \, a^{5} b \sin \left (d x + c\right ) - 84 \, a^{3} b^{3} \sin \left (d x + c\right ) + 100 \, a b^{5} \sin \left (d x + c\right ) + 6 \, a^{6} - 50 \, a^{4} b^{2} + 56 \, a^{2} b^{4}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2} a^{7}} - \frac {25 \, a^{4} \sin \left (d x + c\right )^{4} - 300 \, a^{2} b^{2} \sin \left (d x + c\right )^{4} + 375 \, b^{4} \sin \left (d x + c\right )^{4} + 72 \, a^{3} b \sin \left (d x + c\right )^{3} - 120 \, a b^{3} \sin \left (d x + c\right )^{3} - 12 \, a^{4} \sin \left (d x + c\right )^{2} + 36 \, a^{2} b^{2} \sin \left (d x + c\right )^{2} - 12 \, a^{3} b \sin \left (d x + c\right ) + 3 \, a^{4}}{a^{7} \sin \left (d x + c\right )^{4}}}{12 \, d} \]
1/12*(12*(a^4 - 12*a^2*b^2 + 15*b^4)*log(abs(sin(d*x + c)))/a^7 - 12*(a^4* b - 12*a^2*b^3 + 15*b^5)*log(abs(b*sin(d*x + c) + a))/(a^7*b) + 6*(3*a^4*b ^2*sin(d*x + c)^2 - 36*a^2*b^4*sin(d*x + c)^2 + 45*b^6*sin(d*x + c)^2 + 8* a^5*b*sin(d*x + c) - 84*a^3*b^3*sin(d*x + c) + 100*a*b^5*sin(d*x + c) + 6* a^6 - 50*a^4*b^2 + 56*a^2*b^4)/((b*sin(d*x + c) + a)^2*a^7) - (25*a^4*sin( d*x + c)^4 - 300*a^2*b^2*sin(d*x + c)^4 + 375*b^4*sin(d*x + c)^4 + 72*a^3* b*sin(d*x + c)^3 - 120*a*b^3*sin(d*x + c)^3 - 12*a^4*sin(d*x + c)^2 + 36*a ^2*b^2*sin(d*x + c)^2 - 12*a^3*b*sin(d*x + c) + 3*a^4)/(a^7*sin(d*x + c)^4 ))/d
Time = 6.69 (sec) , antiderivative size = 563, normalized size of antiderivative = 2.55 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {23\,a^5}{4}-172\,a^3\,b^2+272\,a\,b^4\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (27\,a^4\,b-40\,a^2\,b^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (-134\,a^4\,b+200\,a^2\,b^3+128\,b^5\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (106\,a^4\,b-336\,a^2\,b^3+192\,b^5\right )-\frac {a^5}{4}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {5\,a^5}{2}-5\,a^3\,b^2\right )+a^4\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (3\,a^6-276\,a^4\,b^2+768\,a^2\,b^4-352\,b^6\right )}{a}}{d\,\left (16\,a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+16\,a^8\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (32\,a^8+64\,a^6\,b^2\right )+64\,a^7\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+64\,a^7\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^3\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3\,\left (a^2+4\,b^2\right )}{32\,a^5}+\frac {3}{32\,a^3}-\frac {9\,b^2}{8\,a^5}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {6\,b\,\left (\frac {3\,\left (a^2+4\,b^2\right )}{16\,a^5}+\frac {3}{16\,a^3}-\frac {9\,b^2}{4\,a^5}\right )}{a}-\frac {192\,a^2\,b+128\,b^3}{256\,a^6}+\frac {9\,b\,\left (a^2+4\,b^2\right )}{8\,a^6}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^4-12\,a^2\,b^2+15\,b^4\right )}{a^7\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8\,a^4\,d}-\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^4-12\,a^2\,b^2+15\,b^4\right )}{a^7\,d} \]
(tan(c/2 + (d*x)/2)^4*(272*a*b^4 + (23*a^5)/4 - 172*a^3*b^2) - tan(c/2 + ( d*x)/2)^3*(27*a^4*b - 40*a^2*b^3) + tan(c/2 + (d*x)/2)^5*(128*b^5 - 134*a^ 4*b + 200*a^2*b^3) - tan(c/2 + (d*x)/2)^7*(106*a^4*b + 192*b^5 - 336*a^2*b ^3) - a^5/4 + tan(c/2 + (d*x)/2)^2*((5*a^5)/2 - 5*a^3*b^2) + a^4*b*tan(c/2 + (d*x)/2) + (tan(c/2 + (d*x)/2)^6*(3*a^6 - 352*b^6 + 768*a^2*b^4 - 276*a ^4*b^2))/a)/(d*(16*a^8*tan(c/2 + (d*x)/2)^4 + 16*a^8*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^6*(32*a^8 + 64*a^6*b^2) + 64*a^7*b*tan(c/2 + (d*x)/2) ^5 + 64*a^7*b*tan(c/2 + (d*x)/2)^7)) - tan(c/2 + (d*x)/2)^4/(64*a^3*d) + ( tan(c/2 + (d*x)/2)^2*((3*(a^2 + 4*b^2))/(32*a^5) + 3/(32*a^3) - (9*b^2)/(8 *a^5)))/d - (tan(c/2 + (d*x)/2)*((6*b*((3*(a^2 + 4*b^2))/(16*a^5) + 3/(16* a^3) - (9*b^2)/(4*a^5)))/a - (192*a^2*b + 128*b^3)/(256*a^6) + (9*b*(a^2 + 4*b^2))/(8*a^6)))/d + (log(tan(c/2 + (d*x)/2))*(a^4 + 15*b^4 - 12*a^2*b^2 ))/(a^7*d) + (b*tan(c/2 + (d*x)/2)^3)/(8*a^4*d) - (log(a + 2*b*tan(c/2 + ( d*x)/2) + a*tan(c/2 + (d*x)/2)^2)*(a^4 + 15*b^4 - 12*a^2*b^2))/(a^7*d)