Integrand size = 21, antiderivative size = 188 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {4 b \left (a^2-b^2\right ) \csc (c+d x)}{a^5 d}+\frac {\left (2 a^2-3 b^2\right ) \csc ^2(c+d x)}{2 a^4 d}+\frac {2 b \csc ^3(c+d x)}{3 a^3 d}-\frac {\csc ^4(c+d x)}{4 a^2 d}+\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \log (\sin (c+d x))}{a^6 d}-\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \log (a+b \sin (c+d x))}{a^6 d}+\frac {\left (a^2-b^2\right )^2}{a^5 d (a+b \sin (c+d x))} \] Output:
-4*b*(a^2-b^2)*csc(d*x+c)/a^5/d+1/2*(2*a^2-3*b^2)*csc(d*x+c)^2/a^4/d+2/3*b *csc(d*x+c)^3/a^3/d-1/4*csc(d*x+c)^4/a^2/d+(a^4-6*a^2*b^2+5*b^4)*ln(sin(d* x+c))/a^6/d-(a^4-6*a^2*b^2+5*b^4)*ln(a+b*sin(d*x+c))/a^6/d+(a^2-b^2)^2/a^5 /d/(a+b*sin(d*x+c))
Time = 6.11 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.99 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {4 (a-b) b (a+b) \csc (c+d x)}{a^5 d}+\frac {\left (2 a^2-3 b^2\right ) \csc ^2(c+d x)}{2 a^4 d}+\frac {2 b \csc ^3(c+d x)}{3 a^3 d}-\frac {\csc ^4(c+d x)}{4 a^2 d}+\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \log (\sin (c+d x))}{a^6 d}-\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \log (a+b \sin (c+d x))}{a^6 d}+\frac {\left (a^2-b^2\right )^2}{a^5 d (a+b \sin (c+d x))} \] Input:
Integrate[Cot[c + d*x]^5/(a + b*Sin[c + d*x])^2,x]
Output:
(-4*(a - b)*b*(a + b)*Csc[c + d*x])/(a^5*d) + ((2*a^2 - 3*b^2)*Csc[c + d*x ]^2)/(2*a^4*d) + (2*b*Csc[c + d*x]^3)/(3*a^3*d) - Csc[c + d*x]^4/(4*a^2*d) + ((a^4 - 6*a^2*b^2 + 5*b^4)*Log[Sin[c + d*x]])/(a^6*d) - ((a^4 - 6*a^2*b ^2 + 5*b^4)*Log[a + b*Sin[c + d*x]])/(a^6*d) + (a^2 - b^2)^2/(a^5*d*(a + b *Sin[c + d*x]))
Time = 0.41 (sec) , antiderivative size = 173, 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.
| \(\displaystyle \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (c+d x)^5 (a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3200 |
| \(\displaystyle \frac {\int \frac {\csc ^5(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b^5 (a+b \sin (c+d x))^2}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (\frac {\csc ^5(c+d x)}{a^2 b}-\frac {2 \csc ^4(c+d x)}{a^3}+\frac {\left (3 b^4-2 a^2 b^2\right ) \csc ^3(c+d x)}{a^4 b^3}+\frac {4 \left (a^2-b^2\right ) \csc ^2(c+d x)}{a^5}+\frac {\left (a^4-6 b^2 a^2+5 b^4\right ) \csc (c+d x)}{a^6 b}+\frac {-a^4+6 b^2 a^2-5 b^4}{a^6 (a+b \sin (c+d x))}-\frac {\left (a^2-b^2\right )^2}{a^5 (a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {2 b \csc ^3(c+d x)}{3 a^3}-\frac {\csc ^4(c+d x)}{4 a^2}+\frac {\left (a^2-b^2\right )^2}{a^5 (a+b \sin (c+d x))}-\frac {4 b \left (a^2-b^2\right ) \csc (c+d x)}{a^5}+\frac {\left (2 a^2-3 b^2\right ) \csc ^2(c+d x)}{2 a^4}+\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \log (b \sin (c+d x))}{a^6}-\frac {\left (a^4-6 a^2 b^2+5 b^4\right ) \log (a+b \sin (c+d x))}{a^6}}{d}\) |
Input:
Int[Cot[c + d*x]^5/(a + b*Sin[c + d*x])^2,x]
Output:
((-4*b*(a^2 - b^2)*Csc[c + d*x])/a^5 + ((2*a^2 - 3*b^2)*Csc[c + d*x]^2)/(2 *a^4) + (2*b*Csc[c + d*x]^3)/(3*a^3) - Csc[c + d*x]^4/(4*a^2) + ((a^4 - 6* a^2*b^2 + 5*b^4)*Log[b*Sin[c + d*x]])/a^6 - ((a^4 - 6*a^2*b^2 + 5*b^4)*Log [a + b*Sin[c + d*x]])/a^6 + (a^2 - b^2)^2/(a^5*(a + b*Sin[c + d*x])))/d
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 = 3.91 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.91
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{4 a^{2} \sin \left (d x +c \right )^{4}}-\frac {-2 a^{2}+3 b^{2}}{2 a^{4} \sin \left (d x +c \right )^{2}}+\frac {\left (a^{4}-6 b^{2} a^{2}+5 b^{4}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{6}}+\frac {2 b}{3 a^{3} \sin \left (d x +c \right )^{3}}-\frac {4 b \left (a^{2}-b^{2}\right )}{a^{5} \sin \left (d x +c \right )}-\frac {\left (a^{4}-6 b^{2} a^{2}+5 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{6}}+\frac {a^{4}-2 b^{2} a^{2}+b^{4}}{a^{5} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(172\) |
| default | \(\frac {-\frac {1}{4 a^{2} \sin \left (d x +c \right )^{4}}-\frac {-2 a^{2}+3 b^{2}}{2 a^{4} \sin \left (d x +c \right )^{2}}+\frac {\left (a^{4}-6 b^{2} a^{2}+5 b^{4}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{6}}+\frac {2 b}{3 a^{3} \sin \left (d x +c \right )^{3}}-\frac {4 b \left (a^{2}-b^{2}\right )}{a^{5} \sin \left (d x +c \right )}-\frac {\left (a^{4}-6 b^{2} a^{2}+5 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{6}}+\frac {a^{4}-2 b^{2} a^{2}+b^{4}}{a^{5} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(172\) |
| risch | \(\frac {2 i \left (-44 i a^{3} b \,{\mathrm e}^{4 i \left (d x +c \right )}-45 i a \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+15 i a \,b^{3} {\mathrm e}^{8 i \left (d x +c \right )}+44 i a^{3} b \,{\mathrm e}^{6 i \left (d x +c \right )}-15 i a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-18 i a^{3} b \,{\mathrm e}^{8 i \left (d x +c \right )}+45 i a \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+18 i a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}+82 a^{2} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )} b^{4}+3 a^{4} {\mathrm e}^{9 i \left (d x +c \right )}+15 b^{4} {\mathrm e}^{9 i \left (d x +c \right )}-24 a^{4} {\mathrm e}^{7 i \left (d x +c \right )}-60 b^{4} {\mathrm e}^{7 i \left (d x +c \right )}+30 a^{4} {\mathrm e}^{5 i \left (d x +c \right )}+90 b^{4} {\mathrm e}^{5 i \left (d x +c \right )}-24 a^{4} {\mathrm e}^{3 i \left (d x +c \right )}-60 b^{4} {\mathrm e}^{3 i \left (d x +c \right )}-18 a^{2} b^{2} {\mathrm e}^{9 i \left (d x +c \right )}+82 a^{2} b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-128 a^{2} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )} a^{4}-18 \,{\mathrm e}^{i \left (d x +c \right )} b^{2} a^{2}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )} b -b +2 i {\mathrm e}^{i \left (d x +c \right )} a \right ) d \,a^{5}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}-\frac {6 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{a^{4} d}+\frac {5 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{4}}{a^{6} d}-\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^{2} d}+\frac {6 \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^{4} d}-\frac {5 \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^{6} d}\) | \(587\) |
Input:
int(cot(d*x+c)^5/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/4/a^2/sin(d*x+c)^4-1/2*(-2*a^2+3*b^2)/a^4/sin(d*x+c)^2+(a^4-6*a^2* b^2+5*b^4)/a^6*ln(sin(d*x+c))+2/3/a^3*b/sin(d*x+c)^3-4*b*(a^2-b^2)/a^5/sin (d*x+c)-(a^4-6*a^2*b^2+5*b^4)/a^6*ln(a+b*sin(d*x+c))+(a^4-2*a^2*b^2+b^4)/a ^5/(a+b*sin(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 542 vs. \(2 (182) = 364\).
Time = 0.12 (sec) , antiderivative size = 542, normalized size of antiderivative = 2.88 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {21 \, a^{5} - 82 \, a^{3} b^{2} + 60 \, a b^{4} + 12 \, {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (18 \, a^{5} - 77 \, a^{3} b^{2} + 60 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4} + {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5} + {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 12 \, {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4} + {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5} + {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (31 \, a^{4} b - 30 \, a^{2} b^{3} - 6 \, {\left (6 \, a^{4} b - 5 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{7} d \cos \left (d x + c\right )^{4} - 2 \, a^{7} d \cos \left (d x + c\right )^{2} + a^{7} d + {\left (a^{6} b d \cos \left (d x + c\right )^{4} - 2 \, a^{6} b d \cos \left (d x + c\right )^{2} + a^{6} b d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(cot(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="fricas")
Output:
1/12*(21*a^5 - 82*a^3*b^2 + 60*a*b^4 + 12*(a^5 - 6*a^3*b^2 + 5*a*b^4)*cos( d*x + c)^4 - 2*(18*a^5 - 77*a^3*b^2 + 60*a*b^4)*cos(d*x + c)^2 - 12*(a^5 - 6*a^3*b^2 + 5*a*b^4 + (a^5 - 6*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^4 - 2*(a^5 - 6*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^2 + (a^4*b - 6*a^2*b^3 + 5*b^5 + (a^4 *b - 6*a^2*b^3 + 5*b^5)*cos(d*x + c)^4 - 2*(a^4*b - 6*a^2*b^3 + 5*b^5)*cos (d*x + c)^2)*sin(d*x + c))*log(b*sin(d*x + c) + a) + 12*(a^5 - 6*a^3*b^2 + 5*a*b^4 + (a^5 - 6*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^4 - 2*(a^5 - 6*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^2 + (a^4*b - 6*a^2*b^3 + 5*b^5 + (a^4*b - 6*a^2*b ^3 + 5*b^5)*cos(d*x + c)^4 - 2*(a^4*b - 6*a^2*b^3 + 5*b^5)*cos(d*x + c)^2) *sin(d*x + c))*log(-1/2*sin(d*x + c)) - (31*a^4*b - 30*a^2*b^3 - 6*(6*a^4* b - 5*a^2*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a^7*d*cos(d*x + c)^4 - 2*a^7 *d*cos(d*x + c)^2 + a^7*d + (a^6*b*d*cos(d*x + c)^4 - 2*a^6*b*d*cos(d*x + c)^2 + a^6*b*d)*sin(d*x + c))
\[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\cot ^{5}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(cot(d*x+c)**5/(a+b*sin(d*x+c))**2,x)
Output:
Integral(cot(c + d*x)**5/(a + b*sin(c + d*x))**2, x)
Time = 0.04 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.01 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {5 \, a^{3} b \sin \left (d x + c\right ) + 12 \, {\left (a^{4} - 6 \, a^{2} b^{2} + 5 \, b^{4}\right )} \sin \left (d x + c\right )^{4} - 3 \, a^{4} - 6 \, {\left (6 \, a^{3} b - 5 \, a b^{3}\right )} \sin \left (d x + c\right )^{3} + 2 \, {\left (6 \, a^{4} - 5 \, a^{2} b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{5} b \sin \left (d x + c\right )^{5} + a^{6} \sin \left (d x + c\right )^{4}} - \frac {12 \, {\left (a^{4} - 6 \, a^{2} b^{2} + 5 \, b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6}} + \frac {12 \, {\left (a^{4} - 6 \, a^{2} b^{2} + 5 \, b^{4}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{6}}}{12 \, d} \] Input:
integrate(cot(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="maxima")
Output:
1/12*((5*a^3*b*sin(d*x + c) + 12*(a^4 - 6*a^2*b^2 + 5*b^4)*sin(d*x + c)^4 - 3*a^4 - 6*(6*a^3*b - 5*a*b^3)*sin(d*x + c)^3 + 2*(6*a^4 - 5*a^2*b^2)*sin (d*x + c)^2)/(a^5*b*sin(d*x + c)^5 + a^6*sin(d*x + c)^4) - 12*(a^4 - 6*a^2 *b^2 + 5*b^4)*log(b*sin(d*x + c) + a)/a^6 + 12*(a^4 - 6*a^2*b^2 + 5*b^4)*l og(sin(d*x + c))/a^6)/d
Time = 0.14 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.05 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {{\left (a^{4} - 6 \, a^{2} b^{2} + 5 \, b^{4}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{6} d} - \frac {{\left (a^{4} b - 6 \, a^{2} b^{3} + 5 \, b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b d} + \frac {5 \, a^{4} b \sin \left (d x + c\right ) - 3 \, a^{5} + 12 \, {\left (a^{5} - 6 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \sin \left (d x + c\right )^{4} - 6 \, {\left (6 \, a^{4} b - 5 \, a^{2} b^{3}\right )} \sin \left (d x + c\right )^{3} + 2 \, {\left (6 \, a^{5} - 5 \, a^{3} b^{2}\right )} \sin \left (d x + c\right )^{2}}{12 \, {\left (b \sin \left (d x + c\right ) + a\right )} a^{6} d \sin \left (d x + c\right )^{4}} \] Input:
integrate(cot(d*x+c)^5/(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
(a^4 - 6*a^2*b^2 + 5*b^4)*log(abs(sin(d*x + c)))/(a^6*d) - (a^4*b - 6*a^2* b^3 + 5*b^5)*log(abs(b*sin(d*x + c) + a))/(a^6*b*d) + 1/12*(5*a^4*b*sin(d* x + c) - 3*a^5 + 12*(a^5 - 6*a^3*b^2 + 5*a*b^4)*sin(d*x + c)^4 - 6*(6*a^4* b - 5*a^2*b^3)*sin(d*x + c)^3 + 2*(6*a^5 - 5*a^3*b^2)*sin(d*x + c)^2)/((b* sin(d*x + c) + a)*a^6*d*sin(d*x + c)^4)
Time = 17.85 (sec) , antiderivative size = 439, normalized size of antiderivative = 2.34 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (3\,a^4-62\,a^2\,b^2+64\,b^4\right )-\frac {a^4}{4}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {11\,a^4}{4}-\frac {10\,a^2\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (20\,a\,b^3-\frac {62\,a^3\,b}{3}\right )-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (60\,a^4\,b-96\,a^2\,b^3+32\,b^5\right )}{a}+\frac {5\,a^3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{6}}{d\,\left (16\,a^6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+16\,a^6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+32\,b\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a^2\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {\frac {a^2}{16}+\frac {b^2}{8}}{a^4}+\frac {1}{8\,a^2}-\frac {b^2}{2\,a^4}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {b\,\left (32\,a^2+64\,b^2\right )}{64\,a^5}-\frac {b}{4\,a^3}+\frac {4\,b\,\left (\frac {\frac {a^2}{8}+\frac {b^2}{4}}{a^4}+\frac {1}{4\,a^2}-\frac {b^2}{a^4}\right )}{a}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^4-6\,a^2\,b^2+5\,b^4\right )}{a^6\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12\,a^3\,d}-\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^4-6\,a^2\,b^2+5\,b^4\right )}{a^6\,d} \] Input:
int(cot(c + d*x)^5/(a + b*sin(c + d*x))^2,x)
Output:
(tan(c/2 + (d*x)/2)^4*(3*a^4 + 64*b^4 - 62*a^2*b^2) - a^4/4 + tan(c/2 + (d *x)/2)^2*((11*a^4)/4 - (10*a^2*b^2)/3) + tan(c/2 + (d*x)/2)^3*(20*a*b^3 - (62*a^3*b)/3) - (tan(c/2 + (d*x)/2)^5*(60*a^4*b + 32*b^5 - 96*a^2*b^3))/a + (5*a^3*b*tan(c/2 + (d*x)/2))/6)/(d*(16*a^6*tan(c/2 + (d*x)/2)^4 + 16*a^6 *tan(c/2 + (d*x)/2)^6 + 32*a^5*b*tan(c/2 + (d*x)/2)^5)) - tan(c/2 + (d*x)/ 2)^4/(64*a^2*d) + (tan(c/2 + (d*x)/2)^2*((a^2/16 + b^2/8)/a^4 + 1/(8*a^2) - b^2/(2*a^4)))/d - (tan(c/2 + (d*x)/2)*((b*(32*a^2 + 64*b^2))/(64*a^5) - b/(4*a^3) + (4*b*((a^2/8 + b^2/4)/a^4 + 1/(4*a^2) - b^2/a^4))/a))/d + (log (tan(c/2 + (d*x)/2))*(a^4 + 5*b^4 - 6*a^2*b^2))/(a^6*d) + (b*tan(c/2 + (d* x)/2)^3)/(12*a^3*d) - (log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/ 2)^2)*(a^4 + 5*b^4 - 6*a^2*b^2))/(a^6*d)
Time = 0.20 (sec) , antiderivative size = 578, normalized size of antiderivative = 3.07 \[ \int \frac {\cot ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(cot(d*x+c)^5/(a+b*sin(d*x+c))^2,x)
Output:
( - 96*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)* *5*a**4*b + 576*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin( c + d*x)**5*a**2*b**3 - 480*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2) *b + a)*sin(c + d*x)**5*b**5 - 96*log(tan((c + d*x)/2)**2*a + 2*tan((c + d *x)/2)*b + a)*sin(c + d*x)**4*a**5 + 576*log(tan((c + d*x)/2)**2*a + 2*tan ((c + d*x)/2)*b + a)*sin(c + d*x)**4*a**3*b**2 - 480*log(tan((c + d*x)/2)* *2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**4*a*b**4 + 96*log(tan((c + d*x)/2))*sin(c + d*x)**5*a**4*b - 576*log(tan((c + d*x)/2))*sin(c + d*x)** 5*a**2*b**3 + 480*log(tan((c + d*x)/2))*sin(c + d*x)**5*b**5 + 96*log(tan( (c + d*x)/2))*sin(c + d*x)**4*a**5 - 576*log(tan((c + d*x)/2))*sin(c + d*x )**4*a**3*b**2 + 480*log(tan((c + d*x)/2))*sin(c + d*x)**4*a*b**4 + 129*si n(c + d*x)**5*a**4*b - 120*sin(c + d*x)**5*a**2*b**3 + 225*sin(c + d*x)**4 *a**5 - 696*sin(c + d*x)**4*a**3*b**2 + 480*sin(c + d*x)**4*a*b**4 - 288*s in(c + d*x)**3*a**4*b + 240*sin(c + d*x)**3*a**2*b**3 + 96*sin(c + d*x)**2 *a**5 - 80*sin(c + d*x)**2*a**3*b**2 + 40*sin(c + d*x)*a**4*b - 24*a**5)/( 96*sin(c + d*x)**4*a**6*d*(sin(c + d*x)*b + a))