Integrand size = 27, antiderivative size = 147 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\left (2 a^2-3 b^2\right ) \csc (c+d x)}{a^4 d}+\frac {b \csc ^2(c+d x)}{a^3 d}-\frac {\csc ^3(c+d x)}{3 a^2 d}+\frac {4 b \left (a^2-b^2\right ) \log (\sin (c+d x))}{a^5 d}-\frac {4 b \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a^5 d}-\frac {\left (a^2-b^2\right )^2}{a^4 b d (a+b \sin (c+d x))} \] Output:
(2*a^2-3*b^2)*csc(d*x+c)/a^4/d+b*csc(d*x+c)^2/a^3/d-1/3*csc(d*x+c)^3/a^2/d +4*b*(a^2-b^2)*ln(sin(d*x+c))/a^5/d-4*b*(a^2-b^2)*ln(a+b*sin(d*x+c))/a^5/d -(a^2-b^2)^2/a^4/b/d/(a+b*sin(d*x+c))
Time = 1.99 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.86 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {3 a \left (2 a^2-3 b^2\right ) \csc (c+d x)+3 a^2 b \csc ^2(c+d x)-a^3 \csc ^3(c+d x)+12 (a-b) b (a+b) \log (\sin (c+d x))-12 (a-b) b (a+b) \log (a+b \sin (c+d x))-\frac {3 a \left (a^2-b^2\right )^2}{b (a+b \sin (c+d x))}}{3 a^5 d} \] Input:
Integrate[(Cos[c + d*x]*Cot[c + d*x]^4)/(a + b*Sin[c + d*x])^2,x]
Output:
(3*a*(2*a^2 - 3*b^2)*Csc[c + d*x] + 3*a^2*b*Csc[c + d*x]^2 - a^3*Csc[c + d *x]^3 + 12*(a - b)*b*(a + b)*Log[Sin[c + d*x]] - 12*(a - b)*b*(a + b)*Log[ a + b*Sin[c + d*x]] - (3*a*(a^2 - b^2)^2)/(b*(a + b*Sin[c + d*x])))/(3*a^5 *d)
Time = 0.42 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3042, 3316, 27, 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.
| \(\displaystyle \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^4 (a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \frac {\csc ^4(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{(a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\csc ^4(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b^4 (a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (\frac {\csc ^4(c+d x)}{a^2}-\frac {2 b \csc ^3(c+d x)}{a^3}+\frac {\left (3 b^4-2 a^2 b^2\right ) \csc ^2(c+d x)}{a^4 b^2}+\frac {4 b \left (a^2-b^2\right ) \csc (c+d x)}{a^5}+\frac {4 b^2 \left (b^2-a^2\right )}{a^5 (a+b \sin (c+d x))}+\frac {\left (a^2-b^2\right )^2}{a^4 (a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{b d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b^2 \csc ^2(c+d x)}{a^3}-\frac {b \csc ^3(c+d x)}{3 a^2}+\frac {4 b^2 \left (a^2-b^2\right ) \log (b \sin (c+d x))}{a^5}-\frac {4 b^2 \left (a^2-b^2\right ) \log (a+b \sin (c+d x))}{a^5}-\frac {\left (a^2-b^2\right )^2}{a^4 (a+b \sin (c+d x))}+\frac {b \left (2 a^2-3 b^2\right ) \csc (c+d x)}{a^4}}{b d}\) |
Input:
Int[(Cos[c + d*x]*Cot[c + d*x]^4)/(a + b*Sin[c + d*x])^2,x]
Output:
((b*(2*a^2 - 3*b^2)*Csc[c + d*x])/a^4 + (b^2*Csc[c + d*x]^2)/a^3 - (b*Csc[ c + d*x]^3)/(3*a^2) + (4*b^2*(a^2 - b^2)*Log[b*Sin[c + d*x]])/a^5 - (4*b^2 *(a^2 - b^2)*Log[a + b*Sin[c + d*x]])/a^5 - (a^2 - b^2)^2/(a^4*(a + b*Sin[ c + d*x])))/(b*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 1.47 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.95
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{3 a^{2} \sin \left (d x +c \right )^{3}}-\frac {-2 a^{2}+3 b^{2}}{a^{4} \sin \left (d x +c \right )}+\frac {b}{a^{3} \sin \left (d x +c \right )^{2}}+\frac {4 b \left (a^{2}-b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{5}}-\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{a^{4} b \left (a +b \sin \left (d x +c \right )\right )}-\frac {4 b \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{5}}}{d}\) | \(139\) |
| default | \(\frac {-\frac {1}{3 a^{2} \sin \left (d x +c \right )^{3}}-\frac {-2 a^{2}+3 b^{2}}{a^{4} \sin \left (d x +c \right )}+\frac {b}{a^{3} \sin \left (d x +c \right )^{2}}+\frac {4 b \left (a^{2}-b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{5}}-\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{a^{4} b \left (a +b \sin \left (d x +c \right )\right )}-\frac {4 b \left (a^{2}-b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{5}}}{d}\) | \(139\) |
| risch | \(-\frac {2 i \left (-24 i a \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+16 i a^{3} b \,{\mathrm e}^{4 i \left (d x +c \right )}+3 a^{4} {\mathrm e}^{7 i \left (d x +c \right )}-12 a^{2} b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+12 b^{4} {\mathrm e}^{7 i \left (d x +c \right )}+12 i a \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}-12 i a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}-9 a^{4} {\mathrm e}^{5 i \left (d x +c \right )}+44 a^{2} b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-36 b^{4} {\mathrm e}^{5 i \left (d x +c \right )}+12 i a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-12 i a^{3} b \,{\mathrm e}^{6 i \left (d x +c \right )}+9 a^{4} {\mathrm e}^{3 i \left (d x +c \right )}-44 a^{2} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+36 b^{4} {\mathrm e}^{3 i \left (d x +c \right )}-3 a^{4} {\mathrm e}^{i \left (d x +c \right )}+12 a^{2} b^{2} {\mathrm e}^{i \left (d x +c \right )}-12 b^{4} {\mathrm e}^{i \left (d x +c \right )}\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3} b \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right ) d \,a^{4}}+\frac {4 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}-\frac {4 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{5} d}-\frac {4 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{3} d}+\frac {4 b^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{5} d}\) | \(454\) |
Input:
int(cos(d*x+c)*cot(d*x+c)^4/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/3/a^2/sin(d*x+c)^3-(-2*a^2+3*b^2)/a^4/sin(d*x+c)+1/a^3*b/sin(d*x+c )^2+4*b*(a^2-b^2)/a^5*ln(sin(d*x+c))-(a^4-2*a^2*b^2+b^4)/a^4/b/(a+b*sin(d* x+c))-4*b*(a^2-b^2)/a^5*ln(a+b*sin(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 401 vs. \(2 (145) = 290\).
Time = 0.11 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.73 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {5 \, a^{4} b - 6 \, a^{2} b^{3} - 6 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left (a^{2} b^{3} - b^{5} + {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b^{2} - a b^{4} - {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 12 \, {\left (a^{2} b^{3} - b^{5} + {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{3} b^{2} - a b^{4} - {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) - {\left (3 \, a^{5} - 14 \, a^{3} b^{2} + 12 \, a b^{4} - 3 \, {\left (a^{5} - 4 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{5} b^{2} d \cos \left (d x + c\right )^{4} - 2 \, a^{5} b^{2} d \cos \left (d x + c\right )^{2} + a^{5} b^{2} d - {\left (a^{6} b d \cos \left (d x + c\right )^{2} - a^{6} b d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
1/3*(5*a^4*b - 6*a^2*b^3 - 6*(a^4*b - a^2*b^3)*cos(d*x + c)^2 - 12*(a^2*b^ 3 - b^5 + (a^2*b^3 - b^5)*cos(d*x + c)^4 - 2*(a^2*b^3 - b^5)*cos(d*x + c)^ 2 + (a^3*b^2 - a*b^4 - (a^3*b^2 - a*b^4)*cos(d*x + c)^2)*sin(d*x + c))*log (b*sin(d*x + c) + a) + 12*(a^2*b^3 - b^5 + (a^2*b^3 - b^5)*cos(d*x + c)^4 - 2*(a^2*b^3 - b^5)*cos(d*x + c)^2 + (a^3*b^2 - a*b^4 - (a^3*b^2 - a*b^4)* cos(d*x + c)^2)*sin(d*x + c))*log(1/2*sin(d*x + c)) - (3*a^5 - 14*a^3*b^2 + 12*a*b^4 - 3*(a^5 - 4*a^3*b^2 + 4*a*b^4)*cos(d*x + c)^2)*sin(d*x + c))/( a^5*b^2*d*cos(d*x + c)^4 - 2*a^5*b^2*d*cos(d*x + c)^2 + a^5*b^2*d - (a^6*b *d*cos(d*x + c)^2 - a^6*b*d)*sin(d*x + c))
\[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\cos {\left (c + d x \right )} \cot ^{4}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(cos(d*x+c)*cot(d*x+c)**4/(a+b*sin(d*x+c))**2,x)
Output:
Integral(cos(c + d*x)*cot(c + d*x)**4/(a + b*sin(c + d*x))**2, x)
Time = 0.03 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.07 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2 \, a^{2} b^{2} \sin \left (d x + c\right ) - a^{3} b - 3 \, {\left (a^{4} - 4 \, a^{2} b^{2} + 4 \, b^{4}\right )} \sin \left (d x + c\right )^{3} + 6 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{2}}{a^{4} b^{2} \sin \left (d x + c\right )^{4} + a^{5} b \sin \left (d x + c\right )^{3}} - \frac {12 \, {\left (a^{2} b - b^{3}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5}} + \frac {12 \, {\left (a^{2} b - b^{3}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{5}}}{3 \, d} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
1/3*((2*a^2*b^2*sin(d*x + c) - a^3*b - 3*(a^4 - 4*a^2*b^2 + 4*b^4)*sin(d*x + c)^3 + 6*(a^3*b - a*b^3)*sin(d*x + c)^2)/(a^4*b^2*sin(d*x + c)^4 + a^5* b*sin(d*x + c)^3) - 12*(a^2*b - b^3)*log(b*sin(d*x + c) + a)/a^5 + 12*(a^2 *b - b^3)*log(sin(d*x + c))/a^5)/d
Time = 0.18 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.14 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {4 \, {\left (a^{2} b - b^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{5} d} - \frac {4 \, {\left (a^{2} b^{2} - b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b d} + \frac {2 \, a^{3} b^{2} \sin \left (d x + c\right ) - a^{4} b - 3 \, {\left (a^{5} - 4 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \sin \left (d x + c\right )^{3} + 6 \, {\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )^{2}}{3 \, {\left (b \sin \left (d x + c\right ) + a\right )} a^{5} b d \sin \left (d x + c\right )^{3}} \] Input:
integrate(cos(d*x+c)*cot(d*x+c)^4/(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
4*(a^2*b - b^3)*log(abs(sin(d*x + c)))/(a^5*d) - 4*(a^2*b^2 - b^4)*log(abs (b*sin(d*x + c) + a))/(a^5*b*d) + 1/3*(2*a^3*b^2*sin(d*x + c) - a^4*b - 3* (a^5 - 4*a^3*b^2 + 4*a*b^4)*sin(d*x + c)^3 + 6*(a^4*b - a^2*b^3)*sin(d*x + c)^2)/((b*sin(d*x + c) + a)*a^5*b*d*sin(d*x + c)^3)
Time = 22.72 (sec) , antiderivative size = 319, normalized size of antiderivative = 2.17 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (16\,a^2\,b-24\,b^3\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (8\,a\,b^2-\frac {20\,a^3}{3}\right )-\frac {a^3}{3}+\frac {4\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (23\,a^4-44\,a^2\,b^2+16\,b^4\right )}{a}}{d\,\left (8\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+8\,a^5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+16\,b\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {\frac {a^2}{4}+\frac {b^2}{2}}{a^4}+\frac {5}{8\,a^2}-\frac {2\,b^2}{a^4}\right )}{d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,a^2\,b-4\,b^3\right )}{a^5\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{4\,a^3\,d}-\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (4\,a^2\,b-4\,b^3\right )}{a^5\,d} \] Input:
int((cos(c + d*x)*cot(c + d*x)^4)/(a + b*sin(c + d*x))^2,x)
Output:
(tan(c/2 + (d*x)/2)^3*(16*a^2*b - 24*b^3) - tan(c/2 + (d*x)/2)^2*(8*a*b^2 - (20*a^3)/3) - a^3/3 + (4*a^2*b*tan(c/2 + (d*x)/2))/3 + (tan(c/2 + (d*x)/ 2)^4*(23*a^4 + 16*b^4 - 44*a^2*b^2))/a)/(d*(8*a^5*tan(c/2 + (d*x)/2)^3 + 8 *a^5*tan(c/2 + (d*x)/2)^5 + 16*a^4*b*tan(c/2 + (d*x)/2)^4)) - tan(c/2 + (d *x)/2)^3/(24*a^2*d) + (tan(c/2 + (d*x)/2)*((a^2/4 + b^2/2)/a^4 + 5/(8*a^2) - (2*b^2)/a^4))/d + (log(tan(c/2 + (d*x)/2))*(4*a^2*b - 4*b^3))/(a^5*d) + (b*tan(c/2 + (d*x)/2)^2)/(4*a^3*d) - (log(a + 2*b*tan(c/2 + (d*x)/2) + a* tan(c/2 + (d*x)/2)^2)*(4*a^2*b - 4*b^3))/(a^5*d)
Time = 0.19 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.89 \[ \int \frac {\cos (c+d x) \cot ^4(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right )^{4} a^{2} b^{3}+96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right )^{4} b^{5}-96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right )^{3} a^{3} b^{2}+96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right )^{3} a \,b^{4}+96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a^{2} b^{3}-96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} b^{5}+96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a^{3} b^{2}-96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{3} a \,b^{4}-21 \sin \left (d x +c \right )^{4} a^{4} b +24 \sin \left (d x +c \right )^{4} a^{2} b^{3}-45 \sin \left (d x +c \right )^{3} a^{5}+120 \sin \left (d x +c \right )^{3} a^{3} b^{2}-96 \sin \left (d x +c \right )^{3} a \,b^{4}+48 \sin \left (d x +c \right )^{2} a^{4} b -48 \sin \left (d x +c \right )^{2} a^{2} b^{3}+16 \sin \left (d x +c \right ) a^{3} b^{2}-8 a^{4} b}{24 \sin \left (d x +c \right )^{3} a^{5} b d \left (\sin \left (d x +c \right ) b +a \right )} \] Input:
int(cos(d*x+c)*cot(d*x+c)^4/(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)* *4*a**2*b**3 + 96*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*si n(c + d*x)**4*b**5 - 96*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**3*a**3*b**2 + 96*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**3*a*b**4 + 96*log(tan((c + d*x)/2))*sin(c + d *x)**4*a**2*b**3 - 96*log(tan((c + d*x)/2))*sin(c + d*x)**4*b**5 + 96*log( tan((c + d*x)/2))*sin(c + d*x)**3*a**3*b**2 - 96*log(tan((c + d*x)/2))*sin (c + d*x)**3*a*b**4 - 21*sin(c + d*x)**4*a**4*b + 24*sin(c + d*x)**4*a**2* b**3 - 45*sin(c + d*x)**3*a**5 + 120*sin(c + d*x)**3*a**3*b**2 - 96*sin(c + d*x)**3*a*b**4 + 48*sin(c + d*x)**2*a**4*b - 48*sin(c + d*x)**2*a**2*b** 3 + 16*sin(c + d*x)*a**3*b**2 - 8*a**4*b)/(24*sin(c + d*x)**3*a**5*b*d*(si n(c + d*x)*b + a))