Integrand size = 29, antiderivative size = 109 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\csc (c+d x)}{a^2 d}-\frac {2 b \log (\sin (c+d x))}{a^3 d}-\frac {2 \left (a^4-b^4\right ) \log (a+b \sin (c+d x))}{a^3 b^3 d}+\frac {\sin (c+d x)}{b^2 d}-\frac {\left (a^2-b^2\right )^2}{a^2 b^3 d (a+b \sin (c+d x))} \] Output:
-csc(d*x+c)/a^2/d-2*b*ln(sin(d*x+c))/a^3/d-2*(a^4-b^4)*ln(a+b*sin(d*x+c))/ a^3/b^3/d+sin(d*x+c)/b^2/d-(a^2-b^2)^2/a^2/b^3/d/(a+b*sin(d*x+c))
Time = 0.66 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.87 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {\csc (c+d x)}{a^2}+\frac {2 b \log (\sin (c+d x))}{a^3}+2 \left (\frac {a}{b^3}-\frac {b}{a^3}\right ) \log (a+b \sin (c+d x))-\frac {\sin (c+d x)}{b^2}+\frac {\left (a^2-b^2\right )^2}{a^2 b^3 (a+b \sin (c+d x))}}{d} \] Input:
Integrate[(Cos[c + d*x]^3*Cot[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]
Output:
-((Csc[c + d*x]/a^2 + (2*b*Log[Sin[c + d*x]])/a^3 + 2*(a/b^3 - b/a^3)*Log[ a + b*Sin[c + d*x]] - Sin[c + d*x]/b^2 + (a^2 - b^2)^2/(a^2*b^3*(a + b*Sin [c + d*x])))/d)
Time = 0.38 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^5}{\sin (c+d x)^2 (a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \frac {\csc ^2(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{(a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\csc ^2(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b^2 (a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^3 d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (-\frac {2 \csc (c+d x) b^3}{a^3}+\frac {\csc ^2(c+d x) b^2}{a^2}-\frac {2 \left (a^4-b^4\right )}{a^3 (a+b \sin (c+d x))}+\frac {\left (a^2-b^2\right )^2}{a^2 (a+b \sin (c+d x))^2}+1\right )d(b \sin (c+d x))}{b^3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {2 b^4 \log (b \sin (c+d x))}{a^3}-\frac {b^3 \csc (c+d x)}{a^2}-\frac {\left (a^2-b^2\right )^2}{a^2 (a+b \sin (c+d x))}-\frac {2 \left (a^4-b^4\right ) \log (a+b \sin (c+d x))}{a^3}+b \sin (c+d x)}{b^3 d}\) |
Input:
Int[(Cos[c + d*x]^3*Cot[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]
Output:
(-((b^3*Csc[c + d*x])/a^2) - (2*b^4*Log[b*Sin[c + d*x]])/a^3 - (2*(a^4 - b ^4)*Log[a + b*Sin[c + d*x]])/a^3 + b*Sin[c + d*x] - (a^2 - b^2)^2/(a^2*(a + b*Sin[c + d*x])))/(b^3*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 = 5.33 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.97
| method | result | size |
| derivativedivides | \(\frac {\frac {\sin \left (d x +c \right )}{b^{2}}+\frac {\left (-2 a^{4}+2 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{3} b^{3}}-\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{b^{3} a^{2} \left (a +b \sin \left (d x +c \right )\right )}-\frac {1}{a^{2} \sin \left (d x +c \right )}-\frac {2 b \ln \left (\sin \left (d x +c \right )\right )}{a^{3}}}{d}\) | \(106\) |
| default | \(\frac {\frac {\sin \left (d x +c \right )}{b^{2}}+\frac {\left (-2 a^{4}+2 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{3} b^{3}}-\frac {a^{4}-2 a^{2} b^{2}+b^{4}}{b^{3} a^{2} \left (a +b \sin \left (d x +c \right )\right )}-\frac {1}{a^{2} \sin \left (d x +c \right )}-\frac {2 b \ln \left (\sin \left (d x +c \right )\right )}{a^{3}}}{d}\) | \(106\) |
| risch | \(\frac {2 i x a}{b^{3}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {4 i a c}{b^{3} d}-\frac {2 \left (a^{4} {\mathrm e}^{3 i \left (d x +c \right )}-2 a^{2} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+2 b^{4} {\mathrm e}^{3 i \left (d x +c \right )}-a^{4} {\mathrm e}^{i \left (d x +c \right )}+2 a^{2} b^{2} {\mathrm e}^{i \left (d x +c \right )}-2 b^{4} {\mathrm e}^{i \left (d x +c \right )}+2 i a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}\right )}{a^{2} d \,b^{3} \left (2 a \,{\mathrm e}^{3 i \left (d x +c \right )}-i b \,{\mathrm e}^{4 i \left (d x +c \right )}-2 a \,{\mathrm e}^{i \left (d x +c \right )}+2 i b \,{\mathrm e}^{2 i \left (d x +c \right )}-i b \right )}-\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{d \,a^{3}}-\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{3} d}+\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{a^{3} d}\) | \(326\) |
Input:
int(cos(d*x+c)^3*cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(sin(d*x+c)/b^2+(-2*a^4+2*b^4)/a^3/b^3*ln(a+b*sin(d*x+c))-1/b^3*(a^4-2 *a^2*b^2+b^4)/a^2/(a+b*sin(d*x+c))-1/a^2/sin(d*x+c)-2/a^3*b*ln(sin(d*x+c)) )
Time = 0.13 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.96 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {a^{4} b \cos \left (d x + c\right )^{2} - a^{4} b + a^{2} b^{3} + 2 \, {\left (a^{4} b - b^{5} - {\left (a^{4} b - b^{5}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{5} - a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (b^{5} \cos \left (d x + c\right )^{2} - a b^{4} \sin \left (d x + c\right ) - b^{5}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right ) + {\left (a^{3} b^{2} \cos \left (d x + c\right )^{2} + a^{5} - 3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \sin \left (d x + c\right )}{a^{3} b^{4} d \cos \left (d x + c\right )^{2} - a^{4} b^{3} d \sin \left (d x + c\right ) - a^{3} b^{4} d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="frica s")
Output:
(a^4*b*cos(d*x + c)^2 - a^4*b + a^2*b^3 + 2*(a^4*b - b^5 - (a^4*b - b^5)*c os(d*x + c)^2 + (a^5 - a*b^4)*sin(d*x + c))*log(b*sin(d*x + c) + a) - 2*(b ^5*cos(d*x + c)^2 - a*b^4*sin(d*x + c) - b^5)*log(1/2*sin(d*x + c)) + (a^3 *b^2*cos(d*x + c)^2 + a^5 - 3*a^3*b^2 + 2*a*b^4)*sin(d*x + c))/(a^3*b^4*d* cos(d*x + c)^2 - a^4*b^3*d*sin(d*x + c) - a^3*b^4*d)
Timed out. \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*cot(d*x+c)**2/(a+b*sin(d*x+c))**2,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {a b^{3} + {\left (a^{4} - 2 \, a^{2} b^{2} + 2 \, b^{4}\right )} \sin \left (d x + c\right )}{a^{2} b^{4} \sin \left (d x + c\right )^{2} + a^{3} b^{3} \sin \left (d x + c\right )} + \frac {2 \, b \log \left (\sin \left (d x + c\right )\right )}{a^{3}} - \frac {\sin \left (d x + c\right )}{b^{2}} + \frac {2 \, {\left (a^{4} - b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{3} b^{3}}}{d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-((a*b^3 + (a^4 - 2*a^2*b^2 + 2*b^4)*sin(d*x + c))/(a^2*b^4*sin(d*x + c)^2 + a^3*b^3*sin(d*x + c)) + 2*b*log(sin(d*x + c))/a^3 - sin(d*x + c)/b^2 + 2*(a^4 - b^4)*log(b*sin(d*x + c) + a)/(a^3*b^3))/d
Time = 0.17 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.17 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 \, b \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3} d} + \frac {\sin \left (d x + c\right )}{b^{2} d} - \frac {2 \, {\left (a^{4} - b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{3} b^{3} d} - \frac {a b^{2} + \frac {{\left (a^{4} - 2 \, a^{2} b^{2} + 2 \, b^{4}\right )} \sin \left (d x + c\right )}{b}}{{\left (b \sin \left (d x + c\right ) + a\right )} a^{2} b^{2} d \sin \left (d x + c\right )} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="giac" )
Output:
-2*b*log(abs(sin(d*x + c)))/(a^3*d) + sin(d*x + c)/(b^2*d) - 2*(a^4 - b^4) *log(abs(b*sin(d*x + c) + a))/(a^3*b^3*d) - (a*b^2 + (a^4 - 2*a^2*b^2 + 2* b^4)*sin(d*x + c)/b)/((b*sin(d*x + c) + a)*a^2*b^2*d*sin(d*x + c))
Time = 23.13 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.87 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (4\,a^2-b^2\right )}{b}-2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-a+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (4\,a^4-5\,a^2\,b^2+2\,b^4\right )}{a\,b^2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (8\,a^4-9\,a^2\,b^2+4\,b^4\right )}{a\,b^2}}{d\,\left (2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+4\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+2\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+4\,b\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,b\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,a^2\,d}+\frac {2\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b^3\,d}-\frac {2\,b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {2\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^4-b^4\right )}{a^3\,b^3\,d} \] Input:
int((cos(c + d*x)^3*cot(c + d*x)^2)/(a + b*sin(c + d*x))^2,x)
Output:
((2*tan(c/2 + (d*x)/2)^3*(4*a^2 - b^2))/b - 2*b*tan(c/2 + (d*x)/2) - a + ( 2*tan(c/2 + (d*x)/2)^2*(4*a^4 + 2*b^4 - 5*a^2*b^2))/(a*b^2) + (tan(c/2 + ( d*x)/2)^4*(8*a^4 + 4*b^4 - 9*a^2*b^2))/(a*b^2))/(d*(4*a^3*tan(c/2 + (d*x)/ 2)^3 + 2*a^3*tan(c/2 + (d*x)/2)^5 + 2*a^3*tan(c/2 + (d*x)/2) + 4*a^2*b*tan (c/2 + (d*x)/2)^2 + 4*a^2*b*tan(c/2 + (d*x)/2)^4)) - tan(c/2 + (d*x)/2)/(2 *a^2*d) + (2*a*log(tan(c/2 + (d*x)/2)^2 + 1))/(b^3*d) - (2*b*log(tan(c/2 + (d*x)/2)))/(a^3*d) - (2*log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x )/2)^2)*(a^4 - b^4))/(a^3*b^3*d)
Time = 0.48 (sec) , antiderivative size = 369, normalized size of antiderivative = 3.39 \[ \int \frac {\cos ^3(c+d x) \cot ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a^{4} b +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right ) a^{5}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right )^{2} a^{4} b +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right )^{2} b^{5}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right ) a^{5}+8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right ) a \,b^{4}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b^{5}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right ) a \,b^{4}+4 \sin \left (d x +c \right )^{3} a^{3} b^{2}+3 \sin \left (d x +c \right )^{2} a^{2} b^{3}-8 \sin \left (d x +c \right ) a^{5}+11 \sin \left (d x +c \right ) a^{3} b^{2}-8 \sin \left (d x +c \right ) a \,b^{4}-4 a^{2} b^{3}}{4 \sin \left (d x +c \right ) a^{3} b^{3} d \left (\sin \left (d x +c \right ) b +a \right )} \] Input:
int(cos(d*x+c)^3*cot(d*x+c)^2/(a+b*sin(d*x+c))^2,x)
Output:
(8*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a**4*b + 8*log(tan((c + d* x)/2)**2 + 1)*sin(c + d*x)*a**5 - 8*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**2*a**4*b + 8*log(tan((c + d*x)/2)**2*a + 2*t an((c + d*x)/2)*b + a)*sin(c + d*x)**2*b**5 - 8*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)*a**5 + 8*log(tan((c + d*x)/2)**2* a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)*a*b**4 - 8*log(tan((c + d*x)/2) )*sin(c + d*x)**2*b**5 - 8*log(tan((c + d*x)/2))*sin(c + d*x)*a*b**4 + 4*s in(c + d*x)**3*a**3*b**2 + 3*sin(c + d*x)**2*a**2*b**3 - 8*sin(c + d*x)*a* *5 + 11*sin(c + d*x)*a**3*b**2 - 8*sin(c + d*x)*a*b**4 - 4*a**2*b**3)/(4*s in(c + d*x)*a**3*b**3*d*(sin(c + d*x)*b + a))