Integrand size = 27, antiderivative size = 157 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))}{b^6 d}-\frac {4 a \left (a^2-b^2\right ) \sin (c+d x)}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin ^2(c+d x)}{2 b^4 d}-\frac {2 a \sin ^3(c+d x)}{3 b^3 d}+\frac {\sin ^4(c+d x)}{4 b^2 d}+\frac {a \left (a^2-b^2\right )^2}{b^6 d (a+b \sin (c+d x))} \] Output:
(5*a^4-6*a^2*b^2+b^4)*ln(a+b*sin(d*x+c))/b^6/d-4*a*(a^2-b^2)*sin(d*x+c)/b^ 5/d+1/2*(3*a^2-2*b^2)*sin(d*x+c)^2/b^4/d-2/3*a*sin(d*x+c)^3/b^3/d+1/4*sin( d*x+c)^4/b^2/d+a*(a^2-b^2)^2/b^6/d/(a+b*sin(d*x+c))
Time = 0.78 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.20 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {12 a \left (a^2-b^2\right ) \left (a^2-b^2+\left (5 a^2-b^2\right ) \log (a+b \sin (c+d x))\right )+12 b \left (-a^2+b^2\right ) \left (4 a^2+\left (-5 a^2+b^2\right ) \log (a+b \sin (c+d x))\right ) \sin (c+d x)-6 a b^2 \left (5 a^2-6 b^2\right ) \sin ^2(c+d x)+2 b^3 \left (5 a^2-6 b^2\right ) \sin ^3(c+d x)-5 a b^4 \sin ^4(c+d x)+3 b^5 \sin ^5(c+d x)}{12 b^6 d (a+b \sin (c+d x))} \] Input:
Integrate[(Cos[c + d*x]^5*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]
Output:
(12*a*(a^2 - b^2)*(a^2 - b^2 + (5*a^2 - b^2)*Log[a + b*Sin[c + d*x]]) + 12 *b*(-a^2 + b^2)*(4*a^2 + (-5*a^2 + b^2)*Log[a + b*Sin[c + d*x]])*Sin[c + d *x] - 6*a*b^2*(5*a^2 - 6*b^2)*Sin[c + d*x]^2 + 2*b^3*(5*a^2 - 6*b^2)*Sin[c + d*x]^3 - 5*a*b^4*Sin[c + d*x]^4 + 3*b^5*Sin[c + d*x]^5)/(12*b^6*d*(a + b*Sin[c + d*x]))
Time = 0.37 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.88, 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.
| \(\displaystyle \int \frac {\sin (c+d x) \cos ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x) \cos (c+d x)^5}{(a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \frac {\sin (c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{(a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {b \sin (c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{(a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^6 d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (b^3 \sin ^3(c+d x)-2 a b^2 \sin ^2(c+d x)+b \left (3 a^2-2 b^2\right ) \sin (c+d x)-4 \left (a^3-a b^2\right )+\frac {5 a^4-6 b^2 a^2+b^4}{a+b \sin (c+d x)}-\frac {a \left (a^2-b^2\right )^2}{(a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{b^6 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{2} b^2 \left (3 a^2-2 b^2\right ) \sin ^2(c+d x)-4 a b \left (a^2-b^2\right ) \sin (c+d x)+\frac {a \left (a^2-b^2\right )^2}{a+b \sin (c+d x)}+\left (5 a^4-6 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))-\frac {2}{3} a b^3 \sin ^3(c+d x)+\frac {1}{4} b^4 \sin ^4(c+d x)}{b^6 d}\) |
Input:
Int[(Cos[c + d*x]^5*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]
Output:
((5*a^4 - 6*a^2*b^2 + b^4)*Log[a + b*Sin[c + d*x]] - 4*a*b*(a^2 - b^2)*Sin [c + d*x] + (b^2*(3*a^2 - 2*b^2)*Sin[c + d*x]^2)/2 - (2*a*b^3*Sin[c + d*x] ^3)/3 + (b^4*Sin[c + d*x]^4)/4 + (a*(a^2 - b^2)^2)/(a + b*Sin[c + d*x]))/( b^6*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 = 1.48 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.97
| method | result | size |
| derivativedivides | \(\frac {-\frac {-\frac {\sin \left (d x +c \right )^{4} b^{3}}{4}+\frac {2 a \sin \left (d x +c \right )^{3} b^{2}}{3}-\frac {3 a^{2} b \sin \left (d x +c \right )^{2}}{2}+b^{3} \sin \left (d x +c \right )^{2}+4 a^{3} \sin \left (d x +c \right )-4 \sin \left (d x +c \right ) a \,b^{2}}{b^{5}}+\frac {\left (5 a^{4}-6 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}+\frac {a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{b^{6} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(152\) |
| default | \(\frac {-\frac {-\frac {\sin \left (d x +c \right )^{4} b^{3}}{4}+\frac {2 a \sin \left (d x +c \right )^{3} b^{2}}{3}-\frac {3 a^{2} b \sin \left (d x +c \right )^{2}}{2}+b^{3} \sin \left (d x +c \right )^{2}+4 a^{3} \sin \left (d x +c \right )-4 \sin \left (d x +c \right ) a \,b^{2}}{b^{5}}+\frac {\left (5 a^{4}-6 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}+\frac {a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{b^{6} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(152\) |
| parallelrisch | \(\frac {960 \left (a +b \sin \left (d x +c \right )\right ) \left (a^{2}-\frac {b^{2}}{5}\right ) \left (a -b \right ) \left (a +b \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-960 \left (a +b \sin \left (d x +c \right )\right ) \left (a^{2}-\frac {b^{2}}{5}\right ) \left (a -b \right ) \left (a +b \right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (240 a^{3} b^{2}-248 a \,b^{4}\right ) \cos \left (2 d x +2 c \right )+\left (-40 a^{2} b^{3}+33 b^{5}\right ) \sin \left (3 d x +3 c \right )-10 b^{4} a \cos \left (4 d x +4 c \right )+3 b^{5} \sin \left (5 d x +5 c \right )+\left (-960 a^{4} b +1272 a^{2} b^{3}-306 b^{5}\right ) \sin \left (d x +c \right )-240 a^{3} b^{2}+258 a \,b^{4}}{192 b^{6} d \left (a +b \sin \left (d x +c \right )\right )}\) | \(238\) |
| risch | \(\frac {6 i a^{2} x}{b^{4}}+\frac {2 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{b^{5} d}-\frac {10 i a^{4} c}{b^{6} d}-\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 b^{4} d}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )}}{16 b^{2} d}+\frac {12 i a^{2} c}{b^{4} d}-\frac {2 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{b^{5} d}+\frac {7 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{4 b^{3} d}-\frac {i x}{b^{2}}-\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 b^{4} d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{16 b^{2} d}-\frac {2 i c}{b^{2} d}-\frac {5 i a^{4} x}{b^{6}}-\frac {7 i a \,{\mathrm e}^{i \left (d x +c \right )}}{4 b^{3} d}+\frac {2 a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) {\mathrm e}^{i \left (d x +c \right )}}{d \,b^{6} \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {5 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{6} d}-\frac {6 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{4} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d}+\frac {\cos \left (4 d x +4 c \right )}{32 d \,b^{2}}+\frac {a \sin \left (3 d x +3 c \right )}{6 b^{3} d}\) | \(432\) |
| norman | \(\frac {-\frac {\left (150 a^{3}-160 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 b^{4} d}-\frac {\left (150 a^{3}-160 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{3 b^{4} d}-\frac {\left (10 a^{3}-12 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{4} d}-\frac {\left (10 a^{3}-12 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{b^{4} d}-\frac {5 \left (20 a^{3}-20 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{b^{4} d}-\frac {5 \left (20 a^{3}-20 a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{b^{4} d}-\frac {4 \left (45 a^{5}-59 a^{3} b^{2}+15 a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a \,b^{5} d}-\frac {4 \left (45 a^{5}-59 a^{3} b^{2}+15 a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 a \,b^{5} d}-\frac {2 \left (225 a^{5}-310 a^{3} b^{2}+81 a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 b^{5} d a}-\frac {2 \left (225 a^{5}-310 a^{3} b^{2}+81 a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 b^{5} d a}-\frac {8 \left (25 a^{5}-35 a^{3} b^{2}+9 a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{b^{5} d a}-\frac {2 \left (5 a^{5}-6 a^{3} b^{2}+a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \,b^{5} d}-\frac {2 \left (5 a^{5}-6 a^{3} b^{2}+a \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{a \,b^{5} d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {\left (5 a^{4}-6 a^{2} b^{2}+b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \,b^{6}}-\frac {\left (5 a^{4}-6 a^{2} b^{2}+b^{4}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{d \,b^{6}}\) | \(615\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/b^5*(-1/4*sin(d*x+c)^4*b^3+2/3*a*sin(d*x+c)^3*b^2-3/2*a^2*b*sin(d* x+c)^2+b^3*sin(d*x+c)^2+4*a^3*sin(d*x+c)-4*sin(d*x+c)*a*b^2)+1/b^6*(5*a^4- 6*a^2*b^2+b^4)*ln(a+b*sin(d*x+c))+a*(a^4-2*a^2*b^2+b^4)/b^6/(a+b*sin(d*x+c )))
Time = 0.11 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.29 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {40 \, a b^{4} \cos \left (d x + c\right )^{4} - 96 \, a^{5} + 504 \, a^{3} b^{2} - 383 \, a b^{4} - 16 \, {\left (15 \, a^{3} b^{2} - 13 \, a b^{4}\right )} \cos \left (d x + c\right )^{2} - 96 \, {\left (5 \, a^{5} - 6 \, a^{3} b^{2} + a b^{4} + {\left (5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (24 \, b^{5} \cos \left (d x + c\right )^{4} - 384 \, a^{4} b + 392 \, a^{2} b^{3} - 33 \, b^{5} - 16 \, {\left (5 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{96 \, {\left (b^{7} d \sin \left (d x + c\right ) + a b^{6} d\right )}} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
-1/96*(40*a*b^4*cos(d*x + c)^4 - 96*a^5 + 504*a^3*b^2 - 383*a*b^4 - 16*(15 *a^3*b^2 - 13*a*b^4)*cos(d*x + c)^2 - 96*(5*a^5 - 6*a^3*b^2 + a*b^4 + (5*a ^4*b - 6*a^2*b^3 + b^5)*sin(d*x + c))*log(b*sin(d*x + c) + a) - (24*b^5*co s(d*x + c)^4 - 384*a^4*b + 392*a^2*b^3 - 33*b^5 - 16*(5*a^2*b^3 - 3*b^5)*c os(d*x + c)^2)*sin(d*x + c))/(b^7*d*sin(d*x + c) + a*b^6*d)
Timed out. \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)/(a+b*sin(d*x+c))**2,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {12 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )}}{b^{7} \sin \left (d x + c\right ) + a b^{6}} + \frac {3 \, b^{3} \sin \left (d x + c\right )^{4} - 8 \, a b^{2} \sin \left (d x + c\right )^{3} + 6 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{2} - 48 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{b^{5}} + \frac {12 \, {\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{12 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
1/12*(12*(a^5 - 2*a^3*b^2 + a*b^4)/(b^7*sin(d*x + c) + a*b^6) + (3*b^3*sin (d*x + c)^4 - 8*a*b^2*sin(d*x + c)^3 + 6*(3*a^2*b - 2*b^3)*sin(d*x + c)^2 - 48*(a^3 - a*b^2)*sin(d*x + c))/b^5 + 12*(5*a^4 - 6*a^2*b^2 + b^4)*log(b* sin(d*x + c) + a)/b^6)/d
Time = 0.21 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.16 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {{\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6} d} + \frac {a^{5} - 2 \, a^{3} b^{2} + a b^{4}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{6} d} + \frac {3 \, b^{6} d^{3} \sin \left (d x + c\right )^{4} - 8 \, a b^{5} d^{3} \sin \left (d x + c\right )^{3} + 18 \, a^{2} b^{4} d^{3} \sin \left (d x + c\right )^{2} - 12 \, b^{6} d^{3} \sin \left (d x + c\right )^{2} - 48 \, a^{3} b^{3} d^{3} \sin \left (d x + c\right ) + 48 \, a b^{5} d^{3} \sin \left (d x + c\right )}{12 \, b^{8} d^{4}} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")
Output:
(5*a^4 - 6*a^2*b^2 + b^4)*log(abs(b*sin(d*x + c) + a))/(b^6*d) + (a^5 - 2* a^3*b^2 + a*b^4)/((b*sin(d*x + c) + a)*b^6*d) + 1/12*(3*b^6*d^3*sin(d*x + c)^4 - 8*a*b^5*d^3*sin(d*x + c)^3 + 18*a^2*b^4*d^3*sin(d*x + c)^2 - 12*b^6 *d^3*sin(d*x + c)^2 - 48*a^3*b^3*d^3*sin(d*x + c) + 48*a*b^5*d^3*sin(d*x + c))/(b^8*d^4)
Time = 22.70 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {{\sin \left (c+d\,x\right )}^4}{4\,b^2}-{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{b^2}-\frac {3\,a^2}{2\,b^4}\right )+\sin \left (c+d\,x\right )\,\left (\frac {2\,a^3}{b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )-\frac {2\,a\,{\sin \left (c+d\,x\right )}^3}{3\,b^3}+\frac {a^5-2\,a^3\,b^2+a\,b^4}{b\,\left (\sin \left (c+d\,x\right )\,b^6+a\,b^5\right )}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (5\,a^4-6\,a^2\,b^2+b^4\right )}{b^6}}{d} \] Input:
int((cos(c + d*x)^5*sin(c + d*x))/(a + b*sin(c + d*x))^2,x)
Output:
(sin(c + d*x)^4/(4*b^2) - sin(c + d*x)^2*(1/b^2 - (3*a^2)/(2*b^4)) + sin(c + d*x)*((2*a^3)/b^5 + (2*a*(2/b^2 - (3*a^2)/b^4))/b) - (2*a*sin(c + d*x)^ 3)/(3*b^3) + (a*b^4 + a^5 - 2*a^3*b^2)/(b*(a*b^5 + b^6*sin(c + d*x))) + (l og(a + b*sin(c + d*x))*(5*a^4 + b^4 - 6*a^2*b^2))/b^6)/d
Time = 0.18 (sec) , antiderivative size = 490, normalized size of antiderivative = 3.12 \[ \int \frac {\cos ^5(c+d x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right ) a^{4} b +72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right ) a^{2} b^{3}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right ) b^{5}-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{5}+72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{3} b^{2}-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a \,b^{4}+60 \,\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^{4} b -72 \,\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^{2} b^{3}+12 \,\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 ) b^{5}+60 \,\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 ) a^{5}-72 \,\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 ) a^{3} b^{2}+12 \,\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 ) a \,b^{4}+3 \sin \left (d x +c \right )^{5} b^{5}-5 \sin \left (d x +c \right )^{4} a \,b^{4}+10 \sin \left (d x +c \right )^{3} a^{2} b^{3}-12 \sin \left (d x +c \right )^{3} b^{5}-30 \sin \left (d x +c \right )^{2} a^{3} b^{2}+36 \sin \left (d x +c \right )^{2} a \,b^{4}+60 a^{5}-72 a^{3} b^{2}+12 a \,b^{4}}{12 b^{6} d \left (\sin \left (d x +c \right ) b +a \right )} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)/(a+b*sin(d*x+c))^2,x)
Output:
( - 60*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a**4*b + 72*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a**2*b**3 - 12*log(tan((c + d*x)/2)**2 + 1)*s in(c + d*x)*b**5 - 60*log(tan((c + d*x)/2)**2 + 1)*a**5 + 72*log(tan((c + d*x)/2)**2 + 1)*a**3*b**2 - 12*log(tan((c + d*x)/2)**2 + 1)*a*b**4 + 60*lo g(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)*a**4*b - 72*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)*a**2 *b**3 + 12*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d *x)*b**5 + 60*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**5 - 72*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**3*b**2 + 12*l og(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a*b**4 + 3*sin(c + d* x)**5*b**5 - 5*sin(c + d*x)**4*a*b**4 + 10*sin(c + d*x)**3*a**2*b**3 - 12* sin(c + d*x)**3*b**5 - 30*sin(c + d*x)**2*a**3*b**2 + 36*sin(c + d*x)**2*a *b**4 + 60*a**5 - 72*a**3*b**2 + 12*a*b**4)/(12*b**6*d*(sin(c + d*x)*b + a ))