Integrand size = 29, antiderivative size = 235 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {a^2 \left (7 a^4-10 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))}{b^8 d}-\frac {2 a \left (3 a^4-4 a^2 b^2+b^4\right ) \sin (c+d x)}{b^7 d}+\frac {\left (5 a^4-6 a^2 b^2+b^4\right ) \sin ^2(c+d x)}{2 b^6 d}-\frac {4 a \left (a^2-b^2\right ) \sin ^3(c+d x)}{3 b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin ^4(c+d x)}{4 b^4 d}-\frac {2 a \sin ^5(c+d x)}{5 b^3 d}+\frac {\sin ^6(c+d x)}{6 b^2 d}+\frac {a^3 \left (a^2-b^2\right )^2}{b^8 d (a+b \sin (c+d x))} \] Output:
a^2*(7*a^4-10*a^2*b^2+3*b^4)*ln(a+b*sin(d*x+c))/b^8/d-2*a*(3*a^4-4*a^2*b^2 +b^4)*sin(d*x+c)/b^7/d+1/2*(5*a^4-6*a^2*b^2+b^4)*sin(d*x+c)^2/b^6/d-4/3*a* (a^2-b^2)*sin(d*x+c)^3/b^5/d+1/4*(3*a^2-2*b^2)*sin(d*x+c)^4/b^4/d-2/5*a*si n(d*x+c)^5/b^3/d+1/6*sin(d*x+c)^6/b^2/d+a^3*(a^2-b^2)^2/b^8/d/(a+b*sin(d*x +c))
Time = 2.24 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.12 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {60 a^3 \left (a^2-b^2\right ) \left (a^2-b^2+\left (7 a^2-3 b^2\right ) \log (a+b \sin (c+d x))\right )+60 a^2 b \left (a^2-b^2\right ) \left (-6 a^2+2 b^2+\left (7 a^2-3 b^2\right ) \log (a+b \sin (c+d x))\right ) \sin (c+d x)-30 a b^2 \left (7 a^4-10 a^2 b^2+3 b^4\right ) \sin ^2(c+d x)+10 b^3 \left (7 a^4-10 a^2 b^2+3 b^4\right ) \sin ^3(c+d x)+\left (-35 a^3 b^4+50 a b^6\right ) \sin ^4(c+d x)+3 b^5 \left (7 a^2-10 b^2\right ) \sin ^5(c+d x)-14 a b^6 \sin ^6(c+d x)+10 b^7 \sin ^7(c+d x)}{60 b^8 d (a+b \sin (c+d x))} \] Input:
Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + b*Sin[c + d*x])^2,x]
Output:
(60*a^3*(a^2 - b^2)*(a^2 - b^2 + (7*a^2 - 3*b^2)*Log[a + b*Sin[c + d*x]]) + 60*a^2*b*(a^2 - b^2)*(-6*a^2 + 2*b^2 + (7*a^2 - 3*b^2)*Log[a + b*Sin[c + d*x]])*Sin[c + d*x] - 30*a*b^2*(7*a^4 - 10*a^2*b^2 + 3*b^4)*Sin[c + d*x]^ 2 + 10*b^3*(7*a^4 - 10*a^2*b^2 + 3*b^4)*Sin[c + d*x]^3 + (-35*a^3*b^4 + 50 *a*b^6)*Sin[c + d*x]^4 + 3*b^5*(7*a^2 - 10*b^2)*Sin[c + d*x]^5 - 14*a*b^6* Sin[c + d*x]^6 + 10*b^7*Sin[c + d*x]^7)/(60*b^8*d*(a + b*Sin[c + d*x]))
Time = 0.48 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.89, 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.
| \(\displaystyle \int \frac {\sin ^3(c+d x) \cos ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^3 \cos (c+d x)^5}{(a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \frac {\sin ^3(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{(a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {b^3 \sin ^3(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{(a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^8 d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (b^5 \sin ^5(c+d x)-2 a b^4 \sin ^4(c+d x)+b^3 \left (3 a^2-2 b^2\right ) \sin ^3(c+d x)-4 a b^2 \left (a^2-b^2\right ) \sin ^2(c+d x)+b \left (5 a^4-6 b^2 a^2+b^4\right ) \sin (c+d x)-2 a \left (3 a^4-4 b^2 a^2+b^4\right )+\frac {7 a^6-10 b^2 a^4+3 b^4 a^2}{a+b \sin (c+d x)}-\frac {a^3 \left (a^2-b^2\right )^2}{(a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{b^8 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{4} b^4 \left (3 a^2-2 b^2\right ) \sin ^4(c+d x)-\frac {4}{3} a b^3 \left (a^2-b^2\right ) \sin ^3(c+d x)+\frac {1}{2} b^2 \left (5 a^4-6 a^2 b^2+b^4\right ) \sin ^2(c+d x)-2 a b \left (3 a^4-4 a^2 b^2+b^4\right ) \sin (c+d x)+a^2 \left (7 a^4-10 a^2 b^2+3 b^4\right ) \log (a+b \sin (c+d x))+\frac {a^3 \left (a^2-b^2\right )^2}{a+b \sin (c+d x)}-\frac {2}{5} a b^5 \sin ^5(c+d x)+\frac {1}{6} b^6 \sin ^6(c+d x)}{b^8 d}\) |
Input:
Int[(Cos[c + d*x]^5*Sin[c + d*x]^3)/(a + b*Sin[c + d*x])^2,x]
Output:
(a^2*(7*a^4 - 10*a^2*b^2 + 3*b^4)*Log[a + b*Sin[c + d*x]] - 2*a*b*(3*a^4 - 4*a^2*b^2 + b^4)*Sin[c + d*x] + (b^2*(5*a^4 - 6*a^2*b^2 + b^4)*Sin[c + d* x]^2)/2 - (4*a*b^3*(a^2 - b^2)*Sin[c + d*x]^3)/3 + (b^4*(3*a^2 - 2*b^2)*Si n[c + d*x]^4)/4 - (2*a*b^5*Sin[c + d*x]^5)/5 + (b^6*Sin[c + d*x]^6)/6 + (a ^3*(a^2 - b^2)^2)/(a + b*Sin[c + d*x]))/(b^8*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 = 1.83 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.06
| method | result | size |
| derivativedivides | \(\frac {-\frac {-\frac {\sin \left (d x +c \right )^{6} b^{5}}{6}+\frac {2 a \,b^{4} \sin \left (d x +c \right )^{5}}{5}-\frac {3 a^{2} b^{3} \sin \left (d x +c \right )^{4}}{4}+\frac {b^{5} \sin \left (d x +c \right )^{4}}{2}+\frac {4 b^{2} a^{3} \sin \left (d x +c \right )^{3}}{3}-\frac {4 a \,b^{4} \sin \left (d x +c \right )^{3}}{3}-\frac {5 a^{4} b \sin \left (d x +c \right )^{2}}{2}+3 a^{2} b^{3} \sin \left (d x +c \right )^{2}-\frac {b^{5} \sin \left (d x +c \right )^{2}}{2}+6 a^{5} \sin \left (d x +c \right )-8 a^{3} b^{2} \sin \left (d x +c \right )+2 \sin \left (d x +c \right ) a \,b^{4}}{b^{7}}+\frac {a^{3} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{b^{8} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a^{2} \left (7 a^{4}-10 a^{2} b^{2}+3 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{8}}}{d}\) | \(249\) |
| default | \(\frac {-\frac {-\frac {\sin \left (d x +c \right )^{6} b^{5}}{6}+\frac {2 a \,b^{4} \sin \left (d x +c \right )^{5}}{5}-\frac {3 a^{2} b^{3} \sin \left (d x +c \right )^{4}}{4}+\frac {b^{5} \sin \left (d x +c \right )^{4}}{2}+\frac {4 b^{2} a^{3} \sin \left (d x +c \right )^{3}}{3}-\frac {4 a \,b^{4} \sin \left (d x +c \right )^{3}}{3}-\frac {5 a^{4} b \sin \left (d x +c \right )^{2}}{2}+3 a^{2} b^{3} \sin \left (d x +c \right )^{2}-\frac {b^{5} \sin \left (d x +c \right )^{2}}{2}+6 a^{5} \sin \left (d x +c \right )-8 a^{3} b^{2} \sin \left (d x +c \right )+2 \sin \left (d x +c \right ) a \,b^{4}}{b^{7}}+\frac {a^{3} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{b^{8} \left (a +b \sin \left (d x +c \right )\right )}+\frac {a^{2} \left (7 a^{4}-10 a^{2} b^{2}+3 b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{8}}}{d}\) | \(249\) |
| parallelrisch | \(\frac {13440 \left (a^{2}-\frac {3 b^{2}}{7}\right ) \left (a +b \sin \left (d x +c \right )\right ) \left (a -b \right ) \left (a +b \right ) a^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-13440 \left (a^{2}-\frac {3 b^{2}}{7}\right ) \left (a +b \sin \left (d x +c \right )\right ) \left (a -b \right ) \left (a +b \right ) a^{2} \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (3360 a^{5} b^{2}-4240 a^{3} b^{4}+850 a \,b^{6}\right ) \cos \left (2 d x +2 c \right )+\left (-560 a^{4} b^{3}+590 a^{2} b^{5}-45 b^{7}\right ) \sin \left (3 d x +3 c \right )+\left (-140 a^{3} b^{4}+116 a \,b^{6}\right ) \cos \left (4 d x +4 c \right )+\left (42 a^{2} b^{5}-25 b^{7}\right ) \sin \left (5 d x +5 c \right )+14 a \,b^{6} \cos \left (6 d x +6 c \right )-5 b^{7} \sin \left (7 d x +7 c \right )+\left (-13440 a^{6} b +20880 a^{4} b^{3}-7740 a^{2} b^{5}+295 b^{7}\right ) \sin \left (d x +c \right )-3360 a^{5} b^{2}+4380 a^{3} b^{4}-980 a \,b^{6}}{1920 b^{8} d \left (a +b \sin \left (d x +c \right )\right )}\) | \(325\) |
| risch | \(\frac {2 a^{3} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) {\mathrm e}^{i \left (d x +c \right )}}{b^{8} d \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )}-\frac {a \sin \left (5 d x +5 c \right )}{40 b^{3} d}+\frac {3 \cos \left (4 d x +4 c \right ) a^{2}}{32 d \,b^{4}}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{3 b^{5} d}-\frac {5 a \sin \left (3 d x +3 c \right )}{24 b^{3} d}+\frac {7 a^{6} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{8} d}-\frac {10 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{6} d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{4} d}-\frac {7 i a^{6} x}{b^{8}}+\frac {10 i a^{4} x}{b^{6}}-\frac {3 i a^{2} x}{b^{4}}-\frac {5 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{4}}{8 b^{6} d}+\frac {9 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{16 b^{4} d}-\frac {5 \,{\mathrm e}^{-2 i \left (d x +c \right )} a^{4}}{8 b^{6} d}+\frac {9 \,{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{16 b^{4} d}-\frac {5 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{128 b^{2} d}-\frac {5 \,{\mathrm e}^{2 i \left (d x +c \right )}}{128 b^{2} d}-\frac {3 i a^{5} {\mathrm e}^{-i \left (d x +c \right )}}{b^{7} d}+\frac {7 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{5} d}-\frac {5 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{8 b^{3} d}+\frac {3 i a^{5} {\mathrm e}^{i \left (d x +c \right )}}{b^{7} d}-\frac {7 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 b^{5} d}+\frac {5 i a \,{\mathrm e}^{i \left (d x +c \right )}}{8 b^{3} d}-\frac {14 i a^{6} c}{b^{8} d}+\frac {20 i a^{4} c}{b^{6} d}-\frac {6 i a^{2} c}{b^{4} d}-\frac {\cos \left (6 d x +6 c \right )}{192 d \,b^{2}}-\frac {\cos \left (4 d x +4 c \right )}{32 d \,b^{2}}\) | \(601\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/b^7*(-1/6*sin(d*x+c)^6*b^5+2/5*a*b^4*sin(d*x+c)^5-3/4*a^2*b^3*sin( d*x+c)^4+1/2*b^5*sin(d*x+c)^4+4/3*b^2*a^3*sin(d*x+c)^3-4/3*a*b^4*sin(d*x+c )^3-5/2*a^4*b*sin(d*x+c)^2+3*a^2*b^3*sin(d*x+c)^2-1/2*b^5*sin(d*x+c)^2+6*a ^5*sin(d*x+c)-8*a^3*b^2*sin(d*x+c)+2*sin(d*x+c)*a*b^4)+a^3*(a^4-2*a^2*b^2+ b^4)/b^8/(a+b*sin(d*x+c))+a^2/b^8*(7*a^4-10*a^2*b^2+3*b^4)*ln(a+b*sin(d*x+ c)))
Time = 0.13 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.19 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {112 \, a b^{6} \cos \left (d x + c\right )^{6} + 480 \, a^{7} - 3240 \, a^{5} b^{2} + 3185 \, a^{3} b^{4} - 487 \, a b^{6} - 8 \, {\left (35 \, a^{3} b^{4} - 8 \, a b^{6}\right )} \cos \left (d x + c\right )^{4} + 16 \, {\left (105 \, a^{5} b^{2} - 115 \, a^{3} b^{4} + 16 \, a b^{6}\right )} \cos \left (d x + c\right )^{2} + 480 \, {\left (7 \, a^{7} - 10 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + {\left (7 \, a^{6} b - 10 \, a^{4} b^{3} + 3 \, a^{2} b^{5}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (80 \, b^{7} \cos \left (d x + c\right )^{6} - 168 \, a^{2} b^{5} \cos \left (d x + c\right )^{4} + 2880 \, a^{6} b - 3800 \, a^{4} b^{3} + 1007 \, a^{2} b^{5} - 25 \, b^{7} + 16 \, {\left (35 \, a^{4} b^{3} - 29 \, a^{2} b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{480 \, {\left (b^{9} d \sin \left (d x + c\right ) + a b^{8} d\right )}} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/480*(112*a*b^6*cos(d*x + c)^6 + 480*a^7 - 3240*a^5*b^2 + 3185*a^3*b^4 - 487*a*b^6 - 8*(35*a^3*b^4 - 8*a*b^6)*cos(d*x + c)^4 + 16*(105*a^5*b^2 - 11 5*a^3*b^4 + 16*a*b^6)*cos(d*x + c)^2 + 480*(7*a^7 - 10*a^5*b^2 + 3*a^3*b^4 + (7*a^6*b - 10*a^4*b^3 + 3*a^2*b^5)*sin(d*x + c))*log(b*sin(d*x + c) + a ) - (80*b^7*cos(d*x + c)^6 - 168*a^2*b^5*cos(d*x + c)^4 + 2880*a^6*b - 380 0*a^4*b^3 + 1007*a^2*b^5 - 25*b^7 + 16*(35*a^4*b^3 - 29*a^2*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^9*d*sin(d*x + c) + a*b^8*d)
Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**3/(a+b*sin(d*x+c))**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {60 \, {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )}}{b^{9} \sin \left (d x + c\right ) + a b^{8}} + \frac {10 \, b^{5} \sin \left (d x + c\right )^{6} - 24 \, a b^{4} \sin \left (d x + c\right )^{5} + 15 \, {\left (3 \, a^{2} b^{3} - 2 \, b^{5}\right )} \sin \left (d x + c\right )^{4} - 80 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )^{3} + 30 \, {\left (5 \, a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{2} - 120 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{b^{7}} + \frac {60 \, {\left (7 \, a^{6} - 10 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{8}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
1/60*(60*(a^7 - 2*a^5*b^2 + a^3*b^4)/(b^9*sin(d*x + c) + a*b^8) + (10*b^5* sin(d*x + c)^6 - 24*a*b^4*sin(d*x + c)^5 + 15*(3*a^2*b^3 - 2*b^5)*sin(d*x + c)^4 - 80*(a^3*b^2 - a*b^4)*sin(d*x + c)^3 + 30*(5*a^4*b - 6*a^2*b^3 + b ^5)*sin(d*x + c)^2 - 120*(3*a^5 - 4*a^3*b^2 + a*b^4)*sin(d*x + c))/b^7 + 6 0*(7*a^6 - 10*a^4*b^2 + 3*a^2*b^4)*log(b*sin(d*x + c) + a)/b^8)/d
Time = 0.21 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {{\left (7 \, a^{6} - 10 \, a^{4} b^{2} + 3 \, a^{2} b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{8} d} + \frac {a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{8} d} + \frac {10 \, b^{10} d^{5} \sin \left (d x + c\right )^{6} - 24 \, a b^{9} d^{5} \sin \left (d x + c\right )^{5} + 45 \, a^{2} b^{8} d^{5} \sin \left (d x + c\right )^{4} - 30 \, b^{10} d^{5} \sin \left (d x + c\right )^{4} - 80 \, a^{3} b^{7} d^{5} \sin \left (d x + c\right )^{3} + 80 \, a b^{9} d^{5} \sin \left (d x + c\right )^{3} + 150 \, a^{4} b^{6} d^{5} \sin \left (d x + c\right )^{2} - 180 \, a^{2} b^{8} d^{5} \sin \left (d x + c\right )^{2} + 30 \, b^{10} d^{5} \sin \left (d x + c\right )^{2} - 360 \, a^{5} b^{5} d^{5} \sin \left (d x + c\right ) + 480 \, a^{3} b^{7} d^{5} \sin \left (d x + c\right ) - 120 \, a b^{9} d^{5} \sin \left (d x + c\right )}{60 \, b^{12} d^{6}} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c))^2,x, algorithm="giac" )
Output:
(7*a^6 - 10*a^4*b^2 + 3*a^2*b^4)*log(abs(b*sin(d*x + c) + a))/(b^8*d) + (a ^7 - 2*a^5*b^2 + a^3*b^4)/((b*sin(d*x + c) + a)*b^8*d) + 1/60*(10*b^10*d^5 *sin(d*x + c)^6 - 24*a*b^9*d^5*sin(d*x + c)^5 + 45*a^2*b^8*d^5*sin(d*x + c )^4 - 30*b^10*d^5*sin(d*x + c)^4 - 80*a^3*b^7*d^5*sin(d*x + c)^3 + 80*a*b^ 9*d^5*sin(d*x + c)^3 + 150*a^4*b^6*d^5*sin(d*x + c)^2 - 180*a^2*b^8*d^5*si n(d*x + c)^2 + 30*b^10*d^5*sin(d*x + c)^2 - 360*a^5*b^5*d^5*sin(d*x + c) + 480*a^3*b^7*d^5*sin(d*x + c) - 120*a*b^9*d^5*sin(d*x + c))/(b^12*d^6)
Time = 22.48 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.60 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {2\,a^3}{3\,b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{3\,b}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^4\,\left (\frac {1}{2\,b^2}-\frac {3\,a^2}{4\,b^4}\right )}{d}+\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{2\,b^2}+\frac {a^2\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{2\,b^2}-\frac {a\,\left (\frac {2\,a^3}{b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )}{b}\right )}{d}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {a^2\,\left (\frac {2\,a^3}{b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )}{b^2}+\frac {2\,a\,\left (\frac {1}{b^2}+\frac {a^2\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b^2}-\frac {2\,a\,\left (\frac {2\,a^3}{b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )}{b}\right )}{b}\right )}{d}+\frac {{\sin \left (c+d\,x\right )}^6}{6\,b^2\,d}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^5}{5\,b^3\,d}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (7\,a^6-10\,a^4\,b^2+3\,a^2\,b^4\right )}{b^8\,d}+\frac {a^7-2\,a^5\,b^2+a^3\,b^4}{b\,d\,\left (\sin \left (c+d\,x\right )\,b^8+a\,b^7\right )} \] Input:
int((cos(c + d*x)^5*sin(c + d*x)^3)/(a + b*sin(c + d*x))^2,x)
Output:
(sin(c + d*x)^3*((2*a^3)/(3*b^5) + (2*a*(2/b^2 - (3*a^2)/b^4))/(3*b)))/d - (sin(c + d*x)^4*(1/(2*b^2) - (3*a^2)/(4*b^4)))/d + (sin(c + d*x)^2*(1/(2* b^2) + (a^2*(2/b^2 - (3*a^2)/b^4))/(2*b^2) - (a*((2*a^3)/b^5 + (2*a*(2/b^2 - (3*a^2)/b^4))/b))/b))/d - (sin(c + d*x)*((a^2*((2*a^3)/b^5 + (2*a*(2/b^ 2 - (3*a^2)/b^4))/b))/b^2 + (2*a*(1/b^2 + (a^2*(2/b^2 - (3*a^2)/b^4))/b^2 - (2*a*((2*a^3)/b^5 + (2*a*(2/b^2 - (3*a^2)/b^4))/b))/b))/b))/d + sin(c + d*x)^6/(6*b^2*d) - (2*a*sin(c + d*x)^5)/(5*b^3*d) + (log(a + b*sin(c + d*x ))*(7*a^6 + 3*a^2*b^4 - 10*a^4*b^2))/(b^8*d) + (a^7 + a^3*b^4 - 2*a^5*b^2) /(b*d*(a*b^7 + b^8*sin(c + d*x)))
Time = 0.20 (sec) , antiderivative size = 593, normalized size of antiderivative = 2.52 \[ \int \frac {\cos ^5(c+d x) \sin ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^3/(a+b*sin(d*x+c))^2,x)
Output:
( - 420*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a**6*b + 600*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a**4*b**3 - 180*log(tan((c + d*x)/2)**2 + 1 )*sin(c + d*x)*a**2*b**5 - 420*log(tan((c + d*x)/2)**2 + 1)*a**7 + 600*log (tan((c + d*x)/2)**2 + 1)*a**5*b**2 - 180*log(tan((c + d*x)/2)**2 + 1)*a** 3*b**4 + 420*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)*a**6*b - 600*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*s in(c + d*x)*a**4*b**3 + 180*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2) *b + a)*sin(c + d*x)*a**2*b**5 + 420*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**7 - 600*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2) *b + a)*a**5*b**2 + 180*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**3*b**4 + 10*sin(c + d*x)**7*b**7 - 14*sin(c + d*x)**6*a*b**6 + 21*s in(c + d*x)**5*a**2*b**5 - 30*sin(c + d*x)**5*b**7 - 35*sin(c + d*x)**4*a* *3*b**4 + 50*sin(c + d*x)**4*a*b**6 + 70*sin(c + d*x)**3*a**4*b**3 - 100*s in(c + d*x)**3*a**2*b**5 + 30*sin(c + d*x)**3*b**7 - 210*sin(c + d*x)**2*a **5*b**2 + 300*sin(c + d*x)**2*a**3*b**4 - 90*sin(c + d*x)**2*a*b**6 + 420 *a**7 - 600*a**5*b**2 + 180*a**3*b**4)/(60*b**8*d*(sin(c + d*x)*b + a))