Integrand size = 29, antiderivative size = 180 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{b^7 d}-\frac {a \left (a^2-b^2\right )^2 \sin (c+d x)}{b^6 d}+\frac {\left (a^2-b^2\right )^2 \sin ^2(c+d x)}{2 b^5 d}-\frac {a \left (a^2-2 b^2\right ) \sin ^3(c+d x)}{3 b^4 d}+\frac {\left (a^2-2 b^2\right ) \sin ^4(c+d x)}{4 b^3 d}-\frac {a \sin ^5(c+d x)}{5 b^2 d}+\frac {\sin ^6(c+d x)}{6 b d} \] Output:
a^2*(a^2-b^2)^2*ln(a+b*sin(d*x+c))/b^7/d-a*(a^2-b^2)^2*sin(d*x+c)/b^6/d+1/ 2*(a^2-b^2)^2*sin(d*x+c)^2/b^5/d-1/3*a*(a^2-2*b^2)*sin(d*x+c)^3/b^4/d+1/4* (a^2-2*b^2)*sin(d*x+c)^4/b^3/d-1/5*a*sin(d*x+c)^5/b^2/d+1/6*sin(d*x+c)^6/b /d
Time = 0.67 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.85 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {60 \left (a^3-a b^2\right )^2 \log (a+b \sin (c+d x))-60 a b \left (a^2-b^2\right )^2 \sin (c+d x)+30 b^2 \left (a^2-b^2\right )^2 \sin ^2(c+d x)-20 a b^3 \left (a^2-2 b^2\right ) \sin ^3(c+d x)+15 b^4 \left (a^2-2 b^2\right ) \sin ^4(c+d x)-12 a b^5 \sin ^5(c+d x)+10 b^6 \sin ^6(c+d x)}{60 b^7 d} \] Input:
Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
Output:
(60*(a^3 - a*b^2)^2*Log[a + b*Sin[c + d*x]] - 60*a*b*(a^2 - b^2)^2*Sin[c + d*x] + 30*b^2*(a^2 - b^2)^2*Sin[c + d*x]^2 - 20*a*b^3*(a^2 - 2*b^2)*Sin[c + d*x]^3 + 15*b^4*(a^2 - 2*b^2)*Sin[c + d*x]^4 - 12*a*b^5*Sin[c + d*x]^5 + 10*b^6*Sin[c + d*x]^6)/(60*b^7*d)
Time = 0.41 (sec) , antiderivative size = 161, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^5}{a+b \sin (c+d x)}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \frac {\sin ^2(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {b^2 \sin ^2(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^7 d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (b^5 \sin ^5(c+d x)-a b^4 \sin ^4(c+d x)+b^3 \left (a^2-2 b^2\right ) \sin ^3(c+d x)-a b^2 \left (a^2-2 b^2\right ) \sin ^2(c+d x)+b \left (a^2-b^2\right )^2 \sin (c+d x)-a \left (a^2-b^2\right )^2+\frac {\left (a^3-a b^2\right )^2}{a+b \sin (c+d x)}\right )d(b \sin (c+d x))}{b^7 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{2} b^2 \left (a^2-b^2\right )^2 \sin ^2(c+d x)-a b \left (a^2-b^2\right )^2 \sin (c+d x)+a^2 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))+\frac {1}{4} b^4 \left (a^2-2 b^2\right ) \sin ^4(c+d x)-\frac {1}{3} a b^3 \left (a^2-2 b^2\right ) \sin ^3(c+d x)-\frac {1}{5} a b^5 \sin ^5(c+d x)+\frac {1}{6} b^6 \sin ^6(c+d x)}{b^7 d}\) |
Input:
Int[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
Output:
(a^2*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]] - a*b*(a^2 - b^2)^2*Sin[c + d*x ] + (b^2*(a^2 - b^2)^2*Sin[c + d*x]^2)/2 - (a*b^3*(a^2 - 2*b^2)*Sin[c + d* x]^3)/3 + (b^4*(a^2 - 2*b^2)*Sin[c + d*x]^4)/4 - (a*b^5*Sin[c + d*x]^5)/5 + (b^6*Sin[c + d*x]^6)/6)/(b^7*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 = 1.27 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.34
| method | result | size |
| parallelrisch | \(\frac {960 a^{2} \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-960 a^{2} \left (a -b \right )^{2} \left (a +b \right )^{2} \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (-240 a^{4} b^{2}+360 a^{2} b^{4}-75 b^{6}\right ) \cos \left (2 d x +2 c \right )+\left (30 a^{2} b^{4}-30 b^{6}\right ) \cos \left (4 d x +4 c \right )+\left (80 a^{3} b^{3}-100 a \,b^{5}\right ) \sin \left (3 d x +3 c \right )-5 b^{6} \cos \left (6 d x +6 c \right )-12 a \,b^{5} \sin \left (5 d x +5 c \right )+\left (-960 a^{5} b +1680 a^{3} b^{3}-600 a \,b^{5}\right ) \sin \left (d x +c \right )+240 a^{4} b^{2}-390 a^{2} b^{4}+110 b^{6}}{960 b^{7} d}\) | \(242\) |
| derivativedivides | \(\frac {\sin \left (d x +c \right )^{6}}{6 b d}-\frac {a \sin \left (d x +c \right )^{5}}{5 b^{2} d}+\frac {\sin \left (d x +c \right )^{4} a^{2}}{4 d \,b^{3}}-\frac {\sin \left (d x +c \right )^{4}}{2 b d}-\frac {\sin \left (d x +c \right )^{3} a^{3}}{3 d \,b^{4}}+\frac {2 a \sin \left (d x +c \right )^{3}}{3 b^{2} d}+\frac {\sin \left (d x +c \right )^{2} a^{4}}{2 d \,b^{5}}-\frac {\sin \left (d x +c \right )^{2} a^{2}}{d \,b^{3}}+\frac {\sin \left (d x +c \right )^{2}}{2 b d}-\frac {a^{5} \sin \left (d x +c \right )}{d \,b^{6}}+\frac {2 a^{3} \sin \left (d x +c \right )}{d \,b^{4}}-\frac {a \sin \left (d x +c \right )}{b^{2} d}+\frac {a^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{7}}-\frac {2 a^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{5}}+\frac {a^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} d}\) | \(273\) |
| default | \(\frac {\sin \left (d x +c \right )^{6}}{6 b d}-\frac {a \sin \left (d x +c \right )^{5}}{5 b^{2} d}+\frac {\sin \left (d x +c \right )^{4} a^{2}}{4 d \,b^{3}}-\frac {\sin \left (d x +c \right )^{4}}{2 b d}-\frac {\sin \left (d x +c \right )^{3} a^{3}}{3 d \,b^{4}}+\frac {2 a \sin \left (d x +c \right )^{3}}{3 b^{2} d}+\frac {\sin \left (d x +c \right )^{2} a^{4}}{2 d \,b^{5}}-\frac {\sin \left (d x +c \right )^{2} a^{2}}{d \,b^{3}}+\frac {\sin \left (d x +c \right )^{2}}{2 b d}-\frac {a^{5} \sin \left (d x +c \right )}{d \,b^{6}}+\frac {2 a^{3} \sin \left (d x +c \right )}{d \,b^{4}}-\frac {a \sin \left (d x +c \right )}{b^{2} d}+\frac {a^{6} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{7}}-\frac {2 a^{4} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{5}}+\frac {a^{2} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{3} d}\) | \(273\) |
| risch | \(\frac {\cos \left (4 d x +4 c \right ) a^{2}}{32 b^{3} d}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 b^{4} d}+\frac {5 i a \,{\mathrm e}^{i \left (d x +c \right )}}{16 b^{2} d}-\frac {2 i a^{2} c}{d \,b^{3}}-\frac {7 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 b^{4} d}+\frac {7 i a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 b^{4} d}+\frac {4 i a^{4} c}{d \,b^{5}}-\frac {5 a \sin \left (3 d x +3 c \right )}{48 b^{2} d}-\frac {5 i a \,{\mathrm e}^{-i \left (d x +c \right )}}{16 b^{2} d}-\frac {a \sin \left (5 d x +5 c \right )}{80 b^{2} d}+\frac {2 i x \,a^{4}}{b^{5}}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{16 b^{3} d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{16 b^{3} d}-\frac {i a^{2} x}{b^{3}}-\frac {2 a^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \,b^{5}}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \,b^{3}}-\frac {i x \,a^{6}}{b^{7}}+\frac {a^{6} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{7} d}-\frac {{\mathrm e}^{2 i \left (d x +c \right )} a^{4}}{8 b^{5} d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} a^{4}}{8 b^{5} d}+\frac {i a^{5} {\mathrm e}^{i \left (d x +c \right )}}{2 b^{6} d}-\frac {i a^{5} {\mathrm e}^{-i \left (d x +c \right )}}{2 b^{6} d}-\frac {2 i a^{6} c}{b^{7} d}-\frac {\cos \left (6 d x +6 c \right )}{192 b d}-\frac {5 \,{\mathrm e}^{2 i \left (d x +c \right )}}{128 b d}-\frac {5 \,{\mathrm e}^{-2 i \left (d x +c \right )}}{128 b d}-\frac {\cos \left (4 d x +4 c \right )}{32 b d}\) | \(532\) |
| norman | \(\frac {\frac {\left (10 a^{4}-16 a^{2} b^{2}+2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d \,b^{5}}+\frac {\left (10 a^{4}-16 a^{2} b^{2}+2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d \,b^{5}}+\frac {\left (2 a^{4}-4 a^{2} b^{2}+2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{5} d}+\frac {\left (2 a^{4}-4 a^{2} b^{2}+2 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{b^{5} d}+\frac {\left (60 a^{4}-84 a^{2} b^{2}+20 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d \,b^{5}}+\frac {\left (60 a^{4}-84 a^{2} b^{2}+20 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d \,b^{5}}-\frac {2 a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{6} d}-\frac {2 a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{b^{6} d}-\frac {4 a \left (9 a^{4}-16 a^{2} b^{2}+5 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 b^{6} d}-\frac {4 a \left (9 a^{4}-16 a^{2} b^{2}+5 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 b^{6} d}-\frac {8 a \left (25 a^{4}-40 a^{2} b^{2}+13 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 b^{6} d}-\frac {2 a \left (225 a^{4}-370 a^{2} b^{2}+113 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 b^{6} d}-\frac {2 a \left (225 a^{4}-370 a^{2} b^{2}+113 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{15 b^{6} d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}+\frac {a^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{7} d}-\frac {a^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{b^{7} d}\) | \(599\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/960*(960*a^2*(a-b)^2*(a+b)^2*ln(2*b*tan(1/2*d*x+1/2*c)+a*sec(1/2*d*x+1/2 *c)^2)-960*a^2*(a-b)^2*(a+b)^2*ln(sec(1/2*d*x+1/2*c)^2)+(-240*a^4*b^2+360* a^2*b^4-75*b^6)*cos(2*d*x+2*c)+(30*a^2*b^4-30*b^6)*cos(4*d*x+4*c)+(80*a^3* b^3-100*a*b^5)*sin(3*d*x+3*c)-5*b^6*cos(6*d*x+6*c)-12*a*b^5*sin(5*d*x+5*c) +(-960*a^5*b+1680*a^3*b^3-600*a*b^5)*sin(d*x+c)+240*a^4*b^2-390*a^2*b^4+11 0*b^6)/b^7/d
Time = 0.11 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.91 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {10 \, b^{6} \cos \left (d x + c\right )^{6} - 15 \, a^{2} b^{4} \cos \left (d x + c\right )^{4} + 30 \, {\left (a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (d x + c\right )^{2} - 60 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 4 \, {\left (3 \, a b^{5} \cos \left (d x + c\right )^{4} + 15 \, a^{5} b - 25 \, a^{3} b^{3} + 8 \, a b^{5} - {\left (5 \, a^{3} b^{3} - 4 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, b^{7} d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
-1/60*(10*b^6*cos(d*x + c)^6 - 15*a^2*b^4*cos(d*x + c)^4 + 30*(a^4*b^2 - a ^2*b^4)*cos(d*x + c)^2 - 60*(a^6 - 2*a^4*b^2 + a^2*b^4)*log(b*sin(d*x + c) + a) + 4*(3*a*b^5*cos(d*x + c)^4 + 15*a^5*b - 25*a^3*b^3 + 8*a*b^5 - (5*a ^3*b^3 - 4*a*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^7*d)
Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**2/(a+b*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.96 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {10 \, b^{5} \sin \left (d x + c\right )^{6} - 12 \, a b^{4} \sin \left (d x + c\right )^{5} + 15 \, {\left (a^{2} b^{3} - 2 \, b^{5}\right )} \sin \left (d x + c\right )^{4} - 20 \, {\left (a^{3} b^{2} - 2 \, a b^{4}\right )} \sin \left (d x + c\right )^{3} + 30 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (d x + c\right )^{2} - 60 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \sin \left (d x + c\right )}{b^{6}} + \frac {60 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{7}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima" )
Output:
1/60*((10*b^5*sin(d*x + c)^6 - 12*a*b^4*sin(d*x + c)^5 + 15*(a^2*b^3 - 2*b ^5)*sin(d*x + c)^4 - 20*(a^3*b^2 - 2*a*b^4)*sin(d*x + c)^3 + 30*(a^4*b - 2 *a^2*b^3 + b^5)*sin(d*x + c)^2 - 60*(a^5 - 2*a^3*b^2 + a*b^4)*sin(d*x + c) )/b^6 + 60*(a^6 - 2*a^4*b^2 + a^2*b^4)*log(b*sin(d*x + c) + a)/b^7)/d
Time = 0.17 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.39 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{7} d} + \frac {10 \, b^{5} d^{5} \sin \left (d x + c\right )^{6} - 12 \, a b^{4} d^{5} \sin \left (d x + c\right )^{5} + 15 \, a^{2} b^{3} d^{5} \sin \left (d x + c\right )^{4} - 30 \, b^{5} d^{5} \sin \left (d x + c\right )^{4} - 20 \, a^{3} b^{2} d^{5} \sin \left (d x + c\right )^{3} + 40 \, a b^{4} d^{5} \sin \left (d x + c\right )^{3} + 30 \, a^{4} b d^{5} \sin \left (d x + c\right )^{2} - 60 \, a^{2} b^{3} d^{5} \sin \left (d x + c\right )^{2} + 30 \, b^{5} d^{5} \sin \left (d x + c\right )^{2} - 60 \, a^{5} d^{5} \sin \left (d x + c\right ) + 120 \, a^{3} b^{2} d^{5} \sin \left (d x + c\right ) - 60 \, a b^{4} d^{5} \sin \left (d x + c\right )}{60 \, b^{6} d^{6}} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
(a^6 - 2*a^4*b^2 + a^2*b^4)*log(abs(b*sin(d*x + c) + a))/(b^7*d) + 1/60*(1 0*b^5*d^5*sin(d*x + c)^6 - 12*a*b^4*d^5*sin(d*x + c)^5 + 15*a^2*b^3*d^5*si n(d*x + c)^4 - 30*b^5*d^5*sin(d*x + c)^4 - 20*a^3*b^2*d^5*sin(d*x + c)^3 + 40*a*b^4*d^5*sin(d*x + c)^3 + 30*a^4*b*d^5*sin(d*x + c)^2 - 60*a^2*b^3*d^ 5*sin(d*x + c)^2 + 30*b^5*d^5*sin(d*x + c)^2 - 60*a^5*d^5*sin(d*x + c) + 1 20*a^3*b^2*d^5*sin(d*x + c) - 60*a*b^4*d^5*sin(d*x + c))/(b^6*d^6)
Time = 20.07 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.06 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {1}{2\,b}-\frac {a^2\,\left (\frac {1}{b}-\frac {a^2}{2\,b^3}\right )}{b^2}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {1}{2\,b}-\frac {a^2}{4\,b^3}\right )+\frac {{\sin \left (c+d\,x\right )}^6}{6\,b}-\frac {a\,{\sin \left (c+d\,x\right )}^5}{5\,b^2}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (a^6-2\,a^4\,b^2+a^2\,b^4\right )}{b^7}-\frac {a\,\sin \left (c+d\,x\right )\,\left (\frac {1}{b}-\frac {a^2\,\left (\frac {2}{b}-\frac {a^2}{b^3}\right )}{b^2}\right )}{b}+\frac {a\,{\sin \left (c+d\,x\right )}^3\,\left (\frac {2}{b}-\frac {a^2}{b^3}\right )}{3\,b}}{d} \] Input:
int((cos(c + d*x)^5*sin(c + d*x)^2)/(a + b*sin(c + d*x)),x)
Output:
(sin(c + d*x)^2*(1/(2*b) - (a^2*(1/b - a^2/(2*b^3)))/b^2) - sin(c + d*x)^4 *(1/(2*b) - a^2/(4*b^3)) + sin(c + d*x)^6/(6*b) - (a*sin(c + d*x)^5)/(5*b^ 2) + (log(a + b*sin(c + d*x))*(a^6 + a^2*b^4 - 2*a^4*b^2))/b^7 - (a*sin(c + d*x)*(1/b - (a^2*(2/b - a^2/b^3))/b^2))/b + (a*sin(c + d*x)^3*(2/b - a^2 /b^3))/(3*b))/d
Time = 0.18 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.04 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{6}+120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{4} b^{2}-60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a^{2} b^{4}+60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{6}-120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{4} b^{2}+60 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a^{2} b^{4}+10 \sin \left (d x +c \right )^{6} b^{6}-12 \sin \left (d x +c \right )^{5} a \,b^{5}+15 \sin \left (d x +c \right )^{4} a^{2} b^{4}-30 \sin \left (d x +c \right )^{4} b^{6}-20 \sin \left (d x +c \right )^{3} a^{3} b^{3}+40 \sin \left (d x +c \right )^{3} a \,b^{5}+30 \sin \left (d x +c \right )^{2} a^{4} b^{2}-60 \sin \left (d x +c \right )^{2} a^{2} b^{4}+30 \sin \left (d x +c \right )^{2} b^{6}-60 \sin \left (d x +c \right ) a^{5} b +120 \sin \left (d x +c \right ) a^{3} b^{3}-60 \sin \left (d x +c \right ) a \,b^{5}-20 a^{4} b^{2}+40 a^{2} b^{4}-20 b^{6}}{60 b^{7} d} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c)),x)
Output:
( - 60*log(tan((c + d*x)/2)**2 + 1)*a**6 + 120*log(tan((c + d*x)/2)**2 + 1 )*a**4*b**2 - 60*log(tan((c + d*x)/2)**2 + 1)*a**2*b**4 + 60*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**6 - 120*log(tan((c + d*x)/2)** 2*a + 2*tan((c + d*x)/2)*b + a)*a**4*b**2 + 60*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**2*b**4 + 10*sin(c + d*x)**6*b**6 - 12*sin(c + d*x)**5*a*b**5 + 15*sin(c + d*x)**4*a**2*b**4 - 30*sin(c + d*x)**4*b**6 - 20*sin(c + d*x)**3*a**3*b**3 + 40*sin(c + d*x)**3*a*b**5 + 30*sin(c + d* x)**2*a**4*b**2 - 60*sin(c + d*x)**2*a**2*b**4 + 30*sin(c + d*x)**2*b**6 - 60*sin(c + d*x)*a**5*b + 120*sin(c + d*x)*a**3*b**3 - 60*sin(c + d*x)*a*b **5 - 20*a**4*b**2 + 40*a**2*b**4 - 20*b**6)/(60*b**7*d)