Integrand size = 29, antiderivative size = 194 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 a \left (3 a^4-4 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))}{b^7 d}+\frac {\left (5 a^4-6 a^2 b^2+b^4\right ) \sin (c+d x)}{b^6 d}-\frac {2 a \left (a^2-b^2\right ) \sin ^2(c+d x)}{b^5 d}+\frac {\left (3 a^2-2 b^2\right ) \sin ^3(c+d x)}{3 b^4 d}-\frac {a \sin ^4(c+d x)}{2 b^3 d}+\frac {\sin ^5(c+d x)}{5 b^2 d}-\frac {a^2 \left (a^2-b^2\right )^2}{b^7 d (a+b \sin (c+d x))} \] Output:
-2*a*(3*a^4-4*a^2*b^2+b^4)*ln(a+b*sin(d*x+c))/b^7/d+(5*a^4-6*a^2*b^2+b^4)* sin(d*x+c)/b^6/d-2*a*(a^2-b^2)*sin(d*x+c)^2/b^5/d+1/3*(3*a^2-2*b^2)*sin(d* x+c)^3/b^4/d-1/2*a*sin(d*x+c)^4/b^3/d+1/5*sin(d*x+c)^5/b^2/d-a^2*(a^2-b^2) ^2/b^7/d/(a+b*sin(d*x+c))
Time = 1.57 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.16 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-30 a^2 \left (a^2-b^2\right ) \left (a^2-b^2+\left (6 a^2-2 b^2\right ) \log (a+b \sin (c+d x))\right )-30 a b \left (a^2-b^2\right ) \left (-5 a^2+b^2+\left (6 a^2-2 b^2\right ) \log (a+b \sin (c+d x))\right ) \sin (c+d x)+30 b^2 \left (3 a^4-4 a^2 b^2+b^4\right ) \sin ^2(c+d x)+\left (-30 a^3 b^3+40 a b^5\right ) \sin ^3(c+d x)+5 b^4 \left (3 a^2-4 b^2\right ) \sin ^4(c+d x)-9 a b^5 \sin ^5(c+d x)+6 b^6 \sin ^6(c+d x)}{30 b^7 d (a+b \sin (c+d x))} \] Input:
Integrate[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]
Output:
(-30*a^2*(a^2 - b^2)*(a^2 - b^2 + (6*a^2 - 2*b^2)*Log[a + b*Sin[c + d*x]]) - 30*a*b*(a^2 - b^2)*(-5*a^2 + b^2 + (6*a^2 - 2*b^2)*Log[a + b*Sin[c + d* x]])*Sin[c + d*x] + 30*b^2*(3*a^4 - 4*a^2*b^2 + b^4)*Sin[c + d*x]^2 + (-30 *a^3*b^3 + 40*a*b^5)*Sin[c + d*x]^3 + 5*b^4*(3*a^2 - 4*b^2)*Sin[c + d*x]^4 - 9*a*b^5*Sin[c + d*x]^5 + 6*b^6*Sin[c + d*x]^6)/(30*b^7*d*(a + b*Sin[c + d*x]))
Time = 0.45 (sec) , antiderivative size = 172, 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.
| \(\displaystyle \int \frac {\sin ^2(c+d x) \cos ^5(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2 \cos (c+d x)^5}{(a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \frac {\sin ^2(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{(a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {b^2 \sin ^2(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{(a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^7 d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (5 \left (\frac {b^4-6 a^2 b^2}{5 a^4}+1\right ) a^4-2 b^3 \sin ^3(c+d x) a-4 b \left (a^2-b^2\right ) \sin (c+d x) a-\frac {2 \left (3 a^4-4 b^2 a^2+b^4\right ) a}{a+b \sin (c+d x)}+b^4 \sin ^4(c+d x)+b^2 \left (3 a^2-2 b^2\right ) \sin ^2(c+d x)+\frac {\left (a^3-a b^2\right )^2}{(a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{b^7 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-2 a b^2 \left (a^2-b^2\right ) \sin ^2(c+d x)-\frac {a^2 \left (a^2-b^2\right )^2}{a+b \sin (c+d x)}+\frac {1}{3} b^3 \left (3 a^2-2 b^2\right ) \sin ^3(c+d x)+b \left (5 a^4-6 a^2 b^2+b^4\right ) \sin (c+d x)-2 a \left (3 a^4-4 a^2 b^2+b^4\right ) \log (a+b \sin (c+d x))-\frac {1}{2} a b^4 \sin ^4(c+d x)+\frac {1}{5} b^5 \sin ^5(c+d x)}{b^7 d}\) |
Input:
Int[(Cos[c + d*x]^5*Sin[c + d*x]^2)/(a + b*Sin[c + d*x])^2,x]
Output:
(-2*a*(3*a^4 - 4*a^2*b^2 + b^4)*Log[a + b*Sin[c + d*x]] + b*(5*a^4 - 6*a^2 *b^2 + b^4)*Sin[c + d*x] - 2*a*b^2*(a^2 - b^2)*Sin[c + d*x]^2 + (b^3*(3*a^ 2 - 2*b^2)*Sin[c + d*x]^3)/3 - (a*b^4*Sin[c + d*x]^4)/2 + (b^5*Sin[c + d*x ]^5)/5 - (a^2*(a^2 - b^2)^2)/(a + b*Sin[c + d*x]))/(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 = 1.54 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.02
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {\sin \left (d x +c \right )^{5} b^{4}}{5}-\frac {a \sin \left (d x +c \right )^{4} b^{3}}{2}+a^{2} b^{2} \sin \left (d x +c \right )^{3}-\frac {2 b^{4} \sin \left (d x +c \right )^{3}}{3}-2 a^{3} b \sin \left (d x +c \right )^{2}+2 a \,b^{3} \sin \left (d x +c \right )^{2}+5 a^{4} \sin \left (d x +c \right )-6 \sin \left (d x +c \right ) a^{2} b^{2}+b^{4} \sin \left (d x +c \right )}{b^{6}}-\frac {a^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{b^{7} \left (a +b \sin \left (d x +c \right )\right )}-\frac {2 a \left (3 a^{4}-4 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{7}}}{d}\) | \(198\) |
| default | \(\frac {\frac {\frac {\sin \left (d x +c \right )^{5} b^{4}}{5}-\frac {a \sin \left (d x +c \right )^{4} b^{3}}{2}+a^{2} b^{2} \sin \left (d x +c \right )^{3}-\frac {2 b^{4} \sin \left (d x +c \right )^{3}}{3}-2 a^{3} b \sin \left (d x +c \right )^{2}+2 a \,b^{3} \sin \left (d x +c \right )^{2}+5 a^{4} \sin \left (d x +c \right )-6 \sin \left (d x +c \right ) a^{2} b^{2}+b^{4} \sin \left (d x +c \right )}{b^{6}}-\frac {a^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}{b^{7} \left (a +b \sin \left (d x +c \right )\right )}-\frac {2 a \left (3 a^{4}-4 a^{2} b^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{7}}}{d}\) | \(198\) |
| parallelrisch | \(\frac {-2880 \left (a +b \sin \left (d x +c \right )\right ) \left (a^{2}-\frac {b^{2}}{3}\right ) \left (a -b \right ) \left (a +b \right ) a \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+2880 \left (a +b \sin \left (d x +c \right )\right ) \left (a^{2}-\frac {b^{2}}{3}\right ) \left (a -b \right ) \left (a +b \right ) a \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )+\left (-720 a^{4} b^{2}+840 a^{2} b^{4}-125 b^{6}\right ) \cos \left (2 d x +2 c \right )+\left (30 a^{2} b^{4}-22 b^{6}\right ) \cos \left (4 d x +4 c \right )+\left (120 a^{3} b^{3}-115 a \,b^{5}\right ) \sin \left (3 d x +3 c \right )-3 b^{6} \cos \left (6 d x +6 c \right )-9 a \,b^{5} \sin \left (5 d x +5 c \right )+\left (2880 a^{5} b -4200 a^{3} b^{3}+1350 a \,b^{5}\right ) \sin \left (d x +c \right )+720 a^{4} b^{2}-870 a^{2} b^{4}+150 b^{6}}{480 b^{7} d \left (a +b \sin \left (d x +c \right )\right )}\) | \(280\) |
| risch | \(-\frac {5 i {\mathrm e}^{i \left (d x +c \right )} a^{4}}{2 b^{6} d}+\frac {21 i {\mathrm e}^{i \left (d x +c \right )} a^{2}}{8 b^{4} d}+\frac {5 i {\mathrm e}^{-i \left (d x +c \right )} a^{4}}{2 b^{6} d}-\frac {21 i {\mathrm e}^{-i \left (d x +c \right )} a^{2}}{8 b^{4} d}-\frac {a \cos \left (4 d x +4 c \right )}{16 b^{3} d}-\frac {2 a^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) {\mathrm e}^{i \left (d x +c \right )}}{b^{7} d \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )}-\frac {\sin \left (3 d x +3 c \right ) a^{2}}{4 b^{4} d}+\frac {6 i x \,a^{5}}{b^{7}}-\frac {8 i x \,a^{3}}{b^{5}}+\frac {2 i x a}{b^{3}}-\frac {6 a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{7} d}+\frac {8 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{5} d}-\frac {2 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{3} d}+\frac {12 i a^{5} c}{b^{7} d}-\frac {16 i a^{3} c}{b^{5} d}+\frac {4 i a c}{b^{3} d}+\frac {a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{2 b^{5} d}-\frac {3 a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{3} d}-\frac {5 i {\mathrm e}^{i \left (d x +c \right )}}{16 b^{2} d}+\frac {5 i {\mathrm e}^{-i \left (d x +c \right )}}{16 b^{2} d}+\frac {a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{2 b^{5} d}-\frac {3 a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{3} d}+\frac {\sin \left (5 d x +5 c \right )}{80 b^{2} d}+\frac {5 \sin \left (3 d x +3 c \right )}{48 b^{2} d}\) | \(517\) |
| norman | \(\frac {\frac {4 \left (54 a^{4}-66 a^{2} b^{2}+10 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 d \,b^{5}}+\frac {4 \left (54 a^{4}-66 a^{2} b^{2}+10 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{3 d \,b^{5}}+\frac {2 \left (600 a^{4}-680 a^{2} b^{2}+104 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{5 d \,b^{5}}+\frac {4 \left (675 a^{4}-780 a^{2} b^{2}+113 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{15 d \,b^{5}}+\frac {4 \left (675 a^{4}-780 a^{2} b^{2}+113 b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{15 d \,b^{5}}+\frac {4 \left (3 a^{4}-4 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{5} d}+\frac {4 \left (3 a^{4}-4 a^{2} b^{2}+b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{b^{5} d}+\frac {2 \left (126 a^{6}-180 a^{4} b^{2}+58 a^{2} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a \,b^{6} d}+\frac {2 \left (126 a^{6}-180 a^{4} b^{2}+58 a^{2} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 a \,b^{6} d}+\frac {2 \left (1890 a^{6}-2820 a^{4} b^{2}+958 a^{2} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 a \,b^{6} d}+\frac {2 \left (1890 a^{6}-2820 a^{4} b^{2}+958 a^{2} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{15 a \,b^{6} d}+\frac {2 \left (3150 a^{6}-4800 a^{4} b^{2}+1634 a^{2} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 a \,b^{6} d}+\frac {2 \left (3150 a^{6}-4800 a^{4} b^{2}+1634 a^{2} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{15 a \,b^{6} d}+\frac {2 \left (6 a^{6}-8 a^{4} b^{2}+2 a^{2} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{6} d a}+\frac {2 \left (6 a^{6}-8 a^{4} b^{2}+2 a^{2} b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{b^{6} d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7} \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 {2 a \left (3 a^{4}-4 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}-\frac {2 a \left (3 a^{4}-4 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}\) | \(754\) |
Input:
int(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(1/b^6*(1/5*sin(d*x+c)^5*b^4-1/2*a*sin(d*x+c)^4*b^3+a^2*b^2*sin(d*x+c) ^3-2/3*b^4*sin(d*x+c)^3-2*a^3*b*sin(d*x+c)^2+2*a*b^3*sin(d*x+c)^2+5*a^4*si n(d*x+c)-6*sin(d*x+c)*a^2*b^2+b^4*sin(d*x+c))-a^2*(a^4-2*a^2*b^2+b^4)/b^7/ (a+b*sin(d*x+c))-2*a/b^7*(3*a^4-4*a^2*b^2+b^4)*ln(a+b*sin(d*x+c)))
Time = 0.12 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {48 \, b^{6} \cos \left (d x + c\right )^{6} + 240 \, a^{6} - 1440 \, a^{4} b^{2} + 1275 \, a^{2} b^{4} - 128 \, b^{6} - 8 \, {\left (15 \, a^{2} b^{4} - 2 \, b^{6}\right )} \cos \left (d x + c\right )^{4} + 16 \, {\left (45 \, a^{4} b^{2} - 45 \, a^{2} b^{4} + 4 \, b^{6}\right )} \cos \left (d x + c\right )^{2} + 480 \, {\left (3 \, a^{6} - 4 \, a^{4} b^{2} + a^{2} b^{4} + {\left (3 \, a^{5} b - 4 \, a^{3} b^{3} + a b^{5}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (72 \, a b^{5} \cos \left (d x + c\right )^{4} - 1200 \, a^{5} b + 1440 \, a^{3} b^{3} - 293 \, a b^{5} - 16 \, {\left (15 \, a^{3} b^{3} - 11 \, a b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{240 \, {\left (b^{8} d \sin \left (d x + c\right ) + a b^{7} d\right )}} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/240*(48*b^6*cos(d*x + c)^6 + 240*a^6 - 1440*a^4*b^2 + 1275*a^2*b^4 - 12 8*b^6 - 8*(15*a^2*b^4 - 2*b^6)*cos(d*x + c)^4 + 16*(45*a^4*b^2 - 45*a^2*b^ 4 + 4*b^6)*cos(d*x + c)^2 + 480*(3*a^6 - 4*a^4*b^2 + a^2*b^4 + (3*a^5*b - 4*a^3*b^3 + a*b^5)*sin(d*x + c))*log(b*sin(d*x + c) + a) + (72*a*b^5*cos(d *x + c)^4 - 1200*a^5*b + 1440*a^3*b^3 - 293*a*b^5 - 16*(15*a^3*b^3 - 11*a* b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^8*d*sin(d*x + c) + a*b^7*d)
Timed out. \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*sin(d*x+c)**2/(a+b*sin(d*x+c))**2,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {30 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )}}{b^{8} \sin \left (d x + c\right ) + a b^{7}} - \frac {6 \, b^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{3} \sin \left (d x + c\right )^{4} + 10 \, {\left (3 \, a^{2} b^{2} - 2 \, b^{4}\right )} \sin \left (d x + c\right )^{3} - 60 \, {\left (a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{2} + 30 \, {\left (5 \, a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )}{b^{6}} + \frac {60 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2} + a b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{7}}}{30 \, d} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-1/30*(30*(a^6 - 2*a^4*b^2 + a^2*b^4)/(b^8*sin(d*x + c) + a*b^7) - (6*b^4* sin(d*x + c)^5 - 15*a*b^3*sin(d*x + c)^4 + 10*(3*a^2*b^2 - 2*b^4)*sin(d*x + c)^3 - 60*(a^3*b - a*b^3)*sin(d*x + c)^2 + 30*(5*a^4 - 6*a^2*b^2 + b^4)* sin(d*x + c))/b^6 + 60*(3*a^5 - 4*a^3*b^2 + a*b^4)*log(b*sin(d*x + c) + a) /b^7)/d
Time = 0.22 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {2 \, {\left (3 \, a^{5} - 4 \, a^{3} b^{2} + a b^{4}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{7} d} - \frac {a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}}{{\left (b \sin \left (d x + c\right ) + a\right )} b^{7} d} + \frac {6 \, b^{8} d^{4} \sin \left (d x + c\right )^{5} - 15 \, a b^{7} d^{4} \sin \left (d x + c\right )^{4} + 30 \, a^{2} b^{6} d^{4} \sin \left (d x + c\right )^{3} - 20 \, b^{8} d^{4} \sin \left (d x + c\right )^{3} - 60 \, a^{3} b^{5} d^{4} \sin \left (d x + c\right )^{2} + 60 \, a b^{7} d^{4} \sin \left (d x + c\right )^{2} + 150 \, a^{4} b^{4} d^{4} \sin \left (d x + c\right ) - 180 \, a^{2} b^{6} d^{4} \sin \left (d x + c\right ) + 30 \, b^{8} d^{4} \sin \left (d x + c\right )}{30 \, b^{10} d^{5}} \] Input:
integrate(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c))^2,x, algorithm="giac" )
Output:
-2*(3*a^5 - 4*a^3*b^2 + a*b^4)*log(abs(b*sin(d*x + c) + a))/(b^7*d) - (a^6 - 2*a^4*b^2 + a^2*b^4)/((b*sin(d*x + c) + a)*b^7*d) + 1/30*(6*b^8*d^4*sin (d*x + c)^5 - 15*a*b^7*d^4*sin(d*x + c)^4 + 30*a^2*b^6*d^4*sin(d*x + c)^3 - 20*b^8*d^4*sin(d*x + c)^3 - 60*a^3*b^5*d^4*sin(d*x + c)^2 + 60*a*b^7*d^4 *sin(d*x + c)^2 + 150*a^4*b^4*d^4*sin(d*x + c) - 180*a^2*b^6*d^4*sin(d*x + c) + 30*b^8*d^4*sin(d*x + c))/(b^10*d^5)
Time = 22.55 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.31 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {a^3}{b^5}+\frac {a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {2}{3\,b^2}-\frac {a^2}{b^4}\right )}{d}+\frac {\sin \left (c+d\,x\right )\,\left (\frac {1}{b^2}+\frac {a^2\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b^2}-\frac {2\,a\,\left (\frac {2\,a^3}{b^5}+\frac {2\,a\,\left (\frac {2}{b^2}-\frac {3\,a^2}{b^4}\right )}{b}\right )}{b}\right )}{d}+\frac {{\sin \left (c+d\,x\right )}^5}{5\,b^2\,d}-\frac {a\,{\sin \left (c+d\,x\right )}^4}{2\,b^3\,d}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (6\,a^5-8\,a^3\,b^2+2\,a\,b^4\right )}{b^7\,d}-\frac {a^6-2\,a^4\,b^2+a^2\,b^4}{b\,d\,\left (\sin \left (c+d\,x\right )\,b^7+a\,b^6\right )} \] Input:
int((cos(c + d*x)^5*sin(c + d*x)^2)/(a + b*sin(c + d*x))^2,x)
Output:
(sin(c + d*x)^2*(a^3/b^5 + (a*(2/b^2 - (3*a^2)/b^4))/b))/d - (sin(c + d*x) ^3*(2/(3*b^2) - a^2/b^4))/d + (sin(c + d*x)*(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))/d + s in(c + d*x)^5/(5*b^2*d) - (a*sin(c + d*x)^4)/(2*b^3*d) - (log(a + b*sin(c + d*x))*(2*a*b^4 + 6*a^5 - 8*a^3*b^2))/(b^7*d) - (a^6 + a^2*b^4 - 2*a^4*b^ 2)/(b*d*(a*b^6 + b^7*sin(c + d*x)))
Time = 0.18 (sec) , antiderivative size = 543, normalized size of antiderivative = 2.80 \[ \int \frac {\cos ^5(c+d x) \sin ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx =\text {Too large to display} \] Input:
int(cos(d*x+c)^5*sin(d*x+c)^2/(a+b*sin(d*x+c))^2,x)
Output:
(180*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)*a**5*b - 240*log(tan((c + d *x)/2)**2 + 1)*sin(c + d*x)*a**3*b**3 + 60*log(tan((c + d*x)/2)**2 + 1)*si n(c + d*x)*a*b**5 + 180*log(tan((c + d*x)/2)**2 + 1)*a**6 - 240*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 - 180*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)*a* *5*b + 240*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d *x)*a**3*b**3 - 60*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*s in(c + d*x)*a*b**5 - 180*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*a**6 + 240*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 + 6*sin(c + d*x)**6*b**6 - 9*sin(c + d*x)**5*a*b**5 + 15*sin(c + d*x)**4* a**2*b**4 - 20*sin(c + d*x)**4*b**6 - 30*sin(c + d*x)**3*a**3*b**3 + 40*si n(c + d*x)**3*a*b**5 + 90*sin(c + d*x)**2*a**4*b**2 - 120*sin(c + d*x)**2* a**2*b**4 + 30*sin(c + d*x)**2*b**6 - 180*a**6 + 240*a**4*b**2 - 60*a**2*b **4)/(30*b**7*d*(sin(c + d*x)*b + a))