Integrand size = 31, antiderivative size = 202 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\left (a^2-b^2\right )^2 (A b-a B) \log (a+b \sin (c+d x))}{b^6 d}-\frac {\left (a^3 A b-2 a A b^3-a^4 B+2 a^2 b^2 B-b^4 B\right ) \sin (c+d x)}{b^5 d}+\frac {\left (a^2-2 b^2\right ) (A b-a B) \sin ^2(c+d x)}{2 b^4 d}-\frac {\left (a A b-a^2 B+2 b^2 B\right ) \sin ^3(c+d x)}{3 b^3 d}+\frac {(A b-a B) \sin ^4(c+d x)}{4 b^2 d}+\frac {B \sin ^5(c+d x)}{5 b d} \] Output:
(a^2-b^2)^2*(A*b-B*a)*ln(a+b*sin(d*x+c))/b^6/d-(A*a^3*b-2*A*a*b^3-B*a^4+2* B*a^2*b^2-B*b^4)*sin(d*x+c)/b^5/d+1/2*(a^2-2*b^2)*(A*b-B*a)*sin(d*x+c)^2/b ^4/d-1/3*(A*a*b-B*a^2+2*B*b^2)*sin(d*x+c)^3/b^3/d+1/4*(A*b-B*a)*sin(d*x+c) ^4/b^2/d+1/5*B*sin(d*x+c)^5/b/d
Time = 0.47 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {20 (A b-a B) \left (3 b^4 \cos ^4(c+d x)+12 \left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))-12 a b \left (a^2-2 b^2\right ) \sin (c+d x)+6 b^2 \left (a^2-b^2\right ) \sin ^2(c+d x)-4 a b^3 \sin ^3(c+d x)\right )+b^5 B (150 \sin (c+d x)+25 \sin (3 (c+d x))+3 \sin (5 (c+d x)))}{240 b^6 d} \] Input:
Integrate[(Cos[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + b*Sin[c + d*x]),x]
Output:
(20*(A*b - a*B)*(3*b^4*Cos[c + d*x]^4 + 12*(a^2 - b^2)^2*Log[a + b*Sin[c + d*x]] - 12*a*b*(a^2 - 2*b^2)*Sin[c + d*x] + 6*b^2*(a^2 - b^2)*Sin[c + d*x ]^2 - 4*a*b^3*Sin[c + d*x]^3) + b^5*B*(150*Sin[c + d*x] + 25*Sin[3*(c + d* x)] + 3*Sin[5*(c + d*x)]))/(240*b^6*d)
Time = 0.45 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3316, 27, 652, 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.
\(\displaystyle \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^5 (A+B \sin (c+d x))}{a+b \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3316 |
\(\displaystyle \frac {\int \frac {(A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b (a+b \sin (c+d x))}d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{a+b \sin (c+d x)}d(b \sin (c+d x))}{b^6 d}\) |
\(\Big \downarrow \) 652 |
\(\displaystyle \frac {\int \left (b^4 B \sin ^4(c+d x)+b^3 (A b-a B) \sin ^3(c+d x)+b^2 \left (B a^2-A b a-2 b^2 B\right ) \sin ^2(c+d x)-b \left (a^2-2 b^2\right ) (a B-A b) \sin (c+d x)-a^3 A b \left (1-\frac {2 b^2 \left (\frac {B \left (a^2-b^2\right )^2}{2 a A b^3}+1\right )}{a^2}\right )-\frac {\left (a^2-b^2\right )^2 (a B-A b)}{a+b \sin (c+d x)}\right )d(b \sin (c+d x))}{b^6 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {1}{2} b^2 \left (a^2-2 b^2\right ) (A b-a B) \sin ^2(c+d x)+\left (a^2-b^2\right )^2 (A b-a B) \log (a+b \sin (c+d x))-\frac {1}{3} b^3 \left (a^2 (-B)+a A b+2 b^2 B\right ) \sin ^3(c+d x)-b \left (a^4 (-B)+a^3 A b+2 a^2 b^2 B-2 a A b^3-b^4 B\right ) \sin (c+d x)+\frac {1}{4} b^4 (A b-a B) \sin ^4(c+d x)+\frac {1}{5} b^5 B \sin ^5(c+d x)}{b^6 d}\) |
Input:
Int[(Cos[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + b*Sin[c + d*x]),x]
Output:
((a^2 - b^2)^2*(A*b - a*B)*Log[a + b*Sin[c + d*x]] - b*(a^3*A*b - 2*a*A*b^ 3 - a^4*B + 2*a^2*b^2*B - b^4*B)*Sin[c + d*x] + (b^2*(a^2 - 2*b^2)*(A*b - a*B)*Sin[c + d*x]^2)/2 - (b^3*(a*A*b - a^2*B + 2*b^2*B)*Sin[c + d*x]^3)/3 + (b^4*(A*b - a*B)*Sin[c + d*x]^4)/4 + (b^5*B*Sin[c + d*x]^5)/5)/(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[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ )^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + c *x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, 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 = 1.46 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.26
method | result | size |
parallelrisch | \(\frac {\left (a -b \right )^{2} \left (a +b \right )^{2} \left (A b -B a \right ) \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-\left (a -b \right )^{2} \left (a +b \right )^{2} \left (A b -B a \right ) \ln \left (\sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )-b \left (\frac {b \left (a^{2}-\frac {3 b^{2}}{2}\right ) \left (A b -B a \right ) \cos \left (2 d x +2 c \right )}{4}-\frac {b^{2} \left (A a b -a^{2} B +\frac {5}{4} B \,b^{2}\right ) \sin \left (3 d x +3 c \right )}{12}-\frac {b^{3} \left (A b -B a \right ) \cos \left (4 d x +4 c \right )}{32}-\frac {B \,b^{4} \sin \left (5 d x +5 c \right )}{80}+\left (A \,a^{3} b -\frac {7}{4} A a \,b^{3}-B \,a^{4}+\frac {7}{4} B \,a^{2} b^{2}-\frac {5}{8} B \,b^{4}\right ) \sin \left (d x +c \right )-\frac {b \left (a^{2}-\frac {13 b^{2}}{8}\right ) \left (A b -B a \right )}{4}\right )}{b^{6} d}\) | \(254\) |
derivativedivides | \(\frac {-\frac {-\frac {B \sin \left (d x +c \right )^{5} b^{4}}{5}-\frac {A \,b^{4} \sin \left (d x +c \right )^{4}}{4}+\frac {B a \,b^{3} \sin \left (d x +c \right )^{4}}{4}+\frac {A a \,b^{3} \sin \left (d x +c \right )^{3}}{3}-\frac {B \,a^{2} b^{2} \sin \left (d x +c \right )^{3}}{3}+\frac {2 B \,b^{4} \sin \left (d x +c \right )^{3}}{3}-\frac {A \,a^{2} b^{2} \sin \left (d x +c \right )^{2}}{2}+A \,b^{4} \sin \left (d x +c \right )^{2}+\frac {B \,a^{3} b \sin \left (d x +c \right )^{2}}{2}-B a \,b^{3} \sin \left (d x +c \right )^{2}+A \,a^{3} b \sin \left (d x +c \right )-2 A a \,b^{3} \sin \left (d x +c \right )-B \,a^{4} \sin \left (d x +c \right )+2 B \,a^{2} b^{2} \sin \left (d x +c \right )-B \,b^{4} \sin \left (d x +c \right )}{b^{5}}+\frac {\left (A \,a^{4} b -2 A \,a^{2} b^{3}+A \,b^{5}-B \,a^{5}+2 B \,a^{3} b^{2}-B a \,b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}}{d}\) | \(283\) |
default | \(\frac {-\frac {-\frac {B \sin \left (d x +c \right )^{5} b^{4}}{5}-\frac {A \,b^{4} \sin \left (d x +c \right )^{4}}{4}+\frac {B a \,b^{3} \sin \left (d x +c \right )^{4}}{4}+\frac {A a \,b^{3} \sin \left (d x +c \right )^{3}}{3}-\frac {B \,a^{2} b^{2} \sin \left (d x +c \right )^{3}}{3}+\frac {2 B \,b^{4} \sin \left (d x +c \right )^{3}}{3}-\frac {A \,a^{2} b^{2} \sin \left (d x +c \right )^{2}}{2}+A \,b^{4} \sin \left (d x +c \right )^{2}+\frac {B \,a^{3} b \sin \left (d x +c \right )^{2}}{2}-B a \,b^{3} \sin \left (d x +c \right )^{2}+A \,a^{3} b \sin \left (d x +c \right )-2 A a \,b^{3} \sin \left (d x +c \right )-B \,a^{4} \sin \left (d x +c \right )+2 B \,a^{2} b^{2} \sin \left (d x +c \right )-B \,b^{4} \sin \left (d x +c \right )}{b^{5}}+\frac {\left (A \,a^{4} b -2 A \,a^{2} b^{3}+A \,b^{5}-B \,a^{5}+2 B \,a^{3} b^{2}-B a \,b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{6}}}{d}\) | \(283\) |
norman | \(\frac {\frac {\left (8 A \,a^{2} b -12 A \,b^{3}-8 a^{3} B +12 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{b^{4} d}+\frac {\left (8 A \,a^{2} b -12 A \,b^{3}-8 a^{3} B +12 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{b^{4} d}-\frac {2 \left (15 A \,a^{3} b -26 A a \,b^{3}-15 B \,a^{4}+26 B \,a^{2} b^{2}-7 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 b^{5} d}-\frac {2 \left (15 A \,a^{3} b -26 A a \,b^{3}-15 B \,a^{4}+26 B \,a^{2} b^{2}-7 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 b^{5} d}-\frac {4 \left (25 A \,a^{3} b -40 A a \,b^{3}-25 B \,a^{4}+40 B \,a^{2} b^{2}-13 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 b^{5} d}-\frac {4 \left (25 A \,a^{3} b -40 A a \,b^{3}-25 B \,a^{4}+40 B \,a^{2} b^{2}-13 B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 b^{5} d}+\frac {4 \left (3 A \,a^{2} b -4 A \,b^{3}-3 a^{3} B +4 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{b^{4} d}-\frac {2 \left (A \,a^{3} b -2 A a \,b^{3}-B \,a^{4}+2 B \,a^{2} b^{2}-B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{5} d}-\frac {2 \left (A \,a^{3} b -2 A a \,b^{3}-B \,a^{4}+2 B \,a^{2} b^{2}-B \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{b^{5} d}+\frac {2 \left (A \,a^{2} b -2 A \,b^{3}-a^{3} B +2 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{4} d}+\frac {2 \left (A \,a^{2} b -2 A \,b^{3}-a^{3} B +2 B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{b^{4} d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}+\frac {\left (A \,a^{4} b -2 A \,a^{2} b^{3}+A \,b^{5}-B \,a^{5}+2 B \,a^{3} b^{2}-B a \,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^{6} d}-\frac {\left (A \,a^{4} b -2 A \,a^{2} b^{3}+A \,b^{5}-B \,a^{5}+2 B \,a^{3} b^{2}-B a \,b^{4}\right ) \ln \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{b^{6} d}\) | \(714\) |
risch | \(\frac {\sin \left (5 d x +5 c \right ) B}{80 d b}+\frac {\cos \left (4 d x +4 c \right ) A}{32 b d}+\frac {5 \sin \left (3 d x +3 c \right ) B}{48 b d}+\frac {4 i A \,a^{2} c}{b^{3} d}+\frac {2 i B \,a^{5} c}{b^{6} d}-\frac {4 i B \,a^{3} c}{b^{4} d}+\frac {2 i B a c}{d \,b^{2}}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{3}}{2 b^{4} d}+\frac {7 i {\mathrm e}^{-i \left (d x +c \right )} A a}{8 b^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{4}}{2 b^{5} d}-\frac {7 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2}}{8 b^{3} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,a^{3}}{2 b^{4} d}-\frac {7 i {\mathrm e}^{i \left (d x +c \right )} A a}{8 b^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{4}}{2 b^{5} d}+\frac {7 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{2}}{8 b^{3} d}-\frac {2 i A \,a^{4} c}{b^{5} d}-\frac {\cos \left (4 d x +4 c \right ) B a}{32 b^{2} d}+\frac {\sin \left (3 d x +3 c \right ) A a}{12 b^{2} d}-\frac {\sin \left (3 d x +3 c \right ) B \,a^{2}}{12 b^{3} 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 ) A}{d b}-\frac {i A x}{b}+\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )} A}{16 b d}+\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} A}{16 b d}+\frac {i B a x}{b^{2}}+\frac {5 i {\mathrm e}^{-i \left (d x +c \right )} B}{16 b d}-\frac {i x A \,a^{4}}{b^{5}}+\frac {2 i x A \,a^{2}}{b^{3}}-\frac {2 i x B \,a^{3}}{b^{4}}+\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 ) A \,a^{4}}{d \,b^{5}}-\frac {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 ) A \,a^{2}}{d \,b^{3}}-\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 \,a^{5}}{d \,b^{6}}+\frac {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 \,a^{3}}{d \,b^{4}}-\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 a}{d \,b^{2}}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} A \,a^{2}}{8 b^{3} d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} B \,a^{3}}{8 b^{4} d}-\frac {3 \,{\mathrm e}^{-2 i \left (d x +c \right )} B a}{16 b^{2} d}-\frac {{\mathrm e}^{2 i \left (d x +c \right )} A \,a^{2}}{8 b^{3} d}+\frac {{\mathrm e}^{2 i \left (d x +c \right )} B \,a^{3}}{8 b^{4} d}-\frac {3 \,{\mathrm e}^{2 i \left (d x +c \right )} B a}{16 b^{2} d}-\frac {5 i {\mathrm e}^{i \left (d x +c \right )} B}{16 b d}+\frac {i x B \,a^{5}}{b^{6}}-\frac {2 i A c}{d b}\) | \(856\) |
Input:
int(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE )
Output:
((a-b)^2*(a+b)^2*(A*b-B*a)*ln(2*b*tan(1/2*d*x+1/2*c)+a*sec(1/2*d*x+1/2*c)^ 2)-(a-b)^2*(a+b)^2*(A*b-B*a)*ln(sec(1/2*d*x+1/2*c)^2)-b*(1/4*b*(a^2-3/2*b^ 2)*(A*b-B*a)*cos(2*d*x+2*c)-1/12*b^2*(A*a*b-a^2*B+5/4*B*b^2)*sin(3*d*x+3*c )-1/32*b^3*(A*b-B*a)*cos(4*d*x+4*c)-1/80*B*b^4*sin(5*d*x+5*c)+(A*a^3*b-7/4 *A*a*b^3-B*a^4+7/4*B*a^2*b^2-5/8*B*b^4)*sin(d*x+c)-1/4*b*(a^2-13/8*b^2)*(A *b-B*a)))/b^6/d
Time = 0.11 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.10 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {15 \, {\left (B a b^{4} - A b^{5}\right )} \cos \left (d x + c\right )^{4} - 30 \, {\left (B a^{3} b^{2} - A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )^{2} + 60 \, {\left (B a^{5} - A a^{4} b - 2 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + B a b^{4} - A b^{5}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 4 \, {\left (3 \, B b^{5} \cos \left (d x + c\right )^{4} + 15 \, B a^{4} b - 15 \, A a^{3} b^{2} - 25 \, B a^{2} b^{3} + 25 \, A a b^{4} + 8 \, B b^{5} - {\left (5 \, B a^{2} b^{3} - 5 \, A a b^{4} - 4 \, B b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, b^{6} d} \] Input:
integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="fri cas")
Output:
-1/60*(15*(B*a*b^4 - A*b^5)*cos(d*x + c)^4 - 30*(B*a^3*b^2 - A*a^2*b^3 - B *a*b^4 + A*b^5)*cos(d*x + c)^2 + 60*(B*a^5 - A*a^4*b - 2*B*a^3*b^2 + 2*A*a ^2*b^3 + B*a*b^4 - A*b^5)*log(b*sin(d*x + c) + a) - 4*(3*B*b^5*cos(d*x + c )^4 + 15*B*a^4*b - 15*A*a^3*b^2 - 25*B*a^2*b^3 + 25*A*a*b^4 + 8*B*b^5 - (5 *B*a^2*b^3 - 5*A*a*b^4 - 4*B*b^5)*cos(d*x + c)^2)*sin(d*x + c))/(b^6*d)
Timed out. \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\frac {12 \, B b^{4} \sin \left (d x + c\right )^{5} - 15 \, {\left (B a b^{3} - A b^{4}\right )} \sin \left (d x + c\right )^{4} + 20 \, {\left (B a^{2} b^{2} - A a b^{3} - 2 \, B b^{4}\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (B a^{3} b - A a^{2} b^{2} - 2 \, B a b^{3} + 2 \, A b^{4}\right )} \sin \left (d x + c\right )^{2} + 60 \, {\left (B a^{4} - A a^{3} b - 2 \, B a^{2} b^{2} + 2 \, A a b^{3} + B b^{4}\right )} \sin \left (d x + c\right )}{b^{5}} - \frac {60 \, {\left (B a^{5} - A a^{4} b - 2 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + B a b^{4} - A b^{5}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{6}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="max ima")
Output:
1/60*((12*B*b^4*sin(d*x + c)^5 - 15*(B*a*b^3 - A*b^4)*sin(d*x + c)^4 + 20* (B*a^2*b^2 - A*a*b^3 - 2*B*b^4)*sin(d*x + c)^3 - 30*(B*a^3*b - A*a^2*b^2 - 2*B*a*b^3 + 2*A*b^4)*sin(d*x + c)^2 + 60*(B*a^4 - A*a^3*b - 2*B*a^2*b^2 + 2*A*a*b^3 + B*b^4)*sin(d*x + c))/b^5 - 60*(B*a^5 - A*a^4*b - 2*B*a^3*b^2 + 2*A*a^2*b^3 + B*a*b^4 - A*b^5)*log(b*sin(d*x + c) + a)/b^6)/d
Time = 0.37 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.65 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=-\frac {{\left (B a^{5} - A a^{4} b - 2 \, B a^{3} b^{2} + 2 \, A a^{2} b^{3} + B a b^{4} - A b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{b^{6} d} + \frac {12 \, B b^{4} d^{4} \sin \left (d x + c\right )^{5} - 15 \, B a b^{3} d^{4} \sin \left (d x + c\right )^{4} + 15 \, A b^{4} d^{4} \sin \left (d x + c\right )^{4} + 20 \, B a^{2} b^{2} d^{4} \sin \left (d x + c\right )^{3} - 20 \, A a b^{3} d^{4} \sin \left (d x + c\right )^{3} - 40 \, B b^{4} d^{4} \sin \left (d x + c\right )^{3} - 30 \, B a^{3} b d^{4} \sin \left (d x + c\right )^{2} + 30 \, A a^{2} b^{2} d^{4} \sin \left (d x + c\right )^{2} + 60 \, B a b^{3} d^{4} \sin \left (d x + c\right )^{2} - 60 \, A b^{4} d^{4} \sin \left (d x + c\right )^{2} + 60 \, B a^{4} d^{4} \sin \left (d x + c\right ) - 60 \, A a^{3} b d^{4} \sin \left (d x + c\right ) - 120 \, B a^{2} b^{2} d^{4} \sin \left (d x + c\right ) + 120 \, A a b^{3} d^{4} \sin \left (d x + c\right ) + 60 \, B b^{4} d^{4} \sin \left (d x + c\right )}{60 \, b^{5} d^{5}} \] Input:
integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x, algorithm="gia c")
Output:
-(B*a^5 - A*a^4*b - 2*B*a^3*b^2 + 2*A*a^2*b^3 + B*a*b^4 - A*b^5)*log(abs(b *sin(d*x + c) + a))/(b^6*d) + 1/60*(12*B*b^4*d^4*sin(d*x + c)^5 - 15*B*a*b ^3*d^4*sin(d*x + c)^4 + 15*A*b^4*d^4*sin(d*x + c)^4 + 20*B*a^2*b^2*d^4*sin (d*x + c)^3 - 20*A*a*b^3*d^4*sin(d*x + c)^3 - 40*B*b^4*d^4*sin(d*x + c)^3 - 30*B*a^3*b*d^4*sin(d*x + c)^2 + 30*A*a^2*b^2*d^4*sin(d*x + c)^2 + 60*B*a *b^3*d^4*sin(d*x + c)^2 - 60*A*b^4*d^4*sin(d*x + c)^2 + 60*B*a^4*d^4*sin(d *x + c) - 60*A*a^3*b*d^4*sin(d*x + c) - 120*B*a^2*b^2*d^4*sin(d*x + c) + 1 20*A*a*b^3*d^4*sin(d*x + c) + 60*B*b^4*d^4*sin(d*x + c))/(b^5*d^5)
Time = 0.11 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.25 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {{\sin \left (c+d\,x\right )}^4\,\left (\frac {A}{4\,b}-\frac {B\,a}{4\,b^2}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {A}{b}-\frac {a\,\left (\frac {2\,B}{b}+\frac {a\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{b}\right )}{2\,b}\right )}{d}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {2\,B}{3\,b}+\frac {a\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{3\,b}\right )}{d}+\frac {\sin \left (c+d\,x\right )\,\left (\frac {B}{b}+\frac {a\,\left (\frac {2\,A}{b}-\frac {a\,\left (\frac {2\,B}{b}+\frac {a\,\left (\frac {A}{b}-\frac {B\,a}{b^2}\right )}{b}\right )}{b}\right )}{b}\right )}{d}+\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (-B\,a^5+A\,a^4\,b+2\,B\,a^3\,b^2-2\,A\,a^2\,b^3-B\,a\,b^4+A\,b^5\right )}{b^6\,d}+\frac {B\,{\sin \left (c+d\,x\right )}^5}{5\,b\,d} \] Input:
int((cos(c + d*x)^5*(A + B*sin(c + d*x)))/(a + b*sin(c + d*x)),x)
Output:
(sin(c + d*x)^4*(A/(4*b) - (B*a)/(4*b^2)))/d - (sin(c + d*x)^2*(A/b - (a*( (2*B)/b + (a*(A/b - (B*a)/b^2))/b))/(2*b)))/d - (sin(c + d*x)^3*((2*B)/(3* b) + (a*(A/b - (B*a)/b^2))/(3*b)))/d + (sin(c + d*x)*(B/b + (a*((2*A)/b - (a*((2*B)/b + (a*(A/b - (B*a)/b^2))/b))/b))/b))/d + (log(a + b*sin(c + d*x ))*(A*b^5 - B*a^5 - 2*A*a^2*b^3 + 2*B*a^3*b^2 + A*a^4*b - B*a*b^4))/(b^6*d ) + (B*sin(c + d*x)^5)/(5*b*d)
Time = 0.17 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.16 \[ \int \frac {\cos ^5(c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx=\frac {\sin \left (d x +c \right ) \left (3 \sin \left (d x +c \right )^{4}-10 \sin \left (d x +c \right )^{2}+15\right )}{15 d} \] Input:
int(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+b*sin(d*x+c)),x)
Output:
(sin(c + d*x)*(3*sin(c + d*x)**4 - 10*sin(c + d*x)**2 + 15))/(15*d)