Integrand size = 31, antiderivative size = 113 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {\left (2 a A b-3 a^2 B+b^2 B\right ) \log (a+b \sin (c+d x))}{b^4 d}-\frac {(A b-2 a B) \sin (c+d x)}{b^3 d}-\frac {B \sin ^2(c+d x)}{2 b^2 d}+\frac {\left (a^2-b^2\right ) (A b-a B)}{b^4 d (a+b \sin (c+d x))} \]
(2*A*a*b-3*B*a^2+B*b^2)*ln(a+b*sin(d*x+c))/b^4/d-(A*b-2*B*a)*sin(d*x+c)/b^ 3/d-1/2*B*sin(d*x+c)^2/b^2/d+(a^2-b^2)*(A*b-B*a)/b^4/d/(a+b*sin(d*x+c))
Time = 0.33 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.98 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {\left (-a^2+b^2\right ) B \log (a+b \sin (c+d x))}{b}+a B \sin (c+d x)-\frac {1}{2} b B \sin ^2(c+d x)+\left (A-\frac {a B}{b}\right ) \left (2 a \log (a+b \sin (c+d x))-b \sin (c+d x)+\frac {(a-b) (a+b)}{a+b \sin (c+d x)}\right )}{b^3 d} \]
(((-a^2 + b^2)*B*Log[a + b*Sin[c + d*x]])/b + a*B*Sin[c + d*x] - (b*B*Sin[ c + d*x]^2)/2 + (A - (a*B)/b)*(2*a*Log[a + b*Sin[c + d*x]] - b*Sin[c + d*x ] + ((a - b)*(a + b))/(a + b*Sin[c + d*x])))/(b^3*d)
Time = 0.37 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^3 (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \frac {(A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )}{b (a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )}{(a+b \sin (c+d x))^2}d(b \sin (c+d x))}{b^4 d}\) |
| \(\Big \downarrow \) 652 |
| \(\displaystyle \frac {\int \left (-A b+3 a B-B (a+b \sin (c+d x))+\frac {-3 B a^2+2 A b a+b^2 B}{a+b \sin (c+d x)}+\frac {\left (a^2-b^2\right ) (a B-A b)}{(a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{b^4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\left (a^2-b^2\right ) (A b-a B)}{a+b \sin (c+d x)}+\left (-3 a^2 B+2 a A b+b^2 B\right ) \log (a+b \sin (c+d x))-b (A b-3 a B) \sin (c+d x)-\frac {1}{2} B (a+b \sin (c+d x))^2}{b^4 d}\) |
((2*a*A*b - 3*a^2*B + b^2*B)*Log[a + b*Sin[c + d*x]] - b*(A*b - 3*a*B)*Sin [c + d*x] + ((a^2 - b^2)*(A*b - a*B))/(a + b*Sin[c + d*x]) - (B*(a + b*Sin [c + d*x])^2)/2)/(b^4*d)
3.16.54.3.1 Defintions of rubi rules used
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 = 0.93 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04
| method | result | size |
| derivativedivides | \(-\frac {\frac {\frac {B \left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+A \sin \left (d x +c \right ) b -2 a B \sin \left (d x +c \right )}{b^{3}}+\frac {\left (-2 A a b +3 B \,a^{2}-B \,b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}-\frac {A \,a^{2} b -A \,b^{3}-B \,a^{3}+B a \,b^{2}}{b^{4} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(118\) |
| default | \(-\frac {\frac {\frac {B \left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+A \sin \left (d x +c \right ) b -2 a B \sin \left (d x +c \right )}{b^{3}}+\frac {\left (-2 A a b +3 B \,a^{2}-B \,b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{4}}-\frac {A \,a^{2} b -A \,b^{3}-B \,a^{3}+B a \,b^{2}}{b^{4} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(118\) |
| parallelrisch | \(\frac {16 \left (A a b -\frac {3}{2} B \,a^{2}+\frac {1}{2} B \,b^{2}\right ) \left (a +b \sin \left (d x +c \right )\right ) a \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-16 \left (A a b -\frac {3}{2} B \,a^{2}+\frac {1}{2} B \,b^{2}\right ) \left (a +b \sin \left (d x +c \right )\right ) a \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 A a \,b^{3}-6 B \,a^{2} b^{2}\right ) \cos \left (2 d x +2 c \right )+B a \,b^{3} \sin \left (3 d x +3 c \right )+\left (-16 A \,a^{2} b^{2}+8 A \,b^{4}+24 B \,a^{3} b -11 B a \,b^{3}\right ) \sin \left (d x +c \right )-4 A a \,b^{3}+6 B \,a^{2} b^{2}}{8 a d \,b^{4} \left (a +b \sin \left (d x +c \right )\right )}\) | \(219\) |
| risch | \(-\frac {i x B}{b^{2}}-\frac {4 i A a c}{b^{3} d}-\frac {2 i x A a}{b^{3}}+\frac {{\mathrm e}^{2 i \left (d x +c \right )} B}{8 b^{2} d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B a}{b^{3} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} A}{2 b^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B a}{b^{3} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} A}{2 b^{2} d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} B}{8 b^{2} d}+\frac {3 i x B \,a^{2}}{b^{4}}+\frac {6 i B \,a^{2} c}{b^{4} d}-\frac {2 i B c}{b^{2} d}-\frac {2 \left (-A \,a^{2} b +A \,b^{3}+B \,a^{3}-B a \,b^{2}\right ) {\mathrm e}^{i \left (d x +c \right )}}{b^{4} d \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+i b +2 a \,{\mathrm e}^{i \left (d x +c \right )}\right )}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) A a}{b^{3} d}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) B \,a^{2}}{b^{4} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right ) B}{b^{2} d}\) | \(369\) |
| norman | \(\frac {-\frac {2 \left (6 A b -9 B a \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}-\frac {2 \left (6 A b -9 B a \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}-\frac {\left (4 A b -6 B a \right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}-\frac {\left (4 A b -6 B a \right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}-\frac {4 \left (6 A \,a^{2} b -3 A \,b^{3}-9 B \,a^{3}+5 B a \,b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{3} a}-\frac {4 \left (4 A \,a^{2} b -2 A \,b^{3}-6 B \,a^{3}+3 B a \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{3} a}-\frac {4 \left (4 A \,a^{2} b -2 A \,b^{3}-6 B \,a^{3}+3 B a \,b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{3} a}-\frac {2 \left (2 A \,a^{2} b -A \,b^{3}-3 B \,a^{3}+B a \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a \,b^{3} d}-\frac {2 \left (2 A \,a^{2} b -A \,b^{3}-3 B \,a^{3}+B a \,b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a \,b^{3} d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}+\frac {\left (2 A a b -3 B \,a^{2}+B \,b^{2}\right ) \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{4} d}-\frac {\left (2 A a b -3 B \,a^{2}+B \,b^{2}\right ) \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4} d}\) | \(492\) |
-1/d*(1/b^3*(1/2*B*sin(d*x+c)^2*b+A*sin(d*x+c)*b-2*a*B*sin(d*x+c))+(-2*A*a *b+3*B*a^2-B*b^2)/b^4*ln(a+b*sin(d*x+c))-1/b^4*(A*a^2*b-A*b^3-B*a^3+B*a*b^ 2)/(a+b*sin(d*x+c)))
Time = 0.29 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.58 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {4 \, B a^{3} - 4 \, A a^{2} b - 11 \, B a b^{2} + 8 \, A b^{3} + 2 \, {\left (3 \, B a b^{2} - 2 \, A b^{3}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (3 \, B a^{3} - 2 \, A a^{2} b - B a b^{2} + {\left (3 \, B a^{2} b - 2 \, A a b^{2} - B b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (2 \, B b^{3} \cos \left (d x + c\right )^{2} + 8 \, B a^{2} b - 4 \, A a b^{2} - B b^{3}\right )} \sin \left (d x + c\right )}{4 \, {\left (b^{5} d \sin \left (d x + c\right ) + a b^{4} d\right )}} \]
-1/4*(4*B*a^3 - 4*A*a^2*b - 11*B*a*b^2 + 8*A*b^3 + 2*(3*B*a*b^2 - 2*A*b^3) *cos(d*x + c)^2 + 4*(3*B*a^3 - 2*A*a^2*b - B*a*b^2 + (3*B*a^2*b - 2*A*a*b^ 2 - B*b^3)*sin(d*x + c))*log(b*sin(d*x + c) + a) - (2*B*b^3*cos(d*x + c)^2 + 8*B*a^2*b - 4*A*a*b^2 - B*b^3)*sin(d*x + c))/(b^5*d*sin(d*x + c) + a*b^ 4*d)
Timed out. \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (B a^{3} - A a^{2} b - B a b^{2} + A b^{3}\right )}}{b^{5} \sin \left (d x + c\right ) + a b^{4}} + \frac {B b \sin \left (d x + c\right )^{2} - 2 \, {\left (2 \, B a - A b\right )} \sin \left (d x + c\right )}{b^{3}} + \frac {2 \, {\left (3 \, B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{b^{4}}}{2 \, d} \]
-1/2*(2*(B*a^3 - A*a^2*b - B*a*b^2 + A*b^3)/(b^5*sin(d*x + c) + a*b^4) + ( B*b*sin(d*x + c)^2 - 2*(2*B*a - A*b)*sin(d*x + c))/b^3 + 2*(3*B*a^2 - 2*A* a*b - B*b^2)*log(b*sin(d*x + c) + a)/b^4)/d
Time = 0.34 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.66 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {{\left (b \sin \left (d x + c\right ) + a\right )}^{2} {\left (B - \frac {2 \, {\left (3 \, B a b - A b^{2}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} b}\right )}}{b^{4}} - \frac {2 \, {\left (3 \, B a^{2} - 2 \, A a b - B b^{2}\right )} \log \left (\frac {{\left | b \sin \left (d x + c\right ) + a \right |}}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2} {\left | b \right |}}\right )}{b^{4}} + \frac {2 \, {\left (\frac {B a^{3} b^{2}}{b \sin \left (d x + c\right ) + a} - \frac {A a^{2} b^{3}}{b \sin \left (d x + c\right ) + a} - \frac {B a b^{4}}{b \sin \left (d x + c\right ) + a} + \frac {A b^{5}}{b \sin \left (d x + c\right ) + a}\right )}}{b^{6}}}{2 \, d} \]
-1/2*((b*sin(d*x + c) + a)^2*(B - 2*(3*B*a*b - A*b^2)/((b*sin(d*x + c) + a )*b))/b^4 - 2*(3*B*a^2 - 2*A*a*b - B*b^2)*log(abs(b*sin(d*x + c) + a)/((b* sin(d*x + c) + a)^2*abs(b)))/b^4 + 2*(B*a^3*b^2/(b*sin(d*x + c) + a) - A*a ^2*b^3/(b*sin(d*x + c) + a) - B*a*b^4/(b*sin(d*x + c) + a) + A*b^5/(b*sin( d*x + c) + a))/b^6)/d
Time = 11.37 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx=\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (-3\,B\,a^2+2\,A\,a\,b+B\,b^2\right )}{b^4\,d}-\frac {B\,a^3-A\,a^2\,b-B\,a\,b^2+A\,b^3}{b\,d\,\left (\sin \left (c+d\,x\right )\,b^4+a\,b^3\right )}-\frac {\sin \left (c+d\,x\right )\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )}{d}-\frac {B\,{\sin \left (c+d\,x\right )}^2}{2\,b^2\,d} \]