Integrand size = 31, antiderivative size = 57 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {A \sin (c+d x)}{a d}-\frac {(A-B) \sin ^2(c+d x)}{2 a d}-\frac {B \sin ^3(c+d x)}{3 a d} \]
Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\sin (c+d x) \left (6 A-3 (A-B) \sin (c+d x)-2 B \sin ^2(c+d x)\right )}{6 a d} \]
Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 49, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a \sin (c+d x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^3 (A+B \sin (c+d x))}{a \sin (c+d x)+a}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {\int \frac {(a-a \sin (c+d x)) (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^3 d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int (a-a \sin (c+d x)) (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left (-B \sin ^2(c+d x) a^2+A a^2-(A-B) \sin (c+d x) a^2\right )d(a \sin (c+d x))}{a^4 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {-\frac {1}{2} a^3 (A-B) \sin ^2(c+d x)+a^3 A \sin (c+d x)-\frac {1}{3} a^3 B \sin ^3(c+d x)}{a^4 d}\) |
3.11.5.3.1 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[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(-\frac {\frac {B \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )}{d a}\) | \(45\) |
default | \(-\frac {\frac {B \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )}{d a}\) | \(45\) |
parallelrisch | \(\frac {\left (3 A -3 B \right ) \cos \left (2 d x +2 c \right )+B \sin \left (3 d x +3 c \right )+\left (12 A -3 B \right ) \sin \left (d x +c \right )-3 A +3 B}{12 d a}\) | \(58\) |
risch | \(\frac {A \sin \left (d x +c \right )}{a d}-\frac {B \sin \left (d x +c \right )}{4 a d}+\frac {\sin \left (3 d x +3 c \right ) B}{12 a d}+\frac {\cos \left (2 d x +2 c \right ) A}{4 a d}-\frac {\cos \left (2 d x +2 c \right ) B}{4 a d}\) | \(85\) |
norman | \(\frac {\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 A \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 B \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 \left (6 A -B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {2 \left (6 A -B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {2 \left (3 A +2 B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {2 \left (3 A +2 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\) | \(213\) |
Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {3 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (B \cos \left (d x + c\right )^{2} + 3 \, A - B\right )} \sin \left (d x + c\right )}{6 \, a d} \]
Leaf count of result is larger than twice the leaf count of optimal. 588 vs. \(2 (44) = 88\).
Time = 3.58 (sec) , antiderivative size = 588, normalized size of antiderivative = 10.32 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\begin {cases} \frac {6 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {6 A \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {12 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {6 A \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {6 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {6 B \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {8 B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {6 B \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \sin {\left (c \right )}\right ) \cos ^{3}{\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \]
Piecewise((6*A*tan(c/2 + d*x/2)**5/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan( c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) - 6*A*tan(c/2 + d*x/2 )**4/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/ 2 + d*x/2)**2 + 3*a*d) + 12*A*tan(c/2 + d*x/2)**3/(3*a*d*tan(c/2 + d*x/2)* *6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) - 6*A* tan(c/2 + d*x/2)**2/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 6*A*tan(c/2 + d*x/2)/(3*a*d*tan(c/ 2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3* a*d) + 6*B*tan(c/2 + d*x/2)**4/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) - 8*B*tan(c/2 + d*x/2)**3 /(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 6*B*tan(c/2 + d*x/2)**2/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d), Ne(d, 0)), (x*(A + B*sin(c))*cos(c)**3/(a*sin(c) + a), True))
Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {2 \, B \sin \left (d x + c\right )^{3} + 3 \, {\left (A - B\right )} \sin \left (d x + c\right )^{2} - 6 \, A \sin \left (d x + c\right )}{6 \, a d} \]
Time = 0.31 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {2 \, B \sin \left (d x + c\right )^{3} + 3 \, A \sin \left (d x + c\right )^{2} - 3 \, B \sin \left (d x + c\right )^{2} - 6 \, A \sin \left (d x + c\right )}{6 \, a d} \]
-1/6*(2*B*sin(d*x + c)^3 + 3*A*sin(d*x + c)^2 - 3*B*sin(d*x + c)^2 - 6*A*s in(d*x + c))/(a*d)
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\sin \left (c+d\,x\right )\,\left (6\,A-3\,A\,\sin \left (c+d\,x\right )+3\,B\,\sin \left (c+d\,x\right )-2\,B\,{\sin \left (c+d\,x\right )}^2\right )}{6\,a\,d} \]