Integrand size = 21, antiderivative size = 72 \[ \int \cos ^5(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {(A+C) \sin (c+d x)}{d}-\frac {(2 A+3 C) \sin ^3(c+d x)}{3 d}+\frac {(A+3 C) \sin ^5(c+d x)}{5 d}-\frac {C \sin ^7(c+d x)}{7 d} \]
(A+C)*sin(d*x+c)/d-1/3*(2*A+3*C)*sin(d*x+c)^3/d+1/5*(A+3*C)*sin(d*x+c)^5/d -1/7*C*sin(d*x+c)^7/d
Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.40 \[ \int \cos ^5(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {A \sin (c+d x)}{d}+\frac {C \sin (c+d x)}{d}-\frac {2 A \sin ^3(c+d x)}{3 d}-\frac {C \sin ^3(c+d x)}{d}+\frac {A \sin ^5(c+d x)}{5 d}+\frac {3 C \sin ^5(c+d x)}{5 d}-\frac {C \sin ^7(c+d x)}{7 d} \]
(A*Sin[c + d*x])/d + (C*Sin[c + d*x])/d - (2*A*Sin[c + d*x]^3)/(3*d) - (C* Sin[c + d*x]^3)/d + (A*Sin[c + d*x]^5)/(5*d) + (3*C*Sin[c + d*x]^5)/(5*d) - (C*Sin[c + d*x]^7)/(7*d)
Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3492, 290, 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.
\(\displaystyle \int \cos ^5(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^5 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
\(\Big \downarrow \) 3492 |
\(\displaystyle -\frac {\int \left (1-\sin ^2(c+d x)\right )^2 \left (-C \sin ^2(c+d x)+A+C\right )d(-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 290 |
\(\displaystyle -\frac {\int \left (-C \sin ^6(c+d x)+(A+3 C) \sin ^4(c+d x)-(2 A+3 C) \sin ^2(c+d x)+A \left (\frac {C}{A}+1\right )\right )d(-\sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{5} (A+3 C) \sin ^5(c+d x)+\frac {1}{3} (2 A+3 C) \sin ^3(c+d x)-(A+C) \sin (c+d x)+\frac {1}{7} C \sin ^7(c+d x)}{d}\) |
-((-((A + C)*Sin[c + d*x]) + ((2*A + 3*C)*Sin[c + d*x]^3)/3 - ((A + 3*C)*S in[c + d*x]^5)/5 + (C*Sin[c + d*x]^7)/7)/d)
3.1.2.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I nt[ExpandIntegrand[(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d }, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && IGtQ[q, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2 ), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2, 0]
Time = 3.78 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(\frac {\left (700 A +735 C \right ) \sin \left (3 d x +3 c \right )+\left (84 A +147 C \right ) \sin \left (5 d x +5 c \right )+15 \sin \left (7 d x +7 c \right ) C +4200 \left (A +\frac {7 C}{8}\right ) \sin \left (d x +c \right )}{6720 d}\) | \(66\) |
derivativedivides | \(\frac {\frac {C \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}+\frac {A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(74\) |
default | \(\frac {\frac {C \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}+\frac {A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\) | \(74\) |
parts | \(\frac {A \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {C \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7 d}\) | \(76\) |
risch | \(\frac {5 \sin \left (d x +c \right ) A}{8 d}+\frac {35 C \sin \left (d x +c \right )}{64 d}+\frac {\sin \left (7 d x +7 c \right ) C}{448 d}+\frac {\sin \left (5 d x +5 c \right ) A}{80 d}+\frac {7 \sin \left (5 d x +5 c \right ) C}{320 d}+\frac {5 \sin \left (3 d x +3 c \right ) A}{48 d}+\frac {7 \sin \left (3 d x +3 c \right ) C}{64 d}\) | \(101\) |
norman | \(\frac {\frac {2 \left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 \left (A +C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (5 A +3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 \left (5 A +3 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {8 \left (91 A +53 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{35 d}+\frac {2 \left (113 A +129 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {2 \left (113 A +129 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}\) | \(169\) |
1/6720*((700*A+735*C)*sin(3*d*x+3*c)+(84*A+147*C)*sin(5*d*x+5*c)+15*sin(7* d*x+7*c)*C+4200*(A+7/8*C)*sin(d*x+c))/d
Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.88 \[ \int \cos ^5(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {{\left (15 \, C \cos \left (d x + c\right )^{6} + 3 \, {\left (7 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (7 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{2} + 56 \, A + 48 \, C\right )} \sin \left (d x + c\right )}{105 \, d} \]
1/105*(15*C*cos(d*x + c)^6 + 3*(7*A + 6*C)*cos(d*x + c)^4 + 4*(7*A + 6*C)* cos(d*x + c)^2 + 56*A + 48*C)*sin(d*x + c)/d
Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (60) = 120\).
Time = 0.48 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.10 \[ \int \cos ^5(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {8 A \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 A \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {A \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {16 C \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {8 C \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {2 C \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {C \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((8*A*sin(c + d*x)**5/(15*d) + 4*A*sin(c + d*x)**3*cos(c + d*x)** 2/(3*d) + A*sin(c + d*x)*cos(c + d*x)**4/d + 16*C*sin(c + d*x)**7/(35*d) + 8*C*sin(c + d*x)**5*cos(c + d*x)**2/(5*d) + 2*C*sin(c + d*x)**3*cos(c + d *x)**4/d + C*sin(c + d*x)*cos(c + d*x)**6/d, Ne(d, 0)), (x*(A + C*cos(c)** 2)*cos(c)**5, True))
Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.83 \[ \int \cos ^5(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {15 \, C \sin \left (d x + c\right )^{7} - 21 \, {\left (A + 3 \, C\right )} \sin \left (d x + c\right )^{5} + 35 \, {\left (2 \, A + 3 \, C\right )} \sin \left (d x + c\right )^{3} - 105 \, {\left (A + C\right )} \sin \left (d x + c\right )}{105 \, d} \]
-1/105*(15*C*sin(d*x + c)^7 - 21*(A + 3*C)*sin(d*x + c)^5 + 35*(2*A + 3*C) *sin(d*x + c)^3 - 105*(A + C)*sin(d*x + c))/d
Time = 0.28 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int \cos ^5(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {C \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} + \frac {{\left (4 \, A + 7 \, C\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (20 \, A + 21 \, C\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, {\left (8 \, A + 7 \, C\right )} \sin \left (d x + c\right )}{64 \, d} \]
1/448*C*sin(7*d*x + 7*c)/d + 1/320*(4*A + 7*C)*sin(5*d*x + 5*c)/d + 1/192* (20*A + 21*C)*sin(3*d*x + 3*c)/d + 5/64*(8*A + 7*C)*sin(d*x + c)/d
Time = 0.72 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.82 \[ \int \cos ^5(c+d x) \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {\frac {C\,{\sin \left (c+d\,x\right )}^7}{7}+\left (-\frac {A}{5}-\frac {3\,C}{5}\right )\,{\sin \left (c+d\,x\right )}^5+\left (\frac {2\,A}{3}+C\right )\,{\sin \left (c+d\,x\right )}^3+\left (-A-C\right )\,\sin \left (c+d\,x\right )}{d} \]