Integrand size = 19, antiderivative size = 64 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {(a+a \sin (c+d x))^4}{a^3 d}-\frac {4 (a+a \sin (c+d x))^5}{5 a^4 d}+\frac {(a+a \sin (c+d x))^6}{6 a^5 d} \]
Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos ^6(c+d x)}{6 d}+\frac {a \sin (c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \]
-1/6*(a*Cos[c + d*x]^6)/d + (a*Sin[c + d*x])/d - (2*a*Sin[c + d*x]^3)/(3*d ) + (a*Sin[c + d*x]^5)/(5*d)
Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {3042, 3146, 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.
\(\displaystyle \int \cos ^5(c+d x) (a \sin (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^5 (a \sin (c+d x)+a)dx\) |
\(\Big \downarrow \) 3146 |
\(\displaystyle \frac {\int (a-a \sin (c+d x))^2 (\sin (c+d x) a+a)^3d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\int \left ((\sin (c+d x) a+a)^5-4 a (\sin (c+d x) a+a)^4+4 a^2 (\sin (c+d x) a+a)^3\right )d(a \sin (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^2 (a \sin (c+d x)+a)^4+\frac {1}{6} (a \sin (c+d x)+a)^6-\frac {4}{5} a (a \sin (c+d x)+a)^5}{a^5 d}\) |
(a^2*(a + a*Sin[c + d*x])^4 - (4*a*(a + a*Sin[c + d*x])^5)/5 + (a + a*Sin[ c + d*x])^6/6)/(a^5*d)
3.1.3.3.1 Defintions of rubi rules used
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 _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Time = 0.29 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}-\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )\right )}{d}\) | \(63\) |
default | \(\frac {a \left (\frac {\left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}-\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}+\sin \left (d x +c \right )\right )}{d}\) | \(63\) |
risch | \(\frac {5 a \sin \left (d x +c \right )}{8 d}-\frac {a \cos \left (6 d x +6 c \right )}{192 d}+\frac {a \sin \left (5 d x +5 c \right )}{80 d}-\frac {a \cos \left (4 d x +4 c \right )}{32 d}+\frac {5 a \sin \left (3 d x +3 c \right )}{48 d}-\frac {5 a \cos \left (2 d x +2 c \right )}{64 d}\) | \(89\) |
parallelrisch | \(\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (15 \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+78 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+50 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+78 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+15\right )}{15 d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\) | \(135\) |
norman | \(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {14 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {52 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {52 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {14 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {20 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\) | \(169\) |
a/d*(1/6*sin(d*x+c)^6+1/5*sin(d*x+c)^5-1/2*sin(d*x+c)^4-2/3*sin(d*x+c)^3+1 /2*sin(d*x+c)^2+sin(d*x+c))
Time = 0.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.80 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {5 \, a \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right )}{30 \, d} \]
-1/30*(5*a*cos(d*x + c)^6 - 2*(3*a*cos(d*x + c)^4 + 4*a*cos(d*x + c)^2 + 8 *a)*sin(d*x + c))/d
Time = 0.36 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.30 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=\begin {cases} \frac {8 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {a \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + a\right ) \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
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 - a*cos(c + d*x)**6/(6*d), Ne(d , 0)), (x*(a*sin(c) + a)*cos(c)**5, True))
Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.09 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {5 \, a \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} - 15 \, a \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} + 15 \, a \sin \left (d x + c\right )^{2} + 30 \, a \sin \left (d x + c\right )}{30 \, d} \]
1/30*(5*a*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 - 15*a*sin(d*x + c)^4 - 20*a *sin(d*x + c)^3 + 15*a*sin(d*x + c)^2 + 30*a*sin(d*x + c))/d
Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.38 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=-\frac {a \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {a \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {5 \, a \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {5 \, a \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{8 \, d} \]
-1/192*a*cos(6*d*x + 6*c)/d - 1/32*a*cos(4*d*x + 4*c)/d - 5/64*a*cos(2*d*x + 2*c)/d + 1/80*a*sin(5*d*x + 5*c)/d + 5/48*a*sin(3*d*x + 3*c)/d + 5/8*a* sin(d*x + c)/d
Time = 0.06 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.06 \[ \int \cos ^5(c+d x) (a+a \sin (c+d x)) \, dx=\frac {\frac {a\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}-\frac {a\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {a\,{\sin \left (c+d\,x\right )}^2}{2}+a\,\sin \left (c+d\,x\right )}{d} \]