Integrand size = 19, antiderivative size = 60 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \cos ^6(c+d x)}{6 d}+\frac {a \sin (c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \]
Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \cos ^6(c+d x)}{6 d}+\frac {a \sin (c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \]
-1/6*(b*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 = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.23, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 3147, 455, 210, 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.
\(\displaystyle \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (c+d x)^5 (a+b \sin (c+d x))dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {\int (a+b \sin (c+d x)) \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {a \int \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))-\frac {1}{6} \left (b^2-b^2 \sin ^2(c+d x)\right )^3}{b^5 d}\) |
\(\Big \downarrow \) 210 |
\(\displaystyle \frac {a \int \left (\sin ^4(c+d x) b^4-2 \sin ^2(c+d x) b^4+b^4\right )d(b \sin (c+d x))-\frac {1}{6} \left (b^2-b^2 \sin ^2(c+d x)\right )^3}{b^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a \left (\frac {1}{5} b^5 \sin ^5(c+d x)-\frac {2}{3} b^5 \sin ^3(c+d x)+b^5 \sin (c+d x)\right )-\frac {1}{6} \left (b^2-b^2 \sin ^2(c+d x)\right )^3}{b^5 d}\) |
(-1/6*(b^2 - b^2*Sin[c + d*x]^2)^3 + a*(b^5*Sin[c + d*x] - (2*b^5*Sin[c + d*x]^3)/3 + (b^5*Sin[c + d*x]^5)/5))/(b^5*d)
3.4.76.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 )^p, x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 0]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
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*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.55 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.15
method | result | size |
derivativedivides | \(\frac {\frac {b \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {a \left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {b \left (\sin ^{4}\left (d x +c \right )\right )}{2}-\frac {2 a \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+a \sin \left (d x +c \right )}{d}\) | \(69\) |
default | \(\frac {\frac {b \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {a \left (\sin ^{5}\left (d x +c \right )\right )}{5}-\frac {b \left (\sin ^{4}\left (d x +c \right )\right )}{2}-\frac {2 a \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\sin ^{2}\left (d x +c \right )\right ) b}{2}+a \sin \left (d x +c \right )}{d}\) | \(69\) |
risch | \(\frac {5 a \sin \left (d x +c \right )}{8 d}-\frac {b \cos \left (6 d x +6 c \right )}{192 d}+\frac {a \sin \left (5 d x +5 c \right )}{80 d}-\frac {b \cos \left (4 d x +4 c \right )}{32 d}+\frac {5 a \sin \left (3 d x +3 c \right )}{48 d}-\frac {5 b \cos \left (2 d x +2 c \right )}{64 d}\) | \(89\) |
parallelrisch | \(\frac {2 \left (a \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\frac {7 a \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {26 a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{3}+\frac {26 a \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {7 a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\) | \(139\) |
norman | \(\frac {\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {14 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {52 a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {52 a \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {14 a \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {20 b \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 b \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}\) | \(169\) |
1/d*(1/6*b*sin(d*x+c)^6+1/5*a*sin(d*x+c)^5-1/2*b*sin(d*x+c)^4-2/3*a*sin(d* x+c)^3+1/2*sin(d*x+c)^2*b+a*sin(d*x+c))
Time = 0.30 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {5 \, b \cos \left (d x + c\right )^{6} - 2 \, {\left (3 \, a \cos \left (d x + c\right )^{4} + 4 \, a \cos \left (d x + c\right )^{2} + 8 \, a\right )} \sin \left (d x + c\right )}{30 \, d} \]
-1/30*(5*b*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 = 0.33 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.38 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=\begin {cases} \frac {8 a \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {b \cos ^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right ) \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
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 - b*cos(c + d*x)**6/(6*d), Ne(d , 0)), (x*(a + b*sin(c))*cos(c)**5, True))
Time = 0.24 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.17 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {5 \, b \sin \left (d x + c\right )^{6} + 6 \, a \sin \left (d x + c\right )^{5} - 15 \, b \sin \left (d x + c\right )^{4} - 20 \, a \sin \left (d x + c\right )^{3} + 15 \, b \sin \left (d x + c\right )^{2} + 30 \, a \sin \left (d x + c\right )}{30 \, d} \]
1/30*(5*b*sin(d*x + c)^6 + 6*a*sin(d*x + c)^5 - 15*b*sin(d*x + c)^4 - 20*a *sin(d*x + c)^3 + 15*b*sin(d*x + c)^2 + 30*a*sin(d*x + c))/d
Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.47 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=-\frac {b \cos \left (6 \, d x + 6 \, c\right )}{192 \, d} - \frac {b \cos \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {5 \, b \cos \left (2 \, d x + 2 \, c\right )}{64 \, d} + \frac {a \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {5 \, a \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {5 \, a \sin \left (d x + c\right )}{8 \, d} \]
-1/192*b*cos(6*d*x + 6*c)/d - 1/32*b*cos(4*d*x + 4*c)/d - 5/64*b*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 = 4.47 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x)) \, dx=\frac {\frac {b\,{\sin \left (c+d\,x\right )}^6}{6}+\frac {a\,{\sin \left (c+d\,x\right )}^5}{5}-\frac {b\,{\sin \left (c+d\,x\right )}^4}{2}-\frac {2\,a\,{\sin \left (c+d\,x\right )}^3}{3}+\frac {b\,{\sin \left (c+d\,x\right )}^2}{2}+a\,\sin \left (c+d\,x\right )}{d} \]