Integrand size = 21, antiderivative size = 144 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^4}{4 b^5 d}-\frac {4 a \left (a^2-b^2\right ) (a+b \sin (c+d x))^5}{5 b^5 d}+\frac {\left (3 a^2-b^2\right ) (a+b \sin (c+d x))^6}{3 b^5 d}-\frac {4 a (a+b \sin (c+d x))^7}{7 b^5 d}+\frac {(a+b \sin (c+d x))^8}{8 b^5 d} \]
1/4*(a^2-b^2)^2*(a+b*sin(d*x+c))^4/b^5/d-4/5*a*(a^2-b^2)*(a+b*sin(d*x+c))^ 5/b^5/d+1/3*(3*a^2-b^2)*(a+b*sin(d*x+c))^6/b^5/d-4/7*a*(a+b*sin(d*x+c))^7/ b^5/d+1/8*(a+b*sin(d*x+c))^8/b^5/d
Time = 0.37 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.83 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {\frac {1}{4} \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^4-\frac {4}{5} a (a-b) (a+b) (a+b \sin (c+d x))^5+\frac {1}{3} \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^6-\frac {4}{7} a (a+b \sin (c+d x))^7+\frac {1}{8} (a+b \sin (c+d x))^8}{b^5 d} \]
(((a^2 - b^2)^2*(a + b*Sin[c + d*x])^4)/4 - (4*a*(a - b)*(a + b)*(a + b*Si n[c + d*x])^5)/5 + ((3*a^2 - b^2)*(a + b*Sin[c + d*x])^6)/3 - (4*a*(a + b* Sin[c + d*x])^7)/7 + (a + b*Sin[c + d*x])^8/8)/(b^5*d)
Time = 0.33 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3147, 476, 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.
| \(\displaystyle \int \cos ^5(c+d x) (a+b \sin (c+d x))^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^5 (a+b \sin (c+d x))^3dx\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {\int (a+b \sin (c+d x))^3 \left (b^2-b^2 \sin ^2(c+d x)\right )^2d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 476 |
| \(\displaystyle \frac {\int \left ((a+b \sin (c+d x))^7-4 a (a+b \sin (c+d x))^6+2 \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^5-4 \left (a^3-a b^2\right ) (a+b \sin (c+d x))^4+\left (a^2-b^2\right )^2 (a+b \sin (c+d x))^3\right )d(b \sin (c+d x))}{b^5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{3} \left (3 a^2-b^2\right ) (a+b \sin (c+d x))^6-\frac {4}{5} a \left (a^2-b^2\right ) (a+b \sin (c+d x))^5+\frac {1}{4} \left (a^2-b^2\right )^2 (a+b \sin (c+d x))^4+\frac {1}{8} (a+b \sin (c+d x))^8-\frac {4}{7} a (a+b \sin (c+d x))^7}{b^5 d}\) |
(((a^2 - b^2)^2*(a + b*Sin[c + d*x])^4)/4 - (4*a*(a^2 - b^2)*(a + b*Sin[c + d*x])^5)/5 + ((3*a^2 - b^2)*(a + b*Sin[c + d*x])^6)/3 - (4*a*(a + b*Sin[ c + d*x])^7)/7 + (a + b*Sin[c + d*x])^8/8)/(b^5*d)
3.4.100.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[p, 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*(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 = 1.19 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.98
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{3} \left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {3 a \,b^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (3 a^{2} b -2 b^{3}\right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (a^{3}-6 a \,b^{2}\right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (-6 a^{2} b +b^{3}\right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-2 a^{3}+3 a \,b^{2}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {3 \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b}{2}+a^{3} \sin \left (d x +c \right )}{d}\) | \(141\) |
| default | \(\frac {\frac {b^{3} \left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {3 a \,b^{2} \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (3 a^{2} b -2 b^{3}\right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (a^{3}-6 a \,b^{2}\right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (-6 a^{2} b +b^{3}\right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-2 a^{3}+3 a \,b^{2}\right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {3 \left (\sin ^{2}\left (d x +c \right )\right ) a^{2} b}{2}+a^{3} \sin \left (d x +c \right )}{d}\) | \(141\) |
| parallelrisch | \(\frac {105 b^{3} \cos \left (8 d x +8 c \right )-720 a \,b^{2} \sin \left (7 d x +7 c \right )-1680 \cos \left (6 d x +6 c \right ) a^{2} b +280 \cos \left (6 d x +6 c \right ) b^{3}+1344 \sin \left (5 d x +5 c \right ) a^{3}-3024 \sin \left (5 d x +5 c \right ) a \,b^{2}-10080 \cos \left (4 d x +4 c \right ) a^{2} b -420 \cos \left (4 d x +4 c \right ) b^{3}+11200 \sin \left (3 d x +3 c \right ) a^{3}-1680 \sin \left (3 d x +3 c \right ) a \,b^{2}-25200 \cos \left (2 d x +2 c \right ) a^{2} b -2520 \cos \left (2 d x +2 c \right ) b^{3}+67200 a^{3} \sin \left (d x +c \right )+25200 a \,b^{2} \sin \left (d x +c \right )+36960 a^{2} b +2555 b^{3}}{107520 d}\) | \(215\) |
| risch | \(\frac {5 a^{3} \sin \left (d x +c \right )}{8 d}+\frac {15 a \,b^{2} \sin \left (d x +c \right )}{64 d}+\frac {b^{3} \cos \left (8 d x +8 c \right )}{1024 d}-\frac {3 a \,b^{2} \sin \left (7 d x +7 c \right )}{448 d}-\frac {b \cos \left (6 d x +6 c \right ) a^{2}}{64 d}+\frac {b^{3} \cos \left (6 d x +6 c \right )}{384 d}+\frac {\sin \left (5 d x +5 c \right ) a^{3}}{80 d}-\frac {9 \sin \left (5 d x +5 c \right ) a \,b^{2}}{320 d}-\frac {3 b \cos \left (4 d x +4 c \right ) a^{2}}{32 d}-\frac {b^{3} \cos \left (4 d x +4 c \right )}{256 d}+\frac {5 \sin \left (3 d x +3 c \right ) a^{3}}{48 d}-\frac {\sin \left (3 d x +3 c \right ) a \,b^{2}}{64 d}-\frac {15 b \cos \left (2 d x +2 c \right ) a^{2}}{64 d}-\frac {3 b^{3} \cos \left (2 d x +2 c \right )}{128 d}\) | \(241\) |
| norman | \(\frac {\frac {2 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{3} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{2} b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {6 a^{2} b \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (3 a^{2} b +b^{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {4 \left (3 a^{2} b +b^{3}\right ) \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {10 \left (12 a^{2} b +4 b^{3}\right ) \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (39 a^{2} b -8 b^{3}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (39 a^{2} b -8 b^{3}\right ) \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (13 a^{2}+12 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (13 a^{2}+12 b^{2}\right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a \left (163 a^{2}+12 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {2 a \left (163 a^{2}+12 b^{2}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {2 a \left (1883 a^{2}+1032 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d}+\frac {2 a \left (1883 a^{2}+1032 b^{2}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{105 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}\) | \(398\) |
1/d*(1/8*b^3*sin(d*x+c)^8+3/7*a*b^2*sin(d*x+c)^7+1/6*(3*a^2*b-2*b^3)*sin(d *x+c)^6+1/5*(a^3-6*a*b^2)*sin(d*x+c)^5+1/4*(-6*a^2*b+b^3)*sin(d*x+c)^4+1/3 *(-2*a^3+3*a*b^2)*sin(d*x+c)^3+3/2*sin(d*x+c)^2*a^2*b+a^3*sin(d*x+c))
Time = 0.30 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.81 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {105 \, b^{3} \cos \left (d x + c\right )^{8} - 140 \, {\left (3 \, a^{2} b + b^{3}\right )} \cos \left (d x + c\right )^{6} - 8 \, {\left (45 \, a b^{2} \cos \left (d x + c\right )^{6} - 3 \, {\left (7 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{4} - 56 \, a^{3} - 24 \, a b^{2} - 4 \, {\left (7 \, a^{3} + 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{840 \, d} \]
1/840*(105*b^3*cos(d*x + c)^8 - 140*(3*a^2*b + b^3)*cos(d*x + c)^6 - 8*(45 *a*b^2*cos(d*x + c)^6 - 3*(7*a^3 + 3*a*b^2)*cos(d*x + c)^4 - 56*a^3 - 24*a *b^2 - 4*(7*a^3 + 3*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d
Time = 0.68 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.40 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\begin {cases} \frac {8 a^{3} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {a^{3} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {a^{2} b \cos ^{6}{\left (c + d x \right )}}{2 d} + \frac {8 a b^{2} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {4 a b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {a b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {b^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{6 d} - \frac {b^{3} \cos ^{8}{\left (c + d x \right )}}{24 d} & \text {for}\: d \neq 0 \\x \left (a + b \sin {\left (c \right )}\right )^{3} \cos ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((8*a**3*sin(c + d*x)**5/(15*d) + 4*a**3*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + a**3*sin(c + d*x)*cos(c + d*x)**4/d - a**2*b*cos(c + d*x)* *6/(2*d) + 8*a*b**2*sin(c + d*x)**7/(35*d) + 4*a*b**2*sin(c + d*x)**5*cos( c + d*x)**2/(5*d) + a*b**2*sin(c + d*x)**3*cos(c + d*x)**4/d - b**3*sin(c + d*x)**2*cos(c + d*x)**6/(6*d) - b**3*cos(c + d*x)**8/(24*d), Ne(d, 0)), (x*(a + b*sin(c))**3*cos(c)**5, True))
Time = 0.18 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {105 \, b^{3} \sin \left (d x + c\right )^{8} + 360 \, a b^{2} \sin \left (d x + c\right )^{7} + 140 \, {\left (3 \, a^{2} b - 2 \, b^{3}\right )} \sin \left (d x + c\right )^{6} + 168 \, {\left (a^{3} - 6 \, a b^{2}\right )} \sin \left (d x + c\right )^{5} + 1260 \, a^{2} b \sin \left (d x + c\right )^{2} - 210 \, {\left (6 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{4} + 840 \, a^{3} \sin \left (d x + c\right ) - 280 \, {\left (2 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (d x + c\right )^{3}}{840 \, d} \]
1/840*(105*b^3*sin(d*x + c)^8 + 360*a*b^2*sin(d*x + c)^7 + 140*(3*a^2*b - 2*b^3)*sin(d*x + c)^6 + 168*(a^3 - 6*a*b^2)*sin(d*x + c)^5 + 1260*a^2*b*si n(d*x + c)^2 - 210*(6*a^2*b - b^3)*sin(d*x + c)^4 + 840*a^3*sin(d*x + c) - 280*(2*a^3 - 3*a*b^2)*sin(d*x + c)^3)/d
Time = 0.39 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.28 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {b^{3} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {3 \, a b^{2} \sin \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {{\left (6 \, a^{2} b - b^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{384 \, d} - \frac {{\left (24 \, a^{2} b + b^{3}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {3 \, {\left (10 \, a^{2} b + b^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{128 \, d} + \frac {{\left (4 \, a^{3} - 9 \, a b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (20 \, a^{3} - 3 \, a b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{192 \, d} + \frac {5 \, {\left (8 \, a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )}{64 \, d} \]
1/1024*b^3*cos(8*d*x + 8*c)/d - 3/448*a*b^2*sin(7*d*x + 7*c)/d - 1/384*(6* a^2*b - b^3)*cos(6*d*x + 6*c)/d - 1/256*(24*a^2*b + b^3)*cos(4*d*x + 4*c)/ d - 3/128*(10*a^2*b + b^3)*cos(2*d*x + 2*c)/d + 1/320*(4*a^3 - 9*a*b^2)*si n(5*d*x + 5*c)/d + 1/192*(20*a^3 - 3*a*b^2)*sin(3*d*x + 3*c)/d + 5/64*(8*a ^3 + 3*a*b^2)*sin(d*x + c)/d
Time = 0.10 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.98 \[ \int \cos ^5(c+d x) (a+b \sin (c+d x))^3 \, dx=\frac {{\sin \left (c+d\,x\right )}^3\,\left (a\,b^2-\frac {2\,a^3}{3}\right )-{\sin \left (c+d\,x\right )}^5\,\left (\frac {6\,a\,b^2}{5}-\frac {a^3}{5}\right )+{\sin \left (c+d\,x\right )}^6\,\left (\frac {a^2\,b}{2}-\frac {b^3}{3}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {3\,a^2\,b}{2}-\frac {b^3}{4}\right )+a^3\,\sin \left (c+d\,x\right )+\frac {b^3\,{\sin \left (c+d\,x\right )}^8}{8}+\frac {3\,a^2\,b\,{\sin \left (c+d\,x\right )}^2}{2}+\frac {3\,a\,b^2\,{\sin \left (c+d\,x\right )}^7}{7}}{d} \]